Question

In $Ni{(CO)_4}$, the oxidation state of Ni is:
                 (A) 4
                 (B) Zero
                 (C) 2
                 (D) 8

Answer 1

Hint: CO is a neutral ligand and hence its oxidation state will be zero and the overall charge of the complex is also zero. CO binds with these transition metals in such complexes using back bonding. This type of bonding has 3 components which gives it a partial triple bond character.

Complete step by step answer:
-We all know that CO is a neutral ligand. This means that its oxidation state in any metal carbonyl compound will be zero (0).
-Also we can see that the compound $Ni{(CO)_4}$ has an overall charge equal to zero. So, the oxidation state of Ni here is zero.
-Let us see all this mathematically:
For any complex compound its overall charge is always equal to the sum of the oxidation states of all the constituent atoms and ligands.
The overall charge of the complex $Ni{(CO)_4}$ is = 0.
The oxidation state of ligand CO = 0.
Let the oxidation state of Ni be = x.
Oxidation state of $Ni{(CO)_4}$= Oxidation state of Ni + 4(Oxidation state of CO)
                                                               0 = x + 4 (0)
                                          So, x = 0.
Hence the oxidation state of Ni in $Ni{(CO)_4}$ is zero (0).

So, the correct option is: (B) Zero.

Additional information: Such metal carbonyls can prove to be toxic on contact, ingestion or inhalation because they rapidly bind with our hemoglobin to form carboxyhemoglobin thus reducing the ability of hemoglobin to bind with oxygen.

Note: Remember that the oxidation state of CO is zero because it is a neutral ligand and that the overall charge of this complex is 0.