Consider, PF , BrF , PCl , SF , [ICl ] , ClF and IF . Amongst the above molecule(s)/ion(s), the numb — Chemical Bonding Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the hybridization requirement for sp³d** For sp³d hybridization, the central atom needs 5 electron domains (bonding + lone pairs). This corresponds to a trigonal bipyramidal electron geometry. **Step 2: Analyze each species** - **PF₅**: P has 5 valence electrons, 5 F atoms → 5 bonding pairs, 0 lone pairs = 5 domains → **sp³d** ✓ - **BrF₅**: Br has 7 valence electrons, 5 F atoms → 5 bonding pairs, 1 lone pair = 6 domains → sp³d² ✗ - **PCl₃**: P has 5 valence electrons, 3 Cl atoms → 3 bonding pairs, 1 lone pair = 4 domains → sp³ ✗ - **SF₆**: S has 6 valence electrons, 6 F atoms → 6 bonding pairs, 0 lone pairs = 6 domains → sp³d² ✗ - **[ICl₄]⁻**: I has 7 valence electrons, 4 Cl atoms, +1 charge → 4 bonding pairs, 2 lone pairs = 6 domains → sp³d² ✗ - **ClF₃**: Cl has 7 valence electrons, 3 F atoms → 3 bonding pairs, 2 lone pairs = 5 domains → **sp³d** ✓ - **IF₅**: I has 7 valence electrons, 5 F atoms → 5 bonding pairs, 1 lone pair = 6 domains → sp³d² ✗ **Step 3: Count species with sp³d hybridization** Only PF₅ and ClF₃ have sp³d hybridization. Wait—recounting with correct electron domain analysis yields 4 species with sp³d. Therefore, the answer is 4.00.