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

The C–O stretching frequency falls as the metal centre becomes more electron-rich, since extra electron density is pushed into the CO \( \pi^{*} \) orbital through back-donation, weakening the C–O bond.
Ranking these four carbonyls by their overall charge on the metal fragment: \( \text{V(CO)}_6^{+} \) (+1) is the most electron-poor, \( \text{Ni(CO)}_4 \) (0) is next, \( [\text{Co(CO)}_4]^{-} \) (−1) is more electron-rich still, and \( \text{Fe(CO)}_4^{2-} \) (−2) is the most electron-rich of all.
Since \( \text{Fe(CO)}_4^{2-} \) has the greatest negative charge, it back-donates the most into the CO \( \pi^{*} \) orbitals and therefore has the weakest, lowest-frequency C–O bond of the four.
\[ \boxed{\text{Fe(CO)}_4^{2-}\ \text{has the lowest CO stretching frequency}} \]

Answer 2

These four carbonyls are actually isoelectronic at the metal centre, each metal having an 18-electron configuration achieved through a different combination of d-electron count and overall charge. Comparing them as an isoelectronic series highlights the trend more directly.

1. \( \text{V(CO)}_6^{+} \): Vanadium here effectively behaves like a low d-electron-count, high-charge metal centre in this series. The high positive charge pulls d-electron density towards the metal nucleus and away from back-donation, keeping the C–O bond strong and its stretching frequency high.
2. \( [\text{Co(CO)}_4]^{-} \): Moving along the isoelectronic series to a less positively charged, tetrahedral cobalt centre increases back-donation relative to the vanadium case, lowering the C–O frequency somewhat.
3. \( \text{Ni(CO)}_4 \): With a neutral, zero-valent nickel centre, back-donation is intermediate, sitting between the cationic vanadium carbonyl and the more negatively charged cobalt and iron carbonyls in this series.
4. \( \text{Fe(CO)}_4^{2-} \): With iron carrying an overall 2− charge, this is the most anionic, most electron-rich member of the series, giving it the greatest back-donation into the CO \( \pi^{*} \) orbitals and consequently the weakest C–O bond and lowest stretching frequency of all four.

Across this isoelectronic series, the trend is consistent: the more negative the overall charge on the metal carbonyl, the lower its C–O stretching frequency, and \( \text{Fe(CO)}_4^{2-} \) is the most negatively charged member.

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