Question

Which of the following order is correct for the IR vibrational frequency of CO?

• A. ${\left[Fe{\left(CO\right)}_4\right]}^{2-}<{\left[Co{\left(CO\right)}_4\right]}^{-}<\left[Ni{\left(CO\right)}_4\right]$
• B. ${\left[Fe{\left(CO\right)}_4\right]}^{2-}>{\left[Co{\left(CO\right)}_4\right]}^{-}>\left[Ni{\left(CO\right)}_4\right]$
• C. ${\left[Fe{\left(CO\right)}_4\right]}^{2-}>{\left[Co{\left(CO\right)}_4\right]}^{-}<\left[Ni{\left(CO\right)}_4\right]$
• D. ${\left[Fe{\left(CO\right)}_4\right]}^{2-}<{\left[Co{\left(CO\right)}_4\right]}^{-}>\left[Ni{\left(CO\right)}_4\right]$

Answer 1

The correct option is A ${\left[Fe{\left(CO\right)}_4\right]}^{2-}<{\left[Co{\left(CO\right)}_4\right]}^{-}<\left[Ni{\left(CO\right)}_4\right]$

The IR vibrational frequency depends on the strength of CO bond, More is the bond strength more will be the IR vibrational frequency.Since, CO is a pi acceptor ligand and accepts electron in antibonding pi orbital. More electrons in antibonding orbital implies less bond strength of CO. In the given complexes the oxidation state of metals is Fe(2-), Co(1-), Ni(0). Therefore more electrons on metal implies more will be the pi backbonding more stronger will be the Metal-C bond weaker will be the C-O bond hence, the IR vibrational frequency order will be $\left[Fe\right(CO{)}_4{]}^{2-}<\left[Co\right(CO{)}_4{]}^{-}<\left[NiCO{)}_4\right]$