Question

According to molecular orbital theory, the correct ground state electronic configuration for \(\mathrm{[Co(NH_3)_6]^{3+}}\) ion is

• A. \(a_{1g}^{2}\ t_{1u}^{6}\ e_g^{4}\ t_{2g}^{6}\ e_g^{*0}\)
• B. \(a_{1g}^{2}\ t_{1u}^{6}\ t_{2g}^{6}\ e_g^{4}\ e_g^{*0}\)
• C. \(t_{1u}^{6}\ a_{1g}^{2}\ e_g^{4}\ t_{2g}^{6}\ e_g^{*0}\)
• D. \(a_{1g}^{2}\ t_{1u}^{6}\ e_g^{4}\ t_{2g}^{4}\ e_g^{*2}\)

Hint:

For a sigma-only ligand set, the MO energy order is \(a_{1g} < t_{1u} < e_g\) (bonding) \(< t_{2g}\) (non-bonding) \(< e_g^{*}\) (antibonding); fill 12 ligand electrons into the bonding set, then place \(\mathrm{Co(III)}\)'s 6 \(d\) electrons in the low-spin \(t_{2g}^{6}\) configuration, since \(\mathrm{[Co(NH_3)_6]^{3+}}\) is diamagnetic.

Answer 1

The Correct Option is A

Building the molecular orbital picture for an octahedral complex with a pure sigma-donor ligand set like $\mathrm{NH_3}$ comes down to two separate counting jobs: filling the metal-ligand bonding orbitals with the ligand's own electrons, and then figuring out where the metal's own $d$ electrons land.

1. Bonding orbitals come from symmetry matching: the metal's $s$ orbital matches the totally symmetric ligand combination, giving the $a_{1g}$ bonding orbital. The three $p$ orbitals match a triply degenerate ligand combination, giving $t_{1u}$. Two of the five $d$ orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) point straight at the ligands and match the remaining ligand combination, giving $e_g$. The other three $d$ orbitals ($d_{xy}, d_{xz}, d_{yz}$) point between the ligands and have no sigma partner, so they stay non-bonding, labeled $t_{2g}$.
2. Energy order: $a_{1g} < t_{1u} < e_g$ (bonding, filled mostly by ligand electrons) $< t_{2g}$ (non-bonding, purely metal $d$ character) $< e_g^{*}$ (antibonding).
3. Filling with ligand electrons: each of the six $\mathrm{NH_3}$ ligands contributes a lone pair, 12 electrons total, and these exactly fill $a_{1g}^{2}\ t_{1u}^{6}\ e_g^{4}$.
4. Filling with metal electrons: $\mathrm{Co(III)}$ is $d^6$. $\mathrm{NH_3}$ is a strong enough field ligand that $\mathrm{[Co(NH_3)_6]^{3+}}$ is the well known diamagnetic, low-spin case, so all six $d$ electrons pair up in the lower, non-bonding $t_{2g}$ set before touching the antibonding $e_g^{*}$ set: $t_{2g}^{6}\ e_g^{*0}$.

Stacking these in the correct energy order gives $a_{1g}^{2}\ t_{1u}^{6}\ e_g^{4}\ t_{2g}^{6}\ e_g^{*0}$. Any answer that reorders $t_{2g}$ ahead of the bonding $e_g$, or swaps $a_{1g}$ and $t_{1u}$, has the energy ladder wrong, and any answer that splits the six metal electrons across $t_{2g}$ and $e_g^{*}$, like $t_{2g}^4 e_g^{*2}$, is describing a high-spin, paramagnetic complex, which does not match the known diamagnetic $\mathrm{[Co(NH_3)_6]^{3+}}$.

Let's summarize:

• Sigma-only ligands like $\mathrm{NH_3}$ give bonding MOs $a_{1g}, t_{1u}, e_g$ (in that energy order), a non-bonding $t_{2g}$, and an antibonding $e_g^{*}$.
• 12 ligand electrons fill the three bonding MOs completely.
• All 6 of Co(III)'s $d$ electrons pair up in non-bonding $t_{2g}$ because the complex is low-spin, leaving $e_g^{*}$ empty.

The correct configuration is $a_{1g}^{2}\ t_{1u}^{6}\ e_g^{4}\ t_{2g}^{6}\ e_g^{*0}$, option (A).