A different way to count electrons at a metal center is the covalent (neutral atom) method. Here the metal is treated as a neutral atom contributing its own group number of electrons, an X-type ligand (like a halide or H) contributes 1 electron because it forms a normal covalent bond, and an L-type neutral 2-electron donor (like $CO$, $PPh_3$, $NH_3$, or one end of a chelate like $en$) contributes 2 electrons. Any overall ionic charge on the complex is then added (for an anion) or subtracted (for a cation) as extra or missing electrons.
- $\mathrm{[Rh(PPh_3)_3Cl]}$: neutral $Rh$ (group 9) gives 9 electrons, 3 $PPh_3$ give $3\times2=6$, $Cl$ (X-type) gives 1. No overall charge to add. Total $=9+6+1=16$, so this is a 16-electron complex, not 18.
- $\mathrm{[Co(NH_3)_6]^{2+}}$: neutral $Co$ (group 9) gives 9 electrons, 6 $NH_3$ give $6\times2=12$. The complex is a $2+$ cation, so subtract 2 electrons. Total $=9+12-2=19$, an odd, non-18 count.
- $\mathrm{[V(CO)_6]^{-}}$: neutral $V$ (group 5) gives 5 electrons, 6 $CO$ give $6\times2=12$. The complex is a $1-$ anion, so add 1 electron. Total $=5+12+1=18$, exactly the 18-electron count.
- $\mathrm{[Ni(en)_3]^{2+}}$: neutral $Ni$ (group 10) gives 10 electrons, each bidentate $en$ acts as two L-type donors so gives $2\times2=4$ electrons, three $en$ give $3\times4=12$. The complex is $2+$, so subtract 2. Total $=10+12-2=20$, above 18.
Whichever method is used, ionic or covalent, only $\mathrm{[V(CO)_6]^{-}}$ lands exactly on 18 valence electrons, so it is the species that follows the 18-electron rule.
Let's summarize:
- The covalent method always gives the same total electron count as the ionic method for a given real complex, since the two are just different bookkeeping of the same electrons.
- An odd total electron count (like 19 for the $Co(II)$ complex) can never satisfy the 18-electron rule, because it forces an unpaired electron.
The correct option is (C), $\mathrm{[V(CO)_6]^{-}}$.