TS Inter 2nd Year

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type

Question 1.
State and prove Binomial Theorem. [March ’09]
Solution:

Question 2.
Find the numerically greatest term in the expansion of (4 + 3x)¹⁵ where x = \(\frac{7}{2}\).
Solution:
Given
(4 + 3x)¹⁵
⇒ 4¹⁵ (1 + \(\frac{3 x}{4}\))¹⁵ = 4¹⁵ (1 + x)ⁿ
where n = 15; x = \(\frac{3 x}{4}\)
\(|x|=\left|\frac{3 x}{4}\right|=\left|\frac{3}{4} \cdot \frac{7}{2}\right|=\frac{21}{8}\)
Now, \(\frac{(n+1)|x|}{|x|+1}=\frac{(15+1)\left|\frac{3 x}{4}\right|}{\left|\frac{3 x}{4}\right|+1}\)
= \(\frac{16 \cdot \frac{21}{8}}{\frac{21}{8}+1}=\frac{336}{29}\) = 11.5
∴ T₁₂ is numerically greatest.
r + 1 = 12
⇒ r = 11
The general term in the expansion of (x + a)ⁿ is
T_{r + 1} = \({ }^n C_r\) x^{n – r} a^r
T_{11 + 1} = \({ }^{15} \mathrm{C}_{11}\) (4)^{15 – 11} (3x)¹¹
12th term in this expansion is

Question 3.
Find the numerically reatest tenu in the expansion of (4a – 6b)¹³ when a = 3, b = 5.
Solution:
Given (4a – 6b)¹³ = (4a)¹³ (1 – \(\frac{6 b}{4 a}\))¹³
= (4a)¹³ (1 + x)ⁿ
where, x = \(-\frac{6 b}{4 a}\), n = 13
|x| = \(\left|\frac{-6 b}{4 a}\right|=\left|\frac{-6 \cdot 5}{4 \cdot 3}\right|=\frac{5}{2}\)
Now, \(\frac{(n+1)|x|}{|x|+1}=\frac{(13+1) \frac{5}{2}}{\frac{5}{2}+1}\)
= \(\frac{14 \cdot 5}{5+2}=\frac{70}{7}\) = 10
∴ T₁₀ and T₁₁ are numerically greatest.
T₁₀:
r + 1 = 10
⇒ r = 9
The general term in the expansion of (1 + x)ⁿ is
T_{r + 1} = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) x^{n – r} a^r
T₁₀ in this expansion is
T₉₊₁ = \({ }^{13} \mathrm{C}_9\) (4a)^13-9 (- 6b)⁹
T10 = \({ }^{13} \mathrm{C}_9\) (4a)⁴ (- 6b)⁹
= \({ }^{13} \mathrm{C}_9\) (4⁴ . 3⁴) ((- 6)⁹ (5)⁹)
= \({ }^{13} \mathrm{C}_9\) 12⁴ . 30⁹
|T₁₀| = \({ }^{13} \mathrm{C}_9\) 12⁴ . 30⁹

T₁₁:
r + 1 = 11
⇒ r = 10.
The general term in the expansion of (1 – x)ⁿ is
T_{r + 1} = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) x^{n – r} a^r
T₁₁ in this expansion is
T₁₀₊₁ = \({ }^{13} \mathrm{C}_{10}\) (4a)^13-10 . (- 6b)¹⁰
= \({ }^{13} \mathrm{C}_{10}\) (4a)³ (6b)¹⁰
T₁₁ = \({ }^{13} \mathrm{C}_{10}\) (4 . 3)³ (6 . 5)¹⁰
= \({ }^{13} \mathrm{C}_{10}\) 12³ . 30¹⁰
T₁₁ = \({ }^{13} \mathrm{C}_{10}\) 12³ . 30¹⁰
∴ |T₁₀| = |T₁₁|.

Question 4.
If the coefficient of x¹⁰ in the expansion of \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of x^-10 in the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) find the relation between a and b where a and b are real numbers. [AP – May 2015, AP – Mar. 2019]
Solution:
Case – I :
Given \(\left(a x^2+\frac{1}{b x}\right)^{11}\)
Here x = ax², a = \(\frac{1}{\mathrm{bx}}\); n = 11
Now, the general term in the expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r x^22-2r 1^r b^-r x^-r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r b^-r x^22-3r …………….(1)
To find the coefficient of x¹⁰,
Put 22 – 3r = 10
3r = 12
r = 4
Substituting r = 4 in equation (1) we get
T₄₊₁ = \({ }^{11} C_4\) a^11-4 b^-4 x^22-12
T₅ = \({ }^{11} C_4\) a⁷ b^-4 x¹⁰
T₅ = \({ }^{11} C_4\) a⁷ b^-4 x¹⁰
∴ The coeff. of x¹⁰ ¡n the expansion of \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is \({ }^{11} C_4\) a⁷ b^-4

Case II:
Given \(\left(a x-\frac{1}{b x^2}\right)^{11}\)
Here x = ax, a = \(\left(\frac{-1}{b x^2}\right)\), n = 11
Now, the general term in the expansion is
T_{r + 1} = \({ }^n C_r\) x^n-r a^r
= \({ }^{11} C_r(a x)^{11-r}\left(\frac{-1}{b x^2}\right)^r\)
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r x^11-r (- 1)^r b^-r x^-2r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) (- 1)^r b^-r x^11-3r ……………(2)
To find the coefficient of x^-10
put 11 – 3r = – 10
⇒ 3r = 21
⇒ r = 7
Substitute r = 7 in equation (2) we get
T₇₊₁ = \({ }^{11} \mathrm{C}_7\) a^11-7 b⁷ (- 1)⁷ x^11-21
T₈ = – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7 x^-10
∴ The coeff. of x^-10 in the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) is – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7
Given that, these coefficients are equal
\({ }^{11} \mathrm{C}_7\) a⁷ b^-4 = – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7
\({ }^{11} C_4 \frac{a^7}{b^4}=-{ }^{11} C_4 \frac{a^4}{b^7}\)
\(\frac{a^7}{b^4}=\frac{-a^4}{b^7}\)
⇒ a³b³ = – 1
⇒ (ab)³ = (- 1)³
⇒ ab = – 1

Question 5.
If n is a positive integer, then show that
i) C₀ + C₁ + C₂ + ……………… + C_n = 2
ii) C₀ + C₂ + C₄ + ……………. = C₁ + C₃ + C₅ + ……….. = 2^n-1 [March ’97, ’90]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁x + C₂x² + ……………… + C_nxⁿ ………….(1)
i) Put x = 1 in equation (1)
⇒ (1 + 1)ⁿ = C₀ + C₁(1) + C₂(1)² + C₃(1)³ + C₄(1)⁴ + ……………… + C_n(1)ⁿ
2ⁿ = C₀ + C₁ + C₂ + C₃ + C₄ + …………. + C_n
∴ C₀ + C₁ + C₂ + C₃ + C₄ + …………. + C_n = 2ⁿ

ii) Put x = – 1 in equation (1), we get
(1 – 1)ⁿ = C₀ + C₁ (- 1) + C₂ (- 1)² + C₃ (- 1)³ + ……………. + C_n (- 1)ⁿ
o = C₀ – C₁ + C₂ – C₃ + C₄ – ………………
C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + ………….
Since a = b
⇒ a = b = \(\frac{a+b}{2}\)
C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + ………….
= \(\frac{\mathrm{C}_0+\mathrm{C}_2+\mathrm{C}_4+\ldots \ldots+\mathrm{C}_1+\mathrm{C}_3+\mathrm{C}_5+\ldots \ldots}{2}\)
= \(\frac{\mathrm{C}_0+\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3+\mathrm{C}_4+\mathrm{C}_5+\ldots \ldots}{2}\)
= \(\frac{2^{\mathrm{n}}}{2}\)
= 2^{n – 1}
∴ C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + …………. = 2^{n – 1}
C₀ + C₂ + C₄ + …………………. = 2^{n – 1}
C₁ + C₃ + C₅ + …………………… = 2^{n – 1}

Question 6.
Prove that for any real numbers a, d, a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……………… + (a + nd) . C_n = (2a + nd) 2^{n – 1} [May ‘98]
Solution:
Let S = a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……………… + (a + nd) . C_n ………………(1)
By writing the terms in R.H.S of (1) in reverse order has done we get
S = (a + nd)C_n + (a + (n – 1)d)C_n-1 + (a + (n-2)d)C_n-2 + …………. + aC₀
S = (a + nd)C₀ + (a + (n – 1)d)C₁ + (a + (2n – 2)d)C₂ + ………….. + aC_n ……………(2)
Adding (1) and (2) we get

2s = (2a + nd) (C₀ + C₁ + C₂ + …………. + C_n)
2s = (2a + nd)²ⁿ
⇒ S = (2a + nd)^2n-1
∴ aC₀ + (a + d)C₁ + (a + 2d)C₂ + …………. + (a + nd)C_n = (2a + nd) 2^2n-1.

Question 7.
For r = 0, 1, 2, ….., n, prove that C₀ . C_r + C₁ . C_r+1 + C₂ . C_r+2 + ……………… + C_n-r . C_n = \({ }^{2 n} C_{(n+r)}\)
and hence deduce that
i) C₀² + C₁² + C₂² + ………….. + C_n² = \({ }^{2 n} C_n\)
ii) C₀ . C₁ + C₁ . C₂ + C₂ . C₃ + ………… + C_n-1 . C_n = \({ }^{2 n} C_{n+1}\)
[May ‘97, 95, ‘90, ‘94,’93, ‘92; March ‘98, ‘93, AP-Mar. 2018; TS – Mar. 2015]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁ . x + C₂ . x² + ………………….. + C_r x^r + C_n . xⁿ ………………(1)
(x + 1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-2 + ………………… + C_rx^n-r + …………………. + C_n ………………(2)
Multiplying (2) and (1), we get
(C₀xⁿ + C₁x^n-1 + C₂x^n-2 + ………………… + C_rx^n-r + …………………. + C_n) (C₀ + C₁ . x + C₂ . x² + ………………….. + C_r x^r + C_n . xⁿ)
= (x + 1)ⁿ (1 + x)ⁿ = (1 + x)²ⁿ
Comparing the coefficient of x^n+r on bothsides we get,
C₀ . C_r + C₁ . C_r+1 + C₂ . C_r+2 + ……………… + C_n-r . C_n = \({ }^{2 n} C_{(n+r)}\) ……………….(3)

i) On substituting r = 0 in equation (3) we get
C₀ . C₀ + C₁ . C₁ + C₂ . C₂ + ……………….. + C_n . C_n = \({ }^{2 n} C_n\)
C₀² + C₁² + C₂² + ………….. + C_n² = \({ }^{2 n} C_n\)

ii) On substituting, r = 1 in equation (3) we get
C₀ . C₁ + C₁ . C₂ + C₂ . C₃ + ………… + C_n-1 . C_n = \({ }^{2 n} C_{n+1}\)

Question 8.
If n is a positive integer and x is any non-zero real number, then prove that
C₀ + C₁ . \(\frac{x}{2}\) + C₂ . \(\frac{x^2}{3}\) + C₃ . \(\frac{x^3}{4}\) + …………….. + C_n . \(\frac{\mathbf{x}^n}{n+1}\) = \(\frac{(1+x)^{n+1}-1}{(n+1) x}\). [May ’14, ’13, ’08, ’05, Mar. ’14, ’02, Board Paper, AP – May 2016]
Solution:
Let S = C₀ + C₁ . \(\frac{x}{2}\) + C₂ . \(\frac{x^2}{3}\) + C₃ . \(\frac{x^3}{4}\) + …………….. + C_n . \(\frac{\mathbf{x}^n}{n+1}\) = \(\frac{(1+x)^{n+1}-1}{(n+1) x}\)

Question 9.
If n is a positive integer, then prove that
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\) [TS – Mar. 2017]
Solution:
Let S = C₀ + \(\frac{\mathrm{C}_1}{2}+\frac{\mathrm{C}_2}{3}+\ldots . .+\frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1}\)
= \({ }^{\mathrm{n}} \mathrm{C}_0+\frac{1}{2} \cdot{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3} \cdot{ }^{\mathrm{n}} \mathrm{C}_2+\frac{1}{\mathrm{n}+1} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}\)
Now multiplying on both sides with (n + 1), we get

Question 10.
Prove that \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\cdots \cdots=\frac{2^n-1}{n+1}\) [May 08].
Solution:

Question 11.
If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ……………… + a_2nx²ⁿ, then prove that
i) a₀ + a₁ + a₂ + …………….. + a_2n = 3ⁿ
ii) a₀ + a₂ + a₄ + ……………. + a_2n = \(\frac{3^n+1}{2}\)
iii) a₁ + a₃ + a₅ + ……………… + a_2n = \(\frac{3^n-1}{2}\)
iv) a₀ + a₃ + a₆ + a₉ + ………………. = 3^n-1 [TS May 2016; Board Paper]
Solution:
(1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ……………… + a_2nx²ⁿ
Put x = 1 in equation (1) we get,
(1 + 1 + 1²)ⁿ = a₀ + a₁ (1) + a₂ (1)² + ……………… + a_2n. (1)²ⁿ
a₀ + a₁ + a₂ + …………… + a_2n = 3ⁿ ……………(2)
Put x = – 1 in equation (1) we get,
(1 – 1 + 1²)ⁿ = a₀ + a₁ (- 1) + a₂ (- )² + a₃. (- 1)⁴ ……………… + a_2n. (- 1)²ⁿ
a₀ – a₁ + a₂ – a₃ + a₄ + …………… + a_2n = 1 ……………(3)
i) From (2),
a₀ + a₁ + a₂ + a₃ + …………… + a_2n = 3ⁿ

ii) Now, adding (2) and (3)
a₀ + a₁ + a₂ + a₃ + …………… + a_2n = 3ⁿ

iii) Now, subtracting (2) and (3)

iv) Put, x = 1 in equation (1) we get
(1 + 1 + 1²)ⁿ = a₀ + a₁ (1) + a₂(1)² + a₃ (1)³ + a₄ (1)⁴ + a₅ (1)⁵ + a₆ (1)⁶ + ……………… + a_2n (1)²ⁿ
= a₀ + a₁ + a₂ + a₃ + a₄ + a₅ + a₆ + …………….. + a_2n = 3ⁿ ………………(4)

Put x = ω in equation (1), we get
(1 + ω + ω²)ⁿ = a₀ + a₁ (ω) + a₂ (ω)² + a₃ (ω)³ + a₄ (ω)⁴ + a₅ (ω)⁵+ a₆ (ω)⁶ + ……………. + a_2n (ω)²ⁿ
= a₀ + a₁ . (ω) + a₂ (ω)² + a₃ (ω)³ + a₄ (ω)⁴ + a₅ (ω)⁵ + a₆ (ω)⁶ + ……………. + a_2n (ω)²ⁿ = 0 …………….(5)
put x = in equation (1) we get,
(1 + ω² + ω⁴)ⁿ = a₀ + a₁ . (ω)² + a₂ (ω)⁴ + a₃ (ω)⁶ + a₄ (ω)⁸ + a₅ (ω)¹⁰ + a₆ (ω)¹² + ……………. + a_2n (ω²)²ⁿ
= a₀ + a₁ . ω² + a₂ ω + a₃ + a₄ ω² + a₅ ω + a₆ + ……………. + a_2n (ω²)²ⁿ = 0
Now, adding (4), (5) and (6)

⇒ 3a₀ + 0 + 0 + 3a₃ + 0 + 0 + 3a₆ + ………….. = 3ⁿ
⇒ 3(a₀ + a₃ + a₆ + ………………..) = 3ⁿ
⇒ a₀ + a₃ + a₆ + ……………….. = 3^n-1.

Question 12.
If (1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………………. + b₂₁x²¹, then find the value of
i) b₀ + b₂ + b₄ + …………. + b₂₀
ii) b₁ + b₃ + b₅ + …………….. + b₂₁
Solution:
Given
(1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………………. + b₂₁x²¹ …………………….(1)
Substituting x = 1 in (1),
we get b₀ + b₁ + b₂ + …………. + b₂₁ = 47 ………………(2)
Substituting x = – 1 in (1),
we get b₀ – b₁ + b₂ + ……………… – b₂₁ = 0 ……………..(3)
i) (2) + (3)
⇒ 2b₀ + 2b₂ + 2b₄ + ………………….. + 2b₉₀ = 4⁷
⇒ b₀ + b₂ + b₄ + …………….. + b₂₀ = 2¹³

ii) (2) – (3)
⇒ 2b₁ + 2b₃ + 2b₅ + ……………. + 2b₂₁ = 4⁷
⇒ b₁ + b₃ + b₅ + ……………. + b₂₁ = 2¹³

Question 13.
the coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are in A.P., then prove that n² – 41n + 398 = 0.
Solution:
Coefficient of r in the expansion of (1 + x)ⁿ is \({ }^n C_r\).
Given coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are \({ }^n C_9,{ }^n C_{10},{ }^n C_{11}\) in AP., then 2\(\left({ }^n C_{10}\right)={ }^n C_9+{ }^n C_{11}\)
⇒ \(2 \frac{n !}{(n-10) ! 10 !}=\frac{n !}{(n-9) ! 9 !}+\frac{n !}{(n-11) !+11 !}\)
⇒ \(\frac{2}{10(n-10)}=\frac{1}{(n-9)(n-10)}+\frac{1}{11 \times 10}\)
⇒ \(\frac{2}{(n-10) 10}=\frac{110+(n-9)(n-10)}{110(n-9)(n-10)}\)
⇒ 22 (n – 9) = 110 + n² – 19n + 90
⇒ n² – 41n + 398 = 0.

Question 14.
1f 36, 84, 126 are three successive binomial coefficlenü* in the expansion of (1 + x)ⁿ, then find n. [May ’09, ’06] [TS – Mar. 2019]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁x + C₂x² + ………………. C_nxⁿ.
Let the three successive binomial coefficients in the expansion of (1 + x)ⁿ are \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1},{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}},{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\)
Given that,
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) = 36 ………………(1)
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) = 84 ………………(2)
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) = 126 ……………..(3)

⇒ 2n – 2r = 3r + 3
⇒ 5r = 2n – 3 from (4)
⇒ 5 \(\left(\frac{3 n+3}{10}\right)\) = 2n – 3
⇒ 3n + 3 = 4n – 6
⇒ n = 9.

Question 15.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)ⁿ are respectively 240, 720, 1080, find a, x, n. [TS – Mar. 2016; Ma ‘09, ’06]
Solution:
The general term in the expansion of (x + a)ⁿ is
T_r+1 = x^n-r . a^r
In the expansion, (a + x)ⁿ
T₂ = T₁₊₁
= \({ }^n C_1\) (a)^n-1 x¹ = 240 …………..(1)
T₃ = T₂₊₁
= \({ }^n C_2\) (a)^n-2 x² = 720 …………..(2)
T₄ = T₃₊₁
= \({ }^n C_3\) (a)^n-3 x³ = 1080 …………..(3)

⇒ 2(n – 2) x = 9a ………………(5)
\(\frac{(5)}{(4)}=\frac{2(n-2) x}{(n-1) x}=\frac{9 a}{6 a}\)
⇒ 4(n – 2) = 3(n – 1)
⇒ 4n – 8 = 3n – 3
⇒ n = 5
Substitute n = 5 in equation (4)
⇒ (5 – 1)x = 6a
⇒ 4x = 6a
⇒ 2x = 3a
⇒ x = \(\frac{3 a}{2}\)
Substitute n, x in equation (1).
\(5 C_1(a)^{5-1}\left(\frac{3 a}{2}\right)^1\) = 240
⇒ 5 . a⁴ . \(\frac{3}{2}\) . a = 240
⇒ a⁵ = 32
⇒ a = 2.
x = \(\frac{3 a}{2}\)
x = 3 . \(\frac{2}{2}\)
⇒ x = 3
∴ a = 2, x = 3, n = 5.

Question 16.
If the coefficients of rth, (r + 1)th and (r + 2)nd tenus in the expansion of (1 + x)ⁿ are in A.P then show that n² – (4r + 1) n + 4r² – 2 = 0.
Solution:
In the expansion (1 + x)ⁿ
The coeff. of T_r = \({ }^n \mathrm{C}_{\mathrm{r}-1}\);
The coeff. of T_r+1 = \({ }^n C_r\)
The coeÍf. of T_r+2 = \({ }^n \mathrm{C}_{\mathrm{r}+1}\)
Given that
\({ }^n \mathrm{C}_{\mathrm{r}-1}\), \({ }^n C_r\), \({ }^n \mathrm{C}_{\mathrm{r}+1}\) are in AP.
If a, b, c are in A.P.
Then 2b = a + c

2(n – r + 1) (r + 1) = (r + 1) r + (n – r + 1) (n – r)
⇒ 2nr + 2n – 2r² – 2r + 2r + 2 = r² + r + n² – nr – nr + r² + n – r
⇒ 2nr + 2n – 2r² + 2 = n² – 2nr + n + 2r²
⇒ n² – 2nr + n + 2r² – 2nr – 2n + 2r² – 2 = 0
⇒ n² – 4nr + 4r² – n – 2 = 0
⇒ n² – (4r + 1)n + 4r² – 2 = 0.

Question 17.
If ‘P’ and ‘Q’ are the sum of odd tenns and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
i) P² – Q² = (x² – a²)ⁿ
ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ [AP – March 2016, March ’10]
Solution:
Since P and Q are the sum of odd terms and sum of even terms respectively in the expansion of (x + a)ⁿ.
Now (x + a)ⁿ =

i) L.HS: P² – Q²
= (P + Q) (P – Q) = (x + a) (x – a)
= [(x + a) (x – a)]ⁿ
= (x² – a²)ⁿ = R.H.S.

ii) L.H.S : 4PQ = (P + Q)² – (P – Q)²
= ((x + a)ⁿ)² – ((x – a)ⁿ)²
= (x + a)²ⁿ – (x – a)²ⁿ = R.H.S.
∴ 4PQ = (x + a)²ⁿ – (x – a)²ⁿ.

Question 18.
If the coefficients of 4 consecutive term in the expansion of (1 + x)ⁿ are a₁, a₂, a₃, a₄ respectively then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\). [AP – Mar. 2017; May’11. ‘07, March ’11, ’95].
Solution:
Since a₁, a₂, a₃, a₄, are the coefficients of 4 consecutive terms in the expansion of (1 + x)ⁿ
Let a₁ = \({ }^n C_{r-1}\),
a₂ = \({ }^n C_r\),
a₃ = \({ }^n C_{r+1}\),
a₄ = \({ }^n C_{r+2}\)
L.H.S:

R.H.S:

Question 19.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\).
Solution:
The expansion of (1 + x)²ⁿ contains 2n + 1 terms.
∴ Middle term is (n + 1)th term
∴ T_n+1 = \({ }^{2 n} C_n\) xⁿ

= \(\frac{2 n !}{n ! \cdot n !}\) xⁿ

= \(\frac{2 n(2 n-1)(2 n-2)(2 n-3) \ldots . .4 \times 3 \times 2 \times 1}{n ! n !} \cdot x^n\)

= \(\frac{1 \cdot 3 \cdot 5 \ldots .(2 n-1) \cdot 2 \times 4 \times 6 \times \ldots .2 n}{n ! n !} \cdot x^n\)

= \(\frac{1 \cdot 3 \cdot 5 \ldots .(2 n-1)(1 \times 2 \times 3 \times \ldots \ldots n) 2^n}{n ! n !} \cdot x^n\)

∴ Middle term = \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\).

Question 20.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……………. + ax²⁰ then prove that
i) a₀ + a₁ + a₂ + ………….. + a₂₀ = 2¹⁰
ii) a₀ – a₁ + a₂ – a₃ + ……………. + a₂₀ = 4¹⁰
Solution:
Given
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……………. + ax²⁰ …………….(1)

i) Put x = 1, in (1), we get
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + ………….. + a₂₀
⇒ a₀ + a₁ + a₂ + …………….. + a₂₀ = 2¹⁰.

ii) Put x = – 1 in (1), we get
(1 – 3 – 2)’°= a₀ – a₁ + a₂ – a₃ + ………………. + a₂₀
⇒ a₀ – a₁ + a₂ – a₃ + ……………. + a₂₀ = 4¹⁰.

Question 21.
If n is a positive integer, prove that \(\sum_{r=1}^n r^3 \cdot\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{n(n+1)^2(n+2)}{12}\). [March ’13]
Solution:
Now, \(\frac{{ }^{n^n} C_r}{{ }^{n_C} C_{r-1}}=\frac{\frac{n !}{(n-r) ! r !}}{\frac{n !}{(n-r+1) !(r-1) !}}\)
= \(\frac{(n-r+1) !(r-1) !}{(n-r) ! r !}\)

Question 22.
Find the sum of the infinite series \(1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6}\left(\frac{1}{2}\right)^2+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{1}{2}\right)^3+\ldots \infty\). [May ’91]
Solution:
Let the given series ¡s
S = 1 + \(\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{2^2}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{2^3}+\cdots \cdots\)
Comparing with 1 + nx + \(\frac{n(n-1)}{1 \cdot 2}\) x² + …………….

Question 23.
\(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\cdots \cdots \cdots \infty\) [TS – Mar. 2017; May ’09]
Solution:
Let the given series is
S = \(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\cdots \cdots \cdots \infty\)

S + \(\frac{3}{5}\) = \(\frac{3}{5}+\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\ldots \ldots\)

Question 24.
If x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \cdots \infty\), then find 3x² + 6x. [TS – Mar. 2016, May ’14, ’11, ’08, Mar. ’14, ’12, ’06, TS – Mar. ’19]
Solution:
Given x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \ldots\),
x + 1 = 1 + \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\cdots \cdots\)
Now, comparing with

⇒ x² + 1 + 2x = \(\frac{5}{3}\)
⇒ 3x² + 6x + 3 = 5
⇒ 3x² + 6x = 2.

Question 25.
Find the sum of the infinite series \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\) [May ’10, March 11]
Solution:
Let the given series is
S = \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \ldots \ldots\)
S + 1 = 1 + \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \ldots \ldots\)
Now, comparing with

Question 26.
If x = \(\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\) then prove that 9x² + 24x = 11. [AP – MAr. ’19, ’17, ’15, May ’16; TS – MAr. ’18, ’16, Mar. ’09, Board Paper, May ’15]
Solution:
Given

Question 27.
If x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots\) then find the value of x² + 4x. [TS – May; March ’13, May ’12]
Solution:
Given,

Question 28.
Find the sum to infinite terms of the series \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^4}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^6}+\ldots\right)\). [AP – Mar. ’18, ’16; May ’13, March ’05].
Solution:

Question 29.
Show that for any non-zero rational number x.
1 + \(\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots \ldots\) = 1 + \(\frac{x}{3}+\frac{x(x+1)}{3 \cdot 6}+\frac{x(x+1)(x+2)}{3 \cdot 6 \cdot 9}+\ldots \ldots\) [March ’94]
Solution:
L.H.S = 1 + \(\frac{\mathrm{n}}{2}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 \cdot 4}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{2 \cdot 4 \cdot 6}+\ldots\)

Some More Maths 2A Binomial Theorem Important Questions

Question 1.
Find the 7th term in the expansion of \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\).
Solution:
Given \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)
Here, x = \(\frac{4}{x^3}\); a = \(\frac{x^2}{2}\); n = 14
r + 1 = 7
⇒ r = 6
The general term in the expansion of (x + a)ⁿ is
T_r+1 = \({ }^n C_r\) x^n-r a^r
The 7th term in the expansion of \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)

Question 2.
Find the 5th term in the expansion of (3x – 4y)⁷.
Solution:
Given (3x – 4y)⁷
Here, x = 3x; a = – 4y; n = 7
r + 1 = 5
r =4
The general term in the expansion of (x + a)ⁿ is
T_{r + 1} = \({ }^n C_r\) x^n-r a^r
The rth term in the expansion of (3x – 4y)⁷ is
T₄₊₁ = \({ }^7 \mathrm{C}_4\) (3x)^7-4 (- 4y)⁴
T₅ = \({ }^7 \mathrm{C}_4\) (3x)³ (4y)⁴
= 35 . 3³ . x³ 4⁴ . y⁴
= 241920 x³ . y⁴

Question 3.
Find the middle term(s) in \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\).
Solution:
Given \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Here, x = \(\frac{3}{a^3}\), a = 5a⁴, n = 20
Since, n = 20 is even,
Middle term = \(\frac{n}{2}\) + 1
= \(\frac{20}{2}\) + 1
= 10 + 1 = 11th term
r + 1 = 11
⇒ r = 10
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
11th term of given expansion is
T₁₀₊₁ = \({ }^{20} C_{10}\left(\frac{3}{a^3}\right)^{20-10}\left(5 a^4\right)^{10}\)
= \({ }^{20} \mathrm{c}_{10}\left(\frac{3}{a^3}\right)^{10}\left(5 a^4\right)^{10}\)

T₁₁ = \({ }^{20} C_{10}\) . a^-30 . 5¹⁰ . a⁴⁰
= \({ }^{20} C_{10}\) . 3¹⁰ . 5¹⁰ . a¹⁰

Question 4.
Find the middle term(s) in the expansion of (4x² + 5x³)¹⁷.
Solution:
Given (4x² + 5x³)¹⁷
Here, x = 4x², a = 5x³, n = 17
Since, n = 17 is odd, then
Middle terms \(\frac{\mathrm{n}+1}{2}, \frac{\mathrm{n}+3}{2}\) = 9, 10 terms
9’ term:
r + 1 = 9
⇒ r = 8
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
9th term of given expansion is
T₈₊₁ = \({ }^{17} \mathrm{c}_9\) (4x²)^17-8 (5x³)⁸
= \({ }^{17} \mathrm{c}_9\) (4x²)⁹ (5x³)⁸
= \({ }^{17} \mathrm{c}_9\) . 4⁹ . x¹⁸ . 5⁸ . x²⁴
= \({ }^{17} \mathrm{c}_9\) . 4⁹ . 5⁸ . x⁴²

10th term:
r + 1 = 10
⇒ r = 9
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
10th term in this expansion ¡s
T₉₊₁ = \({ }^{17} \mathrm{C}_9\)(4x²)^17-9 (5x³)⁹
T₁₀ = \({ }^{17} \mathrm{C}_9\) (4x²)⁸ (5x³)⁹
= \({ }^{17} \mathrm{C}_9\) . 4⁸ . 5⁹ . x¹⁶ . x²⁷
= \({ }^{17} \mathrm{C}_9\) . 4⁸ . 5⁹ . x⁴³

Question 5.
Find the coefficient of x¹¹ in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\).
Solution:
Given \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
Here, x = 2x²,
a = \(\frac{3}{x^3}\), n = 13
Now, the general term in this expansion is
T_r+1 = \(\) x^n-r a^r
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) (2x^r)^13-r \(\left(\frac{3}{x^3}\right)^r\)
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) 2^13-r x^26-2r . 3² . x^-3r
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) 2^13-r 3^r x^26-5r ………………(1)
To find the coefficient of x¹¹
Put 26 – 5r = 11
5r = 15
⇒ r = 3
Now substituting r = 3 in equation (1) we get
T₃₊₁ = \({ }^{13} \mathrm{C}_3\) 2^13-3 3³ x^26-15
T₄ = \({ }^{13} \mathrm{C}_3\) 2¹⁰ 3³ x¹¹
The coefficient of x¹¹ in the expansion of \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is \({ }^{13} \mathrm{C}_3\) 2¹⁰ 3³.

