Both $[\mathrm{Ni}(\mathrm{CO})_{4}]$ and $[\mathrm{Ni}(\mathrm{CN})_{4}]^{2-}$ are diamagnetic. The hybridisations of nickel in these complexes, respectively, are

(B) $\mathrm{sp}^{3}$, $\mathrm{dsp}^{2}$

(A) $\mathrm{sp}^{3}$, $\mathrm{sp}^{3}$

(C) $\mathrm{dsp}^{2}$, $\mathrm{sp}^{3}$

(D) $\mathrm{dsp}^{2}$, $\mathrm{dsp}^{2}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Coordination Compounds

Both

[Ni (CO)_{4}]

and

[Ni (CN)_{4}]^{2 -}

are diamagnetic. The hybridisations of nickel in these complexes, respectively, are

Answer: (B)

[Ni (CO)_{4}]

, nickel is in zero oxidation state with

3 d^{10}

configuration. CO is a strong field ligand causing

sp^{3}

hybridization, forming a tetrahedral complex. In

[Ni (CN)_{4}]^{2 -}

, nickel is in +2 oxidation state with

3 d^{8}

configuration. CN⁻ is a strong field ligand causing pairing of electrons and

dsp^{2}

hybridization, forming a square planar complex. Both are diamagnetic due to complete pairing of electrons.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option