Q11 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
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.
\(\begin{array}{|l|l|l|l|} \hline & \text { Coordination compounds. } & & \text { Hybridisation \& Magnetic nature. } \\ \hline \text { A } & \mathrm{Ni}(\mathrm{CO})_4 & \text { P } & \mathrm{sp}^3, \quad \mu=5.92 \mathrm{BM} \\ \hline \text { B } & {\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}} & \text { Q } & \mathrm{sp}^3, \quad \mu=2.84 \mathrm{BM} \\ \hline \text { C } & {\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}} & \text { R } & \mathrm{sp}^3, \quad \mu=0 \\ \hline \text { D } & {[\mathrm{MnBr}]^{2-}} & \text { S } & \mathrm{dsp}^2, \quad \mu=0 \\ \hline \end{array}\)
- Correct answerA$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
- 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$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{R} \quad \mathrm{D}=\mathrm{Q}$
Solution
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 $[Ni(CN)_4]^{2-}$, the oxidation state of Ni is +2. The electronic configuration is $[Ar] 3d^8$. CN$^-$ is a strong field ligand, causing pairing of electrons in the $3d$ orbitals, resulting in $dsp^2$ hybridization. All electrons are paired, so $\mu = 0$. This matches S.
For $[Ni(Cl)_4]^{2-}$, the oxidation state of Ni is +2. The electronic configuration is $[Ar] 3d^8$. Cl$^-$ is a weak field ligand, so no pairing occurs. The hybridization is $sp^3$ with two unpaired electrons. The magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$ BM. This matches Q.
For $[MnBr_4]^{2-}$, the oxidation state of Mn is +2. The electronic configuration is $[Ar] 3d^5$. Br$^-$ is a weak field ligand, so no pairing occurs. The hybridization is $sp^3$ with five unpaired electrons. The magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. This matches P.
Answer: $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
For $[Ni(CN)_4]^{2-}$, the oxidation state of Ni is +2. The electronic configuration is $[Ar] 3d^8$. CN$^-$ is a strong field ligand, causing pairing of electrons in the $3d$ orbitals, resulting in $dsp^2$ hybridization. All electrons are paired, so $\mu = 0$. This matches S.
For $[Ni(Cl)_4]^{2-}$, the oxidation state of Ni is +2. The electronic configuration is $[Ar] 3d^8$. Cl$^-$ is a weak field ligand, so no pairing occurs. The hybridization is $sp^3$ with two unpaired electrons. The magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$ BM. This matches Q.
For $[MnBr_4]^{2-}$, the oxidation state of Mn is +2. The electronic configuration is $[Ar] 3d^5$. Br$^-$ is a weak field ligand, so no pairing occurs. The hybridization is $sp^3$ with five unpaired electrons. The magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. This matches P.
Answer: $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
Hence, the correct answer is option: 'A'.