Question

In $Ni{(CO)_4}$, the oxidation state of Ni is:
$(a){\text{ 4}} \\\ (b){\text{ 0}} \\\ (c){\text{ 2}} \\\ (d){\text{ 8}} \\\$

Answer 1

Hint – In this problem let the oxidation of Ni be x and that of CO be y. Use the concept that overall charge on $Ni{(CO)_4}$ is O, thus the oxidation number of Ni in addition to four multiplied by oxidation of CO should be equal to 0.

Complete answer:
Let the oxidation state of Ni be x.
And the oxidation state of CO be y.
Now as we know that complex sum of oxidation states on the overall complex is equal to the sum of the oxidation states of the individual ions of all the atoms involved in making the complex.
Since the overall charge on $Ni{(CO)_4}$ is zero (0).
Therefore, $x + 4\left( y \right) = 0$................. (1)
Now as we know that CO is a neutral ligand, its oxidation state in metal carbonyl is 0.
Or the oxidation state of C and O in CO is (+2 and -2) so the overall oxidation number of CO is zero.
$\Rightarrow y = 0$
Now from equation (1) we have,
$\Rightarrow x + 4\left( 0 \right) = 0$
$\Rightarrow x = 0$
So the oxidation state of Ni in $Ni{(CO)_4}$ is zero.
So this is the required answer.

Hence option (B) is the correct answer.

Note – The oxidation state of Ni is coming out to be zero because Co is a neutral ligand of (L-type) due to this Ni does not have to share any of its electrons with CO to form a bond. In actuality the bond between them is a dative bond. Now $Ni{(CO)_4}$ molecule is tetrahedral, with 4 carbonyl ligands attached to nickel.