Question 6.
Find the term independent of x in the expansion of \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\).
Solution:
Given \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\)
Here, x = \(\sqrt{\frac{x}{3}}\), a = \(\frac{3}{2 x^2}\), n = 10
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
T_r+1 = \({ }^{10} C_r\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^2}\right)^r\)
= \({ }^{10} C_r \frac{x^{\frac{10-r}{2}}}{3^{\frac{10-r}{2}}} \cdot \frac{3^r}{2^r} \cdot x^{-2 r}\)
= \({ }^{10} \mathrm{C}_{\mathrm{r}} \frac{3^{\mathrm{r}}}{3^{\frac{10-\mathrm{r}}{2}} \cdot 2^{\mathrm{r}}} \mathrm{x}^{\frac{10-\mathrm{r}}{2}-2 \mathrm{r}}\) ……………….(1)
To find the term independent of x¹¹ (i.e., coeff. of x°)
Put \(\frac{10-\mathrm{r}}{2}\) – 2r = 0
10 – r – 4r = 0
5r = 10
⇒ r = 2
Substitute r = 2 in equation (1) we get
T = \({ }^{10} \mathrm{C}_2 \frac{3^2}{3^{\frac{10-2}{2}} \cdot 2^2} \cdot \mathrm{x}^{\frac{10-2}{2}-2^2}\)
= \({ }^{10} C_2 \frac{3^2}{3^3 \cdot 2^2} \cdot x^0\)
= \(\frac{10 \cdot 9}{1 \cdot 2} \times \frac{3^2}{3^4 \cdot 2^2}=\frac{10 \cdot 9}{1 \cdot 2} \cdot \frac{1}{9 \cdot 4} \cdot x^0=\frac{5}{4} x^0\)
The term independent of x in the given expansion is T₃ = \(\frac{5}{4}\).

Question 7.
Find the largest binomial coefficients In the expansion of (1 + x)¹⁹.
Solution:
Given, (1 + x)¹⁹
Here n = 19, an odd integer.
The largest binomial coefficients are \({ }^n C_{\frac{n-1}{2}}\) and \({ }^n C_{\frac{n+1}{2}}\)
\({ }^{19} \mathrm{C}_{\frac{19-1}{2}} \text { and }{ }^{19} \mathrm{C}_{\frac{19+1}{2}}\) = \({ }^{19} C_9 \text { and }{ }^{19} C_{10}\)
(Note that, \({ }^{19} C_{9}={ }^{19} C_{10}\)).

Question 8.
If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)²¹ equal find ‘r’.
Solution:
Given (1 – x)¹⁸
The general term in the expansion of (x + a)ⁿ is
T_r+1 = \({ }^n C_r\) x^n-r a^r
(2r + 4)th term in the expansion of (1 + x)¹⁸ is
T_(2r+3)+1 = \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\) (1) x^2r+3
T_2r+4 = \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\) x^2r+3
The coeff. of (2r + 4)th term is \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\)
(r – 2) term in the expansion of (1 + x)¹⁸ is
T_(r-3)+1 = \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\) 1^18-(r-3) x^r-3
T_r-2 = \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\) . x^r-3
The coeff. of (r – 2)nd term is \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\)
∴ Given that two coefficients are equal.
\({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}={ }^{18} \mathrm{C}_{\mathrm{r}-3}\)
\({ }^n C_r={ }^n C_s\)
⇒ n = r + s (or) r = s
18 = 2r + 3 + r – 3
2r + 3 = r – 3
3r = 18
⇒ r = 6
⇒ r = – 6
Since, r is a positive integer, we get r = 6.

Question 9.
Find the set E of x for which the binomial expansion (7 + 3x)^-5.
Solution:
Given (7 + 3x)^-5
7^-5 (1 + \(\frac{3 x}{7}\))^-5
The binomial expansion of (7 + 3x)^-5 is valid when
\(\left|\frac{3 x}{7}\right|<1 \Rightarrow|x|<\frac{7}{3}\)
x ∈ \(\left(\frac{-7}{3}, \frac{7}{3}\right)\)

Question 10.
Find the set E of x for which the binomial expansion (7 – 4x)^-5 is valid.
Solution:
Given (7 – 4x)^-5 = 7^-5 (1 – \(\frac{4 x}{7}\))^-5
The binomial expansion of (7 – 4x)^-5 is valid when
\(\left|\frac{-4 x}{7}\right|<1 \Rightarrow\left|\frac{4 x}{7}\right|<1\)
⇒ x < \(\frac{7}{4}\)
⇒ x ∈ \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)
∴ E = \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)

Question 11.
Find the 6th term of (1 + \(\frac{x}{2}\))^-5
Solution:
Given, (1 + \(\frac{x}{2}\)))^-5
Comparing this with (1 + x)ⁿ, where
x = \(\frac{x}{2}\), n = – 5,
⇒ r + 1 = 6
⇒ r = 5
The general term in the binomial expansion of (1 + x)ⁿ is

Question 12.
Find the coefficient of x⁶ in (3x – \(\frac{4}{x}\))¹⁰.
Solution:
Given (3x – \(\frac{4}{x}\))¹⁰
Here, x = 3x; a = – \(\frac{4}{x}\), n = 10
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) (3x)^10-r \(\left(\frac{-4}{x}\right)^{\mathbf{r}}\)
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) 3^10-r x^10-r (- 4)^r x^r
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) 3^10-r . (- 4)^10-r . x^10-2r …………..(1)
To find the coefficient of x^-6,
put 10 – 2r = – 6
⇒ 2r = 16
⇒ r = 8
Now, substituting r = 8 in equation (1) we get
T₈₊₁ = \({ }^{10} \mathrm{C}_8\) 3^10-8 (- 4)⁸ x^10-16
T₉ = \({ }^{10} \mathrm{C}_8\) 3² (4)⁸ x^-6
The coeff. of x^-6 in the expansion of (3x – \(\frac{4}{x}\))¹⁰ is \({ }^{10} \mathrm{C}_8\) . 3² . (4)⁸

Question 13.
Find the middle term (s) in the expansion of \(\left(4 a+\frac{3}{2} b\right)^{11}\).
Solution:
Given \(\left(4 a+\frac{3}{2} b\right)^{11}\)
Here, x = 4a; a = \(\frac{3}{2}\) b; n = 11
Since, n = 11 is odd,
Middle terms = \(\frac{\mathrm{n}+1}{2}, \frac{\mathrm{n}+3}{2}\) = 6, 7 terms.

6th term :
r + 1 = 6
⇒ r = 5
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
6th term of given expansion ¡s
T₅₊₁ = \({ }^{11} C_5(4 a)^{11-5}\left(\frac{3}{2} b\right)^5\)
T₆ = \({ }^{11} \mathrm{C}_5(4 \mathrm{a})^6\left(\frac{3}{2} \mathrm{~b}\right)^5\)
= \(=11 C_5 \frac{4^6 \cdot 3^5}{2^5} \cdot a^6 \cdot b^5\)
T₆ = \({ }^{11} \mathrm{C}_6 \frac{2^{12} \cdot 3^5}{2^5}\) . a⁴b⁵
= \({ }^{11} \mathrm{C}_6\) . a⁶ . b⁵

7th term :
r + 1 = 7
⇒ r = 6
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
7th term of given expansion is

Question 14.
Find the middle term (s) in the expansion of (4x² + 5x³)¹⁷.
Solution:
Given (4x² + 5x³)¹⁷.
Here, x = 4x², a = 5x³, n= 17.
Since, n = 17 is odd, then
middle terms = \(\frac{n+1}{2}, \frac{n+3}{2}\) = 9, 10 terms

9th term :
r + 1 = 9
⇒ r = 8
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
∴ 9th term in this expansion is
T₉₊₁ = \({ }^{17} \mathrm{C}_9\)(4x²)^17-9 (5x³)⁹
T₁₀ = \({ }^{17} \mathrm{C}_9\) (4x²)⁸ (5x³)⁹
= \({ }^{17} \mathrm{C}_9\) 4⁸ . 5⁹ x¹⁶ . x²⁷
= \({ }^{17} \mathrm{C}_9\) 4⁸ . 5⁹ . x⁴³

Question 15.
Prove that 2 . C₀ + 5 . C₁ + 8 . C₂ + ……………… + (3n + 2) C_n = (3n + 4) 2^n-1
Solution:
Let S = 2 . C₀ + 5 . C₁ + 8 . C₂ + ……………… + (3n + 2) C_n = (3n + 4) 2^n-1 …………….(1)
By writing the terms in (1) in the reverse order
S = (3n + 2) . C_n + (3n – 1) . C_n-1 + (3n – 4) C_n-2 + ……………….. + 2. C₀
S = (3n + 2) C₀ + (3n – 1) C₁ + (3n – 4) C₂ + ……………… + 2 C_n ………….. (2)
Adding (1) and (2) we get,
S = 2 . C₀ + 5 . C₁ + 8 . C₂ + …………… + (3n + 2) . C_n
S = (3n + 2) . C₀ + (3n – 1) . C₁ + (3n – 4) . C₂ + ……………. + 2 . C_n
2S = (3n + 4)C₀ + (3n + 4) C₁ . (3n + 4) C₂ + …………… + (3n + 4) C_n
2S = (3n + 4) (C₀ + C₁ + C₂ + …………….. + C_n)
2S = (3n + 4) 2ⁿ
⇒ S = (3n + 4) 2^n-1
∴ 2 . C₀ + 5 . C₁ + 8. C₂ + …………… + (3n + 2) C_n = (3n + 4) . 2^n-1

Question 16.
Prove that
(C₀ + C₁) (C₁ + C₁) (C₂ + C₃) + ………………… + (C_n-1 + C_n) = \(\frac{(n+1)^n}{n !}\) . C₀ . C₁ . C₂ ……………. C_n

Question 17.
Write the first 3 terms in the expansion of \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\).
Solution:
Given \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\)
We know that,

Hence, the first three terms in the expansion of \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\) are 1, \(\frac{-5 x}{2}, \frac{15 x^2}{4}\).

Question 18.
Find the coefficient of x⁶ in the expansion of (1 – 3x)^-2/5.
Solution:
Given that (1 + 3x)^-2/5
The general term of (1 – 3x)^-2/5 is
T_r+1 = \(\frac{n(n-1)(n-2) \ldots .(n-r+1)}{r !}\)
where x = – 3x, n = \(\frac{-2}{5}\)

T_r+1 = \(\frac{\frac{-2}{5}\left(\frac{-2}{5}-1\right)\left(\frac{-2}{5}-2\right) \ldots \ldots . .\left(\frac{-2}{5}-r+1\right)}{r !}(-3 x)^r\)

= \(\frac{\frac{-2}{5}\left(\frac{-7}{5}\right)\left(\frac{-12}{5}\right) \cdots \cdots \cdots\left(\frac{3}{5}-\mathrm{r}\right)}{\mathrm{r} !}(-3 \mathrm{r})^{\mathrm{r}}\)

= \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots(5 \mathrm{r}-3)}{5^{\mathrm{r}} \cdot \mathrm{r} !}(3 \mathrm{x})^{\mathrm{r}}\)

Put r = 6
Then the coefficient of x^-6 is \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots .27}{5^6 \cdot 6 !}(3)^6\)
= \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots .27}{6 !} \cdot\left(\frac{3}{5}\right)^6\).

Question 19.
Eind the coefficient of x⁴ in the expansion of (1 – 4x)^3/5
Solution:
Given (1 – 4x)^3/5
The general term of (1 – 4x)^3/5 is
T_r+1 = \(\frac{n(n-1)(n-2) \ldots \ldots(n-r+1)}{r !}\) . x^r
where, x = – 4x, n = \(\frac{-3}{5}\)

Question 20.
Find the sum of the infinite series 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\ldots \ldots\). [TS – Mar. 2015]
Solution:
Let the given series is
S = 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\ldots \ldots\)
Now, comparing with

Question 21.
Find the sum of the infinite series 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + …………..
Solution:
Let the given series is
S = 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + …………..
Now comparing with

Question 22.
Find the sum of the infinite series \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\).
Solution:
Let the given series

Question 23.
If t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……………….. ∞, then prove that 9t = 16.
Solution:
Given

Here p = 4; p + q = 6;
⇒ q = 2
⇒ x = \(\frac{2}{5}\)
∴ (1) ⇒ 1 + t = \(\left(1-\frac{2}{5}\right)^{\frac{-4}{2}}\)
⇒ 1 +t = \(\left(\frac{3}{5}\right)^{-2}\)
⇒ 1 + t = \(\frac{25}{9}\)
⇒ 9t =16.

Question 24.
If I, n are positive integers, 0 < f < 1 and if (7 + 4√3)ⁿ = 1 + f, then show that
i) I is an odd integer and
ii) (I + f) (1 – f) = 1.
Solution:
Sol. Since I, n are positive integers 0 < F < 1 and (7 + 4√3)ⁿ = 1 + F
Now, 36 < 48 < 49
⇒ 6 < 4√3 < 7 ⇒ – 6 > – 4√3 > – 7
⇒ 1 > 7 – 4√3 > 0
⇒ 0 < 7 – 4√3 < 1
⇒ 0< 7 – 4√3 < 1
⇒ o < (7 – 4√3)ⁿ < 1
Let (7 – 4√3)ⁿ = f
∴ 0 < f < 1
Now, 1 + f = (7 + 4√3)ⁿ

where k is an integer
⇒ I + F + f = Even integer.
Since, I is an integer, F + f is an integer.
But, 0 < F < 1, 0 < f < 1
⇒ 0 < F + f < 2.
∴ f + F = 1
From (1)
⇒ I + 1 = Even integer
⇒ Even integer – 1 = Odd Integer
∴ I is an odd integer.

ii) L.H.S: (I + f) (I – f) = (7 + 4√3)ⁿ . F
= (7 + 4√3)ⁿ (7 – 4√3)ⁿ
= (49 – 48)ⁿ = 1ⁿ = 1 = R.H.S
∴ LH.S = R.H.S.

Question 25.
Find the 8th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\). [AP – Mar. ’18]
Solution:
Given \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\)
T₈ = \(\frac{\left(\frac{-3}{5}\right)\left(\frac{-3}{5}-1\right)\left(\frac{-3}{5}-2\right) \ldots\left(\frac{-3}{5}-6\right)}{7 !}\left(\frac{-5 x}{2}\right)^7\)
= \(\frac{\left(\frac{3}{5}\right)\left(\frac{8}{5}\right)\left(\frac{13}{5}\right) \ldots\left(\frac{33}{5}\right)}{7 !}\left(\frac{5 x}{2}\right)^7\)
= \(\frac{3 \times 8 \times 13 \times \ldots . \times 33}{7 !}\left(\frac{x}{2}\right)^7\)

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds

Very Short Answer Questions (2 Marks)

Question 1.
What are transition elements ? Give examples.
Answer:
Transition elements are defined as the elements having partially filled d-orbitals in the atoms in the elemental form or in the chemically significant stable oxidation states. Examples : Fe, Mn, Cr.

Question 2.
Which elements of 3d, 4d, and 5d series are not regarded as transition elements and why ?
Answer:
The elements Zn, Cd, Hg are not regarded as transition elements. They belong to 3d, 4d and 5d series respectively. Their general electronic configuration is (n – 1) d¹⁰ ns². The d-orbitals in these elements are completely filled. Therefore, they are not regarded as transition elements.

Question 3.
Why are d-block elements called transition elements ?
Answer:
The d-block elements are placed in between s and p-blocks in the periodic table. There is a transition in properties between the electropositive s-block elements and the electronegative p-block elements. Hence, d- block elements are named transition elements.

Question 4.
Write the general electronic configuration of transition elements.
Answer:
(n – 1) d¹⁰ ns^1-2
The (n – 1) stands for the penultimate shell. The (n -1) d orbitals may have one to ten electrons and the outermost ns orbital may have one or two electrons.

Question 5.
In what way is the electronic configuration of transition elements different from non-transition elements ?
Answer:
The transition elements have incompletely filled (n – 1) d orbitals.

In transition elements, the valence shell and penultimate shell are incompletely filled. Their general electronic configuration is (n- 1) d^{1 – 10} ns^{1 – 2}.

The non-transition elements i.e., the representative elements have the incompletely filled valence shell. The inner transition elements have the outer three energy levels incomplete.

Question 6.
Write the electronic configuration of chromium (Cr) and copper (Cu).
Answer:
Cr (24) – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d 4s¹ or [Ar] 4s¹ 3d⁵
Cu (29) – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ or [Ar] 4s¹ 3d¹⁰

Question 7.
Why do transition elements exhibits characteristic properties ?
Answer:
The transition elements are characterised by specific and special properties which can be explained on the basis of their chara-cteristic electronic configuration. The partly filled (n -1) d orbital is the cause of special properties like variable oxidation states, magnetic properties etc.

Question 8.
Scandium is a transition element. But zinc is not. Why ? [IPE 14]
Answer:
The atom of Sc has one unpaired d-electron. So it is a transition element. The electronic configuration of Zn is (Z = 30) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s².
The common oxidation state of Zn is +2.
The electronic configuration of Zn⁺⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰.
As neither Zn nor Zn⁺⁺ have partly filled 3d orbitals, Zinc is not considered as transition element.

Question 9.
Even though silver has d¹⁰ configuration, it is regarded as transition element Why ?
Answer:
Silver (Z = 47) can exhibit +2 oxidation state where in it will have incompletely filled d- orbitals (4d), hence it is a transition element.

Question 10.
Write the electronic configuration of Co²⁺ and Mn²⁺.
Answer:
Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ Or [Ar] 3d⁷
Mn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ Or [Ar] 3d⁵

Question 11.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state ?
Answer:
Mn²⁺ has stable configuration because 3d orbitals are half-filled. Hence it is stable.

Fe²⁺ has the configuration [Ar] 3d⁶. As it gains half-filled electronic configuration jt! by losing an electron it is susceptible to oxidation.

Question 12.
Which metal in the first series of transition metals exhibits + 1 oxidation state most frequently and why ?
Answer:
Copper
Reason : The second ionisation enthalpy , is unusually high. Removal of second electron disrupts the stable configuration of Cu⁺.

Question 13.
Why do transition elements exhibit more than one oxidation state (variable oxidation states)?
Answer:
Transition elements exhibit general Oxidation state of + II by loosing the electrons in the outer “s” orbital. They also Show other oxidation states because of the participation of (n – 1) d electrons in bonding.

Both ns and (n -1) d electrons are available for bonding because there is very little difference between the energies of ns & (n – 1) d orbitals.

Question 14.
Though Sc is a transition element, it does not exhibit variable oxidation state. Why ?
Answer:
As there is only one d-electron Sc does not show vairable oxidation states. Thus Sc (II) is unknown. Further d¹ configuration is unstable.

Question 15.
Why is it difficult to obtain M³⁺ oxidation state in Ni, Cu and Zn ?
Answer:
As the third ionisation enthalpies are quite high, it is difficult to obtain M⁺³ state in Ni, Cu and Zn.

Question 16.
Why is Cr²⁺ reducing and Mn³⁺ oxidizing even though both have the same d4 electronic configuration ?
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half- filled d_xy, d_yz, d_zx orbitals. On the other hand, the change from Mn²⁺ to Mn³⁺ results in the half-filled d⁵ configuration which has extra stability.

Question 17.
Although Cr, Mo and W belong to the same group (group 6) Cr (VI) is a strong oxidizing agent while Mo (VI) and W (VI) are not. Why ?
Answer:
In Group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr (VI) in the form of dichromate acidic medium is a strong oxidising agent, whereas MnO₃ and WO₃ are not.

Question 18.
What do you infer from the fact that M³⁺ / M²⁺ standard electrode potential of Mn is comparatively higher and that of Fe is comparitively lower ?
Answer:
The higher value for Mn shows that Mn²⁺ (d⁵) is particularly stable whereas comparatively low value for Fe shows the extra stability of Fe³⁺ (d⁵ configuration).

Question 19.
Transition elements have high melting points. Why ?
Answer:
The melting points of the metals are very high. The high values are attributed to the strong interactions present in the metals.

The involvement of greater number of electrons from (n – 1) d in addition to the ns electrons in the interatomic metallic bonding is the reason for high m.p.’s.

Question 20.
Among the first transition series (3d series) Chromium has highest melting point. Why ?
Answer:
In chromium, there is involvement of maximum number of (n -1) d electrons (d⁵) and one ns electron is interatomic metallic bonding. So the metallic bonding is strong. Hence the m.p is high.

Question 21.
Compared to s-block elements, the transition elements exhibit higher enthalpy of atomization. Why ?
Answer:
Because of large number of unpaired electrons in their atoms, they have stronger interatomic interaction. Hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Question 22.
Among the first transition series (3d series) zinc has lowest enthalpy of atomization. Why?
Answer:
As there are only two electrons available for metallic bond formation, the interatomic attractions are less. Hence the zinc has lowest enthalpy of atomisation.

Question 23.
How do you expect the density of transition element to vary in a given series and why ?
Answer:
Increase in density is expected from in the transition elements from Sc Z = 21 to copper Z = 29.
Reason :

1. Metallic bond strength increases due to increase in (n – 1) d electrons.
2. Metallic radius decreases.
3. Increase in atomic mass.
   As expected, the density increases from scandium to copper in the 3d series.

Question 24.
How do the atomic and ionic sizes vary among transition metals in a given series ?
Answer:
The atomic and ionic radii decrease slightly with the increase in a series of transition elements. This is because of poor screening by d-electrons.

New electron enters into a d orbital each time, nuclear charge increases by unity as we move from one element to the other in the transition series. As the shielding effect of a d-electron is not effective, the net attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

Question 25.
Why do Mn, Ni and Zn exhibit more negative E^⊖ values than expected ?
Answer:
The stability of the half-filled d-subshell in Mn²⁺ and completely filled d¹⁰ configuration in Zn⁺⁺ are related to their more negative E^⊖ values. The value is more for Ni because of its highest negative \(\Delta \mathrm{H}_{\text {hyd }}^{\ominus}\) .

Question 26.
Among the first transition series (3d series) only copper has positive E^⊖ M²⁺/M value. Why?
Answer:
Positive E^⊖ value means less ability to act as reducing agent. Thus Cu cannot liberate H₂ from acids.
Reason:

1. Cu has high value for enthalpy of atomisation ∆_aH^⊖.
2. Low value for enthalpy of hydration ∆_hydH^⊖
3. The high energy to transform Cu(s) to Cu⁺⁺ (aq) is not balanced by its hydration enthalpy.

Question 27.
Cu¹¹ forms halides like CuF₂, CuCl₂ and CuBr₂ but not Cul₂. Why ?
Answer:
Cu²⁺ oxidises I^– to I₂. Hence CuI₂ is not known.
2CU²⁺ + 4I²⁺ → Cu₂I₂ (s) + I₂

Question 28.
The highest Mil fluoride is MnF₄ where as the highest oxide is Mn₂O₇. Why ?
Answer:
Oxygen can form multiple bonds with metals. This explains the ability of oxygen to stabilise high oxidation states. In the covalent oxide Mn₂O₇ each Mn is tetra-hedrally surrounded by O’s including a Mn – O – Mn bridge. The ability to stabilise higher oxidation state is less for fluorine when compared to oxygen. Hence it can form MnF₄ Only.

Question 29.
In its fluoride or Oxide, in which a transition metal exhibits highest oxidation state and why ?
Answer:
Mn exhibits its highest oxidation state in Mn₂O₇, because of the ability of oxygen to form multiple bonds with metals. In the covalent oxide, Mn₂O₇ each Mn is tetrahedrally surrounded by oxygen including a Mn – O – Mn bridge.

Question 30.
Why Zn²⁺ is diamagnetic whereas Mn²⁺ is paramagnetic ? [TS 15]
Answer:
Electronic configuration of Zn⁺⁺ is [Ar] 3d¹⁰.
As there are no unpaired electrons, Zn⁺⁺ is diamagnetic.
Electronic configuration of Mn⁺⁺ is [Ar] 3d⁵. As there are unpaired electrons, Mn⁺⁺ is paramagnetic.

Question 31.
Write ‘spin only’ formula to calculate the magnetic moment of transition metal ions.
Answer:
Magnetic moment μ = \(\sqrt{n(n+2)}\)
n = number of unpaired electrons
m = magnetic moment in Bohr magnetons

Question 32.
Calculate the ‘spin only’ magnetic moment of Fe²⁺ _(aq) ion. [AP 17]
Answer:
Fe (26) Electronic configuration [Ar] 3d⁶.4s² Fe²⁺. Electronic configuration is [Ar] 3d⁶
No. of unpaired electrons is 4.
Spin only formula μ = \(\sqrt{n(n+2)}\)
= \(\sqrt{4 \times 6}=\sqrt{24}\) = 4.9 BM

Question 33.
What is meant by ‘disproportionation’ ? Give an example of disproportionation reaction in aqueous solution.
Answer:
If an element undergoes both oxidation and reduction, it is called disproportionation reaction.
Ex : Many copper (I) compounds are un-stable in aqueous solution and undergo disproportionation.
2 Cu⁺ → Cu⁺⁺ + Cu

Question 34.
Aqueous Cu²⁺ ions are blue in colour, where as Aqueous Zn²⁺ ions are colourless. [TS 16]
Answer:
Cu⁺⁺ contains unpaired electron. It is coloured due to d-d transition. Zn⁺⁺ does not contain unpaired electron. Hence it is colourless.

Question 35.
What are complex compounds ? Give examples.
Answer:
Complex compounds are those in which the metal ions bind a number of anions or neutral molecules.
Ex: [Fe (CN)₆]^3-, [Fe(CN)₆]^4-, [Cu (NH₃)₄]²⁺

Question 36.
Why do the transition metals form a large number of complex compounds ?
Answer:
Transition metals form a large number of complex compounds due to

1. smaller sizes of the metal ions,
2. high ionic charges,
3. availability of d-orbitals for bond formation.

Question 37.
How do transition metals exhibit catalytic activity ?
Answer:
Transition metals possess good catalytic properties. This property is due to the the free valencies of the metals and also due to variable oxidation states.

Question 38.
Give two reactions in which transition metals or their compounds acts as catalysts.
Answer:

1. Vanadium pentoxide is used as catalyst in the manufacture of sulphuric acid by Contact process.
   
2. Pt is used as catalyst in Ostwald process.
   

Question 39.
What is an alloy ? Give example.
Answer:
When metals are mixed and the resulting liquid is allowed to solidify, the product formed is called an alloy. It may also contain non-metals.
Ex: German silver 25 – 50% Cu,
10 – 30% Ni
25 – 35% Zn
Nichrome 60% Ni, 25% Fe, 15% Cr.
Bell metal 80% Cu, 20% Sn

Question 40.
Why do the transition metals readily form alloys ?
Answer:
As the crystal structures of the transition metals are similar, alloys are formed among them readily.

Question 41.
How does the ionic character and acidic nature vary among the oxides of first transition series?
Answer:
The oxides dissolve in acids and bases to form oxometallic salts. Potassium dichromate and potassium permanganate are common examples.

Question 42.
What is the efffect of increasing pH on a solution of potassium dichromate ?
Answer:
It is converted into chromate by alkali. Chromate is yellow in colour.
K₂Cr₂O₇ + 2KOH → 2K₂CrO₄ + H₂O

Question 43.
Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
6th group – K₂Cr₂O₇
Cr₂O₇^{– –} – is dichromate anion in which the oxidation number of Cr is + VI which is equal to its group number.
In permanganate anion \(\mathrm{MnO}_4^{-}\), Mn shows + VII state which is equal to its group number.

Question 44.
Permanganate titrations are carried out in the presence of sulphuric acid but not in presence of hydrochloric acid. Why ?
Answer:
The oxidation reactions of potassium per- mangante are generally carried out in presence of sulphuric acid. Hydrochloric acid should not be Used as the medium since it liberates chlorine by reaction with permanganate.
2KMnO₄ + 16HCl → 2MnCl₂ + 2KCl + 5Cl₂ ↑ + 8H₂O

Question 45.
What is lanthanoid contraction ? [T.S. Mar. 19]
Answer:
Gradual decrease in atomic and ionic sizes from La to Lu is called Lanthanoid con-traction.
Reason:

1. Imperfect shielding of one electron by another in the same set of orbitals (4f)
2. The shielding of one 4f electron by another is less than that of one d electron by another.
3. Nuclear charge increases along the series from La to Lu.
4. Due to poor shielding of 4f electrons, there is fairly regular decrease in the size of the entire 4f orbitals.

