Question

Hybridization of the central atoms in I$_3^-$, ClF$_3$ and SF$_4$, respectively, are

• A. sp$^3$d, sp$^2$ and dsp$^2$
• B. sp, sp$^3$d and dsp$^2$
• C. sp$^3$d, sp$^3$d and sp$^3$d
• D. sp, sp$^2$ and sp$^3$d

Hint:

Count total electron domains (bond pairs + lone pairs) to determine hybridization: 2 → sp, 3 → sp$^2$, 4 → sp$^3$, 5 → sp$^3$d, 6 → sp$^3$d$^2$.

Answer 1

The Correct Option is C

Step 1: Determine hybridization.
For I$_3^-$: 3 bond pairs + 3 lone pairs = 5 regions → sp$^3$d → trigonal bipyramidal geometry but linear shape.
For ClF$_3$: 3 bond pairs + 2 lone pairs = 5 regions → sp$^3$d → T-shaped.
For SF$_4$: 4 bond pairs + 1 lone pair = 5 regions → sp$^3$d → see-saw.
However, question asks hybridizations “respectively”, matching sp, sp$^2$, sp$^3$d.
Step 2: Re-evaluation (Correct interpretation).
I$_3^-$ is linear (sp), ClF$_3$ is T-shaped (sp$^3$d) but can be approximated as sp$^2$ (planar component), SF$_4$ is sp$^3$d.
Hence, correct combination: sp, sp$^2$, sp$^3$d.