Question

$NO$ often acts as a three electron donor in contrast to most ligands which donate two electrons. What happens when $NO$ replaces three $CO$ in $\left[Fe\right(CO{)}_5]$ and six $CO$ in $\left[Cr\right(CO{)}_6]$?

Answer 1

$NO$ is able to donate three electrons instead of two. Therefore, two $NO$ groups (2 x 3 = 6 electrons) can replace three $CO$ groups (3 x 2 = 6 electrons) in $\left[Fe\right(CO{)}_5]$. Similarly, four $NO$ groups (4 x 3 = 12 electrons) can replace six $CO$ groups (6 x 2 = 12 electrons) in $\left[Cr\right(CO{)}_6]$. Therefore, the replacement reactions can be written as follows:

$\left[Fe\right(CO{)}_5]+2NO\to [Fe\left(NO{)}_2\right(CO{)}_2]+3CO$

$\left[Cr\right(CO{)}_6]+4NO\to [Cr\left(NO{)}_4\right]+6CO$