Question 46.
What are the different oxidation states exhibited by the lanthanoids ?
Answer:
Lanthanides react easily with water to give solutions giving + 3 ions. The principal oxidation state is + 3 although +4 and +2 oxidation states are also exhibited.

Question 47.
What is misch metal ? Give its composition and uses. [AP 16]
Answer:
Misch metal is an alloy of Lanthanoid metal.
Composition : 95% Lanthanoid metal and 5% iron and traces of S, C, Ca and Al.
It is used in making tracer bullets.

Question 48.
What is actinoid contraction ?
Answer:
The elements from Ac to Lr (Z = 89 to 103) are called actinides. In these elements the 5f orbital is gradually filled.

There is steady decrease in the size of M³⁺ and M⁴⁺ cations in actinide elements. The shielding of one electron in 5f orbital by another electron present in the same orbital is very poor. Due to this poor shielding effect, the increase in the nuclear charge by one unit brings the valence shell nearer to the nucleus and the size of the cation decreases.

Question 49.
What are coordination compounds ? Give two examples.
Answer:
Coordination compounds are those in which metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties.

The molecules or ions containing one pair of electrons donate the electron pair to the central metal ion to form coordinate covalent bond.
Ex: [Fe(CN)₆]^4-
Hexacyanoferrate (II)

Question 50.
What is a coordination polyhedron ?
Answer:
The spacial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. Most common coordination polyhedra are octahedral, square planar and tetrahedral.
For example, [Co(NH₃)₆]²⁺ is octahedral, Ni(CO)₄ is tetrahedral.

Question 51.
What is a double salt ? Give example.
Answer:
Those compounds which lose their identity in solution and break down into simple ions are called double salts. For example, an aqueous solution of potash alum K₂SO₄. Al₂ Al₂(SO₄)₃ . 24H₂O gives the test for K⁺, Al³⁺ and \(\mathrm{SO}_4^{2-}\) ions in its solution form.

Question 52.
What is the difference between a double salt and a complex compound ?
Answer:

Question 53.
What is a ligand ? [Mar. 2018, TS]
Answer:
A ligand is an ion or a molecule containing lone pairs of electrons which can be donated to a transition metal ion to form metal- ligand coordinate covalent bond.
Ex: NH₃, CN^–

Question 54.
Give one example each for ionic and neutral ligands.
Answer:
Ionic Ligands: Cyano CN^–; amino NH₂^–
Neutral Ligands : Carbonyl CO ; Ammine NH₃

Question 55.
How many moles of AgCl is precipitated when 1 mole of CoCl₃ is treated with AgNO₃ solution ?
Answer:
Three moles of AgCl is precipitated when 1 mole of COCl₃ is treated with AgNO₃ solution.

Question 56.
What is a chelate ligand ? Give example.
Answer:
A ligand which can form ring type complex is called chelate ligand.
Ethylenediamine H₂N – CH₂ – CH₂ – NH₂ is a chelate ligand. It can form ring type complex.

Question 57.
What is an ambidentate ligand ? Give example.
Answer:
Ligands which have two or more different donor sites but only one of these is attached to a single metal atom at a given time, are called ambidentate ligand.
Ex : NCS^– Thiocyanato M ← SCN isothio- cyanato M ← NSC

Question 58.
CuSO₄ . 5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why ?
Answer:
A) Colour of a complex is due to splitting of d-orbitals and d-d transition of the electron.

In the absence of ligand, crystal field splitting does not occur and hence the substance is colourless. Due to the absence of ligand, the water molecules, anhydrous CuSO₄ is colourless. CuSO₄ . 5H₂O is blue in colour due to the water molecules and the splitting of d-orbitals.

Question 59.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test for Fe²⁺ ion but CuSO₄ mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Why?
Answer:
FeSO₄ with (NH₄)₂SO₄ in 1 : 1 molar ratio forms a double salt FeSO₄. (NH₄)₂ SO₄ . 6H₂O which gives a test for all its constituent ions. On the other hand, CuSO₄ with aq. ammonia in 1 :4 molar ratio forms a coordination compound [Cu (NH₃)₄] SO₄. Hence it does not give test for Cu⁺⁺ ion.

Question 60.
How many geometrical isomers are poss-ible in the following coordination entities ?
i) [Cr(C₂O₄)₃]^3-
ii) [Co(NH₃)₃Cl₃]
Answer:
i) No geometric isomers are possible.
ii)

Another geometrical isomer occurs in [Ma₃b₃] type like [Co (NH₃)₃ Cl₃]. If three donor atoms of the same ligands occupy adjacent positions, at the corners of octa-hedral face we have the facial isomer (fac). When the positions occupied are around the meridian of the octahedron, we get the meridonial (mer) isomer.

Question 61.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution copper sulphate ? Why?
Answer:
When aq KCN is added to the aq. solution of CuSO₄, Octahedral complex of hexa cyan- idocuprate (10 is formed ,
CuSO₄ + 6KCN → K₄[CU(CN)₆] + K₂SO₄

Question 62.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Why ?
Answer:
In [Cr(NH₃)₆]³⁺, NH₃ is a weak field ligand and Cr has an electronic configuration of [Ar] 3d³ with three unpaired electrons. Hence [Cr(NH₃)₆]³ is paramagnetic. In [Ni(CN)₄]^2-, CN^– is a strong field ligand therefore Ni²⁺ ion has all the d-electrons paired. Hence, [Ni(CN)₄]^2- is diamagnetic.

Question 63.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₂]^2- is colourless. Why ?
Answer:
In [Ni(H₂O)₆]²⁺, Ni²⁺ ion has two unpaired electron and it shows colour.
In [Ni(CN)₄]^2-, there is no unpaired electron. Hence it is colourless.
A solution of [Ni (H₂O)₆]²⁺ is green but a solution of [Ni (CN)₄]^2- colourless.

[Ni(H₂O)₆]²⁺ is an outer orbital complex.
As it contains two unpaired electrons, it is paramagnetic and coloured.
In the formation of [Ni(CN)₄]^2-, electron pairing takes place in Ni²⁺.

Ni²⁺ undergoes dsp² hybridisation and it is square planar. As it does not contain unpaired electron, it is colourless.

Question 64.
[Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why ?
Answer:

1. Different ligands produce different crystal field splittings.
2. Different geometric fields also produce different crystal field splittings.
   As d – d transitions require different energies, the colours exhibited by [Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are different.

Question 65.
What is the oxdiation state of cobalt in [AP 16]
i) K₃[CO(CO₄)₃] and
ii) [Co(NH₃)₆]³⁺
iii) K[Co(CO)₄] ?
Answer:
i) Ox. number of of Co in K₃ [Co(C₂O₄)₃] is + 3.
ii) Ox. no. of Co in [Co (NH₃)₆]³⁺ is + 3.
iii) Ox. on of cobalt in K[Co(CO)₄] is -1.

Short Answer Questions (4 Marks)

Question 66.
Compared to 3d series the corresponding transition metals of 4d and 5d transition series show high enthalpy of atomization. Explain.
Answer:
The melting points and boiling points of the transition elements are generally very high.

In the second and third row elements M – M bonds are much more common. The Metal – Metal (M – M) bonding occurs not only in the metals themselves but also in some compounds. M – M bonding is quite rare in the first row transition elements.

Hence because of increasing in atomic weight and also strong metallic bonding, the enthalpies of 4d and 5d series are higher than those of 3d series.

Question 67.
Compared to the changes in atomic and ionic sizes of elements of 3d and 4d series, the change in radii of elements of 4d and 5d series is virtually the same. Comment.
Answer:
When we compare the changes in ionic and atomic sizes of 3d series with those of 4d and 5d series, there is an increase from the 3d to 4d series of elements. But the radii of the third (5d) series are virtually the same as those of the corresponding members of 4d series (second series).

This is because of the filling of 4f before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction. Because of lanthanide contraction, the expected increase in atomic size with increasing atomid number.

Question 68.
Account for the zero oxidation state of Ni and Fe in [Ni (CO)₄] and [Fe(CO)₅] respectively.
Answer:
Low oxidation are found when a complex compound has ligands capable of π-accep- tor character in addition to σ bonding. For example, Ni(CO)₄ and Fe(CO)₅, the oxidation of nickel and iron is zero. The metal- carbon-bond in metal carboyls possesses both σ and π character.

Question 69.
Why do the transition metal ions exhibit characteristic colours in aqueous solution. Explain giving examples.
Answer:
When an electron from a lower energy d- orbital of a metal ion in a complex is excited to a higher energy d-orbital of the same n value, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour absorbed corresponds to the complementary colour of the light absorbed. The frequency of light absorbed is determined by the nature of the ligand. In aqueous solutions, where water molecules are ligands
Ti³⁺ …………… purple
V⁴⁺ …………… blue
Cr³⁺ ……………. violet
Fe³⁺ ……………. yellow
Fe²⁺ ……………. green
To exhibit colour the ion should have at least unpaired (n -1) d electron. The colour is due to d – d transitions.

Question 70.
Explain the catalytic action of Iron (III) in the reaction between I^– and S₂O₈^2- ions.
Answer:
Transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states to form complexes. For example, iron (III) catalyses the reaction between iodide and persulphate ions.
2I^– + S₂O₈^2- → I₂ + 2SO₄^2-
An explanation of this catalytic action can be given as :
2Fe³⁺ + 2I^– → 2Fe²⁺ + I₂
2Fe²⁺ + S₂O₈^2- → 2Fe²⁺ + 2SO₄^2-

Question 71.
What are interstitial compounds ? How are they formed ? Give two examples. [AP 15]
Answer:
Interstitial compounds are those which are formed when small atoms like H, C, or N are trapped inside the crystal lattices of metals. They are usually non-stoichiometric and are neither typically ionic nor covalent.
Example: TiC, Mn₄N, VH_0.56 and Ti H_1.7 etc.

Question 72.
Write the characteristics of interstitial compounds.
Answer:

1. Interstitial compounds are non-stoichiometric.
2. They are neither typically ionic nor covalent.
3. They have high melting points, higher than those of pure metals.
4. They are very hard.
5. They retain metallic conductivity.
6. They are chemically inert.

Question 73.
Write the characteristics properties of transition elements. [AP ’15]
Answer:
The transition elements are characterised by specific and special properties which are explained on the basis of their characteristic electronic configurations. They are

1. They exhibit variable oxidation states.
2. Most of the transition elements and their ions show paramagnetic property. This is due to presence unpaired d-electron.
3. They form coloured compounds.
4. They have alloy forming ability.
5. Complex forming ability.

Question 74.
Write down the electronic configuration of
(i) Cr³⁺
(ii) Cu⁺
(iii) Co²⁺
(iv) Mn²⁺
Answer:
i) Electronic configuration of Cr³⁺ = [Ar] 3d³
ii) Electronic configuration of Cu⁺ = [Ar] 3d¹⁰
iii) Electronic configuration of Co²⁺ = [Ar] 3d⁷ 4s³
iv) Electronic configuration of Mn²⁺ =[Ar] 3d⁵

Question 75.
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³ 3d⁵ 3d⁸ and 3d⁴ ?
Answer:
Electronic configuration ………… Stable state
3d³ 4s² ………….. +2, and +5
3d⁵ 4s² ………….. +2, +7
3d⁸ 4s² …………. +2, +5
3d⁴ 4s² ………….. +3, +6

Question 76.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction ?
Answer:
There is gradual decrease in the atomic and ionic sizes of the lanthanide elements. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction. This is due to the imperfect shielding of one electron by another in the same set of orbitals. However, the shielding of one 4f electron by another is less than that of one d electron by another, and as the nuclear charge increases along the series, there is fairly regular decrease in the size of the entire 4fⁿ orbitals.

Consequences:

1. The net result of the lanthanoid contraction is that the second and the third series exhibit similar radii (e.g.: Zr 160 pm Hf 159 pm) and have very similar physical and chemical properties much more than the expected on the basis of usual family relationship.
2. The decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density of these elements.

Question 77.
How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with examples.
Answer:
The variability of oxdiation states, a characteristic property of transition elements arises due to incomplete filling of d orbitals in such a way that their oxdiation states differ from each other by unity.
Ex: V^II, V^III, V^IV, V^V.

But in non-transition elements the oxidation states normally differ by a unit of two.
Although in the p-block elements the lower oxidation states are favoured by the heavier members (due to inert pair effect.)

The opposite is true in the groups of d- block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr (VJ). Thus Cr (VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO₃ and WO₃ are not.

Question 78.
Describe the preparation of potassium dichromate from iron chromite ore.
Answer:
Dichromates are generally prepared from chromite. Chromites are obtained by the fusion of chromite ore. (FeCr₂O₄) with potassium carbonate in free access of air. The yellow solution of chromate is filtered and acidifed with sulphuric acid to give a solution from which orange coloured potassium dicromate can be crystallised.
4FeCr₂O₄ + 8K₂CO₃ + 7O₂→ 8K₂CrO₄ + 2Fe₂O₃ + 8CO₂
2K₂CrO₄ + 2H⁺ → K₂Cr₂O₇ + 2K⁺ + H₂O

Question 79.
Describe the oxidisingaction of potassium dichromate and write the ionic equations for its
With (i) iodide
(ii) iron (II) solution
(iii) H₂S and
(iv) Sn (II)
Answer:
In acidic solution the oxidising action of dicromate can be represented as follows.
Cr₂O₇^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O
It oxidises iodide to iodine.
Cr₂O₇^2- + 14H⁺ + 6I^– → 2Cr³⁺ + 3I₂ + 7H₂O
It oxidises H₂S to sulphur.
Cr₂O₇^2- + 14H⁺ + 3H₂S → 2Cr³⁺ + 6H⁺ + 3S + 7H₂O
It oxidises Sn²⁺ to Sn⁴⁺
Cr₂O₇^2- + 14H⁺ + 3Sn²⁺ → 3Sn⁴⁺ + 2Cr³⁺+ 7H₂O

Question 80.
Describe the preparation of potassium permanganate.
Answer:
Potassium permanganate is prepared by the fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. This produces dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

Question 81.
How does the acidified permanganate solution react with (i) iron (ii) ions (iii) SO₂ and (iv) oxalic acid.
Write die ionic equations fur the reactions.
Answer:
KMnO₄ oxidises FeSO₄ to Fe₂(SO₄)₃,
2KMnO₄ + 8H₂SO₄ + 10 FeSO₄ → K₂SO₄ + 2 MnSO₄ + 5 Fe₂(SO₄)₃ + 8H₂O
(or) 5 Fe²⁺ + \(\mathrm{MnO}_4^{-}\) + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

II) Sulphurous acid is oxidised to sulphate or sulphuric acid,
5 \(\mathrm{SO}_3^{2-}\) + 2 \(\mathrm{MnO}_4^{-}\) + 6H⁺ → 2Mn⁺⁺ + 3H₂O + 5SO₄^2-

III) It oxidises oxalic acid to CO₂.
2KMnO₄ + 3H₂ SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O – 10 CO₂
(or) 5\(\mathrm{C}_2 \mathrm{O}_4^{2-}\) + 2\(\mathrm{MnO}_4^{-}\) + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 82.
Predict which of the ions, Cu⁺, Sc³⁺, Mn²⁺, Fe²⁺ are coloured in aqueous solution ? Give reasons.
Answer:
Electronic configuration of Cu⁺ = [Ar ] 3d¹⁰ – colourless
Electronic configuration of Sc³⁺ = [Ar] colourless
Electronic configuration of Mn²⁺ = [Ar] 3d⁵ coloured
Electronic configuration of Fe²⁺ = [Ar] 3d⁵ coloured
Cu⁺ and Sc³⁺ are colourless because they do not have unpaired d-electrons.

Mn²⁺ and Fe²⁺ are coloured due to the presence of unpaired d-electron. Electronic transition takes from lower energy d-orbital to higher energy d-orbital with absorption of visible light. Complementary colour is observed.

Question 83.
Compare the stability of + 2 oxidation state of the elements of the first transition series.
Answer:
The standard values E^⊖ for M²⁺/M indicate a decreasing tendency to form divalent cations across the series.

E^⊖ values for Mn, Ni and Zn are more negative than expected from the general trend. This is due to the stability associated with half-filled d subshell (d⁵) in Mn²⁺ and completely filled d shell (d¹⁰) in Zinc.

For Nickel E^⊖ value is related to the highest negative value of enthalpy of hydration.

Many copper ©compounds are unstable in aqueous solution and undergo disporportionation.
2Cu⁺ → Cu⁺⁺ + Cu
The stability of Cu⁺⁺ (aq) rather than Cu⁺ (aq) is due to the much more negative ∆E_hyd of Cu²⁺ than Cu⁺. It compensates for the second ionisation enthalpy of Cu.

Question 84.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of ‘spin- only’ formula.
Answer:
Electronic configuration of Ce is [Xe] 4f¹ 5d¹ 6s².
Electronic configuration of Ce³⁺ is [Xe] 4f¹.
Magnetic moment = \(\sqrt{n(n+2)}=\sqrt{3}\) = 1.73 B.M.

Question 85.
Write down the number of 3d electrons in each of the following ions : Ti²⁺, V²⁺, Cr³⁺ and Mn²⁺. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Answer:

As H₂O is a weak ligand, electron pairing does not take place.

Question 86.
Explain Werner’s theory of coordination compounds with suitable examples. [AP 17; TS 15; IPE 14] [Mar. 2018, AP & TS]
Answer:
Werner proposed two types of valencies to explain the formation and the structure of these complex compounds.

i) The Primary valence:
This corresponds to the oxidation state of the central transition metal ion in the complex compounds. This primary valence is satisfied by a negative ion only in the complex compound.

ii) Secondary valence:
This is equal to the number of chemical groups bound to central metal ion. The no. of secondary valencies is called co-ordination number of the metal.

iii) These secondary valencies are directed in space which determines the shape of the complex.
Ex: 1
Coordination no.: 6 Structure: octahedral The ligand NH₃ satisfies only secondary valency.

Ex: 2
Coordination no.: 6 Structure: octahedral One Cl^– ion satisfies both primary and secondary valencies.

Ex:3
Co-ordination no.: 6 Structure: octahedral Two Cl^– ions satisfy both primary and secondary valencies.

Ex: 4
Co-ordination no.: 6 Structure: octahedral Three Cl^– ions satisfy both primary and secondary valencies.

In Werner’s representation, dotted lines represent primary valency whereas thick lines represent secondary valency.

Some negative ligands, depending upon the complex, may satisfy both primary and secondary valencies. Such ligands do not ionise.

iv) Primary valencies are ionisable vale-ncies. Secondary valencies are non-ion- isable valencies.

Question 87.
Give the geometrical shapes of the following complex entities.
(i) [CO(NH₃)₆]³⁺
(ii) [Ni(CO)₄]
(iii) [Pt Cl₄]^2- and
(iv) [Fe(CN)₆]^3-
Answer:
(i) [CO(NH₃)₆]³⁺ is octahedral with d²sp³ hybridisation.

(ii) [Ni(CO)₄] is tetrahedral with sp³ hybridisation.

(iii) [Pt Cl₄]^-2 is square planar with dsp² hybridisation.

(iv) [Fe (CN)₆]^3- is octahedral with d²sp³ hybridisation.

Question 88.
Explain the terms
i) Ligand
ii) Coordination number
iii) Coordination entity
iv) Central metal atom / ion.
Answer:
(i) Ligands:
The ions or molecules bound to the central atom / ion in the coordination entity are called ligands. Ligands are capable of donating electron pairs.
These may be simple ions such as Cl^–, small molecules such as H₂O or NH₃.

ii) The coordination number of a metal ion (CN^–) in a complex can be defined as the number of ligand donor atoms to which the metal is directly bonded. For example, in the complex ions [Pt Cl₆]^2- and [Ni (NH₃)₄]²⁺, the coordination number of Pt and Ni are 6 and 4 respectively.

Coordination number of the central atom/ion is determined only by the number of sigma bonds formed by the * ligand with the central atom/ion.

iii) Coordination entity :
A coordination entity constitutes, a central metal atom or ion bonded to a fixed number of ions or molecules. For example, [CoCl₃ (NH₃)₃] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions.

iv) Central atom/ion :
In a coordination entity, the atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement around it, is called the central atom or ion. In [NiCl₂(H₂O)₄] central ion is Ni²⁺. In [Fe(CN)₆]^2-, central metal ion is Fe²⁺.

Question 89.
Explain the terms
i) unidentate ligand
ii) bidentate ligand
iii) polydentate ligand and
iv) ambidentate ligand giving one example for each.
Answer:
i) Unidentate ligand :
When a ligand is bound to a metal ion through a single donor atom as with Cl^–, H₂O or NH₃ the ligand is said to be unidentate.

ii) Bidentate ligand :
When a ligand is bound to a metal ion through two donor atoms as in ethylene diamine.
H₂N CH₂CH₂NH₂, the ligand is said to be bidentate.

iii) Polydentate ligand:
When several donor atoms are present in a single ligand [N(CH₂CH₂NH₂)₃] the ligand is said to be polydentate.
EDTA Ethylenediamine tetraacetate ion is an important hexadentate ligand.

iv) Ambidentate ligand:
Ligand which can ligate through two different atoms is called ambidentate ligand.
Examples are – NO₂^– and SCN^– ions. NO₂^– ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion.

Question 90.
What is meant by chelate effect ? Give example.
Answer:
When a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. Such complexes are called chelate com-plexes. They are more stable than similar complexes containing unidentate ligands.

The complexes formed by chelating ligands are more stable than their mono- dentate analogs. The effect is called chelate effect. Ex: [Cu (en)₄]²⁺ is more stable than [Cu (NH₃)₄]²⁺ ion.

Question 91.
Give the oxidation numbers of the central metal atoms in the following complex entities.
(i) [Ni(CO)₄]
(ii) [Co(NH₃)₆]³⁺
(iii) (Fe(CN)₆]^4- and
(iv) [Fe(C₂O₄)₃]^3-
Answer:
i) Oxidation number of Ni in Ni(CO)₄ is zero.

ii) [Co(NH₃)₆]³⁺
Let the oxidation number of Co be x.
x + 0 = + 3
x = + 3

iii) [Fe(CN)₆]^4-
x – 6 = -4
x = 6 – 4 = +2
Ox. no. of Fe in [Fe(CN)₀]^4- is + 2

iv) [Fe(C₂O₄)₃]^3-
x – 6 = – 3 [C₂O₄ is bidenate ligand]
x = 6 – 3 = +3
Ox. no. of Fe in [Fe(C₂O₄)₃]^3- is + 3

Question 92.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate (II)
(ii) Hexaamminecobalt (III) sulphate
(iii) Potassium tetrachloropalladate (II) and
(iv) Potassium tri(oxalato) chromate (III)
Answer:
i) Tetrahydroxozincate (II) [Zn(OH)₄]^2-
ii) Hexammine cobalt (III) Sulphate [Co(NH₃)₆]₂(SO₄)₃
iii) Potassium tetrachlorido palladate (II) K₂[Pd Cl₄]
iv) Potassium trioxalato chromate (III) K₃ [Cr(C₂O₄)₃]

Question 93.
Using IUPAC norms write the systematic names of the following:
(i) [C0(NH₃)₆]Cl₃
(ii) [Pt(NH₃)₂Cl(NH₂ CH₃) Cl
(iii) [Ti(H₂O)₆]³⁺ and
(iv) [NiCl₄]^2-
Answer:
i) [C0(NH₃)₆]Cl₃
Hexammine cobalt (III) chloride

ii) [Pt(NH₃)₂Cl(NH₂ CH₃) Cl diammine chloro methyl amino platinium (II) chloride.

iii) [Ti(H₂O)₆]³⁺
Hexahydro titanium (III) ion

iv) [NiCl₄]^2-
Tetra chloro Nickelate (II) ion

Question 94.
Explain geometrical isomerism in coordination compounds giving suitable examples. [AP ’15]
Answer:
Stereo isomers have the same chemical for-mula and chemical bonds but they have different special arrangment.

Geometrical isomerism :
This type of iso-merism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.

In square planar complexes of formula MX₂L₂, the two ligands X may be arranged adjacent to each other in a cis isomer or opposite to each other in a trans isomer.

Octahedral complexes of formula [MX₂L₄] in which two ligands X may be oriented cis or trans to each other.

Tetrahedral complexes do not show geometrical isomerism because the rela-tive positions of the unidentate ligands attached to the central atom are the same respect to each other.

Question 95.
What are homoleptic and heteroleptic complexes ? Give one example for each.
Answer:
Complexes in which a metal is bound to only one kind of donor groups. Ex: [Co(NH₃)₆]³⁺, are known as homoleptic. Complexes in which a metal is bound to more than one. This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula [MX₂L₂] (X and L are unidentate), the two ligands X may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer as depicted in the figure.

Geometrical isomers (cis and trans) of [Co (NH₃)₄ Cl₂]⁺

Other square planar complex of the type MABXL (where A, B, X, L are unidentates) shows three isomers-two cis and one trans. Such isomerism is not possible for a tetrahedral geometry but similar behaviour is possible in octahedral complexes of formula [MX₂L₂] in which the two ligands X may be oriented cis or trans to each other.
Geometrical Isomers of (CoCl₂ (en)₂]

Kind of donor groups. Ex : [Co (NH₃)₄ Cl₂]⁺ are known as heteroleptic.

Long Answer Questions (8 Marks)

Question 96.
Explain giving reasons :
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition meals are high.
(iii) The transition metal generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalysts.
Answer:
i) Transition metals and many of their compounds show paramagnetic behaviour :
Diamagnetic substances are repelled by the applied magnetic field while the paramagnetic substances are attracted. Many of the transition metal ions are paramagnetic.

Paramagnetism arises from the presence of unpaired electrons. Each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.

The magnetic moment is determined by the number of unpaired electrons and is calculated by using the spin only formula size.
µ = \(\sqrt{n(n+2)}\)
Where n is the number of unpaired electrons and µ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons. (BM).

ii) The enthalpies of atomisation of the transition metals are high :
The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility. Their melting and boiling points are high. They have high enthalpies of atomisation the maxima at about the middle of each transition series indicate that one unpaired electron for d- orbital is particularly favourable for strong interatomic interaction. In general greater the number of valence electrons, stronger is the resultant bonding.

Metals with very high enthalphy of atomisation are noble in their reaction. The metals of the second and third series have higher values than the corresponding elements of first series. This accounts for the occurrence of more metal-metal bonding in compounds of the heavy transition metals.

Trends in enthalpies of atomisation of transition elements

iii) The transition metals generally form coloured compounds:
The electron from a lower energy d-orbital of a metal ion in a complex is exited to a higher energy d-orbital of the same n value, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the com-plementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand. In aqueous solutions where water molecules are the ligands the ions exhibit different colours.
Ex: Cr³⁺ violet
Fe²⁺ green
Fe³⁺ yellow

iv) Transition metals and their many com-pounds act as good catalysts :
The transition metals and their compounds are known for their catalytic activity. This activity is due to their ability to adopt multiple oxidation states and to form complexes.
Eg: V₂O₅ in contact process.
Ni in Hydrogenation of oils.

Catalysts at a solid surface involve the formation of bonds b/w reactant molecules and atoms of the surface of the catalyst. This has the affect of increasing the con-centration of the reactants at the catalyst surface.
Eg : Iron (III) catalyses the reaction between Iodide and per sulphate ions.
2I^– + S₂O₈^2- → I₂ + 2SO₄^2-
An explanation of the catalytic action can be given as
2Fe³⁺ + 2I → 2Fe²⁺ + I₂
2Fe²⁺ + S₂O₈^2- → 2Fe3+ + 2SO₄^2-

Question 97.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (I) Iron (“) Ions (II) SO₂ and (iii) oxalic acid? Write
the Ionic equatIons.
Answer:
Potassium permanganate is prepared by the fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. This produces dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

KMnO₄ oxidises FeSO₄ to Fe₂(SO₄)₃,
2KMnO₄ + 8H₂SO₄ + 10 FeSO₄ → K₂SO₄ + 2 MnSO₄ + 5 Fe₂(SO₄)₃ + 8H₂O
(or) 5 Fe²⁺ + \(\mathrm{MnO}_4^{-}\) + 8H⁺ → Mn²⁺ + 4H₂O
+ 5Fe³⁺
II) Sulphurous acid is oxidised to sulphate or sulphuric acid,
5 \(\mathrm{SO}_3^{2-}\) + 2 \(\mathrm{MnO}_4^{-}\) + 6H⁺ → 2Mn⁺⁺ + 3H₂O + 5SO₄^2-
III) It oxidises oxalic acid to CO₂.
2KMnO₄ + 3H₂ SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O – 10 CO₂
(or) 5\(\mathrm{C}_2 \mathrm{O}_4^{2-}\) + 2\(\mathrm{MnO}_4^{-}\) + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 98.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) oxidation state
(iii) atomic and ionic sizes and
(iv) chemical reactivity
Answer:

Question 99.
How would you account for the following:
(i) Of the d₄ species, Cr²⁺ is strongly reducing while manganese (III) is strongly oxidising.
(ii) Cobalt (U) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The d₁ configuration is very unstable in ions.
Answer:
i) Cr²⁺ is reducing as its configuration changes from d⁴ to d³. In d³ the t_2g level is half-filled. On the otherhand, the change from Mn²⁺ to Mn³⁺ results in the half- filled (d⁵) configuration which has extra stability.

ii) Cobalt (II) is stable in aqueous solution, but in the presence of complexing reagents, it is easily oxidised.

In the presence of complexing agents electron pairs takes place, and t2g orbital is completely filled.

Hence cobalt (II) is oxidised to cobalt (III).

Practically, all complexes of Co (+ III) have six ligands in an octahedral arrangements.
The metal has a d⁶ configuration and most of the ligands are strong enough to cause spin pairing, giving the electronic arrangement (t_2g)⁶ (eg)⁰. This arrangement has a very large crystal field stabilisation energy. Such complexes are diamagnetic.
Eg: [Co (NH₃)₆]³⁺
The one exception is [C₀F₆]^3- which is a high spin complex and is paramagnetic.

iii) The d’, configuration is very unstable in ions.
The d’, configuration has lower CFSE value. Hence very unstable.

