The number of unpaired electrons in tetrahedral [Ni(CO)4] is

Question

The number of unpaired electrons in tetrahedral [Ni(CO)4] is (A) 0 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

In [N i(C O)4], Ni is present as Ni atom. N i=[A r] 3 d8 4 s2 CO, being a strong field ligand, transfers the 4s-electrons into 3d-orbitals. [N i(C O)4]=[A r] Thus, [N i(C O)4] does not contain any unpaired electron.

Q. The number of unpaired electrons in tetrahedral $[N i (CO)_{4}]$ is

0

2

3

4

In $[N i (CO)_{4}]$ ,
$N i$ is present as $N i$ atom.
$N i = [A r] 3 d^{8} 4 s^{2}$
$CO$ , being a strong field ligand,
transfers the $4 s$ -electrons into $3 d$ -orbitals.
$[N i (CO)_{4}] = [A r]$
Thus, $[N i (CO)_{4}]$
does not contain any unpaired electron.

Q. The number of unpaired electrons in tetrahedral [Ni(CO)4​] is

0

2

3

4

In [Ni(CO)4​], Ni is present as Ni atom. Ni=[Ar]3d84s2 CO, being a strong field ligand, transfers the 4s-electrons into 3d-orbitals. [Ni(CO)4​]=[Ar] Thus, [Ni(CO)4​] does not contain any unpaired electron.

Q. The number of unpaired electrons in tetrahedral $[N i (CO)_{4}]$ is

In $[N i (CO)_{4}]$ ,
$N i$ is present as $N i$ atom.
$N i = [A r] 3 d^{8} 4 s^{2}$
$CO$ , being a strong field ligand,
transfers the $4 s$ -electrons into $3 d$ -orbitals.
$[N i (CO)_{4}] = [A r]$
Thus, $[N i (CO)_{4}]$
does not contain any unpaired electron.