TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements

Very Short Answer Questions (2 Marks)

Question 1.
What inspired Bartlett for carrying out reaction between Xe and PtF6?
Answer:
Neil Bartlett, first prepared a red compound which is formulated as Q2+ PtF6. He, then realised that the first ionisation enthalpy of molecular oxygen was almost identical with that of Xenon. He then made another red colour compound XePtF6 by mixing PtF6 and Xenon.

Question 2.
Which of the following does not exist?
a) XeOF4
b) NeF2
c) XeF2
d) XeF6
Answer:
NeF2 does not exist.

Question 3.
Why do noble gases have comparatively large atomic sizes?
Answer:
The atomic radii of the noble gas elements are all very large. Because these have van der Waals radii which Is larger than Ionic and covalent radii.

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 4.
List out the uses of neons. [Mar. 18 – A.P.]
Answer:

  1. Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
  2. Neon bulbs are used in botanical gardens and in green houses.

Question 5.
Write any two uses of argon.
Answer:

  1. Argon is used to provide an inert atmosphere in high temperature metallurgical processes.
  2. For filling electric bulbs.
  3. Used in laboratory for handling substances that are air sensitive.

Question 6.
In modern diving apparatus, a mixture of He and O2 is used. Why? [AP ’16]
Answer:
Helium has very low solubility in blood. So it is used as a diluent for oxygen in modern diving apparatus.

Question 7.
Helium is heavier than hydrogen. Yet helium is used (instead of H2) in filling balloons for meteorological observations. Why?
Answer:
Helium is non-inflammable and light gas. Hence it is used in filling balloons for mete-orological observations.

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 8.
How is XeO3 prepared ?
Answer:
XeO3 is prepared by the hydrolysis of XeF6.
XeF6 + 3H2O → XeO3 + 6HF

Question 9.
Give the preparation of
a) XeOF4
b) XeO2F2
Answer:
Partial hydrolysis of XeF6 gives oxyfluorides.
a) XeF6 + H2O → XeOF4 + 2HF
b) XeF6 + 2H2O → XeO2F2 + 4HF

Question 10.
Explain the structure of XeO3. [TS ’16 ; IPE ’14]
Answer:
Xe undergoes sp3 hybridisation.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 1
Unpaired electrons in d-orbital form dπ – pπ bonds with oxygen, sp3 orbitals form 3σ bonds with oxygen atoms. Because of the presence of lone pair, the molecule assumes pyramidal shape with a bond angle of 103°.

Question 11.
Noble gases are inert- Explain.
Answer:
The inertness to chemical reactivity is due to –

  1. The noble gases except helium (1 s2) have completely filled ns2np6 electronic configuration in their valence shell.
  2. They have high ionisation enthalpy and high positive electron gain enthalpy.

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 12.
Write the name and formula of the first noble gas compound prepared by Bertlett.
Answer:
Xe+ [PtF6]
Xenon hexa fluoro platinate

Question 13.
Explain the shape of XeF4 on the basis of use VSEPR theory. [Mar. 2018 – AP]
Answer:
XeF4 [Xenon tetrafluoride]:
shape is octahedral. But due to the presence of two lone pairs, the molecule assumes square planar structure.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 2
Total number of electron pairs is 6. The Xe atom undergoes sp3d2 Hybridisation. Hence shape is octahedral. But due to the presence of two lone pairs, the molecule assumes square planar structure.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 3

Question 14.
Give the outer electronic configuration of noble gases.
Answer:
ns2np6

Question 15.
Why do noble gases form compounds with fluorine and oxygen only?
Answer:
Because fluorine and oxygen are highly electronegative, they can form compounds with noble gases.