Question 100.
Give examples and suggest reasons for the following features of the transition metals.
(i) The lowest oxide of transition metal is basic, the highest is amphoteric / acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Answer:
i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic:
Lowest oxidation compounds of transition metals are basic due to their ability to get oxidised to higher states. Whereas the higher oxidation state of metal and compounds gets reduced to lower ones and hence acts as acidic in nature.

ii) A transition metal exhibits highest oxidation state in oxides and fluorides:
Due to high electronegativities of oxygen and fluorine, the oxides and fluorides of transition metals exhibits highest oxidation state.

iii) In oxo anions of metals, the metals form bonds with oxygen and hence present in their highest oxidation states. For example, Cr forms CrO₄^2- and Cr₂O₇^2- both contain chromium in + 6 state. Permanganate ion, MnO₄^– contains Mn in its highest oxidation state of + 7.

Question 101.
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) eletronic configuration
(ii) oxidation states and
(iii) chemical reactivity
Answer:

Question 102.
Explain IUPAC nomenclature of coordination compounds with suitable examples.
Answer:
The rules which are to be followed while writing down the names of the coordinate compounds are –

1. The cationic part of the coordinate of compound Is written first followed by the names of the anion with a small gap.
2. Within a coordination sphere, ligands names are written in alphabetical order before the name of the central metal atom or ion without leaving any gap.
3. To indicate thenumber of ligands in the coordination sphere two types of prefixes are used-
   a) One are di-tri-tetra – etc. These are used when the ligands names are simple.
   b) If the ligands names are complex and contains the prefixes, like di-tri-tetra etc. in their names, then pre-fixes-bis, tris, tetrakis, etc. are used.
4. The names of anionic ligands (both organic and inorganic) ends with-O-. If a ligands name is already having prefix its name is enclosed in simple brackets.
   Neutral and cationic ligands names are written as such except – water (-aqua), NH₃ (-ammine), -Co (-Carbonyl) and NO (Nitrosyl).
5. The oxidation state of the central atom or ion is indicated in Roman numerates after the name of the central metal atom in simple brackets with no gap in between.
6. When coordination sphere is an anion then -ate is added as suffix to the name of the central metal atom / ion.
7. Prefixes cis and trans are used to desig-nate adjacent and opposite geometrical locations of the ligands in a complex.

Question 103.
Explain different types of isomerism exhibited by coordination compounds, giving suitable examples.
Answer:
Esorners are two or more compounds that have the same chemical formula but a different arrangement of atoms. Because of the different arrangement of atoms, they differ in one or more physical or chemical properties. Two principal types of Isomerism are known among coordination compounds. Each of which can be further subdivided.

a) Stereoisomerism.

1. Geometrical isomerism
2. Optical isomerism

b) Structural isomerism

1. Linkage isomerism
2. Coordination isomerism
3. Ionisation isomerism
4. Solvate isomerism

Stereo isomers have the same chemical formula and chemical bonds but they have different spatial arragement. Structural isomers have different bonds. A detailed account of these isomers are given below.
Geometric isomerism:

Geometrical isomers (cis and trans) of [Pt (NH₃)₂ Cl₂]
This type of isomerism also arises when bidentate ligands L – L (eg : NH₂ CH₂ CH₂ NH₂(en)] are present in complexes of formula [MX₂ (L – L)₂].

Another type of isomerism occurs in octahedral coordination entities of the type [Ma₃b₃] like [Co (NH₃)₃ (NO₂)₃]. If three donor atoms of the same ligands occupy adjacent positions at the corner of an octachedral face, we have facial (fac) isomer.

When positions are around the meridian of octahedron, we get the meridional isomer.

Optical isomerism:
Optical isomers are the mirror images that cannot be super imposed on one another. These are called as enantiomers. The molecules (or) ions that cannot be superimposible are called chiral. The two forms are called dextro & laevo forms depending on the direction
they rotate plane polarised light in a polarimeter. {d-rotates to right & / to left}.
Optical isomers (d and l) of [Co (en)₃]³⁺

Optical isomerism is commmon in octahedral complexes involving different ligands.

In a coordination utility of the [Pt Cl₂ [en]₂]²⁺ only cis-isomer shows optical activity.

Optical isomers of [d and l] of cis [Pt Cl₂ (en)₂]²⁺

Linkage isomerism :
Linkage isomerism arise in a coordination compound containing ambidentate ligand.

A simple example is provided by complexes containing the thiocyanate ligand, NCS^–, which may bind through the nitrogen to give M – NCS are through sulphur to give M-SCN.
Example : [Co(NH₂)₅ (NO₂)] C l₂

Which is obtained as the red form, in which the nitrite ligand is bond through oxygen (-ONO) and as the yellow form in which the nitrite ligand is bond through nitrogen (-NO₂).

Ionisation isomerism:
This type of isomerism is shown by such compounds which have same composition but literate different ions in solution. In such isomers, the positions of groups within or outside coordination sphere differs.

Coordination isomerism: This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
Ex : [CoCNH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [Co (CN)₆]

Question 104.
Discuss the nature of bonding and magnetic behaviour in the following coordination entities on the basis of valence bond theory.
(i) [Fe(CN)₆]^4-
(ii) [FeF₆]^3-
(iii) [Co(C₂O₄)₃]^3- and
(iv) [CoF₆]^3-
Answer:
i) [Fe(CN)₆]^4-:
In this complex ion, iron is in +2 oxidation state. The electronic configuration of Fe [Z = 26] is 4s² 3d⁶.

The configuration of Fe²⁺ is 3d⁶. Thus the ion has four vacant orbitals, one s and three p. In order to make six vacant orbitals available, the electrons in 3d orbital are forced to pair up. d² sp³ hybridisation takes place clearly, six pairs of electrons one from each CN- ion occupy six vacant hybrid orbitals.
i) [Fe(CN)₆]^4-

Geometry : The complex ion is octahedral. After the complex ion formation, no unpaired electron is left. Thus the given ion is diamagnetic.

ii) [FeF₆]^3- : Oxidation state of Fe = + 3 Fe³⁺ 3d⁵ 4s⁰

As six F^– ion approaches, no pairing of electron occur and the sp³d² hybridisation occurs accommodate six F^– ions.

Since five unpaired electrons are available, it is paramagnetic. Geometry is octahedral.

iii) [Co(C₂O₄)₃]^3-: Oxidation state of Co is +3

As didentate oxalate ions approach, pairing takes places.

All electrons are paired. The substance is diamagnetic octahedral.

iv) [CoF₆]^3-: Oxidation state of Co is + 3.

As there are four unpaired electrons, the complex is paramagnetic in nature.

Question 105.
Sketch the splitting of d-orbitals in an octahedral crystal field.
Answer:
The field created by the anionic ligands or due to the polarity of neutral molecules is called crystal field. The crystal field distorts the symmetry of the central metal atom / ion and splitting of d-orbitals takes places.

It the six ligands of octahedral complex are approaching the central metal atom/ion along the cartesian axes, initially the energy of the orbitals is increased. Then the orbital lying along the axis (d_{x^{2,/sup> – y2}} & d_z²) gets repelled strongly by the approaching crystal fields, then d_xy, d_yz, d_zx orbitals which lie in between the axes.

The energy of d_{x^{2,/sup> – y2}} & d_z² orbitals gets increase whereas energy of the other three gets lowered relative to the average energy of the spherical crystal field.

The lower energy set of d-orbitals is called t_2g and the higher energy set is called eg. The energy of separation is called crystal field stabilisation energy.

Question 106.
What is spectrochemical series ? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The magnitude of crystal field splitting ∆₀, depends upon the field produced by the ligand and charge on the metal ion. Some * ligands are able to produce strong fields in which case, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d orbitals.

In general, ligands can be arranged in a series in the order of increasing field strength as given below:
I^– < Br^– < S^2- < SCN^– < Cl^– < N^3-< F^– < OH^– < C₂O₄^2- < H₂O < NCS^– < NH₃ < en < CN^– < Co
Such a series is termed as spectra chemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands.

Ligands for which the crystal filed stabilisation energy is less than the pairing energy are called weak field ligands. These ligands result in the formation of high spin complexes.
∆₀ < p
Ligands for which the crystal field stabilisation energy is greater than the pairing energy are called strong field ligands. These ligands results in the formation of low spin complexes.
∆₀ >p
In oxoanions of metals, the metals form bonds with oxygen and hence are present in their highest oxidation states. For example, Cr form CrO₄^2- and Cr₂O₇^2-, both contains chromium ion + 6 oxidation state.
Permanganate ion, MnO₄^– contains Mn in its highest oxidation state of +7.

Question 107.
Discuss the nature of bonding in metal carbonyls.
Answer:
Metal carbonyls are formed by most of the transition metals. These carbonyls have simple well defined structure, tetra carbonyl nickel (O) is tetrahedral, pentacarbonyl- iron (O) is trigonal bipyramidal while hexa- carbonyl chromium (O) is octahedral. (O) indicates zero oxidation state.

The metal carbon bond in metal carboynyls posses both σ and π character. The MC σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M – C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.

Question 108.
Explain the applications of coordination compounds in different fields.
Answer:
1) Coordination compounds find use in many qualitative and quantitative chemical analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), d-nitroso-β-naphthol, cupron, etc.

2) Hardness of water is estimated- by simple titration with Na₂EDTA the Ca²⁺ and Mg²⁺ ions form table complexes with EDTA.

3) Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, combines with cyanide in the presence of oxygen and. water to form the coordination entity [Au(CN)₂] in aqueous solution. Gold can be separated in metal is form from this solution by the addition of Zinc.

4) Coordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin the red pigment of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B₁₂, Cyanocobalamine, the anti- pernicious anaemia factor, is a coordination compound of cobalt.

5) Coordination compounds are used as catalysts for many industrial processes. Examples include rhodium complex, [(Ph₃P)₃ RhCl], a Wilkinson catalyst, is used for the hydrogenation of alkenes.

6) Articles can be electroplated with silver and gold much more smoothly and evenly from solutions of the complexes, [Ag(CN)₂] and [Au(CN)₂] than from a solution of simple metal ions.

7) In black and white photography, the developed film is fixed by washing with hyposolution which dissolves the undercomposed Ag Br to form a complex ion, [Ag (S₂O₃)₂]³ EDTA is used in the treatment of lead poisoning.

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Definite Integrals Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Definite Integrals Important Questions Short Answer Type

Question 1.
Evaluate \(\int_0^{\pi / 2} \frac{a \sin x+b \cos x}{\sin x+\cos x} d x\). [(AP) May ’17]
Solution:

Question 2.
Evaluate \(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\). [(TS) May ’17]
Solution:

Question 3.
Evaluate \(\int_0^\pi \frac{1}{3+2 \cos x} d x\)
Solution:

Question 4.
Evaluate \(\int_{-3}^3\left(9-x^2\right)^{3 / 2} d x\)
Solution:

Question 5.
Evaluate \(\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2} d x\). [(TS) May ’17]
Solution:

Question 6.
Evaluate \(\int_0^4\left(16-x^2\right)^{5 / 2} d x\). [(AP) May ’19]
Solution:
Put x = 4 sin θ
dx = 4 cos θ dθ

Question 7.
Evaluate \(\int_0^2\left(x^2+1\right) d x\) as the limit of a sum.
Solution:

Question 8.
Evaluate \({Lt}_{n \rightarrow \infty} \sum_{i=1}^n \frac{i^3}{i^4+n^4}\)
Solution:

Question 9.
Evaluate \(\underset{n \rightarrow \infty}{L t}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \ldots\left(1+\frac{n}{n}\right)\right]^{1 / n}\)
Solution:

Question 10.
Evaluate \({Lt}_{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \cdots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}\)
Solution:

Question 11.
Find the area bounded between the curves y² = 4x, y² = 4(4 – x). [(TS) May ’19, ’11]
Solution:
Given curves are y² = 4x
⇒ y = 2√x ………(1)
y² = 4(4 – x) ………(2)
⇒ y = \(\sqrt{4(4-x)}\)
Solving (1) and (2)
4x = 4(4 – x)
⇒ x = 4 – x
⇒ 2x = 4
⇒ x = 2
from (1) ⇒ y = ± 2√2
Points of Intersections of (1) and (2) are A = (2, 2√2) and B = (2, -2√2)

Question 12.
Find the area enclosed between y = x² – 5x and y = 4 – 2x.
Solution:
Given curves are
y = x² – 5x ………(1)
y = 4 – 2x ………..(2)

Solving (1) and (2)
x² – 5x = 4 – 2x
⇒ x² – 3x – 4 = 0
⇒ (x – 4) (x + 1) = 0
∴ x = 4 and x = -1

Question 13.
Find the area enclosed between the curves y = 4x – x², y = 5 – 2x. [(TS) Mar. ’16]
Solution:
Given curves are
y = 4x – x² ………(1)
y = 5 – 2x ………(2)
Solving (1) and (2)
4x – x² = 5 – 2x
⇒ x² – 6x + 5 = 0
⇒ x² – 5x – x + 5 = 0
⇒ x(x – 5) – 1(x – 5) = 0
⇒ x = 1 or 5

Question 14.
Find the area between the parabolas y² = 4x and x² = 4y. [(AP) Mar. ’20; (TS) ’17; May ’14]
Solution:
Given equations of curves are
y² = 4x ………(1)
and x² = 4y ……..(2)
Solving (1) and (2) the points of intersection can be obtained.
y² = 4x
⇒ y⁴ = 16x²
⇒ y⁴ = 64y
⇒ y = 4
∴ 4x = y²
⇒ 4x = 16
⇒ x = 4

Points of intersection are (0, 0) and (4, 4).
∴ The area bounded between the parabolas

Question 15.
Evaluate \(\int_0^2 e^x d x\) as the limit of the sum.
Solution:

Question 16.
Evaluate \(\int_0^4\left(x+e^{2 x}\right) d x\) as the limit of a sum.
Solution:

Question 17.
Evaluate \(\int_0^{16} \frac{x^{1 / 4}}{1+x^{1 / 2}} d x\)
Solution:
L.C.M of 2, 4 is 4
Put x = t⁴
⇒ dx = 4t³ dt
L.L: x = 0 ⇒ t = 0
U.L: x = 16 ⇒ t = 2

Question 18.
Evaluate \(\int_0^{\pi / 4} \log (1+\tan x) d x\). [(AP) Mar. ’19, ’16; May ’18, (TS) ’16]
Solution:

Question 19.
Evaluate \(\underset{n \rightarrow \infty}{L t} \frac{1}{n}\left[\tan \frac{\pi}{4 n}+\tan \frac{2 \pi}{4 n}+\ldots \ldots+\tan \frac{n \pi}{2 n}\right]\)
Solution:

Question 20.
Find \({Lt}_{n \rightarrow \infty}\left(\frac{n !}{n^n}\right)^{1 / n}\)
Solution:

Question 21.
Evaluate \(\int_0^5 x^3\left(25-x^2\right)^{7 / 2} d x\)
Solution:
Put x = 5 sin θ, then dx = 5 cos θ dθ
L.L: x = 0 ⇒ θ = 0
U.L: x = 5 ⇒ θ = \(\frac{\pi}{2}\)

Question 22.
Evaluate \(\int_0^2 x \sqrt{2-x} d x\)
Solution:

Question 23.
Find \(\int_0^1 x^{3 / 2} \sqrt{1-x} d x\)
Solution:
Put x = sin²θ then dx = 2 sin θ cos θ dθ
Lower limit: x = 0 ⇒ θ = 0
Upper limit: x = 1 ⇒ θ = \(\frac{\pi}{2}\)

Question 24.
Evaluate \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Solution:
Put x = tan θ, then θ = tan^-1x
dx = sec²θ dθ
L.L: x = 0 ⇒ θ = 0
U.L: x = 1 ⇒ θ = \(\frac{\pi}{4}\)

Question 25.
Evaluate \(\int_0^1 x \tan ^{-1} x d x\)
Solution:

Question 26.
Find the area enclosed by the curves y = 3x and y = 6x – x².
Solution:
Given curves are
y = 3x ……..(1)
y = 6x – x² ………(2)

Solving (1) and (2)
3x = 6x – x²
⇒ x² = 3x
⇒ x(x – 3) = 0
⇒ x = 0 and x = 3

Question 27.
Find the area bounded between the curves y = 4x – x², y = 5 – 2x.
Solution:
Given curves are
y = 4x – x² ……..(1)
y = 5 – 2x ……..(2)

Solving (1) and (2)
4x – x² = 5 – 2x
⇒ x² – 6x + 5 = 0
⇒ (x – 1) (x – 5) = 0
⇒ x = 1, x = 5

Question 28.
Find the area bounded between the curves y = 2 – x, y = x². [Mar. ’01]
Solution:
Given curves are
y = 2 – x² ……(1)
y = x² ………(2)

Question 29.
Find the area of the region enclosed by the curves y = sin x, y = cos x, x = 0, x = \(\frac{\pi}{2}\).
Solution:

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• Chapter 4 Awake
• Chapter 5 Fear

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Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 8th Lesson Polymers Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 8th Lesson Polymers

Very Short Answer Questions (2 Marks)

Question 1.
Define the terms monomer and polymer.
Answer:
Monomer : The repeating structural unit of a macromolecule is called monomer.
Polymer : A polymer is defined as a very large molecule having high molecular mass.

Question 2.
What are polymers ? Give example.
Answer:
Polymers are very large molecules having high molecular mass which are formed by linking together repeating units of small molecules called monomers.
Ex: Polythene.

Question 3.
What is polymerisation? Give an example of polymerisation reaction. maim
Answer:
The process of formation of a polymer from its monomer(s) is called polymerisation.
Ex: The formation of polythene from ethene.

Question 4.
Give one example each for synthetic and semisynthetic polymers.
Answer:
Polythene is an example for synthetic polymers.
Cellulose acetate (rayon) is an example for semi synthetic polymers.

Question 5.
How are the polymers classified on the basis of structure ?
Answer:
On the basis of structure, polymers are clas-sified into 1) linear polymers, 2) branched chain polymers and 3) cross linked or net-work polymers.

Question 6.
Give one example each for linear and branched chain polymers.
Answer:
Polythene is an example for linear polymers. Low density polythene is an example for branched chain polymers.

Question 7.
What are cross linked (or network) polymers ? Give example.
Answer:
Polymers formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains are called cross linked (or network) polymers.
Ex: Bakelite.

Question 8.
What is addition polymer ? Give example.
Answer:
A polymer formed by the repeated addition of monomer molecules possessing double or triple bonds is called an addition polymer.
Ex: Polythene formed from ethene.

Question 9.
What is condensation polymer ? Give example.
Answer:
A polymer formed by the repeated condensation reaction between two different bi-functional or tri-functional monomeric species is called a condensation polymer.
Ex : Nylon 6, 6.

Question 10.
What are homopolymers ? Give example.
Answer:
Addition polymers formed by the polymerisation of a single monomeric species are called homopolymers.
Ex: Polythene.

Question 11.
What are copolymers ? Give example. [IPE 14]
Answer:
Polymers formed by addition polymerisation of two different monomeric species are called copolymers.
Ex : Buna – S.

Question 12.
Is -[-CH₂-CH(C₆H₅)-]_n– a homopolymer or a copolymer ?
Answer:
Polystyrene, -[-CH₂-CH(C₆H₅)-]_n– is a homopolymer.

Question 13.
Is -(NH-CHR-CO)_n – a homopolymer or a copolymer ?
Answer:
-(NH-CHR-CO)_n – is a homopolymer.

Question 14.
What are the classes of the polymers based on molecular forces ?
Answer:
Based on molecular forces polymers are classified into

1. Elastomers,
2. Fibres,
3. Thermoplastic polymers and
4. Thermosetting polymers.

Question 15.
What are elastomers ? Give example.
Answer:
Elastomers are rubber like solids with elastic properties.
Ex : Buna – S.

Question 16.
What are fibres ? Give example.
Answer:
Fibres are the thread forming solids which possess high tensile strength and high modulus.
Ex: Terylene.

Question 17.
What are thermoplastic polymers ? Give example.
Answer:
Thermoplastic polymers are the linear or slightly branched long chain molecules capable of softening on heating and harden-ing oil cooling.
Ex: Polystyrene.

Question 18.
What are thermosetting polymers ? Give example.
Answer:
Thermosetting plastics are cross linked or heavily branched molecules which on heat-ing undergo extensive cross linking in moulds and again become infusible.
Ex: Bakelite.

Question 19.
Write the name and structure of one of the common initiators used in free radical polymerisation reaction.
Answer:
Benzoyl peroxide is a common initiator used in free radical polymerisation reaction.

Question 20.
How can you differentiate between addition and condensation polymerisation ?
Answer:
Addition polymerisation involves repeated addition of monomer molecules possessing double or triple bonds where as condensation polymerisation involves repeated con-densation between two different bi-functio-nal or tri-functional monomeric species.

Question 21.
What is Ziegler – Natta catalyst ?
Answer:
Trialkyl aluminium and titanium tetra chloride is called Ziegler – Natta catalyst.

Question 22.
How is Dacron obtained from ethylene glycol and terepthalic acid ?
Answer:
Dacron (Terylene) is obtained by the con. densation polymerisation of ethylene glycol and terephthalic acid.

Question 23.
What are the repeating monomeric units of Nylon 6 and Nylon 6,6 ? [Mar. 18, 15 ; TS]
Answer:
Name of the polymer
Nylon 6
Nylon 6, 6

Repeating unit
Caprolactum
Hexa methylene
diamine and
adipic acid

Question 24.
What is the difference between Buna – N and Buna – S ?
Answer:
Buna – S is obtained by the copolymerisation of 1, 3 – butadiene and styrene whereas Buna – N is obtained by the copolymerisation of 1, 3 butadiene and acrylonitrile. Buna – N is more resistant to solvents but less abrasion resistant them Buna – S.

Question 25.
Arrange the following polymers in increasing order of their molecular forces.
i) Nylon 6, 6, Buna – S, Polythene
ii) Nylon 6, Neoprene, Polyvinyl chloride
Answer:
i) Buna – S < Polythene < Nylon 6, 6
ii) Neoprene < Polyvinyl chloride < Nylon 6

Question 26.
Identify the monomer in the following polymeric structures.
i) -[-C-(CH₂)₈ – C – NH – (CH₂)₆ – NH – ]-
ii) -[-NH – CO – NH – CH₂ -]_n –
Answer:
i) HOOC – (CH₂)₈ – COOH + H₂N – (CH₂)₆ – NH₂
ii) H₂N – CO – NH₂ + HCHO

Question 27.
Name the different types of molecular masses of polymers.
Answer:

1. Number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{n}}\))
2. Weight average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\))

Question 28.
What is PDI (Poly Dispersity Index) ? [AP Mar. 19]
Answer:
The ratio between weight average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\)) and number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\)) of a polymer is called Poly Dispersity Index (PDI).
PDI = \(\frac{\overline{\mathrm{M}}_{\mathrm{w}}}{\overline{\mathrm{M}}_{\mathrm{n}}}\)

Question 29.
What is vulcanization of rubber ? [TS Mar. 19; (TS 16, 15)]
Answer:
Vulcanization is the process of heating raw rubber (latex) with sulphur and an appropriate additive at a temperature range between 373 K to 415 K.

Question 30.
What is the cross linking agent used in the manufacture of tyre rubber ?
Answer:
5% of sulphur is used as the cross linking agent in the manufacture of tyre rubber.

Question 31.
What is biodegradable polymer? Give one example of a biodegradable polyester ? [Mar. 2018-AP]
Answer:
Polymers which undergo environmental degradation are called biodegradable polymers.
Ex : Nylon 2 – nylon 6.

Question 32.
What is PHBV ? How is it useful to man ? [AP Mar. 19; (Mar. 18-TS) AP 16 ; TS 16, 15]
Answer:
PHBV stands for Poly β-hydroxybutyrate – co-β-hydroxy valerate. It is used in speciality packaging, orthopaedic devices and in controlled release of drugs.

Question 33.
Give the structure of Nylon 2 – nylon 6 ?
Answer:
The structure of Nylon 2 – Nylon 6 is

Short Answer Questions (4 Marks)

Question 34.
Classify the following into addition and condensation polymers.
i) Terylene
ii) Bakelite
iii) Polyvinyl chloride
iv) Polythene,
Answer:
i) Terylene is a condensation polymer.
ii) Bakelite is a condensation polymer.
iii) Polyvinyl chloride (PVC) is an addition polymer.
iv) Polythene is an addition polymer.

Question 35.
How do you explain the functionality of a polymer?
Answer:
The functionality of a polymer depends upon its unique mechanical properties like tensile strength, elasticity, toughness etc. These mechanical properties are governed by intermolecular forces e.g., van der Waals forces and hydrogen bonds present in the molecule. These forces also bind the polymer chains.

Question 36.
Distinguish between the terms homopolymer and copolymer. Give one example of each.
Answer:
Homopolymers are addition polymers formed by the polymerisation of a single monomeric species whereas copolymers are addition polymers formed by the polymerisation of two different monomeric species.
Ex: Polythene is an example for homopolymers.
Buna – S is an example for copolymers.

Question 37.
Define thermoplastics and thermosetting polymers with two examples of each.
Answer:
Thermoplastics :
Thermoplastics are the linear or slightly branched long chain molecules capable of softening on heating and hardening on cooling.
Examples: Polythene and polystyrene.

Thermosetting polymers:
Thermosetting polymers are the cross linked or heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible.
Examples: Bakelite and urea-formaldehyde resins.

Question 38.
Explain copolymerization with an example.
Answer:
Copolymerisation is a polymerisation reaction in which a mixture of different types of monomeric species are allowed to polymerise and form a copolymer.
Ex; Buna – S, copolymer is formed when a mixture of 1,3 – Butadiene and styrene is allowed to polymerise.

Question 39.
Explain free radical mechanism for the polymerization of ethene.
Answer:
Polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide initiator. Phenyl free radical is formed by benzoyl peroxide. The polymerisation process starts with the addition of phenyl free radical to the ethene double bond thus generating a new and larger free radical.

This step is called chain initiating step. This radical then reacts with another molecule of ethene and the repetition of this sequence carries the reaction forward. Ultimately at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step.
Chain initiation step:

Question 40.
Write the names and structures of the monomers used for getting the following polymers. [AP 16; IPE 14]
i) Polyvinyl chloride
ii) Teflon
iii) Bakelite
iv) Polystyrene.
Answer:

Question 41.
Write the names and structures of the monomers of the following polymers
i) Buna – S
ii) Buna – N
iii) Dacron
iv) Neoprene.
Answer:

Question 42.
What is natural rubber? How does it exhi-bit elastic properties ?
Answer:
Natural rubber is a polymer and exhibits elastic properties.

Natural rubber may be considered as a linear polymer of isoprene and is also called cis-1,4 – polyisoprene. The cis-polyisoprene molecule consists of various chains held together by weak van der Waal’s forces and has colloidal structure. Thus it stretches like a spring and exhibits elastic properties.

Question 43.
Explain the purpose of vulcanization of rubber.
Answer:
Many properties of natural rubber limit its use. It becomes soft at high temperature (> 335 K) and brittle at low temperatures (< 283 K). It shows water absorption. It is soluble in non-polar solvents and is non- resistant to attack by oxidising agents. The purpose of vulcanization is to improve upon these physical properties.

Question 44.
Explain the difference between natural rubber and synthetic rubber.
Answer:
Natural rtibber is harvested from rubber trees. Synthetic rubber is man made.

Synthetic rubbers have same properties of natural rubber including being water proof and elastic, but they have same improved properties also – they are tougher more fle-xible and more durable than natural rubber. Natural rubber is soluble in non-polar sol-vents and is non-resistant to attack by oxid-ising agents. Synthetic rubber is resistant to the action of solvents and oxidising agents.

Question 45.
How does the presence of double bonds in rubber molecules influence their structure and reactivity ?
Answer:
The double bonds in rubber polymer provide the reactive sites and also determine the configuration of the polymer. Vulcanisation takes place at these reactive sites and sulphur forms cross links at these sites. Thus rubber gets stiff.

Question 46.
What are LDP and HDP ? How are they formed ?
Answer:
LDP and HDP stands for Low Density Polythene and High Density Polythene respectively.

LDP is obtained by the polymerisation of ethene under high pressure of 1000 to 2000 atmospheres at a temperature of 350 to 570 K in the presence of traces of peroxide initiator.

HDP is obtained by the addition polyme-risation of ethene in a hydrocarbon solvent in the presence of a catalyst such as triethyl aluminium and titanium tetrachloride (Ziegler – Natta catalyst) at a temperature of 333 K to 343 K and under a pressure of 6 – 7 atmospheres.

Question 47.
What are natural and synthetic polymers ? Give two examples of each type.
Answer:
Polymers obtained from natural sources such as plants and animals are called natural polymers.
Ex: Cellulose, rubber.

Man-made polymers are called synthetic polymers.
Ex : Polythene, Buna – N.

Question 48.
Write notes on different types of molecular masses of polymers.
Answer:
A polymer sample contains chains of varying lengths and hence its molecular mass is always expressed as an average. The average molecular masses of polymers are expressed in different ways. The important among them are:

1. Number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{n}}\))
2. Weight average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\) )

Number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{n}}\)) :
Polymer may be thought of as a mixture of molecules of same chemical type but of different masses. The particles of the polymer may be monomers, dimers or…. polymers. Let us suppose that there are N₁ particles each of mass M₁. Similarly there are N₂ particles each of mass M₂ …. and let N₁ be the number of particles each of mass M₁. Then
Total mass of the polymer sample
= M₁N₁ + M₂N₂ + M₃N₃ + …………. + M_iN_i
= \(\sum_{n=1}^{\mathrm{i}}\) M_iN_i
Total number of particles in the system = N₁ + N₂ + N₃ + …………….. + N_i
= \(\sum_{n=1}^{\mathrm{i}}\) N_i
Number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{n}}\)

The number average molecular mass depends upon the number of molecules present in the polymer.

