# CO stretching frequency band in the IR spectrum for tetracarbonylnickel and tetracarbonylferrate(II)

A past exam question says:

With reference to the main bonding interactions between a transition metal and a terminal $\ce{CO}$ ligand, explain whether you would expect to observe a lower $\ce{CO}$ stretching frequency band in the infrared spectrum for $\ce{[Ni(CO)4]}$ or for $\ce{[Fe(CO)4]^{2-}}$

My attempt at an answer is:

There are essentially two bonding interactions at play in the carbonyl ligand: a ligand-to-metal $\ce{n → d\sigma}$ interaction and a metal-to-ligand ${d\pi → \pi^*}$ interaction. The latter interaction is called backbonding, because the metal donates electron density back to the ligand.

Nickel is further right in the periodic table than Iron, so $\ce{[Ni(CO)4]}$ has a greater stretching frequency (a greater $\ce{C–O}$ bond order). This is due to the periodic trend in orbital energy. As we move left to right across the periodic table, the $\ce{d}$ orbital energies decrease and the energies of the $\ce{d\pi}$ and ${d\pi^*}$ orbitals separate. As a result, the backbonding orbital interaction is worse in $\ce{[Fe(CO)4]^{2-}}$ (as strong orbital interactions require well-matched orbital energies)." This means that the ${\pi^*}$-orbitals of the $\ce{CO}$ ligands are significantly more populated in $\ce{[Fe(CO)4]^{2-}}$, so the $\ce{CO}$ bond is accordingly weakened and $\ce{v(CO)}$ is reduced.

Is this correct?

• I can't explain but for a quick answer I'd use the fact that as the charge on the central metal atom increases, more electrons are donated into the anti-bonding MO of $\ce{CO}$. As a result the $\ce{CO}$ bond length increases and its bond strength decreases. – Berry Holmes May 12 '17 at 9:15
• This reads ok to me. The CO molecule frequency is $\approx 2143~\pu{cm^{-1}}$ and is $\approx 2060$ in the Ni carbonyl and $\approx 1790$ in the isoelectronic Fe carbonyl. As the extent of M-C back bonding increases the CO frequencies are lowered, as you suggest, because the CO must accept more $d\pi$ electrons from the metal to prevent accumulation of charge. – porphyrin May 12 '17 at 9:18
• This is correct, it would be a good idea to take the answer out of the question and put it into an answer. – orthocresol May 12 '17 at 10:47