So, $\left(3 \times 2 e ^{-}=6 e ^{-}\right) ; 2$ $NO$ can easily replaced $3 CO$.
$\left(6 \times 2 e ^{-}=12 e ^{-}\right) ; 4 NO$ can easily replace $6 CO$
$Fe ( CO )_{5} \rightarrow Fe ( NO )_{2}( CO )_{2}$
$Cr ( CO )_{5} \rightarrow Cr ( NO )_{4}$