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Consider change in length of the metal-ligand bond in the following reaction:

$\ce{[Cr(CN)6]^3- + e^- -> [Cr(CN)6]^4-}$

I initially had two theories (based on general electrostatics and d-orbital splitting) that contradicted eachother.

General Electrostatics Theory

Since $\ce{Cr}$ is reduced from $\ce{Cr^3+}$ to $\ce{Cr^2+}$, the electrostatic attractions between the $\ce{Cr}$ atom and the $\ce{CN-}$ ligand will decrease. Therefore, according to this explanation, the $\ce{Cr-CN-}$ bond length would increase.

D-orbital Splitting Theory

I examined the change in the d-orbital splitting diagram. Both the oxidized and reduced forms of the compound are in low-spin states due to the strong cyanide ligand. Since the electron change is from ${d^3}$ to ${d^4}$, the diagrams would look like this:

enter image description here

After drawing the diagrams, I concluded that the $\ce{Cr}$ ion experiences a net gain of since the energy loss from the extra electron in the bonding ${t_{2g}}$ orbitals is greater than the energy gain from the paired electrons. Due to this increase in ligand stablization energy, I concluded that the chromium-cyanide bond would decrease in size.

Conclusions

I later saw that my first theory was correct, and that the chromium-cyanide bond increases. However, the answer to this problem also supported the changes in the d-orbital diagrams oultined in the second theory.

Is only the first theory correct, or are both correct with a stronger effect from the first scenario?

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