Question

Consider the following two carbonyl compounds
$\left(i\right)\left[Tc\right(CO{)}_6{]}^{+}$ and $\left(ii\right)\left[Nb\right(CO{)}_6{]}^{-}$

Which of the following is correct?

• A. Only (b) is correct
• B. "$CO$" bond order is greater in $\left[Nb\right(CO{)}_6{]}^{-}$ than in $\left[Tc\right(CO{)}_6{]}^{+}$
• C. Both (a) and (b) are correct
• D. "$Nb-C$" bond order in $\left[Nb\right(CO{)}_6{]}^{-}$ is greater than "$Tc-C$" bond order in $\left[Tc\right(CO{)}_6{]}^{+}$

Answer 1

The correct option is D "$Nb-C$" bond order in $\left[Nb\right(CO{)}_6{]}^{-}$ is greater than "$Tc-C$" bond order in $\left[Tc\right(CO{)}_6{]}^{+}$

We know,
More the extent of back bonding, higher will be the bond order of $M-C$ and lower will be the bond order of $C-O$.

Comparing the oxidation state on central metal in (i) and (ii), $Nb$ has low oxidation so it has higher ability to donate electrons to the carbonyl whereas $Tc$ has high oxidation state so extent of back bonding is lowest.

Therefore, back bonding is efficient in $\left[Nb\right(CO{)}_6{]}^{-}$. So its $M-C$ bond order will be greater than $\left[Tc\right(CO{)}_6{]}^{+}$ and its $C-O$ bond order will be lower than $\left[Tc\right(CO{)}_6{]}^{+}$.

Hence, option (a) is correct.