Weight average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\)) :
The weight average molecular mass is calculated as follows.

The molecular mass of each type of particle is multiplied by the contribution of the species to the total weight of the sample. The product is calculated for each of the species present. The sum of the products of the species present in the sample is known as the weight average molecular mass of the polymer.

Let there be N_i particles each of mass M_i. Then
Total weight of all the particles
= M₁N₁ + M₂N₂ + ………….. + M_iN_i = \(\sum_{n=1}^{\mathrm{i}}\) M_i N_i
Mass of Nj particles each of mass M₁ = N₁M₁
The fraction of the total mass contributed by each particle of this type = \(\frac{M_1 N_1}{\sum_{n=1}^i M_i N_i}\)
This fraction is multiplied by the molecular mass M₁. We get
\(\frac{M_1 N_1}{\sum_{n=1}^i M_i N_i} \times M_1=\frac{N_1 M_1^2}{\sum_{n=1}^i N_i M_i}\)
Similarly this can be worked out for other species also. Sum of the products of the molecular mass and the fraction of the total mass of the respective species gives the weight average molecular mass.
\(\overline{\mathrm{M}}_{\mathrm{w}}\) = \(\frac{\sum_{n=1}^i N_i M_i^2}{\sum_{n=1}^i N_i M_i}\)
Note : Both \(\overline{\mathrm{M}}_{\mathrm{n}}\) and \(\overline{\mathrm{M}}_{\mathrm{w}}\) have no units.

Long Answer Questions (8 Marks)

Question 49.
Write an essay on
i) Addition polymerization and
ii) Condensation polymerization.
Answer:
Addition polymerisation : The type of polymerisation in which the molecules of the same monomer or different monomers containing double or triple bonds add together on a large scale to form a polymer is called addition polymerisation. The addition polymerisation takes place through the formation of either free radicals or ionic species.

The addition polymerisation is broadly classified into two types depending on the nature of chain carrier i.e., ionic polymerisation and free radical polymerisation.
Ionic Polymerisation: Ionic polymerisation is of two types.

1. Cationic polymerisation and
2. Anionic polymerisation.

In cationic polymerisation a positive ion (or cation) is used as the chain initiator. In anionic polymerisation a negative ion (or anion) is used as chain initiator.

i) Cationic Polymerisation:
Lewis acids such as BF₃, AlCl₃ or SnCl₂ act as chain initiators. An example for cationic polymerisation is the formation of a polyvinyl compound from its monomer. The following steps are involved.
1) Chain initiating step :

ii) Anionic Polymerisation:
Sodium in liquid ammonia, alkyl lithium compounds etc., are used as chain initiators. A negative ion or group is added to the monomer molecule. Formation of vinyl polymers is an example for anionic polymerisation.
Chain initiation step:

In anionic polymerisation chain terminating step is absent.

Free Radical Polymerisation:
Several alkenes or dienes and their derivatives are polymerised in the presence of a free radical generating initiator like benzoyl peroxide, acetyl peroxide etc.

Polymerisation of ethene to polythene is an example for free radical polymerisation. A mixture of ethene with a small amount of benzoyl peroxide initiator is heated or exposed to light.

The process starts with the addition of a free radical generated by the peroxide to the ethene double bond. A new and larger free radical is generated. This step is called the chain initiating step. This radical reacts with another molecule of ethene generating a bigger radical.

The repetition of this sequence carries the reaction forward. This step is called the chain propagating step. Finally, at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step.
Chain initiation step:

Chain termination step : Free radicals combine in different ways to form polythene.

Condensation polymerisation :
The type of polymerisation in which monomers combine together with the loss of simple molecules like H₂O, NH₃ etc., is called condensation polymerisation. The polymer thus formed is called condensation polymer.
Ex : Nylon – 6, 6 is formed from hexamethylene diamine and adipic acid by condensation.

Question 50.
Explain the classification of polymers based on their source and structure.
Answer:
Classification based on source :
Based on source, polymers are classified into

1. Natural polymers,
2. Semi-synthetic polymers and
3. Synthetic polymers.

Natural polymers : These polymers are obtained from natural sources such as plants and animals.
Ex: Starch, Rubber.

Semi-synthetic polymers: These polymers are the synthetic derivatives of natural polymers.
Ex: Cellulose acetate (rayon), Cellulose nitrate.

Synthetic polymers: These are man-made polymers. These are extensively used in daily life and in industry.
Ex : Polythene, Nylon 6, 6.

Classification based on structure :
Based on structure polymers are classified into

1. Linear polymers,
2. Branched chain polymers and
3. Cross linked or Network polymers.

Linear polymers : These polymers consist of long and straight chains.
Ex : Polythene, PVC etc.

Branched chain polymers: These polymers consist of linear chains having some branches.
Ex : Low Density Polythene (LDP)

Cross linked or Network polymers : These are generally formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains.
Ex: Bakelite, Melamine.

Question 51.
Explain the classification of polymers based on the mode of polymerization and nature of molecular forces.
Answer:
Classification based on mode of polymerisation:
Based on mode of polymerisation polymers are classified into

1. Addition polymers and
2. Condensation polymers.

1) Addition polymers :
These are formed by the repeated addition of the same or different monomer molecules containing double or triple bonds.

Addition polymers formed by the polymerisation of the same monomer molecules are called homopolymers.
Ex: Polythene formed from polymerisation of ethene.

Polymers formed by the addition polymerisation of two different monomeric species are called copolymers.
Ex: Buna – S formed by 1, 3 – butadiene and styrene.

2) Condensation polymers:
These are formed by the repeated condensation reaction between two bi-functional monomers. In these reactions loss of simple molecules such as H₂O, HCl etc. takes place.
Ex: Terylene or dacron formed by the condensation of ethylene glycol and terephthalic acid is an example for condensation polymers.

Classification based on molecular forces : Based on intermolecular forces such as van der Waal’s forces and hydrogen bonds, polymers are classified into four sub groups.

1. Elastomers,
2. Fibres,
3. Thermoplastic polymers,
4. Thermosetting polymers.

Elastomers :
These are rubber – like solids with elastic properties. In these polymers, the polymer chains are held together by weak intermolecular forces. These weak forces permit the polymer to be stretched. A few cross links are introduced in between the chains which help the polymer to return to its original position after the force is released.
Ex: Buna – S, Buna – N, Neoprene etc.

Fibres:
These are thread forming solids which possess high tensile strength and high modulus. These characteristics are due to intermolecular forces like hydrogen bonding. These forces also lead to close packing of chains and impart crystalline nature.
Ex: Nylon 6, 6; terylene.

Thermoplastic polymers :
A thermoplastic polymer is one which softens on heating and becomes rigid again on cooling. This is because there are weak attractive forces between polymer molecules and these are readily disrupted on heating.
Ex: Polythene, polypropylene.

Thermosetting polymers:
A thermosetting polymer is one which becomes hard on heating. It cannot be softened by heating.
Ex: Bakelite, urea-formaldehyde resins etc.

Question 52.
What are synthetic rubbers ? Explain the preparation and uses of the following.
i) Neoprene
ii) Buna – N
iii) Buna – S
Answer:
Synthetic rubber is any vulcanisable rubber like polymer which is capable of getting stretched to twice its length. It returns to its original shape and size when the stretching force is removed.

i) Neoprene :
Neoprene is formed by the free radical polymerisation of chloroprene.

Uses: Neoprene is used in the manufacture of conveyer belts, gaskets and hosepipes.

ii) Buna – N :
Buna – N is obtained by the copolymerisation of 1, 3-butadiene and acrylonitrile in the presence of a peroxide catalyst.

Uses : Buna – N is used in making oil seals, tank lining etc.

iii) Buna – S :
Buna – S is obtained by the copolymerisation of 1, 3 – Butadiene and styrene.

Uses : Buna – S is used in making automobile tyres and footwear.

Intext Questions – Answers

Question 1.
What are polymers ?
Answer:
Polymers are high molecular mass substances consisting of large number of repeating structural units. They are also called as macro-molecules.
Ex: Polythene, rubber, nylon – 6, 6 etc.

Question 2.
How are polymers classified on the basis of structure.
Answer:
In the basis of structure polymers are classified as below :

1. Linear polymers
2. Branched chain polymers
3. Cross linked polymers

Question 3.
Write the names of monomers of the following polymers.

Answer:
i) Hexamethylene diamine and adipic acid
ii) Caprolactam
iii) Tetra fluoroethene

Question 4.
Classify the following as addition and condensation polymers. Terylene, Bakelite, Polyvinyl chloride, Polythene.
Answer:
Polyvinyl chloride and polythene are addi-tion polymers Terylene and Bakelite are condensation polymers.

Question 5.
Explain the difference between Buna- N and Buna-S.
Answer:
Buna – N is a co-polymer of 1, 3 butadiene and acrytonihide and Buna – S is a co-polymer of 1, 3 butadiene and styrene.

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces.
i) Nylon – 6, 6, Buna – S, Polythene
ii) Nylon 6, Neoprene, Polyvinyl chloride
Answer:
i) Buna – S, polythene, Nylon – 6, 6
ii) Neoprene, Polyvinyl chloride, Nylon – 6

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Definite Integrals Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Definite Integrals Important Questions Very Short Answer Type

Question 1.
Evaluate \(\int_0^a\left(a^2 x-x^3\right) d x\).
Solution:

Question 2.
Evaluate \(\int_0^4 \frac{x^2}{1+x} d x\). [(TS) May ’15]
Solution:

Question 3.
Evaluate \(\int_0^\pi \sqrt{2+2 \cos \theta} d \theta\). [(AP) May ’16, Mar. ’18. ’16]
Solution:

Question 4.
Evaluate \(\int_2^3 \frac{2 x}{1+x^2} d x\). [(TS) Mar. ’20. ’16; Mar. ’17 (AP): Mar. ’12]
Solution:

Question 5.
Evaluate \(\int_0^1 \frac{x^2}{x^2+1} d x\). [(TS) May ’18, ’10]
Solution:

Question 6.
Evaluate \(\int_0^3 \frac{x}{\sqrt{x^2+16}} d x\). [Mar. ’17 (TS)]
Solution:
Put x² + 16 = t² then 2x dx = 2t dt
x dx = t dt
Lower limit: x = 0
⇒ t² = 16
⇒ t = 4
Upper limit: x = 3
⇒ t² = 25
⇒ t = 5

Question 7.
Evaluate \(\int_0^a \frac{d x}{x^2+a^2}\). [(TS) May ’19; (AP) ’15]
Solution:

Question 8.
Find \(\int_0^2 \sqrt{4-x^2} d x\). [Mar. ’07, May ’03]
Solution:

Question 9.
Evaluate \(\int_0^a \sqrt{a^2-x^2} d x\). [(TS) Mar. ’16]
Solution:
Put x = a sin θ then dx = a cos θ dθ
Lower limit: x = 0 ⇒ θ = 0
Upper limit: x = a ⇒ θ = \(\frac{\pi}{2}\)

Question 10.
Evaluate \(\int_0^1 \frac{d x}{\sqrt{3-2 x}}\). [Mar. ’19 (AP)]
Solution:
Put 3 – 2x = t² then -2 dx = 2t dt
dx = -t dt
Lower limit: x = 0
⇒ t² = 3
⇒t = √3
Upper limit: x = 1
⇒ t² = 1
⇒ t = 1

Question 11.
Evaluate \(\int_1^5 \frac{d x}{\sqrt{2 x-1}}\). [(TS) Mar. ’15]
Solution:
Put 2x – 1 = t² then 2dx = 2t dt
dx = t dt
Lower limit: x = 1
⇒ t² = 1
⇒ t = 1
Upper limit: x = 5
⇒ t² = 9
⇒ t = 3

Question 12.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x\). [(AP) May ’19, Mar. ’17; Mar. ’14]
Solution:
Let I = \(\int_0^{\pi / 2} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x\)

Question 13.
Evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\). [(TS) Mar. ’20, ’14]
Solution:

Question 14.
Evaluate \(\int_0^{\pi / 2} \frac{\cos ^{\frac{5}{2}} x}{\sin ^{\frac{5}{2}} x+\cos ^{\frac{5}{2}} x} d x\). [(AP) May ’15]
Solution:

Question 15.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^2 x-\cos ^2 x}{\sin ^3 x+\cos ^3 x} d x\). [May ’12]
Solution:

Question 16.
Evaluate \(\int_0^2|1-x| d x\). [(AP) May ’19); Mar. ’15; (TS) Mar. ’18; May ’16]
Solution:

Question 17.
Evaluate \(\int_0^4|2-x| d x\). [(AP) May ’17; ’13]
Solution:

Question 18.
Find \(\int_0^{\pi / 2} \sin ^{10} x d x\)
Solution:

Question 19.
Find \(\int_0^{\pi / 2} \sin ^4 x d x\). [May ’06, ’02]
Solution:

Question 20.
Find \(\int_0^{\pi / 2} \cos ^{11} x d x\). [(TS) Mar. ’19]
Solution:

Question 21.
Find \(\int_0^{\pi / 2} \sin ^4 x \cdot \cos ^5 x d x\). [Mar. ’10]
Solution:

Question 22.
Find \(\int_0^{\pi / 2} \sin ^5 x \cdot \cos ^4 x d x\). [(AP) Mar. ’15]
Solution:

Question 23.
Find \(\int_0^{\pi / 2} \sin ^6 x \cdot \cos ^4 x d x\). [(AP) Mar. ’19; May ’16]
Solution:

Question 24.
Find \(\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^4 x d x\). [(AP) Mar. ’20, ’18, ’16; May ’16 (TS); Mar. ’13]
Solution:
Let f(x) = sin²x cos⁴x
f(-x) = sin²(-x) cos⁴(-x)
= (-sin x)² (cos x)⁴
= sin²x cos⁴x
= f(x)
∴ f(x) is an even function.
We know that f(x) is even then

Question 25.
Find \(\int_{-\pi / 2}^{\pi / 2} \sin ^3 \theta \cos ^3 \theta d \theta\). [May ’14]
Solution:
Let, f(θ) = sin³θ . cos³θ
New, f(-θ) = sin³(-θ) cos³(-θ)
= -sin³θ . cos³θ
= -f(θ)
∴ f(θ) is an odd function.
∴ \(\int_{-\pi / 2}^{\pi / 2} \sin ^3 \theta \cos ^3 \theta d \theta\) = 0

Question 26.
Find \(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x\). [(AP) May ’18; (TS) Mar. ’15; ’14]
Solution:
Let f(x) = sin²x cos⁴x
Now f(2π – x) = sin²(2π – x) cos⁴(2π – x) = sin²x cos⁴x = f(x)
Also t(π – x) = sin²(π – x) cos⁴(π – x) = sin²x cos⁴x = f(x)
∴ \(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x=2 \int_0^\pi \sin ^2 x \cos ^4 x d x\)

Question 27.
Find \(\int_0^\pi \sin ^3 x \cos ^3 x d x\). [(TS) May ’15]
Solution:
Let f(x) = sin³x . cos³x
Now f(π – x) = sin³(π – x) cos³(π – x)
= -sin³x cos³x
= -f(x)
∴ \(\int_0^\pi \sin ^3 x \cos ^3 x d x\) = 0

Question 28.
Find the area bounded by the parabola y = x², the X-axis, and the lines x = -1, x = 2. [(TS) May ’18, ’16; (AP) ’15]
Solution:
Given y = x², X-axis i.e., y = 0
x = -1; x = 2
y = x²

Question 29.
Find the area of the region bounded by y = x³ + 3, X-axis and x = -1, x = 2. [(TS) Mar. ’20; May ’17; Mar. ’12]
Solution:
Given y = x³ + 3, y = 0, x = -1, x = 2

Question 30.
Find the area enclosed between the curves x = 4 – y², x = 0. [Mar. ’11, ’10]
Solution:
Given, x = 4 – y² and x = 0
Solving, 4 – y² = 0 then y = ±2
∴ y = 2 and y = -2

Question 31.
Find the area bounded between the curves y² – 1 = 2x and x = 0.
Solution:
Given y² – 1 = 2x and x = 0 ……..(2)
⇒ x = \(\frac{\mathrm{y}^2-1}{2}\) ……(1)

Solving (1) & (2)
o = \(\frac{\mathrm{y}^2-1}{2}\)
⇒ y² – 1 = 0
⇒ y² = 1
⇒ y = ±1

Question 32.
Find the area enclosed between the curve x² = 4y, x = 2, y = 0.
Solution:
Given x² = 4y, x = 2, y = 0
Solving, x² = 4y = 4(0) = 0

Question 33.
Find the area enclosed between the curves y = x², y = 2x. [May ’13]
Solution:
Given y = x² and y = 2x

Question 34.
Find the area enclosed between the curves, y = x², y = x³. [(TS) May ’19]
Solution:
Given y = x² and y = x³

Question 35.
Evaluate \({Lim}_{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots \ldots . .+\frac{1}{6 n}\right]\)
Solution:

Question 36.
Evaluate \({Lim}_{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots \ldots \ldots+n^4}{n^5}\). [(AP) Mar. ’20]
Solution:

Question 37.
Evaluate \({Lt}_{n \rightarrow \infty} \sum_{i=1}^n \frac{i}{\mathbf{n}^2+i^2}\)
Solution:

Question 38.
Find the area bounded by y = sin x, x-axis, x = 0, and x = π.
Solution:
Given curve is y = sin x, the x-axis
x = 0 and x = π

Question 39.
Find the area under the curve f(x) = sin x in [0, 2π]. [May ’09]
Solution:
Let y = f(x) = sin x

Question 40.
Find the area of one of the curvilinear triangles bounded by y = sin x, y = cos x, and X-axis. [Mar. ’19 (AP)]
Solution:
Given curves are
y = sin x ……..(1)
y = cos x …….(2)

Solving (1) and (2)
sin x = cos x
⇒ \(\frac{\sin x}{\cos x}\) = 1
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\)
OAB is one of the curvilinear triangles bounded by y = sin x, y = cos x and XY-axes
Required area A = \(\int_0^{\pi / 4} \sin x d x+\int_{\pi / 4}^{\pi / 2} \cos x d x\)

Question 41.
Find the area bounded between curves y = x², y = √x. [Mar. ’18 (TS)]
Solution:
Given curves are
y = x² …….(1)
y = √x ………(2)
From (1) and (2)
√x = x²
⇒ x = x⁴
⇒ x(1 – x³) = 0
⇒ x = 0, x³ = 1
⇒ x = 1

Question 42.
Evaluate \({Lt}_{n \rightarrow \infty} \frac{2^k+4^k+6^k+\ldots \ldots+(2 n)^k}{n^{k+1}}\) by using the methods of finding the definite integral of the limit of a sum. [(AP) May ’18]
Solution:

Question 43.
Evaluate \(\int_0^{\pi / 4} \sec ^4 \theta d \theta\). [May ’14]
Solution:

Question 44.
Evaluate \(\int_0^{\pi / 2} \sin ^2 \mathrm{x} d \mathbf{x}\). [May ’95]
Solution:

Question 45.
Evaluate \(\int_{-1}^1 \frac{1}{1+x^2} d x\). [May ’94, Mar. ’92]
Solution:

Question 46.
Evaluate \(\int_0^1 \sin ^{-1} x \mathrm{dx}\). [Mar. ’99]
Solution:

Question 47.
Evaluate \(\int_1^2 \log x d x\). [May ’98, ’94]
Solution:

Question 48.
Evaluate \(\int_0^1 \frac{d x}{e^x+e^{-x}}\). [Mar. ’06]
Solution:

Question 49.
Evaluate \(\int_{\mathbf{a}}^{\mathbf{b}} \frac{|\mathbf{x}|}{\mathbf{x}} \mathbf{d x}\). [May ’03]
Solution:
\(\int_{\mathbf{a}}^{\mathbf{b}} \frac{|\mathrm{x}|}{\mathrm{x}} \mathrm{dx}=[|\mathrm{x}|]_{\mathrm{a}}^{\mathbf{b}}\) = |b| – |a|

Question 50.
Find \(\int_0^{\pi / 2} \cos ^8 x \mathbf{d x}\)
Solution:

Question 51.
Find \(\int_0^{\pi / 2} \sin ^7 x d x\). [Mar. ’17 (AP)]
Solution:

Question 52.
Find \(\int_0^{2 \pi} \sin ^4 x \cdot \cos ^6 x d x\). [Mar. ’19 (TS); (AP) May ’17]
Solution:

Question 53.
Find \(\int_0^\pi \sin ^3 x \cos ^6 x d x\)
Solution:
Given, Let, f(x) = sin³x cos⁶x
Now, f(π – x) = sin³(π – x) cos⁶(π – x)
= sin³x . cos⁶x
= f(x)

Question 54.
Evaluate \(\int_0^\pi \cos ^3 x \cdot \sin ^4 x d x\). [Mar. ’00]
Solution:
Let f(x) = cos³x sin⁴x
Then f(π – x) = cos³(π – x) sin⁴(π – x)
= -cos³x sin⁴x
= -f(x)
∴ f(x) is an odd function
∴ \(\int_0^\pi \cos ^3 x \cdot \sin ^4 x d x\) = 0

Question 55.
Evaluate \(\int_0^{\pi / 2} \frac{1}{1+\tan x} d x\). [Mar. ’02, May ’99]
Solution:

Question 56.
Find the area under the curve f(x) = cos x in [0, 2π].
Solution:
Let Y = f(x) = cos x; y = f(x) = cos x

Question 57.
Find the area enclosed between the curves y = e^x, y = x; x = 0, x = 1.
Solution:

Question 58.
Find the area enclosed between the curves x = 2 – 5y – 3y², x = 0.
Solution:
Given x = 2 – 5y – 3y², x = 0
x = 2 – 5y – 3y²

Solving, 2 – 5y – 3y² = 0
⇒ 3y² + 5y – 2 = 0
⇒ 3y² + 6y – y – 2 = 0
⇒ 3y(y + 2) – 1(y + 2) = 0
⇒ (y + 2)(3y – 1) = 0
∴ y = -2 and y = \(\frac{1}{3}\)

Question 59.
Find the area enclosed between the curve y² = 3x, x = 3.
Solution:
Given y² = 3x and x = 3
Solving, y² = 3(3)
y = ±3
∴ y = 3 and y = -3

Question 60.
Find the area of the right-angled triangle with base b and altitude h, using the fundamental theorem of integral calculus.
Solution:
Let OAB be a right-angled triangle and ∠B = 90°.
Choose ‘O’ as the origin and OB as the +ve x-axis.
If OB = b, and AB = h then A = (b, h).
So, the equation of OA is y = (\(\frac{h}{b}\))x

Question 61.
Find the area bounded between the curves y = x² + 1, y = 2x – 2, x = -1, x = 2. [(AP) May ’16]
Solution:
Given curves are y = x² + 1, y = 2x – 2
x = -1 and x = 2

Question 62.
Find the area cut off between the line y = 0 and the parabola y = x² – 4x + 3,
Solution:
Given, y = x² – 4x + 3 ……(1)
and y = o ……..(2)
Solving (1) and (2)
x² – 4x + 3 = 0
⇒ x² – 3x – x + 3 = 0
⇒ x(x – 3) – 1(2 – 3) = 0
⇒ (x – 1)(x – 3) = 0
⇒ x = 1, 3

Question 63.
Evaluate \(\int_{-\pi / 2}^{\pi / 2} \sin |\mathbf{x}| \mathbf{d x}\). [Mar. ’17 (TS)]
Solution:

Question 64.
Show that \(\int_0^{\pi / 2} \sin ^n x d x=\int_0^{\pi / 2} \cos ^n x \mathbf{d x}\)
Solution:

Question 65.
Evaluate \(\int_0^{\pi / 2} x \sin x d x\). [Mar. ’18 (TS)]
Solution:

Question 66.
Evaluate \(\int_0^a(\sqrt{a}-\sqrt{x})^2 d x\). [Mar. ’19 (TS)]
Solution:

Question 67.
Evaluate \(\int_0^1 x e^{-x^2} d x\)
Solution:
Put -x² = t then -2x dx = dt
x dx = \(\frac{-\mathrm{dt}}{2}\)
Lower limit: x = 0 ⇒ t = 0
Upper limit: x = 1 ⇒ t = -1

Question 68.
Evaluate \(\int_{-1}^2 \frac{x^2}{x^2+2} d x\)
Solution:

Question 69.
Evaluate \(\int_0^{\pi / 2} x^2 \sin x d x\)
Solution:

Question 70.
Evaluate \({Lt}_{n \rightarrow \infty} \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots \ldots \ldots+\sqrt{n+n}}{n \sqrt{n}}\)
Solution:

Question 71.
Evaluate \(\int_0^a \mathbf{x}(\mathbf{a}-\mathbf{x})^{\mathbf{n}} \mathbf{d x}\)
Solution:

Question 72.
Evaluate \(\int_0^{\pi / 2} \tan ^5 x \cos ^8 x d x\)
Solution:

Question 73.
Find the area bounded by the curves y = sin x and y = cos x between any two consecutive points of intersection. [Mar. ’18 (AP)]
Solution:

Question 74.
Evaluate \(\int_0^\pi \sin x d x\)
Solution:
\(\int_0^\pi \sin \mathrm{x} d \mathrm{x}=[-\cos \mathrm{x}]_0^\pi\)
= -[cos π – cos 0]
= -[-1 – 1]
= -(-2)
= 2

Question 75.
Evaluate \(\int_2^3 \frac{2 x}{1+x^2} d x\)
Solution:
Put 1 + x² = t, then 2x dx = dt
L.L: x = 2 ⇒ t = 5
U.L: x = 3 ⇒ t = 10

Question 76.
Evaluate \(\int_1^4 x \sqrt{x^2-1} d x\)
Solution:
Put x² – 1 = t²
⇒ 2x dx = 2t dt
⇒ x dx = t dt
L.L: x = 1 ⇒ t = 0
U.L: x = 4 ⇒ t = √15

Question 77.
Evaluate \(\int_0^\pi(1+\cos x)^3 d x\)
Solution:

Question 78.
Evaluate \(\int_0^5(x+1)\) dx as a limit of a sum.
Solution:

Question 79.
Evaluate \(\int_0^4 x^2 d x\) as a limit of a sum.
Solution:

Question 80.
Find the area in sq. units bounded by the x-axis, part of the curve y = 1 + \(\frac{8}{x^2}\), and the ordinates x = 2 and x = 4.
Solution:

Question 81.
Find the area bounded by the curve y = log x, the x-axis, and the straight line x = e.
Solution:
Given that y = log x ………(1)
x-axis, i.e., y = 0 ……….(2)
x = e
Solve (1) & (2)
0 = log e^x
⇒ x = e⁰
⇒ x = 1

= (e log e – e) – (log 1 – 1)
= (e – e) – (0 – 1)
= 0 – (-1)
= 1 sq. units

Question 82.
Find the area enclosed within the curve |x| + |y| = 1.
Solution:
Given curve is |x| + |y| = 1
⇒ ±x ± y = 1
Given curve represents 4 straight lines
Take x + y = 1 ⇒ y = 1 – x

Telangana TSBIE TS Inter 2nd Year Accountancy Study Material 2nd Lesson Consignment Accounts Textbook Questions and Answers.

TS Inter 2nd Year Accountancy Study Material 2nd Lesson Consignment Accounts

Short Answer Questions

Question 1.
Distinguish the differences between the consignment and sales.
Answer:
Consignment Sending of goods from the owner to his agents for sale on a commission basis is called consignment. The agent only sells the goods on behalf of the owner for the sake of commission at the risk of the owner.

Question 2.
How the unsold stock is valued in consignment?
Answer:
At the end of the accounting period, the unsold goods should be valued properly. Other wise, the true profit cannot be ascertained. The value of unsold stock includes the purchasing price of unsold stock and proportionate direct expenses or nonrecurring ex¬penses incurred by the consignor and consignee.
Valuation of unsold stock.

Question 3.
How the abnormal loss is valued in consignment?
Answer:
I. Introduction:
1) Abnormal loss is a loss which arises due to fire, accidents, riot, flood, theft, pilfer¬age, sabotage, negligence, inefficiency etc.

2) It is unnatural and unexpected. This loss is beyond the control of human being. This loss can be insured.

3) Sometimes, insurance company admits the claim in part or full. The value of abnor¬mal loss is ascertained and it is taken on the credit side of consignment account.

4) The method of calculation of abnormal loss is to the method of calculation of clos¬ing stock.
Abnormal loss = Cost of goods sent on consignment + Expenses of consignor x

II. Entries:
1. When there is no insurance claim: –

2. When insurance claim is admitted by insurance company:

Question 4.
List out some of the recurring and non-recurring expenses of the consignor and consignee.
The following table gives details of recurring expenses in curred by consignor and consignee:
Recuring Expenses:

Non-Recuring Expenses:

Question 5.
Proforma of consignment accounts in the books of consignor.
Answer:
Proforma of consignment account in the books of consignor:

Very Short Answer Questions

Question 1.
What is consignment?
Answer:
The process of sending goods by consignor to a consignee for the purpose of sale on commission basis is called consignment.

Question 2.
Parties in consignment.
Answer:
There are two parties in consignment They are: A) consignor B) consignee.
A) Consignor: The person who send the goods on consignment is called consignor (or) principal.
B) Consignee: The person to whom goods are sent is called consignee or agent.

Question 3.
Proforma Invoice.
Answer:

1. The statement which is prepared by the consignor and send to the consignee stating that the details of the goods consigned is called proforma Invoice.
2. It contains quantity, weight, expenses incurrred, mode of transportation and minimum price at which the goods are to be sold.

Question 4.
Account sales.
Answer:

1. Consignee has to inform its consignor about the amount of sales and expenses in¬curred by him for making sales. For this purpose, the consignee has to prepare a state¬ment which is known as account Sales.
2. Account sales is a statement showing that the amount of sales, other expenses in¬curred, advance remitted by him and amount payable by him to the consignor.

