Question

Match the coordination complexes given in Column I with the most appropriate properties in Column II. (Given: Atomic numbers of Mn: 25; Co: 27; Ni: 28) 

• A. E-1, F-2, G-3, H-4
• B. E-2, F-1, G-4, H-3
• C. E-4, F-2, G-1, H-3
• D. E-1, F-4, G-3, H-2

Hint:

To solve such matching problems, first determine the oxidation state and $d$-electron count of the central metal. Then use the ligand field strength (weak field vs strong field) to decide high-spin or low-spin configuration, which directly helps predict magnetic moment, CFSE, and hybridisation.

Answer 1

The Correct Option is A

We analyze each complex one by one using electronic configurations and ligand field theory. Step 1: For [Mn(H$_2$O)$_6$]$^{2+$}
Mn atomic number = 25 $⇒$ Mn$^{2+}$ has configuration $3d^5$.
All ligands are weak field ($H_2O$) $⇒$ high spin $d^5$ $⇒$ 5 unpaired electrons.
Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.
So, E $⇒$ 1. Step 2: For [CoF$_6$]$^{3-$}
Co atomic number = 27 $⇒$ Co$^{3+}$ has configuration $3d^6$.
F$^-$ is a weak field ligand $⇒$ high spin complex.
High-spin $d^6$ in octahedral field gives CFSE = $0.4 \Delta_o$.
So, F $⇒$ 2. Step 3: For [NiCl$_4$]$^{2-$}
Ni atomic number = 28 $⇒$ Ni$^{2+}$ has configuration $3d^8$.
Cl$^-$ is weak field $⇒$ does not cause pairing.
In tetrahedral field, complex is high spin with hybridisation $sp^3$.
So, G $⇒$ 3. Step 4: For [Ni(CN)$_4$]$^{2-$}
Ni$^{2+}$ $⇒$ $3d^8$.
CN$^-$ is a strong field ligand $⇒$ causes pairing of $3d$ electrons.
Configuration: $dsp^2$ (square planar), with all electrons paired $⇒$ Diamagnetic.
So, H $⇒$ 4. Therefore, the correct matching is: E-1, F-2, G-3, H-4. \[ \boxed{(A)} \]