Question 16.
How is XeOF4 prepared? Describe its molecular shape.
Answer:
Partial hydrolysis of XeF6 gives oxyfluorides
XeF6 + H2O → XeOF4 + 2HF
XeOF4 has square pyramidal.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 4

Question 17.
What is the major source of helium?
Answer:
Natural gas

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 18.
Which noble gas is radioactive? How is it formed?
Answer:
Radon. Radon is obtained as a decay product of Ra226.
\({ }_{88}^{226} \mathrm{Ra}\) → \({ }_{86}^{222} \mathrm{Rn}\) + \({ }_2^4 \mathrm{He}\)

Question 19.
Name the following.
a) most abundant noble gas in atmosphere
b) radioactive noble gas
c) noble gas with least boiling point
d) noble gas forming large number of compounds
e) noble gas not present in atmosphere.
Answer:
a) Argon
b) Radon
c) Helium
d) Xenon
e) Radon

Short Answer Questions (4 Marks)

Question 20.
How are xenon fluorides XeF2, XeF4 and XeF6 obtained ? [AP ’17 ; IPE ’14] [Mar. 2018 – TS]
Answer:
Xenon fluorides are obtained by the direct combination of elements under appropriate conditions.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 5

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 21.
How are XeO3 and XeOF4 prepared?
Answer:
Hydrolysis of XeF4 and XeF6 gives XeO3.
XeF6 + 3H2O → XeO3 + 6HF
Partial hydrolysis of XeF6 gives XeOF4
XeF6 + H2O → XeOF4 + 2HF

Question 22.
Give the formulae and describe the struc-tures of noble gas sp>ecies, isoelectronic with
a) ICl4
b) IBr2
c) BrO3
Answer:
a) XeF4
b) XeF2
c) XeF6

a) ICl4 is isoelectronic with XeF4. XeF4 is square planar in which Xe undergoes sp3d2 hybridisation. If has two lone pairs.

Structure of XeF4: Square planar.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 6

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

b) IBr2 is isoelectronic with XeF2. XeF2 is linear. Xe undergoes sp3d hybridisation. Three lone pairs occupy three vertices of triangle.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 7

c) BrO3 is isoelectronic with XeF6. In XeF6, Xe undergoes sp3d3 hybridisation. XeF6 has seven electron pairs (6 bonding pairs and one lone pair) XeF6 thus has a distorted octahedral structure
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 8

Question 23.
Explain the reaction of the following with water. [AP ’15 ; IPE ’14]
a) XeF2
b) XeF4
c) XeF6
Answer:
a) XeF2 reacts with water and give Xe, HF and O2
2XeF2 + 2H2O → 2Xe + 4HF + O2

b) XeF4 reacts with water and give Xe, XeO3, HF and O2
6XeF4 + 12H2O → 4 Xe + 2XeO3 + 24 HF + 3O2

c) XeF6 reacts with water and give XeO3, HF
XeF6 + 3H2O → XeO3 + 6 HF

Question 24.
Explain the structures of [(AP ’17) (Mar. 2018 – T.S)]
a) XeF2 and
b) XeF4
Answer:
a) In XeF2, Xe undergoes sp3d hybridisation.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 9
There are three bond pairs and two lone pairs. The two bond pairs overlap end-end with 2pz orbitals of two F atoms. Thus XeF2 molecule has linear structure.

b) XeF4 is square planar. Xe undergoes sp3d2 hybridisation. [Mar. 2018 – A.P.]
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 10
There are two lone pairs and four bond pairs. The four bond pairs overlap end-end with 2pz orbitals of four F atoms.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 11

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 25.
Explain the structure of [TS ’15]
a) XeF6 and
b) XeOF4.
Answer:
a) XeF6 has distorted octohedral shapes. XeF6 has seven electron pairs. (6 bonding and one lone pair). It has a distorted octahedral structure in the gas phase.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 12

b) XeOF4 has square pyramidal structure.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 13
Xe in excited state undergoes sp3d2 hybridisation. One hybrid orbital is occupied by a lone pair. Other five orbitals are occupied by unpaired electrons. Five orbitals form five σ bonds. The unpaired electrons in unhybridised’d’ orbital forms a π bond with oxygen. As the lone pair occupies axial position, XeOF4 assumes square pyramidal structure.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 14

