Chemistry - Coordination Chemistry Question with Solution | TestHub

There is synergic bonding present in $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$.

As $\mathrm{CO}$ is neutral ligand, so oxidation state of nickel is zero i.e.${\mathrm{Ni}}^0$.

So, electronic configuration of ${\mathrm{Ni}}^0$ $=\left[\mathrm{Ar}\right]3{\mathrm{d}}^8 4{\mathrm{s}}^2$

As $\mathrm{CO}$ is strong field ligand, it will back paired the 4s electrons to $3\mathrm{d}$ orbitals.

So in complex electronic configuration of ${\mathrm{Ni}}^0$ $=\left[\mathrm{Ar}\right] 3{\mathrm{d}}^{10} 4{\mathrm{s}}^0$ here inner d orbital is filled.

In ligand $\mathrm{CO}$ (carbonyl) the anti-bonding orbital is vacant.

$\mathrm{CO}: \mathrm{\sigma }1{\mathrm{s}}^2, \mathrm{\sigma }*1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \mathrm{\sigma }*2{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2, {\mathrm{\pi p}}_{\mathrm{x}}^2={\mathrm{\pi p}}_{\mathrm{y}}^2, \mathrm{\pi }*{\mathrm{p}}_{\mathrm{x}}^0=\mathrm{\pi }*{\mathrm{p}}_{\mathrm{y}}^0$

The fully filled inner d-orbital of central metal atom gives it's electrons to vacant $\mathrm{\pi }$ anti-bonding orbital, this is called back $\mathrm{\pi }$ bonding, due to this back bonding bond between metal and ligand is very strong.

This back $\mathrm{\pi }$ bonding is also called synergic bonding and this synergic bonding is explained by molecular orbital theory.

Carbon monoxide $\mathrm{CO}$  made synergic bonding with Haemoglobin in are blood, as synergic bonding is very strong, so oxygen taking capacity of haemoglobin decrease and it cause death.