The number of paramagnetic complex among $[\mathrm{FeF}_{6}]^{3-},[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-},[\mathrm{Mn}(\mathrm{CN})_{6}]^{3-},[\mathrm{Co}(\mathrm{C}_{2} \mathrm{O}_{4})_{3}]^{3-}$, $[\mathrm{MnCl}_{6}]^{3-}$ and $[\mathrm{CoF}_{6}]^{3-}$, which involved $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization is ______

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Coordination Compounds

The number of paramagnetic complex among

[FeF_{6}]^{3 -}, [Fe (CN)_{6}]^{3 -}, [Mn (CN)_{6}]^{3 -}, [Co (C_{2} O_{4})_{3}]^{3 -}

[MnCl_{6}]^{3 -}

and

[CoF_{6}]^{3 -}

, which involved

d^{2} sp^{3}

hybridization is ______

Answer: 2

An inner-orbital

d^{2} s p^{3}

set needs a field strong enough to pair electrons and leave two inner

d

orbitals vacant.

[F e (C N)_{6}]^{3 -}

has

F e^{3 +}

(

d^{5}

) with the strong cyanide field, giving a low-spin

d^{2} s p^{3}

ion with one unpaired electron, hence paramagnetic.

[M n (C N)_{6}]^{3 -}

has

M n^{3 +}

(

d^{4}

) with cyanide, again low-spin

d^{2} s p^{3}

, with two unpaired electrons, hence paramagnetic.

[C o (C_{2} O_{4})_{3}]^{3 -}

C o^{3 +}

(

d^{6}

) low-spin

d^{2} s p^{3}

but with no unpaired electrons, so diamagnetic. The fluoride and chloride complexes are weak-field outer-orbital

s p^{3} d^{2}

ions. The number that are both

d^{2} s p^{3}

and paramagnetic is 2.

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Integer answer