TS Inter 2nd Year Chemistry Study Material Chapter 6(a) Group-15 Elements

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements

Very Short Answer Questions (2 Marks)

Question 1.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Nitrogen has unique ability to form pπ – pπ multiple bonds with itself and with other elements having small size and high electro-negativity. Nitrogen exists as a diatomic molecule with a triple bond (N ≡ N) between the two atoms. Consequently its bond enthalpy (941.4 kJ mol^-1) is very high. On the contrary phosphorus forms single bonds (p – p.) Hence, the reactivity of Nitrogen differ from phosphorus.

White phosphorus is less stable and therefore more reactive, because of angular strain in the P₄ molecule where the angles are only 60°. Hence phosphorous is much more reactive than nitrogen.

Question 2.
How is nitrogen prepared in the laboratory ? Write the chemical equations of the reactions involved.
Answer:
In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
NH₄Cl (aq) + NaNO₂(aq) → N₂(g) + 2H₂O(l) + NaCl (aq)
It can also be obtained by thermal decomposition of ammonium dichromate.

Question 3.
Nitrogen exists as diatomic molecule and phosphorus as P₄. Why? [TS 15]
Answer:
Nitrogen has unique ability to form pπ – pπ multiple bonds with itself and with other elements having small size and high electronegativity. Thus nitrogen exists as a diatomic molecule with a triple bond between the two atoms. Phosphorus forms p-p single bonds and exists as P₄ molecule.

As it has large size and lower eiectro- negavitity it does not form pπ – pπ multiple bonds with itself.

Question 4.
Why does nitrogen show catenation properties less them phosphorus ?
Answer:
Single N – N bond is weaker than the single P – P bond because of high electronic repulsion of the non-bonding electrons, owing to small bond length. Hence nitrogen shows less catenation property than phosphorus.

Question 5.
Nitrogen molecule is highly stable. Why? [IPE ’14]
Answer:
Nitrogen exists as a diatomic molecule with a triple bond between the two atoms (N ≡ N), The bond enthalpy of N₂ is very high (941.4 kj. mol^-1). Hence, nitrogen molecule is highly stable.

Question 6.
Why are the compounds of bismuth more stable in +3 oxidation state?
Answer:
The common oxidation states of Bismuth are +3 and +5. The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect.

Question 7.
What is allotropy? Explain the different allotropic forms of phosphorus.
Answer:
Occurrence of an element in different physical forms with similar chemical properties but different physical properties is called allotropy.
Phosphorus is found in many allotropic forms, they are :
a) White or Yellow – P
b) Scarlet – P
c) Red – P
d) Violet – P
e) α – black – P and β – black – P

Question 8.
How do you account for the inert character of dinitrogen ?
Answer:
Nitrogen exists as a diatomic molecule with a triple bond N ≡ N.
The bond dissociation enthalpy of N ≡ N is very high. (941.4kJ mol^-1).
As, such a high energy is not available at room temperature nitrogen is inert.

Question 9.
Explain the difference in the structures of white and red phosphorus.
Answer:
White phosphorus is less stable and there fore more reactive because of angular strain in the P₄ molecule. The bond angle in P₄ is 60° only.
It consists of discrete tetrahedral P₄ molecules. These molecules are held by van der Waal’s forces.
Red phosphorus is polymeric consisting of chains of P₄ tetrahedra linked through covalent bonds. Hence less reactive.

Question 10.
How is a-black phosphorus prepared from red phosphorus ?
Answer:
When red phosphorus is heated in a sealed tube at 803K, a-black phosphorus is obtained.

Question 11.
Write the difference between the properties of white and red phosphorus.
Answer:

White Phosphorus	Red Phosphorus
1. Highly reactive.	1. Less reactive.
2. Consists of discrete P4 units held by van der Waal’s forces.	2. Consists of P4 units linked through covalent bonds.
3. Glows in dark.	3. Does not glow in dark.
4. Poisonous.	4. Non -poisonous.
5. Insoluble in water but soluble in carbon disulphide.	5. Insoluble in water as well as in carbon disulphide.

Question 12.
What is inert pair effect?
Answer:
The reluctance of electron pair in the outermost s-orbital to uncouple and take part in bonding is called inert pair effect.

