The hybridization and magnetic nature of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$, respectively are:

(D) $\mathrm{d}^{2} \mathrm{sp}^{3}$ and paramagnetic

(A) $\mathrm{sp}^{3} \mathrm{d}^{2}$ and diamagnetic

(B) $\mathrm{sp}^{3} \mathrm{d}^{2}$ and paramagnetic

(C) $d^{2} sp^{3}$ and diamagnetic

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Coordination Compounds

The hybridization and magnetic nature of

[Mn (CN)_{6}]^{4 -}

and

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

, respectively are:

Answer: (D)

Cyanide is a strong-field ligand, so both complexes are low spin and the pairing it forces empties two inner

3 d

orbitals for hybridisation.

M n^{2 +}

3 d^{5}

, so in

[Mn (CN)_{6}]^{4 -}

the five electrons crowd into the

t_{2 g}

set as

t_{2 g}^{5} e_{g}^{0}

, leaving one unpaired electron.

F e^{3 +}

is also

3 d^{5}

, so

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

has the same

t_{2 g}^{5} e_{g}^{0}

arrangement and likewise one unpaired electron. With two vacant

3 d

orbitals available in each, both are inner-orbital octahedral complexes. Both are

d^{2} s p^{3}

hybridised and paramagnetic.

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option