JEE Advanced · Chemistry · 23. Coordination Compounds
Both \(\left[\mathrm{Ni}(\mathrm{CO})_4\right]\) and \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\) are diamagnetic. The hybridisations of nickel in these complexes, respectively, are
- A \(s p^3, s p^3\)
- B \(s p^3, d s p^2\)
- C \(d s p^2, s p^3\)
- D \(d s p^2, d s p^2\)
Answer & Solution
Correct Answer
(B) \(s p^3, d s p^2\)
Step-by-step Solution
Detailed explanation
In \(\left[\mathrm{Ni}(\mathrm{CO})_4\right]\), the oxidation state of \(\mathrm{Ni}\) is zero \((0)\).
\(\mathrm{Ni}(28)[\mathrm{Ar}] 4 \mathrm{~s}^2, 3 d^8\)

\(\mathrm{CO} \text { is a strong ligand, causes coupling, thus }\)

In \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\), the oxidation state of \(\mathrm{Ni}\) is \(+2\) \(\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8, 4 s^0\)

\(\mathrm{CN}^{-} \text {is strong ligand causes coupling. }\)

\(\mathrm{Ni}(28)[\mathrm{Ar}] 4 \mathrm{~s}^2, 3 d^8\)

\(\mathrm{CO} \text { is a strong ligand, causes coupling, thus }\)

In \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\), the oxidation state of \(\mathrm{Ni}\) is \(+2\) \(\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8, 4 s^0\)

\(\mathrm{CN}^{-} \text {is strong ligand causes coupling. }\)

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