Question 13.
Explain why is NH₃ basic while BiH₃ is only feebly basic.
Answer:
Ability to donate electron pair decreases as the atomic size increases from NH₃ to BiH₃. This is due to decrease in electron density. Hence NH₃ is basic while BiH₃ is feebly basic.

Question 14.
Arrange the hydrides of group-15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:
Basic strength
NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
Reducing character
NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃

Question 15.
PH₃ is a weaker base than NH₃ – Explain.
Answer:
Atomic size of phosphorous is more than that of Nitrogen. Ability to donate electron pair decreases from NH₃ to PH₃. Hence PH₃ is weaker base than NH₃.

Question 16.
A hydride of group-15 elements dissolves in water to form a basic solution. This solution dissolves the AgCl precipitate. Name the hydride. Write the chemical equations involved.
Answer:
The hydride is NH₃.
NH₃ + H₂O > NH₄OH
AgCl + 2NH₃ > [Ag(NH₃)⁺Cl^–

Question 17.
What happens when white phosphorus is heated with cone. NaOH solution in an inert atmospnere of CO₂ ? [AP Mar. ’19 ; AP ’15]
Answer:
Phosphine is formed.
P₄ + 3NaOH + 3H₂O → PH₃ ↑ + 3NaH₂PO₂ sodium hypophosphite

Question 18.
NH₃ forms hydrogen bonds but PH₃ does not. Why ? [TS ’15]
Answer:
N-H bond is more polar while P-H bond, is less polar. Nitrogen is a small atom with high electronegativity. Hence N-H bond is more polar. So, NH₃ can form hydrogen bonds.

Question 19.
The HNH angle is higher than HPH, HAsH and HSbH angles- Why?
Answer:
The NH₃ molecule is trigonal pyramidal. This shape results from the sp³ hybridisation of central atom. The decrease in the bond angle in HPH, HAsH and HSbH is due to increase in the size of the central atoms.

Question 20.
How do calcium phosphide and heavy water react?
Answer:
Calcium phosphide reacts with heavy water forming Calcium Deuteroxide and Deuterophosphine.
Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PD₃

Question 21.
Ammonia is a good complexing agent. Explain with an example. [IPE ’14]
Answer:
Ammonia can donate an electron pair and form dative bond. Hence ammonia can form complex compounds.

Question 22.
A mixture of Ca₃P₂ and CaC₂ is used in making Holme’s signal – Explain. [AP ’16]
Answer:
The spontaneous combustion of phosphine is technically used in Holme’s signals. Containers containing calcium phosphide and calcium carbide are pierced and thrown in the sea when the gases evolved burn and serve as a signal.

Question 23.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
Fe²⁺ reduces \(\mathrm{NO}_3^{-}\) to NO.
No forms brown coloured complex with Fe²⁺
[Fe(H₂O)₆]²⁺ + NO > [Fe(H₂O)₅NO]²⁺ (brown) + H₂O

Question 24.
Give the resonating structures of NO, and N₂O₅.
Answer:

Question 25.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?
Answer:
Nitrogen cannot form R₃N = O because of absence of d-orbitals in its valence shell. Nitrogen cannot form dπ – pπ bond as the heavier elements can e.g. R₃P = O.

Question 26.
How is nitric oxide (NO) prepared?
Answer:
Nitric oxide is obtained by the reduction of dilute HNO₃ with copper.
3Cu + 8HNO₃ → 2NO + SCu(NO₃)₂ + 4H₂O

Question 27.
Give one example each of normal oxide and mixed oxide of nitrogen.
Answer:
Normal oxide : NO₂, NO.
Mixed oxide : N₂O₃

Question 28.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why?
Answer:
NO contains odd number of valence electrons. It exists as a monomer in gaseous state and hence paramagnetic. It dimerises in liquid and solid states and is converted to N₂O₂ molecule with even number of electrons. Hence NO is diamagnetic in liquid and solid states.

Question 29.
Give an example of [Mar. 2018 – TS]
a) acidic oxide of phosphorus
b) neutral oxide of nitrogen.
Answer:
a) P₂O₃ and P₂O₅ are acidic oxides of phosphorus.
b) NO and N₂O are neutral oxides of nitrogen.

