Question

Match the coordination compounds in Column I having the given type of hybridisation of \(M^{n+}\) ion and magnetic moment as given in Column II.

• A. A = R, B = S, C = Q, D = P
• B. A = S, B = R, C = P, D = Q
• C. A = S, B = P, C = R, D = Q
• D. A = Q, B = P, C = S, D = R

Hint:

Strong field ligands cause pairing and \(dsp^2\) hybridisation, while weak field ligands give \(sp^3\) hybridisation with unpaired electrons.

Answer 1

The Correct Option is A

Step 1: Analyse \(Ni(CO)_4\).
Nickel is in oxidation state 0. 
Electronic configuration: 
\[ Ni: [Ar]3d^8 4s^2 \Rightarrow d^{10} \text{ after bonding} \]
CO is a strong field ligand, pairing all electrons. 
Hybridisation: 
\[ sp^3 \]
No unpaired electrons: 
\[ \mu = 0 \]
So, 
\[ A \rightarrow R \]

Step 2: Analyse \([Ni(CN)_4]^{2-}\)
Ni is in +2 oxidation state: 
\[ Ni^{2+} = d^8 \]
CN- is a strong field ligand → pairing occurs. 
Hybridisation: 
\[ dsp^2 \ (\text{square planar}) \]
All electrons paired: 
\[ \mu = 0 \]
So, 
\[ B \rightarrow S \]

Step 3: Analyse \([NiCl_4]^{2-}\). 
\([Ni]^{2+}\) = \(\ d^8\). 
\([Cl]^{-}\)is a weak field ligand With no pairing. 
Hybridisation: 
\[ sp^3 \]
Two unpaired electrons: 
\[ \mu \approx 2.84\ \text{BM} \]
So, 
\[ C \rightarrow Q \]

Step 4: Analyse \([MnBr_4]^{2-}\). 
\([Mn]^{2+}\) = \( \ d^5\). 
Br- is weak field → high spin. 
Hybridisation: 
\[ sp^3 \]
Five unpaired electrons: 
\[ \mu \approx 5.92\ \text{BM} \]
So, 
\[ D \rightarrow P \]

Step 5: Compile matching. 
\[ A = R,\quad B = S,\quad C = Q,\quad D = P \]

Step 6: Match with options. 
This corresponds to option (A). 

Step 7: Final conclusion. 
\[ \boxed{A = R,\ B = S,\ C = Q,\ D = P} \]