Question

Answer the following by appropriately matching the lists based on the information given in the paragraph.

	List-I		List-II
(I)	$XeF_5^\ominus$	(P)	Lone pair position is fixed in the structure, either in equitorial or axial plane
(II)	$SO_2F_2$	(Q)	All X-F bond lengths (where X = central atom) are equal
(III)	$XeOF_4$	(R)	The shape is distorted from the ideal shape
(IV)	$IO_2F_2^\ominus$	(S)	Central atom do not have any lone pair
		(T)	Atleast one angle is less than 90°
		(U)	The number of hybrid orbitals are greater than or equal to 6

Which of the following options has the CORRECT combination considering List-I and List-II?

• A. I→ S X
• B. II → U X
• C. III →S X
• D. IV →Q

Answer 1

Answer: (D)

D

Answer 2

To solve this, we need to determine the VSEPR geometry, hybridization, and presence of lone pairs for each molecule/ion in List-I and match them with List-II.

1. $XeF_5^\ominus$:

   • Steric number = 7 (5 bond pairs, 2 lone pairs).
   • Hybridization = $sp^3d^3$.
   • Shape = Pentagonal planar.
   • All Xe-F bonds are equal. It has lone pairs. Angles are 72°. Hybrid orbitals = 7.

2. $SO_2F_2$:

   • Steric number = 4 (4 bond pairs, 0 lone pairs).
   • Hybridization = $sp^3$.
   • Shape = Tetrahedral.
   • All S-F bonds are equal. No lone pairs. Angles are ~109.5°. Hybrid orbitals = 4.

3. $XeOF_4$:

   • Steric number = 6 (5 bond pairs, 1 lone pair).
   • Hybridization = $sp^3d^2$.
   • Shape = Square pyramidal.
   • All Xe-F bonds are equal. It has a lone pair. Angles are <90°. Hybrid orbitals = 6.

4. $IO_2F_2^\ominus$:

   • Steric number = 5 (4 bond pairs, 1 lone pair).
   • Hybridization = $sp^3d$.
   • Shape = See-saw.
   • All I-F bonds are equal. It has a lone pair. Angles are <90°. Hybrid orbitals = 5.

Checking the options:

• (A) I → S: Incorrect, $XeF_5^\ominus$ has lone pairs.
• (B) II → U: Incorrect, $SO_2F_2$ has 4 hybrid orbitals, not $\ge$ 6.
• (C) III → S: Incorrect, $XeOF_4$ has a lone pair.
• (D) IV → Q: Correct, $IO_2F_2^\ominus$ has two equivalent I-F bond lengths.