Question: 49. Match List I with List II.
| List I (Compound) | List II (Shape/geometry) |
|---|---|
| (A) \mathrm{NH}_{3} | (I) Trigonal Pyramidal |
| (B) \mathrm{BrF}_{5} | (II) Square Planar |
| (C) \mathrm{XeF}_{4} | (III) Octahedral |
| (D) \mathrm{SF}_{6} | (IV) Square Pyramidal |
TEST PAPER WITH ANSWER
Choose the correct answer from the options given below:
(1) A-I, B-IV, C-II, D-III
(2) A-II, B-IV, C-III, D-I
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-III, C-IV, D-I
Answer: Option (1)
Explanation:
The shape of a molecule is determined using VSEPR theory based on the number of bond pairs and lone pairs around the central atom.
\mathrm{NH}_{3} has 3 bond pairs and 1 lone pair on nitrogen,
giving a trigonal pyramidal shape, so A matches with I.
\mathrm{BrF}_{5} has 5 bond pairs and 1 lone pair on bromine,
resulting in a square pyramidal geometry, so B matches with IV.
\mathrm{XeF}_{4} has 4 bond pairs and 2 lone pairs on xenon arranged opposite to each other, giving a square planar shape,
so C matches with II.
\mathrm{SF}_{6} has 6 bond pairs and no lone pairs on sulfur,
leading to an octahedral geometry, so D matches with III.
Thus, the correct matching is A-I, B-IV, C-II, D-III,
which corresponds to Option (1).