Consider the following species: BrF5, XeF5-, BF4-, ICl4-, XeF4, SF4, NH4+, ClF3, XeF2, ICl2- Number of species having sp3d hybridized central atom is .
Answer
4
Solution
Steric number = bond pairs + lone pairs on the central atom; 5 ⇒ sp3d.
BrF5: 5 bp + 1 lp = 6 ⇒ sp3d2
XeF5-: 5 bp + 2 lp = 7 ⇒ sp3d3
BF4-: 4 bp = 4 ⇒ sp3
ICl4-: 4 bp + 2 lp = 6 ⇒ sp3d2
XeF4: 4 bp + 2 lp = 6 ⇒ sp3d2
SF4: 4 bp + 1 lp = 5 ⇒ sp3d ✓
NH4+: 4 bp = 4 ⇒ sp3
ClF3: 3 bp + 2 lp = 5 ⇒ sp3d ✓
XeF2: 2 bp + 3 lp = 5 ⇒ sp3d ✓
ICl2-: 2 bp + 3 lp = 5 ⇒ sp3d ✓
∴ Number of sp3d species = 4
Answer: 4