Question

Which of the following sequences of hybridisation, geometry and magnetic nature are correct for the given coordination compounds?}

• [A.] \([NiCl_4]^{2-}\) -- \(sp^3\), tetrahedral, paramagnetic
• [B.] \([Ni(NH_3)_6]^{2+}\) -- \(sp^3d^2\), octahedral, paramagnetic
• [C.] \([Ni(CO)_4]\) -- \(sp^3\), tetrahedral, paramagnetic
• [D.] \([Ni(CN)_4]^{2-}\) -- \(dsp^2\), square planar, diamagnetic

Choose the correct answer.

• A. A, B, C and D
• B. B, C and D only
• C. A, C and D only
• D. A, B and D only

Answer 1

The Correct Option is D

Concept: Electronic configuration of \(Ni^{2+}\): \[ Ni: [Ar]\,3d^8\,4s^2 \] \[ Ni^{2+}: [Ar]\,3d^8 \] Nature of ligand determines hybridisation and geometry. Step 1: {Compound A} \[ [NiCl_4]^{2-} \] Cl\(^-\) is a weak field ligand. Hybridisation: \[ sp^3 \] Geometry: tetrahedral. Two unpaired electrons present \(\Rightarrow\) paramagnetic. Thus A is correct. Step 2: {Compound B} \[ [Ni(NH_3)_6]^{2+} \] \(NH_3\) is a moderate field ligand. Hybridisation: \[ sp^3d^2 \] Geometry: octahedral. Unpaired electrons present \(\Rightarrow\) paramagnetic. Thus B is correct. Step 3: {Compound C} \[ [Ni(CO)_4] \] Nickel oxidation state \(=0\). Electronic configuration: \[ 3d^{10} \] Hybridisation: \[ sp^3 \] Geometry: tetrahedral. All electrons paired \(\Rightarrow\) diamagnetic. Thus statement C is incorrect. Step 4: {Compound D} \[ [Ni(CN)_4]^{2-} \] \(CN^-\) is a strong field ligand. Hybridisation: \[ dsp^2 \] Geometry: square planar. All electrons paired \(\Rightarrow\) diamagnetic. Thus D is correct. Hence correct statements: \[ A,\; B,\; D \]