COMEDK · Chemistry · 24. Coordination Compounds
Match the coordination compounds in Column I having the given type of hybridisation of \(\mathrm{M}^{\mathrm{n}+}\) ion and magnetic moment as given in Column II.
\(\begin{array}{|l|l|l|l|} \hline & \text { Coordination compounds. } & & \text { Hybridisation \& Magnetic nature. } \\ \hline \text { A } & \mathrm{Ni}(\mathrm{CO})_4 & \text { P } & \mathrm{sp}^3, \quad \mu=5.92 \mathrm{BM} \\ \hline \text { B } & {\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}} & \text { Q } & \mathrm{sp}^3, \quad \mu=2.84 \mathrm{BM} \\ \hline \text { C } & {\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}} & \text { R } & \mathrm{sp}^3, \quad \mu=0 \\ \hline \text { D } & {[\mathrm{MnBr}]^{2-}} & \text { S } & \mathrm{dsp}^2, \quad \mu=0 \\ \hline \end{array}\)
- A \(\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}\)
- B \(\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}\)
- C \(\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{R}\)
- D \(\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{R} \quad \mathrm{D}=\mathrm{Q}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}\)
Step-by-step Solution
Detailed explanation
For \(\mathrm{Ni(CO)_4}\), the oxidation state of Ni is 0. The electronic configuration is \([Ar] 3d^8 4s^2\). Since CO is a strong field ligand, it causes pairing of electrons, leading to \(sp^3\) hybridization. All electrons are paired, so \(\mu = 0\). This matches R. For…
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