The ion that has $\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridization for the central atom is:

(B) $\left[\mathrm{ICl}_{4}\right]^{-}$

(A) $\left[\mathrm{IF}_{6}\right]^{-}$

(C) $\left[\mathrm{ICl}_{2}\right]^{-}$

(D) $\left[\mathrm{BrF}_{2}\right]^{-}$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Chemical Bonding and Molecular Structure

The ion that has

sp^{3} d^{2}

hybridization for the central atom is:

(A) $[IF_{6}]^{-}$
(B) $[ICl_{4}]^{-}$ ✓ Correct
(C) $[ICl_{2}]^{-}$
(D) $[BrF_{2}]^{-}$

Answer: (B)

Solution

Hybridisation follows from the steric number, the count of bonding atoms plus lone pairs on the central atom. In

[ICl X_{4}] X^{-}

iodine brings 7 valence electrons plus 1 for the charge; four go into bonds with the chlorines and the remaining four form two lone pairs, so there are

4 + 2 = 6

electron domains, which is

s p^{3} d^{2}

. For comparison

[IF X_{6}] X^{-}

has seven domains (

s p^{3} d^{3}

), while

[ICl X_{2}] X^{-}

and

[BrF X_{2}] X^{-}

each have five (

s p^{3} d

). So the ion with

s p^{3} d^{2}

hybridisation at the central atom is

[ICl X_{4}] X^{-}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Chemical Bonding and Molecular Structure

Solution

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