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

Answer 1

The Correct Option is C

In coordination chemistry, the Crystal Field Stabilization Energy (CFSE) is determined by the splitting of d-orbitals in transition metal complexes. The general formula for CFSE is \((P-m)\times\Delta-\text{pairing energy}\), where \(P\) and \(m\) are the numbers of electrons in the \(t_{2g}\) and \(e_g\) orbitals respectively.

Let's analyze the CFSE for the given ions in their octahedral complexes:

1. Mn²⁺ vs. Mn³⁺:
   Mn²⁺ has an electronic configuration of \([\text{Ar}]3d^5\). In an octahedral complex, these 5 electrons occupy both the \(t_{2g}\) and \(e_g\) orbitals equally, distributing as \(t_{2g}^3e_g^2\). Therefore, CFSE is approximately zero as the stabilization energy gained by placing electrons in the lower energy \(t_{2g}\) orbital is offset by electron-electron repulsion.
   Mn³⁺ has an electronic configuration of \([\text{Ar}]3d^4\), with \(t_{2g}^3e_g^1\) distribution in an octahedral field. Each electron in \(t_{2g}\) contributes negatively to CFSE since they are stabilized.
   This makes \(\mathrm{CFSE}\left(\text{Mn}^{2+}\right)<\mathrm{CFSE}\left(\text{Mn}^{3+}\right)\).
2. Fe²⁺ vs. Fe³⁺:
   Fe²⁺ has an electronic configuration of \([\text{Ar}]3d^6\). These distribute as \(t_{2g}^4e_g^2\), contributing negatively to CFSE due to more electrons in the stabilized \(t_{2g}\) orbitals.
   Fe³⁺ has an electronic configuration of \([\text{Ar}]3d^5\), distributing equally as \(t_{2g}^3e_g^2\), resulting in zero CFSE (same logic as Mn²⁺ earlier).
   This results in \(\mathrm{CFSE}\left(\text{Fe}^{2+}\right)>\mathrm{CFSE}\left(\text{Fe}^{3+}\right)\).

Thus, the correct answer is: \(\mathrm{Mn}^{2+}<\mathrm{Mn}^{3+}\) and \(\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\).