# Which of the following has the smallest bond order for "CO"? ["Mn"("CO")_6]^(+), "Cr"("CO")_6, ["Ti"("CO")_6]^(2-), ["V"("CO")_6]^(-)

Feb 19, 2016

The ${\text{Ti}}^{2 -}$ complex, involving the most negative oxidation state out of all these ${d}^{6}$ complexes (${\text{Ti}}^{2 -}$ being the most electron-rich metal center), is actually the one with the smallest bond order within the $\text{C"-"O}$ bond because it will involve the most $\pi$ back-bonding.

The more $\pi$ backbonding into the ${\pi}^{\text{*}}$ of $\text{CO}$, the more weakened the $\text{C"-"O}$ bond is, within itself.

CARBONYL LIGANDS ARE BOTH SIGMA DONORS AND PI ACCEPTORS

One of the most easily forgotten concepts here to remember is that $\text{CO}$ is a $\setminus m a t h b f \left(\sigma\right)$ donor AND $\setminus m a t h b f \left(\pi\right)$ acceptor, since it has both a $\sigma$ HOMO and two ${\pi}^{\text{*}}$ LUMOs.

• Its $\sigma$ HOMO is the one contributed to by the $2 s$ of carbon and the $2 {p}_{z}$ of oxygen (that's why there's only one HOMO, not two or three).
• Its two ${\pi}^{\text{*}}$ LUMOs are the second-highest-energy ones with contribution from the $2 {p}_{x}$ and $2 {p}_{y}$ of carbon and of oxygen.

METAL CHARGE, AND COMPLEX FORMED

1) Each $\text{CO}$, in acting as a $\setminus m a t h b f \left(\sigma\right)$ donor and $\setminus m a t h b f \left(\pi\right)$ acceptor, contributes ZERO charge (remember that it is $\text{^((-)) :"C"-="O} {:}^{\left(+\right)}$), so ${\text{Mn}}^{+}$ is our cation here.

This is really saying that we have a ${d}^{6}$ metal, because ${\text{Mn}}^{+}$ has a valence configuration of $\left[A r\right] 3 {d}^{5} 4 {s}^{1}$.

This is an octahedral complex because there are six ligands. In fact, all four of these complexes are octahedral.

2) Same situation, but with ${\text{Cr}}^{0}$, which is a ${d}^{6}$ metal ($\left[A r\right] 3 {d}^{5} 4 {s}^{1}$).
3) Same situation, but with ${\text{Ti}}^{2 -}$, which is a ${d}^{6}$ metal ($\left[A r\right] 3 {d}^{4} 4 {s}^{2}$).
4) Same situation, but with ${\text{V}}^{-}$, which is a ${d}^{6}$ metal ($\left[A r\right] 3 {d}^{4} 4 {s}^{2}$).

Now, since we are talking about bond order, which correlates with the effective degree of the overall bonding (such as triple bonding being a bond order of 3, or $s {p}^{2}$ aromatic bonding being a bond order of 1.5), we should expect that the strongest bond WITHIN $\text{CO}$ has the highest bond order.

CHECKING TRANSITION METAL CHARGE AND RADIUS

Something definitely to consider here is the charge and radius of the metal. Here are the metals in order of ascending positive charge, and also decreasing radius:

${\text{Ti"^(2-) > "V"^(-) > "Cr"^(0) > "Mn}}^{+}$

Now, what exactly would this do? At a glance, I would note that titanium is the largest and most negatively charged. That would imply that the internuclear distance is longer. Generally that implies a weaker $\text{M"-"CO}$ bond, and a stronger $\text{C"-"O}$ bond!

However, the largest anionic charge makes the metal the most electron-rich, and the most able to $\pi$ back-donate, which weakens the $\text{C"-"O}$ bond more so than the larger radius strengthens it.

Therefore, the $\boldsymbol{{\text{Ti}}^{2 -}}$ complex contains $\text{CO}$ with the weakest $\text{C"-"O}$ bond.