The number of species from the following that are involved in $\mathrm{sp}^3\mathrm{d}^2$ hybridization is $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$, $\mathrm{SF}_6$, $[\mathrm{CrF}_6]^{3-}$, $[\mathrm{CoF}_6]^{3-}$, $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$ and $[\mathrm{MnCl}_6]^{3-}$

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Coordination Compounds

The number of species from the following that are involved in

sp^{3} d^{2}

hybridization is

[Co (NH_{3})_{6}]^{3 +}

SF_{6}

[CrF_{6}]^{3 -}

[CoF_{6}]^{3 -}

[Mn (CN)_{6}]^{3 -}

and

[MnCl_{6}]^{3 -}

Answer: (D)

Octahedral

s p^{3} d^{2}

hybridisation is the outer-orbital arrangement, used when the metal keeps its

3 d

electrons unpaired, that is, under weak-field ligands or a configuration that cannot free two inner

d

orbitals. In

S F_{6}

sulphur hybridises as

s p^{3} d^{2}

. In

[Co F_{6}]^{3 -}

the

C o^{3 +}

3 d^{6}

centre with weak-field fluoride, and in

[MnC l_{6}]^{3 -}

the

M n^{3 +}

3 d^{4}

centre with weak-field chloride, are both outer-orbital

s p^{3} d^{2}

. By contrast

[Co (N H_{3})_{6}]^{3 +}

[Cr F_{6}]^{3 -}

and

[Mn (CN)_{6}]^{3 -}

use inner

d^{2} s p^{3}

orbitals. The number of species using

s p^{3} d^{2}

hybridisation is three.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option