Question

Select the correct order of C-O bond length :

• A. $\left[M\right(CO{)}_3\left(P{F}_3\right)]<[M\left(CO{)}_3\right(PC{l}_3\left)\right]<\left[M\right(CO{)}_3\left(PM{e}_3\right)]$
• B. $\left[M\right(CO{)}_3\left(P{F}_3\right)]>[M\left(CO{)}_3\right(PC{l}_3\left)\right]>\left[M\right(CO{)}_3\left(PM{e}_3\right)]$
• C. $\left[M\right(CO{)}_3\left(P{F}_3\right)]=[M\left(CO{)}_3\right(PC{l}_3\left)\right]=\left[M\right(CO{)}_3\left(PM{e}_3\right)]$
• D. $\left[M\right(CO{)}_3\left(P{F}_3\right)]=[M\left(CO{)}_3\right(PC{l}_3\left)\right]>\left[M\right(CO{)}_3\left(PM{e}_3\right)]$

Answer 1

The correct option is A $\left[M\right(CO{)}_3\left(P{F}_3\right)]<[M\left(CO{)}_3\right(PC{l}_3\left)\right]<\left[M\right(CO{)}_3\left(PM{e}_3\right)]$

In the options, we can see that the metal atom and carbonyl Ligands are the same in all. Thus, the metal carbonyl bond strength and bond length will depend on the other Ligands present. The tendency to donate lone pair is maximum in $PM{e}_3$ due to $+I$ effect of Me group which increases electron density on P making it a good donor. This results in less effective bonding between metal and $CO.$ Therefore synergic bond strength is least and bond length is maximum. Minimum tendency to donate electron pair is in $P{F}_3$ followed by $PC{l}_3$ as $-I$ effect of $-F>-Cl$ . So synergic bonding is maximum where $P{F}_3$ ligand is present and hence $M-CO$ bond strength is maximum and bond length is minimum.