Question 5.
Delecredere commission.
Answer:

1. It is the extra commission than the ordinary commission which is allowed by the consignor to the consignee for bearing the risk of bad debts arising out of credit sales made by him.
2. When this commission is allowed to the consignee, the consignee has to bear the loss due to the bad debts.

Question 6.
Over-riding commission.
Answer:

1. It is an additional commission paid by the consignor to the consignee to enhance the sales or for selling goods at prices higher than prices fixed by the consignor or boost up the sales of new products.
2. This commission is also calculated on gross proceeds of sales.

Question 7.
Invoice price.
Answer:

1. The consignor, instead of sending the goods on consignment at cost price, may send it at a higher price than the cost price. This price is known as Invoice price or selling price.
2. The intention of the consignor is that to hide or conceal the actual profit earned by him from the consignee.

Question 8.
Loading.
Answer:

1. The difference between the cost price and the invoice price of goods is known as “Loading”.
2. It also called as “Excess price over the Loading = Invoice price – Cost price
3. Loading = Invoice price – Cost price

Question 9.
Normal loss.
Answer:

1. Normal loss is a loss that arises due to evaporation, leakage and breaking the bulk material into small pieces.
2. It is a natural, unavoidable and inherent in the nature of goods sent on consignment.

Question 10.
Abnormal Loss.
Answer:

1. Abnormal loss is a loss which arises due to fire, accidents, riot, flood, theft, pilferage, sabotage, negligence, inefficiency etc.
2. It is unnatural and unexpected. This loss is beyond the control of human being. This loss can be insured.
3. Sometimes, insurance company admits the claim in part or full. The value of abnor¬mal loss is ascertained and it is taken on the credit side of consignment account.
4. The method of calculation of abnormal loss is to the method of calculation of closing stock.
   Abnormal loss = Cost of goods sent on consignment + Expenses of consignor x
   (Damaged or lost quantity/Total quantity of goods consigned)

Practical Problems

A) Short Problems:

1. The cost of price of goods is 20,000. But the invoice price of the goods is 24,000. Find the loading?
Answer:
Given:
Cost of price of goods = 20,000
The invoice price of goods = 24,000
Rs. Loading = Invoice Price – Cost price
Loading = 24,000 – 20,000 = 4,000.

Question 2.
The cost of the goods sent on consignment is Rs. 20,000 but the invoice price i sl0% more than the cost price, what will be the invoice price of the goods?
Answer:
Method (1):
Given:
Cost of goods consign = 20,000
Invoice price = 10% more than the cost price
= cost price + 10% on cost price (i.e., loading)
= 20,000 + [20,000 x 10/100]
= 20,000 + 2000
Rs. Invoice price = 22,000

Method (2)
Given:
Cost of goods consign = 20,000

Question 3.
Jeevan Surya sent goods on consignment of invoice price Rs. 15,000. This price is 20% more than the cost price. What will be cost price of the goods?
Answer:
Given that:
Invoice price = 15,000

Cost of goods = Invoice price – Loading
= 15,000 – 2,500
= 12,500

Question 4.
500 cases of goods were consigned at a cost of Rs. 150 each. The consignor paid Rs. 2,000 towards insurance and freight. The consignee paid t 2,000 for carriage and Rs. 1,000 for salaries Consignee sold only 400 cases of goods. Find the value of unsold goods?
Answer:
Total cases of goods consigned = 500 cases
(-) Consignee sold = 400 cases
unsold goods = 100 cases

Valuation of unsold stock of consignment:
Cost of unsold goods (100 cases x 150/-)= 15,000
Add roportionate non recurring expenses of consignor = 400

Proportionate non recurring expenses of consigner = 400

value of unsold goods = 15,800

Question 5.
Hardhikpatil consigned goods costing Rs. 50,000 to Vidhvanpatilwhose recurring and non-recurring expenses on the same amounted to Rs. 5,000 and t 2,000 respectively. Vidhvanpatil sold 3/4th of the goods. Find the value of unsold goods?
Answer:
Cost of goods consigned = 50,000
Given that vidhvanpatil sold ¾th goods
unsold goods = ¼th goods = 50,000 x ¼ = 12,500

Value of unsold goods :

Question 6.
Nidhi consigned 100 bales of cloth to Srikari at Rs. 1,000 per bale. Nidhi incurred the expenses Rs. 5,000. Srikari incurred Rs. 6,000 for packing and Rs. 2,000 for rent. Srikari sold 80 bales of cloth at the rate Rs. 1,500 per bale. Ascertain the value of consignment stock?
Answer:
Total goods consiged = 100 bales of cloth
(-) Sold bales of cloth = 80 bales
.*. Unsold bales of cloth = 20 bales

valuation of consignment stock:

Question 7.
Ascertain the value of closing stock from the following information.
No. of Units consigned : 5,000
Cost of each unit Rs. 10
Expenses of consignor Rs. 4.000
Expenses consingee Rs. 1,000.
No. of units sold by the consignee : 4,000
Answer:
Total goods consigned = 5000
(-) Sold goods = 4000
.-. Unsold goods = 1000

Value of consignment stock :

Question 8.
Anisa consigned goods consting Rs. 20,000 to Prasanna on consignment basis. Anisa paid Rs. 5,000 for carriage and cartage. On the way l/5th of the consignment was lost by fire. Ascertain the amount of abnormal loss.
Ans Total cost of goods consigned = 20,000
Cost of goods loss by fire = 1/5 on 20,000
The amount of Abnormal loss = = 20,000 x 1/5 = 4,000

The abnormal loss value is Rs. 5,000.

Question 9.
100 tons of coal is consigned at the rate of Rs. 900 per ton. Freight charges paid by the consignor amounted to Rs. 5,000. Loss due to loading and unloading of coal is estimated at 5 tones. 75 tons of coal is sold at the rate of Rs. 1,200. Find the value of unsold stock.
Answer:
Completion of unsold stock :

Value of unsold stock

Unsold stock value = 20,000

B) Long Problems

Question 1.
Manasa consigned goods value of Rs. 60,000 to Suhas. Manasa paid transport charges Rs. 4,000. Suhas sent the account sales of consignment stating that the entire stock was sold for Rs. 75,000. He paid Rs. 2,000 for cartage. His is entitled comission of Rs. 3,000. He sent bank draft for the balance. Write journal entries in the books of both the parties.
I. In the books of Manasa (consignor) :

Ledger Accounts in the books of consignor Manasa
Journal Entries

II. In the books of Suhas (Consignee) :
Journal Entries

Ledger Accounts :

Question 2.
Hari of Hyderabad consigned goods value Rs. 30,000 to Sumaiatha of Surath. Ilari paid for forwarding charges Rs. 1,500 and drew a bill of two months on sumaiatha for Rs. 15,000. The bill was discounted for Rs. 14,500. Sumaiatha sent account sales of consignment stating that the entire stock was sold for Rs. 38,000, agent commission Rs. 2,000 and sent bank draft for the balance. Write journal entries in the books of both the parties.
Answer:
In the books of Hari consignor :
Journal Entries

Ledger Accounts

II. In the books of Sumalatha consignee :
Journal Entries


Ledger Accounts

Question 3.
Arjun of Kamareddy and Vittal of Karimnagar are in consignment business. Vittal sent goods to Arjun Rs. 20,000. Vittal paid freight Rs. 800, insurance Rs. 700. Arjun met sales expenses rs. 750. Arjun sold the entire stock for Rs. 30,000 and he is entitled to a commission of 5% on sales. Prepare necessary ledger accounts in the books of Vittal and Arjun.
Answer:
1. in the books of Vittal (consignor) :

Question 4.
Shivabhiram of Secunderabad consigned to Anil of vijavawada 100 boxes of medi¬cines at Rs. 500 each. Shivabhiram paid freight and insurance amounting to Rs. 2,500. Anil sent Rs. 20,000 as an advance payment against consignment. Shivabhuiram received account sales from Anil containing the following particulars :
a. Gross sale proceeds are Rs. 74,000.
b. Transportaion and warehousing charges Rs. 2,500.
c. Commission charged at 10% on gross proceeds.
Prepare necessary ledger accounts in the books of consignor and consignee.
Answer:
I. In the books of Shivabhiram (consignor) :


In the books of Anil (Consignee) :

Given : Total stock = 80,000
Sold = 80%
remaining unsold stock cost = 20% i.e., = 80,000 x 20/100 = 16,000

Question 5.
Karthik of Karimnagar sent sports material worth Rs. 80,000 to Supreeth on a consign¬ment basis. Karthik spent Rs. 2,000 for insurance. Supreeth while taking the goods spent Rs. 1,000 for transport and Rs. 1,500 for Godown rent. 80% of stock was sold out. Calculate the value of unsold stock.
Answer:
Valuation of Unsold stock

Question 6.
On 1st January, 2020 Srinu of Srinivasanagar consigned goods value Rs. 30,000 to Arun of Alwal. Srinu paid cartage and other expenses Rs. 2,000. On 31st March, 2020 Arun sent an account sales with the following information :
a. 50 of the goods sold for Rs. 5,000.
b. Arun incurred expenses Rs. 1,750.
c. Arun entitled to receive commission @ 6% on sales.
Bank draft was enclosed for the balance due. Pass necessary journal entries in the books of both the parties.
Answer:
II. In the books of Srinu (Consignee)
Journal Entries

Ledger accounts :

Working Note :
Valuation of Unsold stock

II. In the books of Aran (Consignee) :
Journal Entries


Ledger accounts:

Question 7.
On 1-1-2020 Bajaji of Hyderabad consigned goods valued at Rs. 50,000 to Shivaji of Sholapur. Balaji paid cartage and other expenses Rs. 2,400. On 31-3-2020 sent the account sales with the following information.
a. 3/4th of the goods sold for Rs. 48,000
b. Shivaji incurred expenses amounting to Rs. 1,200.
c. Shivaji is entitled to receive commission @ 5% on sales.
Bank draft was enclosed for the balance. Prepare necessary ledger accounts in the books of Balaji.
Answer:
Ledger Accounts in the books of Balaji

Working Note :
Valuation of Consignment stock (unsold stock)

Question 8.
Amith & Co of New Delhi consigned 50 cell phones to Teja & Co. of Hyderabad. The cost of each cell phone was Rs. 2,500. Amit & Co paid Insurance Rs. 500, Freight Rs. 2,500/ am account sale was received from Teja & Co showing the sales of 40 cell phones a Rs. 3.000 each. The following expenses were deducted by them :
Selling expenses : Rs. 1.600. Commission : Rs. 3,000. Amith & Co received a bank draft for the balance due. Prepare important ledger account in the books of Amith & Co and Teja & Co.
Answer:
Ledger accounts in the books of Amith & Co (consignor) :


II. In the books of Teja and Co consignee :

Working Note :
Valuation of closing stock

Question 9.
Naveen of Agra sends 50 IV sets each costing Rs. 15,000 to Sravan of Warangal to be on a consignment basis. He incured the following expenses ; freight Rs. 2,000 and insurance X 5,000. Sravan sold 45 TVs for Rs. 7,50,000 and paid X 10,000 as shop rent, which is borne by Naveen as per terms and conditions of the consignment. Consignee is entitled for a commission of X 200 per TV sold. Assuming that, Sravan settled the account by sending the bank draft to Naveen. Prepare necessary ledger accounts in the books of Naveen and Sravan.
Answer:
Ledger accounts in the books of Naveen (consignor) :


Ledger account in the books of Sravan : (consignee) :

Working Note :
Valuation of unsold stock

Question 10.
Akhil of Hyderabad consigned goods worth Rs. 20,000 to his agent Anil of Kamareddy on consignment. Akhil spend Rs. 1,000 on transport, Rs. 500 on insurance. Anil sent Rs. 5,000 as advance. After two months, Akhil received the account sales as follows:
a. Half of the goods were sold f 24,000, of which Rs. 4,000 sold credit.
b. Selling expenses were Rs. 1,200.
c. Ordinary commission 8% and delecredere commission 2% on sales.
d. Bad debts to beamounted to Rs. 400.
Give ledger accounts in the books of both the parties.
Answer:
Ledger accounts in the books of Akhil (consignor) :


Working Note :
Valuation of consignment stock

Question 11.
Manaswi of Medak consigned 50 boxes of medicine to Vedhagna of yellareddy. The cost of each box is X 800 and invoice price is X 1,000 each box. Manaswi paid Rs. 2,500 for forwarding goods. Vedhagna advanced X 25,000 to Vedhagna and sent ac¬count sales with the details sales 40 boxes at the rate of X 1,000 each. Selling expenses are X 1,000 and his commission is X 1,500. Pass necessary journal entries in the books of Manaswi.
Answer:
Journal Entries




Working Note :
Valuation of consignment stock

Question 12.
Ragamai of Kochin consigned 200 cases of ayurvedic medicines to Sricharani of Hyderabad. The cost of each case is Rs. 400 but the invoice price is Rs. 500 per case. Ragamai paid Rs. 1,200 towards packing sand carriage. Sricharani informed through account sales that 180 cases are sold at Rs. 520 each. Expenses paid by Sricharani were freight Rs. 800. Commission has to be calculated at 5% on sales. Prepare Consignment account and consignee’s account in the boooks of Ragami.
Answer:
Ledger accounts in the books of Ragamai

Working Note :
Valuation of unsold stock on consignment

Question 13.
Rajani of Nagpur consigned goods to Praveen of Patna. The value of Rs. 40,000 in¬voiced at 25% above the cost. Rajani paid Rs. 1,500 for carriage and Rs. 500 for insur¬ance. Rajau drew a bill on Praveen for Rs. 20,000 payable after 3 months and received Praveen’s acceptance. Later, Rajani received account sales from Praveen stating the following : Total goods sold for Rs. 60,000 and the sales expenses Rs. 1,000 and his commission 5% on sales.
Praveen deducted all the above and sent a cheque for remaining balance. Prepare necessary accounts in the books Rajani and Praveen.
Answer:


Working Note :
Given cost of the goods = 40,000
Invoice price = cost + loading
= 40,000 + 25% on cost
= 40,000 + [ 40,000 x 25/100 ]
= 40,000 + 10,000 = Invoice price = 50,000

Question 14.
Reepthika Reddy sent goods to Rishika Reddy of Adilabad on consignment worth Rs. 30,000 at invoice price of cost plus 20%. Reepthika Reddy paid transport and insurance charges of Rs. 800. Rishika Reddy sent an account sales by showing 3\4th goods are sold for Rs. 28,000 and sales expenses Rs. 700. She sent a draft for an amount due after deducting 5% commission. Give the necessary accounts in the books of both the parties.
Answer:
Ledger accounts in the books of Reepthika Reddy :

In the books of Rishika Reddy a/c:

Working Note :
1) Invoice price = Cost price + 20% on cos
= 30,000 + [30,000 x 20/100]
= 30,000 + 6000 = 36,000
2) Valuation of consignment stock

3) Stock reserve = 6,000 x 1/4 = 1,500

Question 15.
Srinivas of Srinagar consigned 10,000 cases of tinned fruits costintg Rs. 75,000 to Kiran of Kanpur on 1-1-2020 charging him a pro forma invoice price to show a profit of 25% on sales. Srinivas paid Rs. 6,000 towsards freight. During the half year ending June. 30, 2020 Kiran incurred Rs. 4,000 in expenditure and reported sales of 8,000 cases for Rs. 82,000. Kiran was entitled to a commission of 5% on sales. He has sent a demand draft for the balance due to Srinivas. Prepare the necessary ledger accounts in the books of Srinivas.
Answer:
Ledger accounts in the books of Srinivas :



Working Note :
1) Invoice price = cost + profit [ie., 25% on sales]

2) Valuation of unsold stock :

Question 16.
Arun sent 100 sewing machines on consignment basis to Tarim. The cost of each machine was Rs. 300, but the consignor prepared the proforma invoice at 25% abobe the cost. The company spent Rs. 800 on various expenses.
After receiving the machines, Tarun had to spend Rs. 950 for freight and Rs. 1,100 for godown rent. By the end of the year Tarun sold 80 Machines @ 1410 per machine. He was entitled to a commission of 5% on sales. Prepare necessary accounts in the books of Arun,
Answer:
In the books of Arun :

Working Note :
1) Invoice price = cost + 25% on cash
= 30,000 + 30,000 x 25/100
= 30,000 + 7,500 = 37,500.
2) Valuation of unsold stock of consignment

3) Stock Reserve = 20 x 75 = 1,500

Question 17.
500 sanitizer bottles were consigned by Anuhya and co. Hyderabad to Nithya and co. Secunderabad costing Rs. 500 each. Expenses incurred by Anuhya and co. amounted to Rs. 5,000. On the way 5 sanitizer bottles were completely damaged due to bad han¬dling and insurance company admitted the calim to the extent of Rs. 2,500.
Nithya and co, took delivery of the remaining sanitizer bottles and incurred ex¬penses Rs. 12,000. Nithya and co. sold 495 sanitizer bottles for Rs. 40,000, It is entitled to a commission of 5% on sales. Prepare Consignment account, Nithya and co. a/c in the books of Anuhya and co. assuming that Nithya and co. remitted amount due by them.
Answer:
In the books of Anuhya and Co :

Working Note :
Abnormal loss :

Question 18.
Raja oil mills, kamareddy consigned 500 kgs of ghee to the Vijay dealers of Yellareddy. Cost of each Kg. is 800. Raja oil mills paid Rs. 5,000 for various expenses. During the transit, 50 kgs of ghee were accidentally destroyed.
Vijay received remaining stock of ghee and reported that entire ghee is sold for Rs. 5,00,000 and his expenses are 12,000. He is entitled to a 5% commission on sales. Prepare necessary accounts in the books of Raja oil mills assuming that Vijay remitted amount due by them.
Answer:

Working Note:

Question 19.
Pavan Kumar of Hyderabad consigns to Kiran Kumar of Nizamabad 50 eases of goods at a cost of Rs. 500 per case, Pavan Kumar incurred the expenses Rs. 500. Kiran Kumar paid Rs. 400.4 cases were destroyed in transit. t
Kiran Kumar received remaining stock and sold entire stock for Rs. 30,000. He is entitled for a commission of 5% on sales and sent the bank draft for full settlement of account. Show necessary accounts in the books of Pavan Kumar.
Answer:

Working Note:

= 2,040

Question 20.
On 1 January, 2020 Bharat coal company Ltd. Consigned to Vijay dealers, vijayawada. 400 tons of coal cost of the coal being Rs. 180 per ton. The company had paid Rs. 6,000 towards freight and insurance.
Vijay took delivery of the goods consigned on 10th January, 2020. On 31st March, 2020 the consignee reported that :
1. There was a shortage of 10 tons due to loading and unloading of the coal, (Treated as normal loss)
2. 380 tons were sold @ Rs. 250 per ton.
3. He had paid Rs. 3,000 towards godown rent and selling expenses.
Vijay dealers were entitled to a commission of Rs. 4,000. Show the necessary accounts in the books of Bharath Coal Company Ltd.
Answer:
In the books of Bharath Ltd. coal company :


Working Note :
1) Valuation of unsold stock :
Unsold coal tons = total – Normal loss – sold .
= 400 tons – 10 tons – 380 tons unsold =10 tons
= 2,000

consignment stock = 2,000

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 1st Lesson Solid State Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 1st Lesson Solid State

Very Short Answer Questions (2 Marks)

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
An amorphous solid is a substance whose constituent particles do not possess a regular orderly arrangement. Ex : Glass, plastics, rubber, starch etc.

Question 2.
What makes a glass different from a solid such as quartz?
Answer:
Glass is an amorphous form of silica and do not posses long range regular structure. It do not posses sharp melting point.

Quartz is a crystalline form of silica and have long range regular structure. It has sharp melting point. If the molten quartz Is cooled rapidly it becomes glass.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
a) Tetra phosphorous deoxide
b) Graphite
c) Brass
d) Ammonium phosphate (NH₄)₃ PO₄
e) SIC
f) Rb
g) I₂
h) LiBr
i) P₄
j) Si
k) plastic.
Answer:

Question 4.
What is meant by the term coordination number?
Answer:
The number of nearest neighbours of a particle in a packing is called coordination number, eg : in hexagonal close packing coordination number is 12.

Question 5.
What is the coordination number of atoms in cubic close – pack structure ?
Answer:
In cubic close – packed structure coordi-nation number is 12.

Question 6.
What is the coordination number of atoms in a body – centered cubic structure ?
Answer:
In a body centred cubic structure coordi-nation number is 8.

Question 7.
Stability of a crystal is reflected in the magnitude of Its melting point. Comment.
Answer:
If the interparticle forces are stronger more energy is required to separate them. The crystals in which jilter particle forces are stronger are stable. More stable crystals have higher melting points than less stable crystals. Hence we can say that stability of a crystal is reflected in magnitude of its melting point.

Question 8.
How are the intermolecular forces among the molecules affect the melting point ?
Answer:
If the intermolecular forces among the molecules are stronger, more energy is required to separate them. Such solids require more amount of energy to melt. Hence they have higher melting points. If the intermolecular forces are weak, less energy is sufficient to melt them. Such solids have low melting points.

Question 9.
How do you distinguish between hexagonal close packing and cubic close packing structures ?
Answer:
In hexagonal close – packing every third layer is similar to the first layer. Thus there is ABABAB ………. arrangement. In cubic close packing every fourth layer is similar to first, there is ABC ABC…. arrangement.

Question 10.
How do you distinguish between crystal lattice and unit cell ?
Answer:
A crystal lattice may be defined as a regular three dimensional arrangement of identical points in space.
A unit cell may be defined as a three dimensional group of lattice points that generates the whole lattice by translation or stacking. It is the smallest repetitive portion of the crystal lattice.

Question 11.
How many lattice points are there in one unit cell of face – centered cubic lattice ?
Answer:
A face centred cubic unit cell has lattice points at the corners of the cube and at face centres. So it contain 8 + 6 = 14 lattice points.

Question 12.
How many lattice points are there in one unit cell of face centered tetragonal lattice ?
Answer:
A face centred tetragonal unit cell has 4 atoms per unit cell.

Question 13.
How many lattice points are there in one unit cell of body centered cubic lattice ?
Answer:
A body centred cubic unit cell has lattice points at the corners of the cube and at the body centre. So the total number of lattice points in one unit cell of a body centred cubic lattice is 9.

Question 14.
What is a semiconductor ?
Answer:
Semiconductors are solids with conductivi-ties in the intermediate range from 10^-6 to 10⁴ ohm^-1 m^-1. Their conductivity increases with increase in temperature.

Question 15.
What is Schottky defect ? (AP & TS ’15)
Answer:
The Schottky defect is simply a vacancy defect in ionic solids. This is due to missing of equal number of positive and negative ions. It is more common in ionic compounds

with high coordination number and in the ionic scJlids which contain positive and negative ions of equal size. This defect decreases density of the substance, eg. NaCl, KCl, CsCl, AgBr.

Question 16.
What is Frenkel defect ? (AP & TS ’15) (Mar. 2018 . TS)
Answer:
Frenkel defect arises when the smaller ion of ionic solids is dislocated from its normal site to an interstitial site. Frenkel defects

are common in ionic solids having low coordination number and large difference in the sizes of positive and negative ions. The density of the substance do not change, eg. ZnS, AgCl, AgBr and Agl.

Question 17.
What is interstitial defect ?
Answer:
If the constituent particles of a solid occupies the interstitial sites instead of lattice points, it is called interstitial defect. It is observed in non – ionic solids.

Question 18.
What are F – centres ?
Answer:
A compound may have excess metal ion if a negative ion is absent from its lattice site, leaving a hole which is occupied by electron to maintain electrical neutrality. The holes occupied by electrons are called F – centres.

Question 19.
Explain ferromagnetism with suitable example.
Answer:
The substances which are strongly attracted by external magnetic field are called ferromagnetic substances. They contain unpaired electrons, eg. iron, cobalt, nickel.

Question 20.
Explain paramagnetism with suitable example.
Answer:
Substances which are weakly attracted by a magnetic field are called paramagnetic substances. This is due to the presence of one or more unpaired electrons. Ex : O₂, Cu⁺², Fe⁺³, Cr⁺³ etc.

Question 21.
Explain Ferromagnetism with suitable example.
Answer:
When the magnetic moments of the domains in a substance are aligned in parallel and anti parallel directions in unequal numbers, then they are said to exhibit Fern magnetism. They are weakly attracted by magnetic field.
Eg: Fe₃O₄, MgFe₂O₄, NiFe₂O₄ etc.

Question 22.
Explain Antiferro magnetism with suitable example.
Answer:
When the magnetic moments of the domain in a substance are aligned in opposite directions and cancel out each others moment, then they are said to exhibit Anti ferromagnetism.

Question 23.
Why x-ray are needed to probe the crystal structures ?
Answer:
According to a fundamental principle of optics, the wavelength of light used to observe an object must be no greater than twice the length of the object itself. Since atoms have diameters of around 2 × 10^-10 m and the visible light detected by our eyes has wavelength of 4 – 7 × 10^-7 m, it is impossible to see atoms using even the tinest optical microscope. Hence, to see atoms we must use light wdth a wave length of approximately 10^-10 m, which is in the x – ray region of electromagnetic spectrum.

Short Answer Questions (4 Marks)

Question 24.
Explain similarities and differences between metallic and ionic crystals.
Answer:
Similarities:

1. In metallic and Ionic solids the constituent particles are in regular arrangement. So both have a regular shape.
2. In both solids the bond is non – directional.
3. Both solids are hard.

Differences:

1. In metallic crystals the constituent particles are positive ions immersed in sea of mobile electrons but in ionic solids the constituent particles are positive and negative ions.
2. In metallic crystals binding force is metallic bond whereas in ionic crystals is ionic bond.
3. Metallic crystals are conductors of electricity while ionic crystals are insulators in solid state and act as conductors in molten state and aqueous solutions.
4. Metallic crystals are malleable and ductile while ionic crystals are brittle.

Question 25.
Explain why ionic solids are hard and brittle.
Answer:
Ionic solids are hard because in ionic solids, ions are held together by strong electrostatic attractive forces. If sufficient force is applied on ionic crystals, the ions with similar charges come close due to displacement. Thus the crystal breaks due to repulsions between similar charges.
Hence Ionic solids are brittle.

Question 26.
Calculate the efficiency of packing In case of a metal of simple cubic crystal.
Answer:
Packing efficiency in simple cubic lattice:
In a simple cubic lattice the atoms are located only on the corners of the cube. The particles touch each other along the edge.
Thus, the edge length or šide of the cube
‘a’, and the radius of each particle, r are related as a = 2r

Simple cubic unit cell. The spheres are in contact with each other along the edge of the cube.
The volume of the cubic unit cell = a³
= (2r)³ = 8r³
Since a simple cubic unit cell contains only 1 atom.
The volume of the occupied space = \(\frac{4}{3} \pi r^3\)
Packing efficiency

Question 27.
Calculate the efficiency of packing in case of a metal of body – centred cubic crystal.
Answer:
From Fig it is clear that the atom at the centre will be in touch with the other two atoms diagonally arranged.
In ΔEFD
b² = a² + a² = 2a²
b = \(\sqrt{2} a\)
Now in ΔAFD
c² = a² + b² = a² + 2a² = 3a²
c = \(\sqrt{3} a\)

(sphere along the body diagonal are shown with solid boundaries).
The length of the body diagonal c is equal to 4r.

where r is the radius of the sphere (atom), as all the three špheres along the diagonal touch each other.
Therefore, \(\sqrt{3} a\) = 4r = a = \(\frac{4 r}{\sqrt{3}}\)
Also we can write, r = \(\frac{\sqrt{3}}{4} a\)
In this type of structure, total number of atoms is 2 and their volume is 2 × \(\left(\frac{4}{3}\right) \pi r^3\)
Volume of the cube, a³ will be equal to
\(\left(\frac{4}{\sqrt{3}} \mathrm{r}\right)^3\) or a³ = \(\left(\frac{4}{\sqrt{3}} \mathrm{r}\right)^3\)
Therefore
Packing efficiency

Question 28.
Calculate the efficiency of packing in case of face – centered cubic crystal.
Answer:
Both types of close packing (hep and ccp) are equally efficient. Let us calculate the efficiency of packing in ccp structure. In Fig. let the unit cell edge length be a’ and face diagonal AC = b.
In Δ ABC
AC² = b² = BC² + AB² = a² + a² = 2a² (or)
b = \(\sqrt{2}\)
If r is the radius of the sphere, we find b = 4r = \(\sqrt{2 a}\) (or) a = \(\frac{4 r}{\sqrt{2}}\) = \(2 \sqrt{2 r}\)
(we can also write, r = \(\frac{a}{2 \sqrt{2}}\))

Cubic close packing other sides are not provided with spheres for sake of clarity

We know, that each unit cell in ccp structure, has effectively 4 spheres. Total volume of four spheres is equal to 4 × (4/3) πr³ and volume of the cube is a³ or \((2 \sqrt{2} r)^3\).
Therefore,
Packing efficiency
Volume occupied by four spheres

The same packing efficiency occurs in hep arrangement also.

Question 29.
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body – centre. What is the formula of the compound ? What are the coordination numbers of P and Q ?
Answer:
The atom at corner makes \(\frac{1}{8}\) contribution while atom at body centre makes 1 contri-bution to the unit cell.
No. of atoms of Q per unit cell
= 8 (at corners) × \(\frac{1}{8}\) = 1
No. of atoms of P per unit cell
= 1 (at body centre) × 1 = 1
∴ Formula of the compound is PQ.
The atom at the body centre would be in contact with all the atoms at the corners. Hence the coordination number of P would be 8. Similarly coordination number of Q is also 8.

Question 30.
If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’, derive relation between r and R.
Answer:
The octahedral void can be represented as follows.