Question 30.
Explain the following.
a) reaction of alkali with red phosphorous.
b) reaction between PCl₃ and H₃PO₃.
Answer:
a) Red phosphorous is less reactive than white phosphorous, It does not react with alkali.
b) PCl₃ + 5H₃PO₃ → 3H₄P₂O₅ + 3HCl
Pyrophosphorous acid formed

Question 31.
How does PCl₃ react with
a) CH₃COOH
b)C₂H₅OH and
c) water
Answer:
a) PCl₃ reacts with CH₃COOH to give acetyl chloride.
3CH₃COOH + PCl₃ → 3CH₃COCl + H₃PO₃

b) Reacts with ethyl alcohol to give ethyl chloride.
3CH₃CH₂OH + PCl₃ → 3C₂H₅Cl + H₃PO₃

c) Hydrolysis of PCl₃ gives phosphorous acid.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 32.
PCl₃ can act as an oxidising as well as a reducing agent – Justify.
Answer:
1) PCl₃ oxidises Sb to SbCl₃.
Sb + PCl₃ → SbCl₃ + P
2) It reduces PCl₃ to PCl₅.
PCl₃ + Cl₂ → PCl₅

Question 33.
Which of the following are not known ?
PCl₃, ASCl₃, SbCl₃, NCl₅, BiCl₅, PH₅
Answer:
NCl₅, BiCl₅, PH₅.

Question 34.
Which of the following is more covalent – SbCl₅ or SbCl₃?
Answer:
SbCl₅

Question 35.
Write the oxidation states of phosphorous in solid PCl₅.
Answer:
In the solid state PCl₅ exists as an ionic solid [PCl₄]⁺ [PCl₆]^– in which the cation is [PCl₄]⁺ and [PCl₆]^– is anion.
Oxidation state of phosphorous in PCl₄⁺ is + 5 and in PCl₆^– is + 5

Question 36.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:
With dilute HNO₃, it gives NO.
3Cu + 8HNO₃ (dilute) → 3Cu (NO₃)₂ + 2NO + 4H₂O
With cone. HNO₃, NO₂ is obtained.
Cu + 4HNO₃ (conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Question 37.
Which oxide of nitrogen has oxidation number of N same as that in nitric acid?
Answer:
Nitrogen in N₂O₅ has same oxidation number as in Nitric acid.

Question 38.
Write the chemical reactions, that occur in the manufacture of nitric acid.
Answer:
Nitric acid is manufactured by Ostwald’s process and the following chemical reactions will occur.
4NH₃(g) + 5O₂(g) 4NO(g) + 6H₂O(g)
2NO (g) + O₂(g) ⇌ 2NO₂ (g)
3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)

Question 39.
Iron becomes passive in cone HNO₃. Why ?
Answer:
Iron becomes passive in cone. HNO₃ because of formation of a passive film of oxide on the surface.

Question 40.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
a) Uses of nitric acid :

1. It is used in the manufacture of NH₄NO₃ for fertilisers and other nitrates for use in explosives,
2. It is used in the preparation of nitro-glycerin, trinitrotoluene and other nitro compounds.

b) Uses of Ammonia:

1. Ammonia is used to produce various nitrogeneous ferti- Users, (amonium nitrate, urea etc),
2. In the Ostwald process for preparation of HNO₃.
3. Liquid ammonia is used as refrigerant.

Question 41.
What are the oxidation states of phosphorus in the following ?
i) H₃PO₃
ii) PCl₃
iii) Ca₃P₂
iv)Na₃PO₄
v) POF₃
Answer:
i) H₃PO₃ ; 3 + x – 6 = 0; x = +3
ii) PCl₃ ; x – 3 = 0 ; x = +3
iii) + 6 + 2x = 0 ; 2x = -6 ; x = – 3
iv) + 3 + x – 8 = 0 ; x – 5 = 0 ; x = +5
v) x – 2 – 3 = 0 ; x = + 5

Question 42.
H₃PO₃ is diprotic while H₃PO₂ is monoprotic – Why?
Answer:

The H atoms which are attached with oxygen in P – OH form are ionisable and cause basicity. As per the structure, H₃PO₃ is diprotic while H₃PO₂ is monoprotic.

Question 43.
Give the disproportionation reaction of H₃PO₃.
Answer:
Orthophosphorus acid on heating disproportionates to give orthophosphoric acid and phosphine.

