Question: 64. Match the List-I with List-II.
| List-I | List-II |
|---|---|
| A. \mathrm{XeO}_{3} | I. s p^{3} d; linear |
| B. \mathrm{XeF}_{2} | II. \mathrm{sp}^{3}; pyramidal |
| C. \mathrm{XeOF}_{4} | III. \quad \mathrm{sp}^{3} \mathrm{~d}^{3}: \quad distorted octahedral |
| D. \mathrm{XeF}_{6} | IV. \quad \mathrm{sp}^{3} \mathrm{~d}^{2} ; \quad square pyramidal |
Choose the correct answer from the options given below.
(1) A-II, B-I, C-IV, D-III
(2) A-II, B-I, C-III, D-IV
(3) A-IV, B-II, C-III, D-I
(4) A-IV, B-II, C-I, D-III
Answer: Option (1)
Explanation:
The hybridization and shape of xenon compounds can be determined using the number of bond pairs and lone pairs around the xenon atom according to VSEPR theory.
A. \mathrm{XeO}_{3}
Xenon has three sigma bonds with oxygen and one lone pair.
Total electron pairs around xenon = 4.
Hybridization is \mathrm{sp}^{3} and due to one lone pair, the shape is pyramidal.
Thus, A matches with II.
B. \mathrm{XeF}_{2}
Xenon forms two sigma bonds with fluorine and has three lone pairs.
Total electron pairs = 5.
Hybridization is \mathrm{sp}^{3} \mathrm{d}.
Three lone pairs occupy equatorial positions, giving a linear shape.
Thus, B matches with I.
C. \mathrm{XeOF}_{4}
Xenon forms five sigma bonds (one with oxygen and four with fluorine) and has one lone pair.
Total electron pairs = 6.
Hybridization is \mathrm{sp}^{3} \mathrm{d}^{2}.
Due to one lone pair, the shape becomes square pyramidal.
Thus, C matches with IV.
D. \mathrm{XeF}_{6}
Xenon forms six sigma bonds with fluorine and has one lone pair.
Total electron pairs = 7.
Hybridization is \mathrm{sp}^{3} \mathrm{d}^{3}.
The presence of a lone pair causes distortion from regular octahedral geometry.
Thus, D matches with III.
Therefore, the correct matching is A-II, B-I, C-IV, D-III,
corresponding to Option (1).