Question

The magnitudes of CFSE in \([\mathrm{M}(\mathrm{H}_2\mathrm{O})_6]^{n+}\) for Mn and Fe ions satisfy the relations

• A. \(\mathrm{Mn}^{2+}<\mathrm{Mn}^{3+}\) and \(\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}\)
• B. \(\mathrm{Mn}^{2+}>\mathrm{Mn}^{3+}\) and \(\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)
• C. \(\mathrm{Mn}^{2+}<\mathrm{Mn}^{3+}\) and \(\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)
• D. \(\mathrm{Mn}^{2+}>\mathrm{Mn}^{3+}\) and \(\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}\)

Hint:

For high-spin octahedral complexes: \(d^5 \Rightarrow \text{CFSE}=0\); \(d^4 \Rightarrow 0.6\Delta_o\) (magnitude); \(d^6 \Rightarrow 0.4\Delta_o\).
With weak-field ligands (e.g., H\(_2\)O), assume high spin unless stated otherwise.

Answer 1

The Correct Option is C

Step 1: \(\mathrm{H_2O}\) is a weak-field ligand, so octahedral complexes are high-spin. Compute CFSE using \[ \text{CFSE} = (-0.4\Delta_o)\,n_{t_{2g}} + (0.6\Delta_o)\,n_{e_g}. \]
Step 2: Determine \(d^n\) counts and high-spin configurations. \[ \begin{aligned} \mathrm{Mn}^{2+}:&\ d^5 \Rightarrow t_{2g}^3e_g^2 \ \Rightarrow \ \text{CFSE}=3(-0.4)+2(0.6)=0,
\mathrm{Mn}^{3+}:&\ d^4 \Rightarrow t_{2g}^3e_g^1 \ \Rightarrow \ \text{CFSE}=3(-0.4)+1(0.6)=-0.6\Delta_o,
\mathrm{Fe}^{2+}:&\ d^6 \Rightarrow t_{2g}^4e_g^2 \ \Rightarrow \ \text{CFSE}=4(-0.4)+2(0.6)=-0.4\Delta_o,
\mathrm{Fe}^{3+}:&\ d^5 \Rightarrow t_{2g}^3e_g^2 \ \Rightarrow \ \text{CFSE}=0. \end{aligned} \] Step 3: Compare magnitudes. \(|\text{CFSE}|\) values: \(\mathrm{Mn}^{2+}=0\), \(\mathrm{Mn}^{3+}=0.6\Delta_o\), \(\mathrm{Fe}^{2+}=0.4\Delta_o\), \(\mathrm{Fe}^{3+}=0\). Hence, \[ \mathrm{Mn}^{2+}<\mathrm{Mn}^{3+} \quad \text{and} \quad \mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}. \] Therefore, option (C) is correct.