Question

Formation of [M(en)\(_3\)]\(^{2+}\) from [M(H\(_2\)O)\(_6\)]\(^{2+}\) and three equivalents of ethylenediamine (en) is LEAST favored when M is

• A. Co
• B. Ni
• C. Cu
• D. Fe

Answer 1

The Correct Option is D

Step 1: Apply the Irving--Williams series for M\(^{2+}\) (3d) complexes.
For high-spin octahedral divalent first-row transition metals, the thermodynamic stabilities of complexes (including chelates like en) follow:
\[ \text{Mn}^{2+}<\text{Fe}^{2+}<\text{Co}^{2+}<\text{Ni}^{2+}<\text{Cu}^{2+}>\text{Zn}^{2+}. \] Thus, among Co, Ni, Cu, and Fe, the \emph{least} stable chelate is for Fe\(^{2+}\). Step 2: Rationalize using LFSE and additional stabilization.
The chelate effect favors \([M(en)_3]^{2+}\) for all M, but relative stabilities are governed by metal-dependent factors. Approximate octahedral LFSE (high spin) values:
\[ \begin{aligned} \text{Fe}^{2+} (d^6\ \text{HS}) &: -0.4\,\Delta_o,
\text{Co}^{2+} (d^7\ \text{HS}) &: -0.8\,\Delta_o,
\text{Ni}^{2+} (d^8) &: -1.2\,\Delta_o,
\text{Cu}^{2+} (d^9) &: \text{JT stabilization (very strong)}. \end{aligned} \] Smaller (less negative) LFSE for Fe\(^{2+}\) leads to a less stable octahedral complex than Co\(^{2+}\) and Ni\(^{2+}\). Cu\(^{2+}\) is exceptionally stabilized overall despite Jahn--Teller distortion, consistent with the Irving--Williams series. Step 3: Conclusion.
Therefore, the formation of \([M(en)_3]^{2+}\) from \([M(H_2O)_6]^{2+}\) is \underline{least favored} when \(M=\text{Fe}\).