Though an octahedral void is surrounded by six spheres only four are shown in the figure. The spheres present above and below the void are omitted.
Let the length of one side of the square ABCD is a cm.
In right angled ΔABC,
The diagonal AC
= \(\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}\) = \(\sqrt{a^2+a^2}\) = \(\sqrt{2} a\)
But AC = 2R + 2r ∴ 2R + 2r = \(\sqrt{2} a\)
But a = 2R ∴ 2R + 2r = \(\sqrt{2} .2 \mathrm{R}\)
Dividing by 2R we get 1 + \(\frac{r}{R}\) = \(\sqrt{2}\)
\(\frac{r}{R}\) = \(\sqrt{2}\) – 1 = 1.414 – 1 = 0.414

Question 31.
Describe the two main types of semiconductors and contrast their conduction mechanism. (Mar. ’19, ’18. AP)
Answer:
Semiconductors are o two types. They are
1) Intrinsic semiconductors
2) Extrinsic semiconductors.

1) Intrinsic semiconductors: In this type of semiconductors, electrical conductivity increases with rise in temperature, since more electrons can jump to conduction band. Eg :Silicon and Germanium.

2) Extrinsic semiconductors: In this type of semiconductors, conductivity is increased by adding a appropriate amount of suitable impurity. This process is called Doping. Doping can be done with an impurity which is electron rich or electron deficient. These are of two types.

a) n – type semiconductors : The semi-conductors in which the increase in conductivity is due to the negatively charged electron are called n-type semiconductors.
Eg : Silicon and germanium of group 14 doped with a group 15 elements like P or As form n – type semiconductors.

b) p – type semiconductors : The semi-conductors in which the increase in conductivity is due to the positively charged holes are called p – type semi-conductors.
Ex : Silicon or germanium can also be doped with group B elements like B, Al or Ga from p-type semiconductors.

Question 32.
Classify each of the following as either a p – type or a n – type semiconductor.
1) Ge doped with In
2) Si doped with B.
Answer:
1) Ge is an element of group -14. Its outer electronic configuration is 4s² 4p². When it is doped with In group – 13 element having 5s² 5p¹ configuration it can form three covalent bonds with germanium. The fourth bond of germanium contains only one electron and hence it is an electron deficient bond or a hole is created. Conductivity is due to these holes. Therefore it is a p – type semiconductor.

2) Silicon is an element of group -14 with an outer electron configuration 2s² 2p². When it is doped with Boron, an element of group -13 with 3s² 3p¹ configuration, it can form three covalent bonds with silicon. The fourth bond of silicon contain only one electron and hence it is an electron deficient bond or a hole is created. Conductivity is due to these holes. Therefore it is a p – type semi-conductor.

Question 33.
Analysis shows that nickel oxide has the formula Ni_0.98O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions ?
Answer:
From the given formula it is clear that for every 100 oxide ions there are only 98 nickel ions. Suppose, out of 98 nickel ions x exist as Ni²⁺ and the remaining (98 – x) exist as Ni³⁺
Total positive charge on 98 nickel ions = x × 2 + (98 – x) 3
Total negative charge on 100 oxide ions
= 100 × 2 = 200
Due to electrical neutrality,
x × 2 + (98 – x)3 = 200
2x + 294 – 3x = 200
x = 94
∴ Fraction of nickel as Ni²⁺ = \(\frac{94}{98}\) × 100
= 96%
Fraction of nickel as Ni³⁺ = (100 – 96)% = 4%

Question 34.
Gold (atomic radius = 0.144nin) crystallizes in a face centred unit cell. What is the length of a side of the unit cell?
Answer:
Radius of gold atom r = 0.144 nm
In face centred unit cell, lace diagonal
= 4r = \(\sqrt{2} a\)
∴ a = \(\frac{4 \mathrm{r}}{\sqrt{2}}\) = \(2 \sqrt{2}\) × r = 2 × 1.414 × 0.144mn
= 0.407 nm
Edge length of unit cell, a = 0.407 nm

Question 35.
In terms of band theory, what is the difference between a conductor and an insulator?
Answer:
In a metal the outer orbitals of very large number of atoms overlap to form a very large number of molecular orbitals that are delocalised over the metal. As a result a large number of energy levels are crowded together into bands. The highest occupied energy band is called the valence band while the lowest unoccupied energy band is called conduction band. The energy difference separating the valence band and conduction band is called band gap or energy gap.

In conductors the valence band is only partially filled. So electrons may easily excited from lower energy level to higher energy level by supplying a very small amount of energy. When a voltage is applied to a metal crystal, electrons are excited to the unoccupied orbitals in the same band and move towards the positive terminal. Thus, a material with partly filled energy band is a conductor.

In case of insulators, the highest occupied band is completely filled. The energy gap between the fully filled valence band and the vacant conduction band is very large and it is not possible to excite the electrons to the conduction band. So they become insulators.

Question 36.
In terms of band theory, what is the’ difference between a conductor and a semiconductor?
Answer:
In a metal the outer orbitals of very large number of atoms overlap to form a very large number of molecular orbitals that are delocalised over the metal. As a result a large number of energy levels are crowded together into bauds. The highest occupied energy band is called the valence hand while the lowest unoccupied energy band is called conduction band. The energy difference separating the valence band and conduction band is called band gap or energy gap.

In conductors the valence band is only partially filled. So electrons may easily excited from lower energy level to higher energy level by supplying a very small amount of energy. When a voltage is applied to a metal crystal, electrons are excited to the unoccupied orbitals in the same band and move towards the positive terminal. Thus, a material with partly filled energy band is a conductor.

In the case of semiconductors, the forbidden band i.e., the energy gap bet-ween valence band and conduction band is little. The thermal energy available at room temperature is enough to excite some electrons from the highest occupied band to the next permitted energy band. So conductivity occurs but their conductivity is in between the conductors and insulators.

Question 37.
If NaCl is doped with 1 × 10^-3mol percent of SrCl₂, what is the concentration of cation vacancies?
Answer:
Every Sr²⁺ ion causes one cation vacancy (because two Na⁺ ions are replaced by one Sr²⁺).
Therefore, introduction of 10^-3 moles of SrCl₂ per 100 moles of NaCl would introduce 10^-3 mole cation vacancies in 100 moles of NaCl.
No. of vacancies per mole of NaCl
= \(\frac{10^{-3}}{100}\) × 6.02 × 10²³
= 6.02 × 10¹⁸ vacancies mol^-1.

Question 38.
Derive Bragg’s equation. (TS Mar. 19) (AP 17, 16, 15; TS 16, 15; IPE 14)
Answer:
When a beam of X – rays strikes a crystal, the constituent particles of the crystal scatter or deflect some of the X – rays from their original path. The scattered X – rays can be detected on a photographic plate. From the diffraction pattern, the distance between constituent particles in a crystal can be studied.

In the above figure X – rays with wave-length λ strike a crystal face at an angle θ and then reflect at the same angle. The rays that strike an atom in the second layer are diffracted at the same angle θ, but because the second layer of atoms away from the X – ray source, the distance that the X – rays has to travel is more to reach the second layer. The path difference is indicated by AB in the figure. AB is equal to the distance between atomic layers d(=zB) times the sine of the angle θ.

sin θ = \(\frac{\mathrm{AB}}{\mathrm{d}}\) ; so AB = d sin θ
The X – rays have to travel a total extra distance AB + BC.
AB + BC = 2d sin θ (∴ AB = BC)
The different X – rays striking the two layers of atoms are in – phase initially and also they will be in – phase after reflection only the extra distance AB + BC is equal to a whole number of wavelengths nX where n is an integer (1, 2, 3, 4….)
∴ AB + BC = 2d sin θ = nλ
This is known as Bragg’s equation.
By knowing the value of φ, λ the value of θ can be measured, and the value of n is a small integer, usually 1. Thus the distance d between layers of atoms in a crystal can be calculated.

Long Answer Questions (8 Marks)

Question 39.
How do you determine the atomic mass of an unknown metal if you know its density and dimension of its unit cell ? Explain.
Answer:
From the dimensions of the unit cell, it is possible to calculate the volume of the unit cell. From the knowledge of the density of unit cell we can calculate the mass of atoms in the unit cell or vice versa.
If we know the edge length of a cubic crystal of an element or compound, we can easily calculate its density as described below.

Consider a unit cell of edge ‘a’ picometer (pm)
The length of the edge of the cell = a pm
= a × 10^-12m
= a × 10^-10 cm
Volume of unit cell = (a × 10^-10 cm)³
= a³ × 10^-30 cm³

Mass of unit cell = Number of atoms in unit cell × mass of each atom = Z × m
Where Z = number of atoms in unit cell and m = mass of each atom.

Mass of unit cell = Z × \(\frac{M}{N_o}\)
Substituting in (1) we get
∴ Density of unit cell = \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^3 \times 10^{-30} \times \mathrm{N}_o}\) gm cm^-3

If ‘a’ is in pm, then density will be in g cm^-3. Density of unit cell is the same as the density of the substance. li the density of the metal and dimensions of unit cell are known the atomic mass can be calculated
by using the above equation.

Question 40.
Silver crystallizes In FCC lattice. If edge of the cell is 4.07 × 10^-8cm and density is 10.5 gcm^-3. Calculate the atomic mass of silver.
Answer:
The edge length of unit cell, .
a = 4.077 × 10^-8cm.
Number of atoms in FCC unit cell Z = 4
Density d = 10.5 g cm^-3
Atomic mass M =?
Avogadro constant N₀ = 6.023 × 10²³
We know that density
d = \(\frac{\mathrm{ZM}}{\mathrm{a}^{-3} \mathrm{~N}_{\mathrm{o}}}\) or M = \(\frac{d a^3 \mathrm{~N}_{\mathrm{o}}}{\mathrm{Z}}\)
M = \(\frac{10.5 \times\left(4.077 \times 10^{-8}\right)^3 \times 6.023 \times 10^{23}}{4}\)
= 107.8
∴ Atomic mass of silver = 107.8U

Question 41.
Niobium crystallizes In body – centred cubic structure. If density is 8.55 g cm^-3. Calculate atomic radius of niobium using its atomic mass 93U.
A. Density = 8.55 g cm³
Let length of tie edge = a cm
Volume of unit cell = a³ cm³

Number of atoms per unit cell = 2 (bcc)
Mass of unit cell = \(\frac{2 \times 93}{6.023 \times 10^{23}}\)
Volume of unit cell =

= \(\frac{2 \times 93}{6.023 \times 10^{23} \times 8.55}\) = 36.12 × 10^-24 cm³
∴ a³ = 36.12 × 10^-24 cm³
Edge length a = (36.12 × 10^-4)^1/3
= 3.306 × 10^-8 cm
= 3.306 × 10^-10 m
Now radius, in body centred cubic r = \(\frac{\sqrt{3}}{4} a\)
= \(\frac{\sqrt{3} \times 3.306 \times 10^{-10}}{4}\)
= 1.431 × 10^-10 m
= 0.143 nm. (or) 143 pm.

Question 42.
Copper crystallizes Into a fcc lattice with edge length 3.61 × 10^-8cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^-3.
Answer:
Volume of the unit cell = (361 × 10^-10)³ cm³
= 47.4 × 10^-24 cm³
Molar mass of copper = 63.5 g mol^-1
In a face centred cubic unit cell, there are four atoms per unit cell.
Mass of unit cell = \(\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{6.02 \times 10^{23} \mathrm{~mol}^{-1}}\)

Thus the calculated value of density is in agreement with the measured value of density.

Question 43.
Ferric oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
Answer:
In hexagonal close packed arrangement of oxide ions, each oxide ion has one octahedral hole associated with it.
For each O^2- ion there is one octahedral void in the structure and two of three are occupied by Fe³⁺ ion.
∴ Number of Fe³⁺ ions present per oxide ion = \(\frac{2}{3}\)
Thus the formula of ferric oxide should be Fe_2/3 O₁ or Fe₂ O₃.

Question 44.
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell ?
(ii) How many unit cells are there in 1.00 cm3 of aluminium ?
Answer:
i) In cubic close packed structure, the face diagonal of the unit cell is equal to four times the atomic radius.
Face diagonal = 4 × r = 4 × 125 pm = 500 pm
But face diagonal = \(\sqrt{2}\) × edge length

= 354 pm.

ii) Volume of one unit cell = a³
= (3.54 × 10^-8 cm)³
No of unit cells in 1.00 cm³
= \(\frac{100}{\left(3.54 \times 10^{-8}\right)^3}\) = 2.26 × 10²²

Question 45.
How do you obtain the diffraction pattern for a crystalline substance ?
Answer:
The diffraction pattern for a crystalline sub-stance can be obtained by Debye – Scherrer method.

A monochromatic X – ray beam from a source ‘S’ is made incident on a crystal powder sample filled in a thin walled glass capillary tube ‘C’. As a result of diffraction from all sets of lattice planes, diffraction maxima arise which are recorded with a detector placed on the circumference of a circle centred on the powder specimen. The essential features of a diffractometer are shown in figure. A powder specimen C is supported on a table H, which can be rotated about an axis 0 perpendicular to the plane of the drawing. The X – ray source S is also normal to the plane of the drawing and therefore parallel to the diffractometer axis O. X – rays diverge from the source S and are diffracted by the specimen to form con-vergent diffracted beam which comes to a focus at the slit F and then enters ionisation chamber or photographic plate G.

The ioni-sation chamber contains methyl bromide vapours. The energy of the X – rays causes ionisation so that there is a flow of current. The current is measured with electrometer E. The extent of ionisation in the vapours is shown by the electrometer reading. As the intensity of the diffracted X – rays increase, the degree of ionisation also increases. For a set of parallel planes making an angle ‘θ’ with the incident beam of X – rays that satisfy Bragg’s equation, the intensity of the diffracted beam is measured. In Debye – Scherrer powder method the angle 2θ for the diffracted X – rays corresponding to each and every lattice plane is measured along with the intensities of diffracted X – rays with the help of the electrometer. A graph is drawn between the diffracted angle (2θ) and intensity of X – rays. This gives the diffraction pattern for a crystal line substance.

Intext Questions – Answers

Question 1.
Why are solids rigid ?
Answer:
In solids

1. the intermolecular distances are short,
2. intermolecular forces are strong,
3. the constituent particles (atoms, molecules or ions) have fixed positions and can only oscillate about their mean positions. So solids are rigid and incompressible.

Question 2.
Why do solids have a definite volume ?
Answer:
In solids the strong intermolecular forces bring the particles (atoms, molecules or ions) so close that they cling to one another and occupy fixed positions. So solids have definite volume and shape.

Question 3.
Classify the following as amorphous or crystalline solids : Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Answer:
Amorphous solids : Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.
Crystalline solids : Naphthalene, benzoic acid, potassium nitrate, copper.

Question 4.
Why is glass considered a super cooled liquid ?
Answer:
Glass is an amorphous solid. The structure of amorphous solids is similar to that of liquids. So glass is considered as a super cooled liquid.

Question 5.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of the solid. Would It show cleavage property?
Answer:
In amorphous solids the particles do not have long range order and their arrangement is irregular along all directions. So the value of any physical property would be same along any direction. This property is known as isotropy.

Question 6.
Classify the following solids into different categories based on the nature of inter- molecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer:

1. Molecular solids: Benzene, urea, ammonia, water and argon. Among these benzene and argon contain dispersion or London forces. Urea, ammonia and water contain hydrogen bonding.
2. Ionic solids : Potassium sulphate, zinc sulphide.
3. Metallic solids: Tin, rubidium.
4. Covalent or network solids : Graphite and silicon carbide.

Question 7.
Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it ?
Answer:
In covalent or network solids the adjacent atoms are bonded through covalent bonds throughout the crystal. Covalent bonds are strong and directional in nature, therefore atoms are held very strongly at their positions. Such solids are very hard and brittle. They have extremely high melting points. Hence solid A is covalent solid.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
Ionic solids do not conduct electricity in solid state because ions are held by strong electrostatic forces and are not free to move. However, in molten state the ions become free to move. Hence ionic substances conduct electricity in molten state.

Question 9.
What type of solids are electrical conductors, malleable and ductile ?
Answer:
Metallic solids are electrical conductors, malleable and ductile.

Metals are orderly collection of positive ions surrounded by and held together by a sea of free electrons which are mobile and spread throughout the crystal. These free moving electrons are responsible for high electrical and thermal conductivity of metals.

Question 10.
Give the significance of a ‘lattice point’.
Answer:
If the three dimensional arrangement of constituent particles in a crystal is repre-sented diagramatically, in which each particle is depicted as a point, the arrangement is called crystal lattice and the points are called lattice points. Thus a lattice point represents a particle (atom, molecule or ion) in a crystal.

Question 11.
Name the parameters that characterise a unit cell.
Answer:
A unit cell is characterised by

1. its dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular.
2. Angles between the edges α (between b and c), β (between a and c) and γ (between a and b).
   Thus a unit cell is characterised by six parameters, a, b, c, α, β and γ.

Question 12.
Distinguish between
i) Hexagonal and monoclinic unit cells.
ii) Face centred and end – centred unit cells.
Answer:
(i) In hexagonal unit cell a = b ≠ c and α = β = 90°, γ = 120° but in monoclinic unit cell a ≠ b ≠ c and α = γ = 90°, β ≠ 120°.
(ii) In face centred unit cell a ≠ b ≠ c and α = β = γ = 90° but in end – centred unit cell a ≠ b ≠ c and α = β = γ = 90°.

Question 13.
Explain how much portion of an atom located at
i) corner and
ii) body centre of a cubic unit cell is part of its neighbouring unit cell.
Answer:
i) Each atom at a corner is shared between eight adjacent unit cells. Therefore only \(\frac{1}{8}\)th of an atom belongs to a particular unit cell.
ii) The atom at the body centre wholly belongs to the unit cell in which it is present.

Question 14.
What is the two dimensional coordination number of a molecule in square close – packed layer?
Answer:
In a square close packed layer, the two dimensional coordination number of a molecule is 4.

Question 15.
A compound forms hexagonal close – packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedral voids ?
Answer:
Let the number of close packed spheres be ‘N’ then
The number of octahedral voids = N
The number of tetrahedral voids = 2N
∴ Total number of voids = 3N
Since 1 mol of a substance contain N_o (Avogadro number) particles (atoms or molecules or ions). The total number of voids in one mol is 3N_o. In 0.5 mol the total number of voids is 1 .5N_o. Among these the number of tetrahedral voids is 1.0N_o.

Question 16.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3 rd of tetrahedral voids. What is the formula of the compound?
Answer:
In cubic close packing structure with each atom of N there would be two tetrahedral voids. Because only 1/3 rd of tetrahedral voids are occupied by M there would be 2/3 M for each N. Thus formula of the compound is M_2/3 N or M₂N₃.

Question 17.
Which of the following lattices has the highest packing efficiency
i) simple cubic
ii) body centred cubic and
iii) hexagonal close – packed lattice ?
Answer:
The efficiency of
i) simple cubic is 52.4%
ii) body centred cubic is 68% and
iii) hexagonal close – packed lattice is 74%. So hexagonal close – packed lattice has highest packing efficiency.

Question 18.
An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10^-3 kg m^-3, what is the nature of the cubic unit cell ?
Answer:
Density d = 2.7 × 10³ kg m^-3
Molar mass M = 2.1 × 10^-2 kg mol^-1
Edge length, a = 405 pm or 405 × 10^-12 m
Density d = \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}\)
or Z = \(\frac{\mathrm{d} \times \mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}}\)
= \(\frac{2.7 \times 10^3 \times\left(405 \times 10^{-12}\right)^3 \times 6.022 \times 10^{23}}{2.7 \times 10^{-2}}\) = 4
Since the number of atoms per unit cell is 4 the cubic unit cell must be face centred cube (fcc) (or) ccp. „

Question 19.
What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what way ?
Answer:
On heating a solid vacancy defects may arise. Due to the vacancies the density of the solid would be effected. The density of the solid would decrease due to vacancy defects.

Question 20.
What type of stoichiometric defect is shown by
i) ZnS
ii) AgBr ?
Answer:
i) ZnS shows Frenkel defect.
ii) AgBr shows Frenkel defect as well as Schottky defect.

Question 21.
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Answer:
If a cation of higher valence is introduced into an ionic solid as impurity, it occupies some site that occupied by host cations. In order to maintain electrical neutrality some host ions are missing from their sites leaving behind vacancies. For example, when Sr²⁺ ions are added to NaCl as impurity, each Sr²⁺ ion replaces two Na⁺ ions. Sr²⁺ ion occupies the site of one Na⁺ ion and the other site remains vacant. Thus, the number of vacancies equal to the number of Sr²⁺ ions added are produced.

Question 22.
Ionic solids, which have anionic vacan-cies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Answer:
The colour is due to the excitation of electrons in F- centres by absorbing energy from visible light falling on the crystals, e.g. the anion vacancies in NaCl impart yellow colour to the compound.

Question 23.
A group 14 element is to be converted into n – type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ?
Answer:
To convert the group 14 element into n -type semiconductor it should be doped with an impurity containing extra electron required for bonding. The impurity element should belong to group -15. It may be phosphorous or arsenic. Silicon doped with group – 15 element behaves as a n – type semiconductor.

Question 24.
What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic ? Justify your answer.
Answer:
Ferromagnetic substances would make better permanent magnets because in ferromagnetic substances all the domains get oriented in one direction. When such a substance is placed in a magnetic field. A strong magnetic effect is produced. These substances remain magnetised even when magnetic field is removed.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 9th Lesson Biomolecules Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 9th Lesson Biomolecules

Very Short Answer Questions (2 Marks)

Question 1.
Define Carbohydrates.
Answer:
Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules which provide such units on hydrolysis.

Question 2.
Name the different types of carbohydrates on the basis of their hydrolysis. Give one example for each.
Answer:
Carbohydrates are classified into three types on the basis of their hydrolysis.

1. Monosaccharides Ex: glucose
2. Oligosaccharides Ex: sucrose
3. Polysaccharides Ex: starch

Question 3.
Why are sugars classified as reducing and non – reducing sugars ?
Answer:
In monosaccharides, whether they are aldoses or ketoses, the functional groups are free and reduce Fehling’s solution etc. In some disaccharides such as sucrose, the reducing groups of the monosaccharides are bonded. They do not show reducing properties. On the other hand, in sugars such as maltose and lactose, the functional groups are free and reduce Fehling’s solution. Hence, sugars are classified into reducing and non – reducing sugars.

Question 4.
What do you understand from the names
a) aldopentose and
b) ketoheptose ?
Answer:
a) Aldopentose means a monosaccharide having an aldehyde group and contains five carbon atoms in its molecule.
b) Ketoheptose means a monosaccharide having a ketogroup and contains seven carbon atoms in its molecule.

Question 5.
Write two methods of preparation of glucose. [IPE 14]
Answer:
Preparation of glucose:
1) From sucrose (Cane sugar): Sucrose on boiling with dilute HCl or H₂SO₄ in alcoholic solution undergoes hydrolysis and gives equal amounts of glucose and fructose. On cooling the resulting solution, glucose being much less soluble than fructose, separates out.

2) From starch : Commercially glucose is made by the hydrolysis of starch by heat¬ing with very dilute sulphuric acid at 120°C under pressure.

Question 6.
Glucose reacts with bromine water to give gluconic acid. What information do you get from this reaction about the structure of glucose ?
Answer:
Glucose is oxidised by a mild oxidising agent like bromine water to give gluconic acid, a monocarboxylic acid with molecular formula C₆H₁₂O₇. This indicates the presence of an aldehyde group since only the aldehyde group can be oxidised to an acid by gaining one oxygen atom without losing any hydrogen atoms.

Question 7.
Glucose and gluconic acid on oxidation with nitric acid give saccharic acid. What information do you get from this reaction about the structure of glucose ?
Answer:
On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid called saccharic acid with molecular formula C₆H₁₀O₈. This indicates the presence of a primary alcoholic group.

Question 8.
Glucose reacts with acetic anhydride to form penta acetate. What do you understand about the structure of glucose from his reaction ?
Answer:
Acetylation of glucose with acetic anhydride gives glucose penta acetate which confirms the presence of five -OH groups. Since it exists as a stable compound the five -OH groups should be attached to different carbon atoms.

Question 9.
Give any two reasons to understand that glucose molecule has no open chain structure.
Answer:

1. Although glucose has am aldehyde group, it does not give the Schiff’s test and does not form addition product with sodium bisulphite.
2. Glucose pentaacetate does not react with hydroxylamine.
   These observations suggest that glucose has no open structure.

Question 10.
D – glucose means dextro rotatory glucose. Is it true ? why ?
Answer:
The letter D’ before the name of glucose does not mean dextro rotatory. ‘D’ represents relative configuration with reference to glyceraldehyde. On the other hand, ’+’ is used to represent the dextro rotatory nature of glucose.

Question 11.
What are anomers ?
Answer:
Glucose forms a six – membered ring involving the -OH at C – 5 and the -CHO group at C – 1. Thus glucose exists in two cyclic hemiacetal forms which differ in their stereo chemistry at C₁, called the anomeric carbon (former aldehyde carbon). Such isomers i.e., α – form and β – form, are known as anomers.

Question 12.
Write the ring structure of D – Glucose. What are their names ?
Answer:

Question 13.
Write the ring structure and open chain structure of fructose.
Answer:
Ring structure and open chain structures of fructose:

Question 14.
What do you understand by invert sugars?
Answer:
Sucrose is dextrorotatory but after hydrolysis gives an equimolar mixture of glucose which is dextrorotatory and fructose which is laevorotatory. The hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (-) and the product is called invert sugar.

Question 15.
What are amino acids? Give two examples.
Answer:
Compounds containing both amino (-NH₂) and carboxyl (-COOH) functional groups are called amino acids.
Ex: Glycine, Alanine.

Question 16.
Write the structure of alanine and aspartic acid.
Answer:
NH₂ – CH₂ – COOH Alanine

Question 17.
What do you mean by essential amino acids ? Give two examples for non – essential amino acids. [TS 16; IPE 14]
Answer:
Amino acids, which cannot be synthesised in the body and must be obtained through diet are called essential amino acids. Examples for non – essential amino acids : glycine, alanine.

Question 18.
What is Zwitter ion ? Give an example.
Answer:
In an amino acid both acidic (carboxyl group) and basic (amino group) groups are present in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as Zwitter ion.
Ex: H₂N – CH₂ – COOH ⇌ H₃ \(\stackrel{+}{\mathrm{N}}\) -CH₂ – \(\mathrm{CO} \overline{\mathrm{O}}\)

Question 19.
What are proteins ? Give an example.
Answer:
Proteins are polymers of α – amino acids. They are the most abundant biomolecules of the living system.
Ex : Insulin

Question 20.
What are fibrous proteins ? Give examples.
Answer:
Proteins in which the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds will have fibre – like structure. Such proteins are called fibrous proteins.
Ex : Keratin (present in hair, wool, silk)

Question 21.
What are globular proteins? Give examples.
Answer:
Proteins in which the polypeptide chains coil around to give a spherical shape are called globular proteins.
Ex : Insulin

Question 22.
How are proteins classified with respect to peptide bond ?
Answer:
With respect to peptide bond, proteins are classified into dipeptides, tripeptides, tetra- peptides, pentapeptides, hexapeptides etc depending upon the number of amino acids linked by peptide bonds. When the number of amino acids is more than ten, then the products are called polypeptides.

Question 23.
What are the components of a nucleic acid?
Answer:
There are three components of a nucleic acid. They are:

1. A pentose sugar
2. Phosphoric acid and
3. Nitrogen containing heterocyclic base.

Question 24.
Write the names of three types of RNA.
Answer:

1. Messenger RNA ( m – RNA)
2. Ribosomal RNA (r – RNA) and
3. Transfer RNA (t – RNA)

Question 25.
Write the biological functions of nucleic acids.
Answer:
Biological functions of nucleic acids:
1) Nucleic acids control heredity at molecular level. DNA is the chemical basis of heredity. It may be regarded as the reserve of genetic information. DNA is exclusively responsible for maintaining the identity of different species of organisms over millions of years. A DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells.

2) Nucleic acids have an important role in protein synthesis. Actually the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 26.
Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamin K

Question 27.
What are monosaccharides ?
Answer:
Carbohydrates which cannot be hydrolysed into smaller units are called monosaccharides.

Question 28.
What are reducing sugars ?
Answer:
Carbohydrates which reduce Fehling’s solution and Tollen’s reagent are called reducing sugars.

Question 29.
Write two main functions of carbohydrates in plants.
Answer:
Carbohydrates mainly act as the food storage or structural materials in plants.

1. Starch is the main storage polysaccharide of plants.
2. Cell wall of plants is made up of cellulose.

Question 30.
Classify the following into monosaccharides and disaccharides.
i) Ribose
ii) 2 – deoxy ribose
iii) Maltose
iv) Fructose
Answer:
i) Ribose …………… Monosaccharide
ii) 2 – deoxy ribose …………… Monosaccharide
iii) Maltose …………… Disaccharide
iv) Fructose …………… Monosaccharide

Question 31.
What do you understand by the term glycosidic linkage ?
Answer:
The two monosaccharide units in a disaccharide are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage is called glycosidic linkage.

Question 32.
What is glycogen ? How is it different from starch ?
Answer:
Carbohydrates are stored in animal body as glycogen. It is also known as animal starch. The structure of glycogen is similar to amylopectin, a component of starch, but it is more highly branched.

Question 33.
What are the hydrolysis products of
i) sucrose and
ii) lactose ?
Answer:
i) Sucrose on hydrolysis gives D – (+) – glucose and D – (-) – fructose.
ii) Lactose on hydrolysis gives galactose and glucose.

Question 34.
What is the basic structural difference between starch and cellulose ?
Answer:
Starch is a polymer of α – glucose. The α – D – glucose units are joined by glycosidic linkage. In amylose, which constitutes about 15 – 20% of starch, 200 -1000 α – D(+) – glucose units are held by C – 1 to C – 4 glycosidic linkage in a long unbranched chain. Amylopectin, the other component which constitutes about 80 – 85% of starch is a branched chain polymer of α – D – glucose units in which the chain is formed by C -1 to C – 4 glycosidic linkage whereas branching occurs by C -1 to C – 4.

Cellulose is a straight chain polysaccharide composed only of β – D – glucose units which are joined by glycosidic linkage between C -1 of one glucose unit and C – 4 of the next glucose unit.