Question 44.
H₃PO₂ is a good reducing agent – Explain with an example.
Answer:
H₃PO₂ contains two P – H bonds and reduces AgNO₃ to metallic silver.
4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag + 4HNO₃ + H₃PO₄

Question 45.
Draw the structures of
a) Hypo phosphoric acid
b) Cyclic meta phosphoric acid.
Answer:

Short Answer Questions (4 Marks)

Question 46.
Discuss the general characteristics of Group-15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Answer:
Nitrogen, Phosphorous, Arsenic, Antimony and Bismuth are included in Group-15.
Metallic Nature: Nitrogen and Phosphorus are non-metals, Arsenic, Antimony are metalloids and Bismuth is a metal.
Electronic configuration : The valence shell electronic configuration of these elements is
Nitrogen 7 → 1s²2s²2p³
Phosphorus 15 → 1s²2s²2p⁶3s²3p³

Oxidation states:
The common oxidation states of Group 15 elements are -3, +3 and +5. The tendency to exhibit -3 oxidation state decreases down the group. This is due to increase in size and metallic character.

The stability of +5 oxidation state decreases down the group. The stability of +5 oxidation state decreases and that of +3 state increases down the group due to inert pair effect.

Nitrogen exhibits +1, +2, +4 oxidation states also. Phosphorus also shows +1 and +4 oxidation states in some oxoacids.

Nitrogen is restricted to maximum cova-lency of 4 since only four (one s and three p) orbitals are available for bonding. The heavier elements have vaccant d – orbitals which can be used for bonding.

Atomic size :
Covalent and ionic radii increase in size down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and / or f – orbitals in heavier members.

Ionisation Enthalpy :
The ionisation enthalpy decreases down the group due to gradual increase in atomic size. The ionisation enthalpy of group 15 elements is much greater than that of group 14 elements in the corresponding periods. This is due to extra stable half-filled p-orbitals and small size of group 15 elements.

Electronegativity :
The electronegativity decreases down the group with increasing atomic size from N to Bi. However, amongst the heavier elements the difference is less.

Question 47.
Discuss the trends in chemical reactivity of group-15 elements.
Answer:
Nitrogen differs from the rest of the group 15 elements. This is due to its (i) small size (ii) high electronegativity (iii) high ionisation enthalpy and (iv) non-availability of d-orbitals.

The general oxidation states of group 15 elements are -3, +3 and +5. In Bismuth +3 is more stable than +5 oxidation state due to inert pair effect.

Hydrides: All the elements N to Bi form hydrides of the type EH₃ where E = N to Bi.
The stability decreases from NH₃ to BiH₃.
Ammonia is mild reducing agent while BiH3 is the strongest reducing agent. Reducing character increases from NH₃ to BiH₃.
Basic character decreases from NH₃ to BiH₃. NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

Oxides:
These elements form two types of oxides E₂O₃ and E₂O₅. However nitrogen forms number of oxides due to pn – pn multiple bonding between nitrogen and oxygen atoms.

These oxides are acidic and acidic character decreases from N₂O₃ to Bi₂O₃. Similarly from N₂O₅ to Bi₃O₅ the acidic character decreases.

Pentoxides are more acidic than trioxides.

N₂O₃ and P₂O₃ are purely acidic. As₂O₃ and Sb₂O₃ are amphoteric. Bi₂O₃ is basic.

Halides :
Trihalides and pentahalidies are formed. Nitrogen does not form penta- halide due to non-availability of d – orbitals in its valence shell.

Pentahalides are more covalent than trihalides because the elements in the higher oxidation state exert more polarising power. All the trihalides of these elements except those of nitrogen are stable. Trihalides except BiF₃ are covalent in nature.

Reactivity towards metals : All these elements react with metals to form their binary compounds exhibiting – 3 oxidation state.
Example: Ca₃N₂ (Calcium nitride)
Ca₃P₂ (Calcium phosphide)

Question 48.
How does P₄ react with the following ?
a) SOCl₂
b) SO₂Cl₂
Answer:
a) P₄ reacts with thionyl chloride SOCl₂ to give PCl₃.
P₄ + 8 SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂

b) PCl₅ is formed when P₄ reacts with SO₂Cl₂.
P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂

Question 49.
Explain the anomalous nature of nitrogen in group -15.
Answer:
Nitrogen differs from the rest of the elements of Group-15 due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals.

