Question

Match the Xenon compounds in List-I with their molecular geometries in List-II: [h] {|l|l|} List - I (Compound) & List - II (Geometry)
(A) XeF\(_2\) & (I) Trigonal pyramidal
(B) XeO\(_3\) & (II) Distorted octahedral
(C) XeF\(_6\) & (III) Linear
(D) XeOF\(_4\) & (IV) Square pyramidal

• A. A-III, B-I, C-II, D-IV
• B. A-IV, B-III, C-II, D-I
• C. A-III, B-I, C-IV, D-II
• D. A-IV, B-III, C-II, D-I

Hint:

To quickly find the geometry of Xenon compounds, count valence electrons (8), subtract electrons used for bonds (1 per F, 2 per O), and divide the remainder by 2 to get lone pairs. Ex: XeF\(_2\) \(\rightarrow\) (8 - 2)/2 = 3 lone pairs.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Molecular geometry is determined using the **VSEPR Theory** (Valence Shell Electron Pair Repulsion).
Xenon is a noble gas with 8 valence electrons. To find the shape, we calculate the number of Bond Pairs (BP) and Lone Pairs (LP) around the central Xe atom.
Step 2: Detailed Explanation:
(A) **XeF\(_2\):** Xe has 8 electrons. 2 are used for bonds with F. Remaining 6 electrons form 3 lone pairs. Steric number = 2 BP + 3 LP = 5 (\(sp^3d\)). The three LPs occupy equatorial positions to minimize repulsion, leaving the F atoms at axial positions. Geometry: **Linear**. (Matches III).
(B) **XeO\(_3\):** 3 Xe-O double bonds are formed. 6 electrons are used. 1 lone pair remains. Steric number = 3 BP + 1 LP = 4 (\(sp^3\)). Geometry: **Trigonal Pyramidal**. (Matches I).
(C) **XeF\(_6\):** 6 BP + 1 LP = 7 (\(sp^3d^3\)). The presence of the lone pair distorts the regular octahedron. Geometry: **Distorted Octahedral**. (Matches II).
(D) **XeOF\(_4\):** 4 BP with F and 1 BP with O. Total 5 BP + 1 LP = 6 (\(sp^3d^2\)). The oxygen and lone pair take trans positions. Geometry: **Square Pyramidal**. (Matches IV).
Step 3: Final Answer:
The matching pairs are A-III, B-I, C-II, D-IV.
This matches option (A).