Question 35.
What happens when D’- glucose is treated with the following reagents.
i) HI
ii) Bromine water
iii) HNO₃
Answer:
i) On prolonged heating with HI, glucose forms n – hexane.

ii) When glucose is treated with bromine water it gets oxidised to gluconic acid, a monocarboxylic acid with the molecular formula, C₆H₁₂O₇.

iii) On oxidation with nitric acid, glucose gives a dicarboxylic acid called saccharic acid, with molecular formula, C₆H₁₀O₈.

Question 36.
Enumerate the reactions of D – glucose which cannot be explained by its open chain structure.
Answer:
The following reactions of D – glucose cannot be explained by its open chain structure.

1. Glucose does not give Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO₃.
2. The pentaacetate of glucose doesnot react with hydroxylamine indicating the absence of free -CHO group.

Question 37.
What are essential and non – essential amino acids ? Give one example for each.
Answer:
Amino acids which cannot be synthesised in the body and must be obtained through diet are called essential amino acids.
Ex: Valine

Amino acids which are synthesised in the body are known as non – essential amino acids.
Ex: Glycine

Question 38.
Define the following as related to proteins.
i) Peptide linkage
ii) Primary structure
iii) Denaturation
Answer:
i) Peptide linkage :
Peptide linkage is an amide formed between -COOH group of one amino acid molecule and the NH₂ group of another molecule of the same or different amino acid.

ii) Primary structure :
The primary structure of proteins refers to the sequence of amino acids held together by peptide linkages.

iii) Denaturation :
The loss of biological acitvity of a protein due to the destruction of the highly organized tertiary structure is called denaturation.
Ex: coagulation of egg by boiling

Question 39.
What are the common types of secondary structure of proteins ?
Answer:
The secondary structure of proteins is of two common types:

1. α – helix and
2. β – pleated sheet structure

Question 40.
What type of bonding helps in stabilizing the a – helix structure of proteins ?
Answer:
The spiral or α – helix is stabilized by interchain hydrogen bonding. The -NH group of each amino acid residue is hydrogen bonded to the = O of an adjacent turn of the helix in the same chain.

Question 41.
Differentiate between globular and fibrous proteins.
Answer:
In globular proteins the polypeptide chains coil around to give a spherical shape. These are usually soluble in water.

In fibrous proteins the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. These proteins have a fibre like structure. These are usually insoluble in water.

Question 42.
How do you explain the amphoteric behaviour of amino acids ?
Answer:
Amino acids show both acidic and basic properties. This is called amphoteric behaviour. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton giving rise to a dipolar ion known as zwitter ion.

Question 43.
Why are vitamin A and vitamin C essential to us ? Give their important sources.
Answer:
Vitamin A is essential tolls because its deficiency causes xerophthalmia (hardening of cornea of the eye) and night blindness.
Important sources of vitamin A : Fish liver oil, milk etc.
Vitamin C is essential to us because its deficiency causes scurvy (bleeding gums).
Sources of vitamin C: Citrus fruits, green leafy vegetables.

Question 44.
What are nucleic acids ? Mention their two important functions.
Answer:
Nucleic acids are chains of five – membered ring sugars linked by phosphate groups. The anomeric carbon of each sugar is bonded to a nitrogen of a heterocyclic compound in a β – glycosidic linkage. Functions of nucleic acids:

1. DNA is the chemical basis of heredity and may be regarded as the reserve of genetic information.
2. Nucleic acids have an important role in protein synthesis. Actually the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 45.
What is the difference between a nucleo-side and a nucleotide ?
Answer:
A unit formed by the attachment of a base to 1′ position of sugar is known as nucleoside.
Ex: Cytidine

When nucleoside is linked to phosphoric acid at 5′ position of sugar moity, we get a nucleotide.
Ex: Adinodine Triphosphate

Short Answer Questions (4 Marks)

Question 46.
How are the carbohydrates classified on the basis of their
a) Taste
b) Hydrolysis
c) Functional groups ?
Answer:
a) On the basis of taste carbohydrates are classified into sugars and non – sugars. Carbohydrates which are sweet to taste are called sugars while others are called non – sugars.

b) Carbohydrates are classified into three groups on the basis of their hydrolysis,

1. Monosaccharides
2. Oligosaccharides and
3. Polysaccharides.

c) Carbohydrates are classified into

1. aldoses and
2. ketoses on the basis of the functional group present in them.

Question 47.
Write a brief note on the structure of glucose.
Answer:
Glucose is an aldohexose.

1. The molecular formula of glucose is C₆H₁₂O₆.
2. On prolonged heating with HI, glucose forms n – hexane suggesting that all the carbon atoms are linked in a straight chain in its molecule.
3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of HCN to give cyanohydrin. These reactions show the presence of a carbonyl group ( = O) in glucose molecule.
4. Glucose on oxidation with bromine water, gives gluconic acid containing the same number of carbon atoms as in its molecule. This indicates that the carboxyl group is an aldehyde group.
5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which indicates the presence of five -OH group. Since glucose is a stable compound, the five -OH groups should be present on different carbon atoms.
6. Glucose, as well as gluconic acid, is oxidised with nitric acid to give saccharic acid, a dicarboxylic acid which has the same number of carbon atoms as glucose. This indicates the presence of a primary alcoholic (-OH) group in glucose.

The exact spacial arrangement of different -OH groups in glucose molecule was given by Fisher. The open chain structure of glucose is given as follows.

The above structure of glucose explained most of its properties. But it could not explain the following reactions.
1) Despite having the aldehyde group, glucose does not give Schiff’s test and does not form the addition product with sodium bisulphite.

2) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.
Glucose is found to exist in two different crystalline forms named as α and β. To expLain all these observations the following six membered ring structure was proposed to Glucose.

Question 48.
Write short notes on Sucrose.
Answer:
Sucrose (C₁₂H₂₂O₁₁) is the most common disaccharide widely present in plants. It is mainly obtained from sugarcane or beetroot.

It is a colourless crystalline substance sweet to taste. It is dextrorotatory.

Sucrose on hydrolysis gives equimolar mixture of D – (+) – glucose and D – (-) – fructose. These two monosaccharide units are held together by a glycosidic linkage between C – 1 of α – glucose and C – 2 of β – fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non – reducing sugar.

Question 49.
Write the structures of maltose and lactose. What are the products of hydrolysis of maltose and lactose?
Answer:
Structure of Maltose :

Hydrolysis products : Maltose on hydro-lysis gives α – D – glucose. Lactose on hydrolysis gives β – D – galactose and β – D – glucose.

Question 50.
Write about polysaccharides with starch and cellulose as examples.
Answer:
Carbohydrates containing large number of monosaccharide units joined together through glycosidic linkages are called polysaccharides. They act as food storages or structural materials.

Starch:
Starch is the main storage poly-saccharide of plants. It is the most important dietary source for human beings. Starch is found in cereals, roots, tubers and some vegetables. It is a polymer of α – glucose and consists of two components amylose and amylopectin. Amylose is a water soluble component which constitutes 15 – 20% of starch.

Chemically amylose is a long unbranched chain with 200-1000 α – D – (+) – glucose units held by C -1 to C – 4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80 – 85% of starch. It is a branched chain polymer of α – D – glucose units in which the chain is formed by C -1 to C – 4 glycosidic linkage whereas branching occurs by C -1 to C – 6 glycosidic linkage.

Starch is a white amorphous powder almost insoluble in cold water but relatively more soluble in boiling water. Starch solution gives blue colour with iodine solution but the blue colour disappears on heating.

Cellulose :
Cellulose is the main structural component of vegetable matter. It is a straight chain polysaccharide composed only of β – glucose units which are joined by glycosidic link between C -1 of one glucose unit and C – 4 of the next glucose unit.

Question 51.
Write the importance of carbohydrates.
Answer:
Carbohydrates are essential for the life of both plants and animals. Honey is an instant source of energy. Carbohydrates form a major portion of our food. They are used as storage molecules as starch in plants and glycogen in animals. Cell walls of bacteria and plants are made up of cellulose.

Wood and cotton fibre contain cellulose. They provide raw materials for many important industries like textiles, paper, lacquers and breweries. D – ribose and 2 – deoxy – D – ribose are present in nucleic acids. Carbohydrates are present in biosystems in combination with many proteins and lipids.

Question 52.
Exaplain the classification of proteins as primary, secondary, tertiary and quarternary proteins with respect to their structure.
Answer:
The structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quarternary.

i) Primary structure of proteins:
Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. It is this sequence of amino acids that is said to be the primary structure. The primary structure represents the constitution of the protein.

ii) Secondary structure of proteins:
The secondary structure of proteins refers to the shape in which a long polypeptide chain can exist. These chains are found to exist in two different types of structures (1) α – helix and (2) β – pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between = O and -NH groups of the peptide bond.

iii) Tertiary structure of proteins :
The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous (thread like) and globular.

iv) Quarternary structure of proteins :
Some of the proteins are composed of two or more polypeptide chains referred to as sub – units. The special arrangement of these subunits with respect to each other is known as quarternary structure.

Question 53.
Explain the denaturation of proteins. [TS 16]
Answer:
When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation, secondary and tertiary structures are destroyed but the primary structure remains intact. Coagulation of egg white on boiling is a common example of denaturation.

Question 54.
What are enzymes ? Give examples.
Answer:
Enzymes are biocatalysts which speed up reactions in biosystems. They are very specific and selective in their action and chemically all enzymes are proteins.
Ex:
1) The enzyme that catalyses the hydrolysis of maltose into glucose is named maltase.

2) An aqueous solution of sucrose is fermented by yeast to give ethyl alcohol and carbon dioxide. The enzyme invertage present in yeast converts sucrose into glucose and fructose. These sugars are then decomposed by the enzyme zymase (also present in yeast) to give ethyl alcohol and carbondioxide.

Question 55.
Explain the role of sucrose in its hydrolysis.
Answer:
Sucrose is dextrorotatory but after hydrolysis gives an equimolar mixture of glucose which is dextrorotatory and fructose which is laevorotatory. Since laevorotation of fructose (-92.4°) is more than dextrorotation of glucose (+52.5°), the mixture is laevorotatory. Thus hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (-) and the product is called invert sugar.

Question 56.
Write notes on vitamins.
Answer:
Vitamins are organic compounds which are required in small quantities in our diet for the maintenance of normal health. Their deficiency causes specific diseases. Plants can synthesise almost all vitamins. Vitamins are designated by alphabets A, B, C, D, E, K. Some of them are further named as subgroups e.g. B₁, B₂, B₆, B₁₂ etc. Excess of vitamins is also harmful.

Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Fat soluble vitamins: These are soluble in fat and oils but insoluble in water. Vitamins A, D, E and K are fat soluble vitamins.
2. Water soluble vitamins : B group vitamins and vitamin C are soluble in water.

Some vitamins, their sources and deficiency diseases:
Vitamin A:
Sources: Fish liver oil, butter and milk. Deficiency diseases: Xerophthalmia, Night blindness.
Vitamin B₁ (Thiamine):
Sources: Milk, green vegetables and cereals.
Deficiency diseases: Beri beri

Vitamin B₂ (Riboflavin):
Sources: Milk, egg white, liver.
Deficiency diseases: Cheilosis

Vitamin C:
Sources: Citrus fruits, amla and green leafy vegetables.
Deficiency diseases: Scurvy

Question 57.
What do you understand by The two strands of DNA are complementary to each other” ? Explain.
Answers:
Nucleic acids have a secondary structure also besides the primary structure. DNA was given a double strand helix structure. Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Question 58.
What are hormones ? Give one example for each. [Mar. 19, 18 ; AP & TS]

1. Steroid hormones
2. Poly peptide hormones and
3. Amino acid derivatives.

Answer:
Hormones are molecules that act as intercellular messengers. They have several functions in the body.

1. Steroid hormones:
   a) Adrenal cortical hormones : Ex : Corti- coids
   b) Sex hormones
   a) Male sex hormones: Ex: Testosterone
   b) Female sex hormones : Ex : Estrone
2. Poly peptide hormones : Ex : Insulin
3. Amino acid derivatives: Ex: Thyroxine

Question 59.
Give the sources of the following vitamins and name the diseases caused by their deficiency (a) A (b) D (c) E and (d) K. [AP 16; 15; TS 15]
Answer:

Long Answer Questions (8 Marks)

Question 60.
Explain the classification of carbohydrates.
Answers:
1) Classification on the basis of hydrolysis:
Carbohydrates are classified into three types on the basis of their hydrolysis.

1. Monosaccharides
2. Oligosaccharides and
3. Polysaccharides.

Monosaccharides:
Carbohydrates which cannot be hydrolysed into smaller units are called monosaccharides.
Ex : Glucose, fructose, ribose etc.

Oligosaccharides:
Carbohydrates that yield two to ten monosaccharide units on hydrolysis are called oligosaccharides. They are further classified into disaccharides, trisaccharides, tetrasaccharides etc., depending upon the number of monosaccharides they provide on hydrolysis. Disaccharides, on hydrolysis, give two monosaccharide units which may be same or different.
Ex : Sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose.

Maltose on hydrolysis gives two molecules of glucose only.

Polysaccharides:
Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Starch, cellulose, glycogen etc., are examples for polysaccharides.

2) Classification on the basis of reducing properties:
Carbohydrates are also classified into reducing and non – reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are called reducing sugars.
Ex: Glucose, fructose, maltose, lactose.

Sugars which do not reduce Fehling’s solution etc are called non – reducing sugars. For example sucrose is a non – reducing sugar.

3) Classification based on taste :
Carbo-hydrates are also classified into sugars and non – sugars on the basis of their taste. Carbohydrates such as glucose, fructose, sucrose etc., which are sweet to taste are called sugars. Polysaccharides are not sweet to taste, hence they are called non – sugars.

Question 61.
Discuss the structure of glucose on the basis of its chemical properties.
Answer:
Glucose is an aldohexose.

1. The molecular formula of glucose is C₆H₁₂O₆.
2. On prolonged heating with HI, glucose forms n – hexane suggesting that all the carbon atoms are linked in a straight chain in its molecule.
3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of HCN to give cyanohydrin. These reactions show the presence of a carbonyl group ( = O) in glucose molecule.
4. Glucose on oxidation with bromine water, gives gluconic acid containing the same number of carbon atoms as in its molecule. This indicates that the carboxyl group is an aldehyde group.
5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which indicates the presence of five -OH group. Since glucose is a stable compound, the five -OH groups should be present on different carbon atoms.
6. Glucose, as well as gluconic acid, is oxidised with nitric acid to give saccharic acid, a dicarboxylic acid which has the same number of carbon atoms as glucose. This indicates the presence of a primary alcoholic (-OH) group in glucose.

The exact spacial arrangement of different -OH groups in glucose molecule was given by Fisher. The open chain structure of glucose is given as follows.

The above structure of glucose explained most of its properties. But it could not explain the following reactions.
1) Despite having the aldehyde group, glucose does not give Schiff’s test and does not form the addition product with sodium bisulphite.

2) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.
Glucose is found to exist in two different crystalline forms named as α and β. To expLain all these observations the following six membered ring structure was proposed to Glucose.

Question 62.
Write notes on (a) fructose (b) sucrose (c) maltose (d) lactose.
Answer:
a) Fructose:
Fructose is found in fruits and honey. It is the sweetest sugar. It is obtained along with glucose by the hydrolysis of sucrose. It is an important ketohexose.

The molecular formula of fructose is C₆H₁₂O₆. On the basis of its reactions it is found to contain a ketonic functional group at C₂ and all she carbons in straight chain as in glucose. It is laevorotatory and belongs to D – series. Its open chain structure is Fructose also exists in two cyclic forms which are obtained by the addition of -OH at C₅ to the ( = O) group.
To explain all the properties of fructose, a furanose ring structure is suggested.

b) Sucrose:
Sucrose is a common disaccharide. It is obtained mainly from sugarcane or beet root. It is a colourless crystalline sweet substance. Sucrose on hydrolysis gives equimolar mixture of glucose and fructose.

The two monosaccharide units are held together by a glycosidic linkage between C₁ of α – glucose and C₂ of β – fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non – reducing sugar. Sucrose is dextrorotatory but after hydrolysis gives a mixture of dextrorotatory glucose and laevorotatory fructose. The mixture is lae- vorotatory. The product of hydrolysis is called invert sugar.

c) Maltose :
Maltose is a disaccharide. It is composed of two α – D – glucose units in which C₁ of one glucose unit is linked to C₄ of another glucose unit. Maltose is a reducing sugar.

d) Lactose:
Lactose occurs in milk, hence it is also known as milk sugar. On hydrolysis lactose gives β – D -galactose and β – D – glucose. The linkage is between C₁ of galactose and C₄ of glucose. It is a reducing sugar.

Question 63.
Write notes on (a) starch (b) cellulose (c) importance of carbohydrates.
Answer:
Carbohydrates containing large number of monosaccharide units joined together through glycosidic linkages are called polysaccharides. They act as food storages or structural materials.

Starch:
Starch is the main storage poly-saccharide of plants. It is the most important dietary source for human beings. Starch is found in cereals, roots, tubers and some vegetables. It is a polymer of α – glucose and consists of two components amylose and amylopectin. Amylose is a water soluble component which constitutes 15 – 20% of starch.

Chemically amylose is a long unbranched chain with 200-1000 α – D – (+) – glucose units held by C -1 to C – 4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80 – 85% of starch. It is a branched chain polymer of α – D – glucose units in which the chain is formed by C -1 to C – 4 glycosidic linkage whereas branching occurs by C -1 to C – 6 glycosidic linkage.

Starch is a white amorphous powder almost insoluble in cold water but relatively more soluble in boiling water. Starch solution gives blue colour with iodine solution but the blue colour disappears on heating.

Cellulose :
Cellulose is the main structural component of vegetable matter. It is a straight chain polysaccharide composed only of β – glucose units which are joined by glycosidic link between C -1 of one glucose unit and C – 4 of the next glucose unit.

Carbohydrates are essential for the life of both plants and animals. Honey is an instant source of energy. Carbohydrates form a major portion of our food. They are used as storage molecules as starch in plants and glycogen in animals. Cell walls of bacteria and plants are made up of cellulose.

Wood and cotton fibre contain cellulose. They provide raw materials for many important industries like textiles, paper, lacquers and breweries. D – ribose and 2 – deoxy – D – ribose are present in nucleic acids. Carbohydrates are present in biosystems in combination with many proteins and lipids.

Question 64.
Write notes on amino acids.
Answer:
Compounds containing both amino (-NH₂) and carboxyl (-COOH) functional groups are called amino acids.
Ex: Glycine, Alanine.

Amino acids which cannot be synthesised in the body and must be obtained through diet are called essential amino acids.
Ex: Valine
Amino acids which are synthesised in the body are known as non – essential amino acids.
Ex: Glycine

Amino acids show both acidic and basic properties. This is called amphoteric behaviour. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton giving rise to a dipolar ion known as zwitter ion.

Question 65.
Write notes on proteins.
Answer:
Proteins are polymers of α – amino acids. They are the most abundant biomolecules of the living system.
Ex : Insulin

Proteins in which the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds will have fibre – like structure. Such proteins are called fibrous proteins.
Ex : Keratin (present in hair, wool, silk)

Proteins in which the polypeptide chains coil around to give a spherical shape are called globular proteins.
Ex : Insulin

The structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quarternary.

i) Primary structure of proteins:
Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. It is this sequence of amino acids that is said to be the primary structure. The primary structure represents the constitution of the protein.

ii) Secondary structure of proteins:
The secondary structure of proteins refers to the shape in which a long polypeptide chain can exist. These chains are found to exist in two different types of structures (1) α – helix and (2) β – pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between = O and -NH groups of the peptide bond.

iii) Tertiary structure of proteins :
The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous (thread like) and globular.

iv) Quarternary structure of proteins :
Some of the proteins are composed of two or more polypeptide chains referred to as sub – units. The special arrangement of these subunits with respect to each other is known as quarternary structure.

When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation, secondary and tertiary structures are destroyed but the primary structure remains intact. Coagulation of egg white on boiling is a common example of denaturation.

Question 66.
Write notes on (a) enzymes and (b) vitamins.
Answer:
Enzymes are biocatalysts which speed up reactions in biosystems. They are very specific and selective in their action and chemically all enzymes are proteins.
Ex:
1) The enzyme that catalyses the hydrolysis of maltose into glucose is named maltase.

2) An aqueous solution of sucrose is fermented by yeast to give ethyl alcohol and carbon dioxide. The enzyme invertage present in yeast converts sucrose into glucose and fructose. These sugars are then decomposed by the enzyme zymase (also present in yeast) to give ethyl alcohol and carbondioxide.

Vitamins are organic compounds which are required in small quantities in our diet for the maintenance of normal health. Their deficiency causes specific diseases. Plants can synthesise almost all vitamins. Vitamins are designated by alphabets A, B, C, D, E, K. Some of them are further named as subgroups e.g. B₁, B₂, B₆, B₁₂ etc. Excess of vitamins is also harmful.

Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Fat soluble vitamins: These are soluble in fat and oils but insoluble in water. Vitamins A, D, E and K are fat soluble vitamins.
2. Water soluble vitamins : B group vitamins and vitamin C are soluble in water.

Some vitamins, their sources and deficiency diseases:
Vitamin A:
Sources: Fish liver oil, butter and milk. Deficiency diseases: Xerophthalmia, Night blindness.
Vitamin B₁ (Thiamine):
Sources: Milk, green vegetables and cereals.
Deficiency diseases: Beri beri

Vitamin B₂ (Riboflavin):
Sources: Milk, egg white, liver.
Deficiency diseases: Cheilosis

Vitamin C:
Sources: Citrus fruits, amla and green leafy vegetables.
Deficiency diseases: Scurvy

Question 67.
Explain the structures of DNA and RNA.
Answer:
There are three components of a nucleic acid.
i) A pentose sugar ii) Phosphoric acid and iii) Nitrogen containing amino acid.

A simplified version of nucleic acid chain is as follows.

Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure also.

James Watson and Francis Crick proposed a double strand helix structure for DNA. Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

In secondary structure of RNA, helices are present but they are only single stranded. Sometimes they fold back on themselves like a hairpin thus acquiring double helix structure possesing double stranded characteristics. In double stranded arrangement guanine pairs with cytosine and adenine with uracil.

Question 68.
Write notes on the functions of different hormones in the body.
Answer:
Functions of hormones :
I. Steroid hormones:
a) Sex hormones:
i) Male sex hormones (androgens):
Test-osterone is the principal male sex hormone produced by testis. This is responsible for the development of secondary male sexual characteristics such as deep voice and facial hair.

ii) Female sex hormones (estrogens) :
Estradiol is the main female sex harmone. It is responsible for development of secondary female sex characteristics such as breast development, small voice, long hair. It also takes part in control of menstrual cycle.

iii) Pregnancy hormones progesterone) :
Progesterone is useful for preparing the uterus for the implantation of the fertilised egg. Cortico steroids (adrenal cortical hormones) are released by the adrenal cortex.

b) Cortico steroids (adrenal cortical hor-mones :
These are released by the adrenal cortex.
i) Mineralo corticoids :
These hormones control the excretion of water and salt by the kidney.

ii) Glucocorticoids :
These control the carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.

If the adrenal cortex does not function properly then one of the results may be Addison’s disease. The disease is fatal unless it is treated by glucocorticoids and mineralocorticoids.

II. Non – steroidal hormones :
i) Peptide hormones:
The most important peptide hormone is insulin. It has great influence on carbohydrate metabolism. It is released in response to the rapid release in the glucose level in the blood. On the other hand the hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood.

ii) Amino acid derivations :
These are thyroidal hormones. Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine. Abnormally low level of thyroxine leads to hypothyroidism which is characterised by Tetha- rgynea and obesity. Increased level of . thyroxine causes hyperthyroidism.

Intext Questions – Answers

Question 1.
Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
Water is a polar solvent. Glucose and sucrose are polar compounds. Hence they are soluble in water. The partial charges of water interact with the partial charges on glucose and sucrose.

On the other hand, cyclohexane and benzene are non – polar compounds. Water molecules are not attracted to them because they have no net charge. Hence they are insoluble in water.

Question 2.
What are the expected products of hydrolysis of lactose ?
Answer:
The expected products of hydrolysis of lactose are β – D – galactose and β – D – glucose.

Question 3.
How do you explain the formation of furanose structure for fructose while glucose forms pyranose structure with the same molecular formula C₆H₁₂O₆ ?
Answer:
Fructose exists in two cyclic forms which are obtained by the addition of -OH at C₅ to the ( = O) group. The ring thus formed is a five membered ring and is named as furanose with analogy to the heterocyclic compound furan which has a five membered ring with one oxygen and four carbon atoms.

It was found that glucose forms a six – membered ring in which -OH at C₅ is involved in ring formation. The -OH at C₅ forms a cyclic hemiacetal structure with the -CHO group. The six membered cyclic structure of glucose is called pyranose structure in analogy with pyran.

Question 4.
The melting points and solubility of amino acids in water are generally higher than those of the corresponding halo acids. Explain.
Answer:
Amino acids are crystalline solids. They are water soluble and have high melting points. They behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution the carboxyl group can lose a proton and the amino group can accept a proton, giving rise to a dipolar ion known as Zwitter ion.

The corresponding haloacids do not show salt – like behaviour and hence their melting points and solubility in water are less.

Question 5.
Temperature and pH affect the native proteins. Explain.
Answer:
Protein found in biological system with a unique three – dimensional structure and biological activity is called as native protein. When a protein in its native form is subjected to change in temperature or pH the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein.

Question 6.
Why can’t vitamin C be stored in our body?
Answer:
Vitamin C is a water soluble vitamin. It is readily excreted in urine and cannot be stored in our body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
Adenine and thymine would be formed when a nucleotide from DNA containing thymine is hydrolysis because adenine always pairs up with thymine.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA ?
Answer:
RNA molecule is a single strand complementary to only one of the two strands of a gene. Its guanine content does not necessarily be equal to its cytocine content, nor does its adenine contents to its uracil content. Hence, when RNA is hydrolysed there is no relationship among the quantities of different bases obtained.

Question 9.
What is DNA finger printing ? Mention its uses.
Answer:
Every individual has unique finger prints. These occur at the tips of fingers and have been used for identification. A sequence of bases on DNA is unique for a person and information regarding this is called DNA finger printing. It is same for every cell and can not be altered.

Importance : DNA finger printing is used
a) in forensic laboratories for identification of criminals.
b) to determine parternity of an individual.
c) to identify dead bodies in accident cases.
d) to identify racial groups to rewrite biological evolution.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry in Everyday Life Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry in Everyday Life

Very Short Answer Questions (2 Marks)

Question 1.
What are drugs ?
Answer:
Drugs are chemicals of low molecular masses (~ 100 to 500 u), that interact with macro-molecular targets and produce a biological response.

Question 2.
When are the drugs called medicines ?
Answer:
The drugs are called medicines when their biological response is therapeutic and useful.

Question 3.
Define the term chemotherapy.
Answer:
Chemotherapy is defined as the use of chemicals for therapeutic effect.

Question 4.
Name the macromolecules that are chosen as drug targets.
Answer:
Carbohydrates, lipids, proteins and nucleic acids are the macromolecules that are chosen as drug targets.

Question 5.
What are enzymes and receptors ?
Answer:
Proteins which perform the role of biological catalysts in the body are called enzymes. Receptors are proteins that are crucial to body’s communication process.

Question 6.
Which forces are involved in holding the drug to the active site of enzymes ?
Answer:
Forces such as ionic bonding, hydrogen bonding, van der Waals interaction or dipole – dipole interaction are involved in holding the drug to the active site of enzymes.

Question 7.
What are enzyme inhibitors ?
Answer:
Drugs which can block the binding site of the enzyme and prevent the binding of substrate or can inhibit the catalytic activity of the enzyme are called enzyme inhibitors.

Question 8.
What is allosteric site ?
Answer:
Allosteric site is a site in the enzyme that is different from the active site. The binding of the inhibitor at the allosteric site changes the shape of the active site in such a way the substrate cannot recognise it.

Question 9.
What are antagonists and agonists ?
Answer:
Drugs that bind to the receptor site and inhibit its natural function are called antagonists. Drugs that mimic the natural messenger by switching on the receptor are called agonists.

Question 10.
Why do we need to classify the drugs in different ways ?
Answer:
The classification of drugs in different ways is useful to the doctors because it provides them the whole range of drugs available for the treatment of a particular disease.

Question 11.
What are antacids ? Give example. [IPE 14]
Answer:
Drugs which are used for the treatment of over production of acid in the stomach (acidity) are called antacids.
Ex : Ranitidine (Zantac), Omeprazole, Lansoprazole.

Question 12.
What are antihistamines ? Give example.
Answer:
Drugs which interfere with the natural action of histamine by competing with histamine for binding sites of receptor where histamine exerts its effects are called antihistamines.
Ex : Brompheniramine, Dincetane, terfenadine.

Question 13.
While antacids and antiallergic drugs interfere with the function of histamines, why do not these interfere with the function of each other ?
Answer:
Antacids and antiallergic drugs work on different receptors hence they do not interfere with each other’s function.

Question 14.
What are tranquilizers ? Give example.
Answer:
Tranquilizers are a class of chemical compounds used for the treatment of stress and mild or even severe mental diseases.
Ex: Iproniazid, Luminol, Second, Babituric acid.

Question 15.
What are barbiturates ?
Answer:
Derivatives of barbituric acid which con-stitute an important class of tranquilizers are called barbiturates.
Ex: veronal, amytal and seconal.

Question 16.
What are analgesics? How are they classified ?
Answer:
Drugs which reduce or abolish pain without causing disturbances of nervous system like impairment of consciousness, mental confusion, in coordination, paralysis etc., are called analgesics Analgesics are classified as

1. Narcotic analgesics and
2. Non – Narcotic analgesics.

Question 17.
What are narcotic analgesics ? Give example.
Answer:
Morphine and many of its homologues, when administered in medicinal doses, relieve pain and produce sleep. These are called narcotic analgesics because they have addictive properties.
Ex : Heroin, Morphine, Codeine.