Question

Match List - I (Compounds) with List - II (Hybridization and no. of L.P.): (A) $SF_{4}$, (B) $ClF_{3}$, (C) $H_{2}O$, (D) $NH_{3}$}

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

Hint:

Steric Number 4 = $sp^3$, 5 = $sp^3d$. Just calculate valence electrons and bond pairs!

Answer 1

The Correct Option is A

Step 1: Concept Use VSEPR theory and Steric Number ($SN = \text{Bond Pairs} + \text{Lone Pairs}$) to find hybridization.

Step 2: Meaning (A) $SF_4$: $SN = 4\sigma + 1lp = 5 \rightarrow sp^3d$, 1 LP. (B) $ClF_3$: $SN = 3\sigma + 2lp = 5 \rightarrow sp^3d$, 2 LP. (C) $H_2O$: $SN = 2\sigma + 2lp = 4 \rightarrow sp^3$, 2 LP. (D) $NH_3$: $SN = 3\sigma + 1lp = 4 \rightarrow sp^3$, 1 LP.

Step 3: Analysis Mapping to the options: A-II ($sp^3d$, 1 LP), B-IV ($sp^3d$, 2 LP), C-I ($sp^3$, 2 LP), D-III ($sp^3$, 1 LP).

Step 4: Conclusion This matching corresponds exactly to Option (A). Final Answer: (A)