Question:medium
The pair of complex ions that exhibits the slowest outer-sphere electron-exchange reaction at 25 \(^{\circ}\mathrm{C}\) is:
The pair of complex ions that exhibits the slowest outer-sphere electron-exchange reaction at 25 \(^{\circ}\mathrm{C}\) is:
Show Hint
Outer-sphere self-exchange rate drops when the metal-ligand bond length (and spin state) changes a lot between the two oxidation states; check which pair has the biggest jump in \(e_g\) electron occupancy.
Updated On: Jul 20, 2026
- \(\mathrm{[Co(NH_3)_6]^{3+}}\) and \(\mathrm{[Co(NH_3)_6]^{2+}}\)
- \(\mathrm{[Fe(H_2O)_6]^{3+}}\) and \(\mathrm{[Fe(H_2O)_6]^{2+}}\)
- \(\mathrm{[Ru(NH_3)_6]^{3+}}\) and \(\mathrm{[Ru(NH_3)_6]^{2+}}\)
- \(\mathrm{[Mn(CN)_6]^{3-}}\) and \(\mathrm{[Mn(CN)_6]^{4-}}\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Frame the question as a bond-reorganization comparison.
Outer-sphere electron transfer needs both partner ions to reach a matching geometry before the electron hops. So the slowest pair is the one whose two oxidation states look most different structurally: biggest change in $M-L$ bond length, and worse, a change in $e_g$ electron occupancy.
Step 2: Work out the $d$-electron count and spin state for each metal in each option.
$\mathrm{Co^{3+}}$ ($d^6$) with $\mathrm{NH_3}$ is low spin, $t_{2g}^6e_g^0$; $\mathrm{Co^{2+}}$ ($d^7$) is high spin, $t_{2g}^5e_g^2$. $\mathrm{Fe^{3+}}$ ($d^5$) and $\mathrm{Fe^{2+}}$ ($d^6$) with water are both high spin. $\mathrm{Ru^{3+}}$ ($d^5$) and $\mathrm{Ru^{2+}}$ ($d^6$) with $\mathrm{NH_3}$ are both low spin, since second/third-row metals have a much larger field splitting. $\mathrm{Mn(III)}$ $d^4$ and $\mathrm{Mn(II)}$ $d^5$ with cyanide are both low spin.
Step 3: Score each pair on how many $e_g$ electrons change.
Cobalt pair: $e_g$ occupancy jumps from 0 to 2, stretching the bond by roughly 0.2 angstrom. Iron pair: both stay high spin with $e_g$ occupancy unchanged at 2; only $t_{2g}$ changes by one electron, a smaller effect. Ruthenium pair: both low spin, $e_g$ stays empty in both; only $t_{2g}$ changes by one, minimal bond-length change. Manganese cyanide pair: both low spin, $e_g$ stays empty in both, a small change.
Step 4: Rank the reorganization energies.
The cobalt pair has by far the biggest jump in $e_g$ occupancy among all four options, needing the most nuclear reorganization before electron transfer, giving the largest activation barrier. All other pairs keep $e_g$ occupancy fixed and change only $t_{2g}$ population, a much smaller perturbation.
Final Answer:
The cobalt ammine pair, option (A), has the largest reorganization energy and the slowest outer-sphere self-exchange rate. \[\boxed{\text{[Co(NH}_3\text{)}_6\text{]}^{3+}\text{ and [Co(NH}_3\text{)}_6\text{]}^{2+}}\]
Outer-sphere electron transfer needs both partner ions to reach a matching geometry before the electron hops. So the slowest pair is the one whose two oxidation states look most different structurally: biggest change in $M-L$ bond length, and worse, a change in $e_g$ electron occupancy.
Step 2: Work out the $d$-electron count and spin state for each metal in each option.
$\mathrm{Co^{3+}}$ ($d^6$) with $\mathrm{NH_3}$ is low spin, $t_{2g}^6e_g^0$; $\mathrm{Co^{2+}}$ ($d^7$) is high spin, $t_{2g}^5e_g^2$. $\mathrm{Fe^{3+}}$ ($d^5$) and $\mathrm{Fe^{2+}}$ ($d^6$) with water are both high spin. $\mathrm{Ru^{3+}}$ ($d^5$) and $\mathrm{Ru^{2+}}$ ($d^6$) with $\mathrm{NH_3}$ are both low spin, since second/third-row metals have a much larger field splitting. $\mathrm{Mn(III)}$ $d^4$ and $\mathrm{Mn(II)}$ $d^5$ with cyanide are both low spin.
Step 3: Score each pair on how many $e_g$ electrons change.
Cobalt pair: $e_g$ occupancy jumps from 0 to 2, stretching the bond by roughly 0.2 angstrom. Iron pair: both stay high spin with $e_g$ occupancy unchanged at 2; only $t_{2g}$ changes by one electron, a smaller effect. Ruthenium pair: both low spin, $e_g$ stays empty in both; only $t_{2g}$ changes by one, minimal bond-length change. Manganese cyanide pair: both low spin, $e_g$ stays empty in both, a small change.
Step 4: Rank the reorganization energies.
The cobalt pair has by far the biggest jump in $e_g$ occupancy among all four options, needing the most nuclear reorganization before electron transfer, giving the largest activation barrier. All other pairs keep $e_g$ occupancy fixed and change only $t_{2g}$ population, a much smaller perturbation.
Final Answer:
The cobalt ammine pair, option (A), has the largest reorganization energy and the slowest outer-sphere self-exchange rate. \[\boxed{\text{[Co(NH}_3\text{)}_6\text{]}^{3+}\text{ and [Co(NH}_3\text{)}_6\text{]}^{2+}}\]
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