Answer
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Hint:Bond order affects the position of absorption bands. Higher the bond order—larger is the band frequency. So, the stronger the C-O-bond or higher is the bond order; higher will be the stretching vibrational frequency. As it is given in the statement that \[(P{{F}_{3}})\] is better \[\pi \]-acceptor than CO it means that the bond of Ni with \[(P{{F}_{3}})\] is more stronger than the bond of Ni with CO when both are together as in the molecule$\left[ Ni(P{{F}_{3}}){{(CO)}_{3}} \right]$.
Complete answer:
In $\left[ Ni{{(CO)}_{4}} \right]$ Ni forms a bond with CO only. So, this bond is strong because there is no competition as only one pi-acceptor ligand is present. Due to the Ni-CO bond being stronger, the C-O bond becomes weak because there is more back donation which reduces its bond order so correspondingly there is reduction in stretching vibration frequency as it increases with the increasing bond order.
Whereas in $Ni(P{{F}_{3}}){{(CO)}_{4}}$, Ni forms two bonds, one with \[(P{{F}_{3}})\] and one with CO. Ni forms a strong bond with\[(P{{F}_{3}})\] because it is better \[\pi \]- acceptor than CO as compared to the bond formed by Ni with CO. Consequently this weakens the Ni-CO bond resulting in the increase of bond order in C-O bond because there is less back donation of CO as compared to\[(P{{F}_{3}})\]. So, due to the increase in bond order of C-O bond it has high stretching vibrational frequency. So, correct order of stretching vibration frequency of C-O bond in the given molecules is$\left[ Ni{{(CO)}_{4}} \right]<\left[ Ni{{(CO)}_{3}}(P{{F}_{3}}) \right]$.
So, option-(B) is correct.
Additional Information:
\[\pi \]back bonding, also called \[\pi \] back donation, is a concept in which electrons move from an atomic orbital on one atom to an appropriate symmetry anti-bonding orbital on a π-acceptor ligand. Ligands having empty orbitals which can interact with metal d-orbitals for the formation of \[\pi \]-bond are called \[\pi \]-acceptor ligands.
Note:
CO is a dative, L-type ligand that does not affect the oxidation state of the metal center upon binding, but does increase the total electron count by two units. There are two bonding interactions at play in the carbonyl ligand: a ligand-to-metal n → $d\sigma $ interaction and a metal-to-ligand $d\pi \to {{\pi }^{*}}$ interaction. The latter interaction is called back-bonding, because the metal donates electron density back to the ligand.
Complete answer:
In $\left[ Ni{{(CO)}_{4}} \right]$ Ni forms a bond with CO only. So, this bond is strong because there is no competition as only one pi-acceptor ligand is present. Due to the Ni-CO bond being stronger, the C-O bond becomes weak because there is more back donation which reduces its bond order so correspondingly there is reduction in stretching vibration frequency as it increases with the increasing bond order.
Whereas in $Ni(P{{F}_{3}}){{(CO)}_{4}}$, Ni forms two bonds, one with \[(P{{F}_{3}})\] and one with CO. Ni forms a strong bond with\[(P{{F}_{3}})\] because it is better \[\pi \]- acceptor than CO as compared to the bond formed by Ni with CO. Consequently this weakens the Ni-CO bond resulting in the increase of bond order in C-O bond because there is less back donation of CO as compared to\[(P{{F}_{3}})\]. So, due to the increase in bond order of C-O bond it has high stretching vibrational frequency. So, correct order of stretching vibration frequency of C-O bond in the given molecules is$\left[ Ni{{(CO)}_{4}} \right]<\left[ Ni{{(CO)}_{3}}(P{{F}_{3}}) \right]$.
So, option-(B) is correct.
Additional Information:
\[\pi \]back bonding, also called \[\pi \] back donation, is a concept in which electrons move from an atomic orbital on one atom to an appropriate symmetry anti-bonding orbital on a π-acceptor ligand. Ligands having empty orbitals which can interact with metal d-orbitals for the formation of \[\pi \]-bond are called \[\pi \]-acceptor ligands.
Note:
CO is a dative, L-type ligand that does not affect the oxidation state of the metal center upon binding, but does increase the total electron count by two units. There are two bonding interactions at play in the carbonyl ligand: a ligand-to-metal n → $d\sigma $ interaction and a metal-to-ligand $d\pi \to {{\pi }^{*}}$ interaction. The latter interaction is called back-bonding, because the metal donates electron density back to the ligand.
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