Question

Which of the following will have the lower CO stretching frequency?

• A. \( \text{V(CO)}_6^+ \)
• B. \( [\text{Co(CO)}_4]^- \)
• C. \( \text{Ni(CO)}_4 \)
• D. \( \text{Fe(CO)}_4^{2-} \)

Hint:

A higher negative charge on the metal results in more electron density on the CO ligands, leading to a lower CO stretching frequency.

Answer 1

The Correct Option is D

Step 1: Understanding CO stretching frequency.
The CO stretching frequency is influenced by the electron density on the metal. Higher electron density on the metal leads to a lower stretching frequency because the metal-carbon bond becomes weaker.
Step 2: Analyzing the options.
- (1) \( \text{V(CO)}_6^+ \): Vanadium has a positive charge, which attracts electron density from the CO ligands, increasing the CO stretching frequency. - (2) \( [\text{Co(CO)}_4]^- \): Cobalt in this complex is in a negative oxidation state, which increases electron density on the CO ligands, slightly decreasing the CO stretching frequency. - (3) \( \text{Ni(CO)}_4 \): Nickel in this complex is neutral, and the electron density on the CO ligands is balanced, leading to a moderate CO stretching frequency. - (4) \( \text{Fe(CO)}_4^{2-} \): Iron in this complex is in a negative oxidation state, which greatly increases electron density on the CO ligands, leading to the lowest CO stretching frequency.
Step 3: Conclusion.
The correct answer is (4) \( \text{Fe(CO)}_4^{2-} \), as the increased electron density on the CO ligands lowers the CO stretching frequency.

Answer 2

The C–O stretching frequency in a metal carbonyl depends on how much electron density the metal back-donates into the antibonding \( \pi^{*} \) orbitals of CO. More back-donation weakens the C–O bond and lowers its stretching frequency, so the more electron-rich (more negatively charged) the metal, the lower the C–O frequency.

1. \( \text{V(CO)}_6^{+} \): This carbonyl carries an overall positive charge, so the metal is comparatively electron-poor. Less electron density is available for back-donation into the CO \( \pi^{*} \) orbitals, so the C–O bond stays closer to a normal triple-bond character and the stretching frequency is the highest among these four.
2. \( [\text{Co(CO)}_4]^{-} \): This carbonyl carries a single negative charge, making the metal somewhat electron-rich. This increases back-donation compared to a neutral or cationic carbonyl, lowering the C–O frequency somewhat, but less than in a more highly charged anion.
3. \( \text{Ni(CO)}_4 \): This carbonyl is neutral overall. Its back-donation, and hence its C–O frequency, sits in between the cationic and the anionic carbonyls discussed here.
4. \( \text{Fe(CO)}_4^{2-} \): This carbonyl carries a double negative charge, making the metal the most electron-rich of the four. This drives the strongest back-donation into the CO \( \pi^{*} \) orbitals, weakening the C–O bond the most and giving the lowest C–O stretching frequency of the four.

Increasing negative charge on the metal carbonyl consistently increases back-donation and lowers the C–O stretching frequency, and \( \text{Fe(CO)}_4^{2-} \) carries the largest negative charge here.

Therefore, the correct answer is \( \text{Fe(CO)}_4^{2-} \).