Question 26.
Complete following:
a) XeF2 + H2O →
b) XeF2 + PF5
c) XeF4 + SbF5
d) XeF6 + AsF5
e) XeF4 + O2F2
f) NaF + XeF6
(Hint: NaF + XeF6 → Na+[XeF7])
Answer:
a) XeF2 is hydrolysed to give Xe, HF and O2.
2XeF2(s) + 2H2O (l) → 2Xe (g) + 4HF (aq) + O2(g)

b) Xenon fluorides react with fluoride ion acceptors to form cationic species.
XeF2 + PF5 → [XeF]+ [PF6]

c) Xenon tetrafluoride reacts with fluoride ion to form cationic species.
XeF4 + SbF5 → [XeF3]+ [SbF6]

d) XeF6 + ASF5 → [XeF5]+ [AsF6]

e) XeF6 is prepared by the interaction of XeF6 and O2F2 at 143 K.
XeF4 + O2F2 → XeF6 + O2

f) XeF6 reacts with fluoride ion donors to form fluoroanions.
XeF4 + NaF → Na+ [XeF7]

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 27.
How are XeF2 and XeF4 prepared? Give their structures. [AP ’17]
Answer:
Xenon fluorides are obtained by the direct combination of elements under appropriate conditions.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 5

Hydrolysis of XeF4 and XeF6 gives XeO3.
XeF6 + 3H2O → XeO3 + 6HF
Partial hydrolysis of XeF6 gives XeOF4
XeF6 + H2O → XeOF4 + 2HF

Long Answer Questions (8 Marks)

Question 28.
How are XeF2, XeF2. Explain their reaction with water. Discuss their structures.
Answer:
XeF2 is prepared by heating Xenon and Fluorine in a seated Nickel tube in the ratio 1 : 2 at 673K for 8hrs.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 15
Reaction with water:
XeF2 is hydrolysed to give Xe, HF and O2
2XeF2 (s) + 2H2O (l) → 2Xe(g) + 4HF (aq) + O2(g)

Structure:
Xe in the first excited state
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 16
Xe undergoes sp3d hybridisation. The sp3d orbitals assume trigonal bipyramidal geometry. Two fluorine atoms form 2π bonds with Xe.

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 17
The bond pairs occupy axial positions. XeF4: XeF4 is formed when Xe and F2 mixed in the ratio 1 : 5, heated to 873K at 7 bar.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 18
Reaction with water: XeF4 hydrolysed with water to XeO3.
6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2
Structure of XeF4 : XeF4 is square planar Xe in the second excited state
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 19
Xe undergoes sp3 d2 hybridisation. Xe forms four o bonds with 4 fluorine atoms. Two lone pairs occupy axial positions. The four fluorine atoms occupy corners of a square.
TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements 20

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Additional Questions – Answers

Question 1.
Why are the elements of group 18 known as noble gases?
Answer:
The group 18 elements have completely filled s and p orbitals in their valence shell. They react with a few elements only under certain conditions. Therefore they are known as noble gases.

Question 2.
Noble gases have very low boiling points. Why?
Answer:
Noble gases are monoatomic. There are no inter atomic forces except weak dispersion forces. Hence they are liquified at very low temperatures. Hence they have low boiling points.

Question 3.
Does the hydrolysis of XeF6 lead to a redox reaction?
Answer:
No, the products of hydrolysis are XeF4 and XeO2F2 where the oxidation states of all the elements remain the same.

Intext Questions – Answers

Question 1.
Why is Helium used in diving apparatus ?
Answer:
Helium acts as a diluent for oxygen. Hence it is used in diving apparatus.

TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements

Question 2.
Balance the following equation.
XeF6 + H6O > XeO2 F2 + HF
Answer:
XeF6 + 2H2O > XeO2F2 + 4HF

Question 3.
Why has it been difficult to study the chemistry of Radon?
Answer:
Radon is radioactive element with very short Half-life period. Hence its study is difficult.

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