Question

In compound $\left[M\right(CO{)}_n{]}^z$, the correct match for highest ${}^{\prime }C-O^{\prime }$ bond length for given $M$, $n$ and $z$ respectively, is:

• A. $M=Ti;n=6;z=-2$
• B. $M=Cr;n=6;z=0$
• C. $M=V;n=6;z=-1$
• D. $M=Mn;n=6;z=+1$

Answer 1

The correct option is A $M=Ti;n=6;z=-2$

In metal carbonyls, the bond order of $M-C$ increases and $C-O$ decreases.

Metals with lowest oxidation state shows more back bonding as they have high tendency to donate. And the $M-C$ bond order will be highest and the $C-O$ bond order will be lowest. So the bond length of $C-O$ will be highest in option (b).

Hence, option (b) is correct.