Question

The molecule `X' has see-saw shape with central atom in \(sp^3d\) hybridization. What is `X'?

• A. \(\text{ClF}_3\)
• B. \(\text{XeF}_4\)
• C. \(\text{SF}_4\)
• D. \(\text{BrF}_5\)

Hint:

For \(sp^3d\) hybridization: - \(BP=5, LP=0\): Trigonal Bipyramidal (\(\text{PCl}_5\)) - \(BP=4, LP=1\): See-saw (\(\text{SF}_4\)) - \(BP=3, LP=2\): T-shape (\(\text{ClF}_3\)) - \(BP=2, LP=3\): Linear (\(\text{XeF}_2\))

Answer 1

The Correct Option is C

Step 1: Analyze the conditions - Hybridization: \(sp^3d\) implies 5 electron domains (Steric Number = 5). - Shape: See-saw implies 4 bonding pairs and 1 lone pair (\(AX_4E_1\) type).
Step 2: Check each option 1. \(\text{ClF}_3\): Cl has 7 valence \(e^-\). 3 bonded to F. Total valence \(e^-\) on central atom = \(7 + 3 = 10\). Pairs = 5. Bonding pairs = 3, Lone pairs = 2. Shape: T-shaped. 2. \(\text{XeF}_4\): Xe has 8 valence \(e^-\). 4 bonded to F. Total = \(8 + 4 = 12\). Pairs = 6. \(sp^3d^2\). Shape: Square Planar. 3. \(\text{SF}_4\): S has 6 valence \(e^-\). 4 bonded to F. Total = \(6 + 4 = 10\). Pairs = 5. Bonding pairs = 4, Lone pairs = 1. Hybridization: \(sp^3d\). Shape: See-saw. 4. \(\text{BrF}_5\): Br has 7 valence \(e^-\). 5 bonded to F. Total = \(7 + 5 = 12\). Pairs = 6. \(sp^3d^2\). Shape: Square Pyramidal. Final Answer:
\(\text{SF}_4\).