Question

Correct sequence of $CO$ bond order in given compounds is:
$\left(P\right)Fe(CO{)}_5$
$\left(Q\right)CO$
$\left(R\right)H_{3}B\leftarrow CO$
$\left(S\right)\left[Mn\right(CO{)}_5{]}^{-}$

• A. $R>Q>P>S$
• B. $S>P>R>Q$
• C. $Q>S>P>R$
• D. $P>R>S>Q$

Answer 1

The correct option is A $R>Q>P>S$

$CO\text{Bond order}\propto \frac{1}{\text{Extent of back bonding (}M\to CO\text{)}}$

In free carbonyls ($CO$), there will be no back bonding so $C-O$ bond order will be 3.

Comparing $Fe(CO{)}_5$ and $\left[Mn\right(CO{)}_5{]}^{-}$, back bonding is more in $\left[Mn\right(CO{)}_5{]}^{-}$ because the $Mn$ is in -1 oxidation state, so it donates more electron density and the extent of back bonding will be higher. So bond order of $CO$ will be more.

$H_{3}B\leftarrow CO$, the electrons are donated from non-bonding molecular orbital of carbonyl carbon.
So bond order of $CO$ is more than 3.

Hence, correct order will be
$R>Q>P>S$