Nitrogen has unique ability to form pπ – pπ multiple bonds with itself and with other elements (e.g.: C, O). Other elements cannot form pit – pn bonds as their atomic orbitals are large.

Thus nitrogen exists as N₂ with triple bond. Consequently bond enthalpy is high (225 kcal / mole, or 941.4 kJ/mole). Hence it is i~ert at room temperature.

Catenation tendency is weaker in Nitrogen as the N – N single bond is weaker because of high repulsion of the non-bonding electrons owing to small bond length.

Nitrogen cannot form NCl₅ because of the absence of d-orbitals in its valence shell.

Question 50.
Complete the following reactions:
a) Ca₃P₂ + H₂O →
b) P₄ + KOH →
c) CuSO₄ + NH₃ →
d) Mg + N₂ →
e) (NH₄)₂ Cr₂O₇
f) Decomposition of nitrous acid
Answer:
a) Phosphine is formed when calcium phosphide reacts with water.
Ca₃P₂ + H₂O → 3Ca(OH)₂ + 2PH₃ ↑

b) P₄ + KOH + 3H₂O → ↑ PH₃ T + 3KH₂PO₂ (Potassium hypophosphite)

c) CuSO₄(aq) + NH₃ (aq) ⇌ [Cu(NH₃)₄]²⁺(aq) + SO₄^-2 deep blue

d) At high temperatures N₂ directly combines with Mg to form Mg₃N₂.
3Mg + N₂ Mg₃N₂

e) Nitrogen is evolved when Ammonium dichromate decomposes on heating.

Question 51.
How does PCl₅ react with the following ?
a) Water
b) C₂H₅OH
c) CH₃COOH
d) Ag
Answer:
a) PCl₅ hydrolyses to give POCl₃ and H₃PO₄.
PCl₅ + H₂O → POCl₃ + 2HCl
POCl₃+ 3H₂O → H₃PO₄ + 3HCl

b) Ethyl chloride is formed.
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl

c) Acetyl chloride is formed.
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

d) Finely divided metals on heating with
PCl₅ to give corresponding chlorides.
2Ag + PCl₅ → 2AgCl + PCl₃

Question 52.
Complete the following:
a) NH₄NO₃
b) HNO₃ + P₄O₁₀ →
c) Pb(NO₃)₂
d) Zn + dil. HNO₃ →
e) P₄ + cone. HNO₃
f) HgCl₂ + PH₃ →
Answer:
a) Nitrous oxide forms.
NH₄NO₃ N₂O + 2H₂O
b) 4HNO₃ + P₄O₁₀ → 4HPO₃ + 2N₂O₅
c) 2Pb(NO₃)₂ 2PbO + 4NO₂ + O₂
d) 4Zn + 10 HNO₃ (dilute) → 4Zn(NO₃)₂ + 5H₂O + N₂O
e) P₄ + 20HNO₃ → 4H₃PO₄ + 20 NO₃ + 4H₂O
f) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl

Long Answer Questions (8 Marks)

Question 53.
How is ammonia manufactured by Haber’s process ? Explain the reactions of ammonia with [Mar. 2018 . AP]
a) ZnSO_{4 (aq)}
b) CuSO_{4 (aq)}
c) AgCl_(s)
Answer:
On a large scale, ammonia is manufactured by Haber’s process.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ∆_fH°= -46.1. kJmol^-1
The optimum conditions for production of ammonia are a pressure of about 2000 × 10⁵ Pa and a temperature of 700K. Iron oxide is the catalyst with small amounts of K₂O and Al₂O₃.

