JEE Main 2020 — Chemistry Inorganic Chemistry

There is synergic bonding present in $[\mathrm{Ni}{(\mathrm{CO})}_{4}]$[Ni(CO)4​].

As $\mathrm{CO}$CO is neutral ligand, so oxidation state of nickel is zero i.e.${\mathrm{Ni}}^{0}$Ni0.

So, electronic configuration of ${\mathrm{Ni}}^{0}$Ni0 $=[\mathrm{Ar}]3{d}^{8}4{s}^{2}$=[Ar]3d84s2

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

So in complex electronic configuration of ${\mathrm{Ni}}^{0}$Ni0 $=[\mathrm{Ar}]3{d}^{10}4{s}^{0}$=[Ar]3d104s0 here inner d orbital is filled.

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

$$\mathrm{CO}:\sigma 1{s}^{2},\sigma *1{s}^{2},\sigma 2{s}^{2},\sigma *2{s}^{2},\sigma 2{p}_{z}^{2},{\pi p}_{x}^{2}={\pi p}_{y}^{2},\pi *{p}_{x}^{0}=\pi *{p}_{y}^{0}$$CO:σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2pz2​,πpx2​=πpy2​,π∗px0​=π∗py0​

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

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

Carbon monoxide $\mathrm{CO}$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.