Compressed mixture of N₂ and H₂ in the volume ratio is heated to 700K at a pressure of 200 atm and passed over finely divided iron oxide mixed with small amounts of K₂O and Al₂O₃. Ammonia formed is liquified and unreacted mixture of N₂ and H₂ again pumped into catalytic chamber.

a) With ZnSO₄, Zinc hydroxide precipitate is formed.
ZnSO₄ (aq) + 2NH₄OH (aq) → Zn(OH)₂ (s) + (NH₄)₂SO₄ (aq)

b) When ammonia reacts with CuSO₄, deep blue colouration is obtained.
CuSO₄(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺ (aq) + SO₄^-2 deep blue

c) AgCl dissolves in NH₃.
AgCl + 2NH₃ (aq) → [Ag(NH₃)₂]Cl (aq)

Question 54.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? [TS Mar. 19 ; (AP ’17)]
a) Copper
b) Zn
c) S₈
d) P₄
Answer:
Ostwald’s process:
Step (1) : A mixture of ammonia and excess of atmospheric oxygen (1 : 10) is passed through a tower containing pt/Rh guaze catalyst maintained at 500 K temperature and 9 bar pressure. Then catalytic oxidation of NH₃ takes place, forming Nitric oxide.

Step (2) : The nitric oxide thus obtained is made to combine with oxygen. Then NO₂ is obtained.
2NO_(g) + O₂(g) ⇌ 2NO_2(g)

Step (3) :The nitrogen dioxide, is dissolved in water to give nitric acid
3NO₂(g) + H₂O_(l) → 2HNO₃(aq) + NO(g)
The NO formed is recycled.

Reactions:
a) Copper: N02 gas is liberated
Cu + 4 HNO₃ (conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

b) Zinc : NO₂ gas is liberated
Zn + 4 HNO₃ (cone) → Zn(NO₃)₂ + 2H₂O + 2NO₂

c) Sulphur (S₈) : Oxidation takes place forming sulphuric acid
S₈ + 48 HNO₃(conc) → 8H₂SO₄ + 48 NO₂ + 16 H₂O

d) Phosphorous (P₄) : Oxidation takes place forming H₃PO₄
P₄ + 20 HNO₃ → 4H₃PO₄ + 20 NO₂ + 4H₂O

Intext Questions – Answers

Question 1.
Why are pentahalides more covalent than trihalides?
Answer:
Pentahalides are more covalent than trihalides because the elements in the higher oxidation state exert more polarising power.

Question 2.
Why is BiH₃ the strongest reducing agent amongst the hydrides of Group 15 elements?
Answer:
Because BiH₃ is least stable.

Question 3.
Why is N₂ less reactive at room temperature ?
Answer:
Because of high bond enthalpy of N ≡ N.

Question 4.
Mention the conditions required to maximise the yield of ammonia.
Answer:
In Haber’s process
N₂(g) + 3H₂ (g) ⇌ 2NH₃(g)
Conditions required for higher yield are

1. High pressure = 200 atm
2. Optimum temperature – 700 k
3. Catalyst: Iron oxide mixed with K₂O + Al₂O₃.

Question 5.
How does ammonia react with a solution of Cu⁺⁺ (aq)?
Answer:
Cu⁺⁺ (ag) + 4NH₃ (aq) ⇌ \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]_{(\mathrm{aq})}^{2+}\) deep blue
NH₃ forms complexion with Cu⁺⁺ by donating its lone pair of electrons, forming Tetraamine copper (II) ion.

Question 6.
What is the covalence of N in N₂O₅ structure ?
Answer:

From the structure of N₂O₅, it is evident that the covalence of N in N₂O₅ is four.

Question 7.
Bond angle in \(\mathrm{PH}_4^{+}\) is higher than that in PH₃. Why?
Answer:
In both PH₄⁺ and PH₃ the central P atom is sp3 hybridised. In \(\mathrm{PH}_4^{+}\) all the four orbitals are bonded whereas in PH₃ there is a lone pair on P. Due to lone pair-bond pair repulsion in PH₃ bond angle is less than 109°28′.

Question 8.
What happens when PCl₅ is heated?
Answer:
It decomposes to give Cl₂.
PCl₅ → PCl₃ + Cl₂

Question 9.
What is the basicity of H₃PO₄?
Answer:
3

Question 10.
What happens when H₃PO₃ is heated? (or write the disproportion of H₃PO₃)
Answer:
Orthphosphoric acid and Phosphine are formed.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 11.
What happen when white phosphorus is heated with conc. NaOH solution in an inert atmosphere of CO₂?
Answer:
Phosphine is formed.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ sodium hypophosphite

Question 12.
Write a balanced equation for the hydrolytic reaction of PCl₅ in Heavy water ?
Answer:
PCl₅ + 4D₂O → D₃PO₄ + 5DCl