Question

The oxidation state of chromium in $Cr{{(CO)}_{6}}$ is:
(A) $0$
(B) $+2$
(C) $-2$
(D) $+6$

Answer 1

The oxidation state of the metal is related to the overall charge of the complex. So, knowing the charge of the ligands which are attached to the metal atom and the charge on the complex. The oxidation state of the metal is obtained.

Complete step by step solution:
The $Cr{{(CO)}_{6}}$ is a coordination compound, in which the chromium central atom is surrounded by six identical carbonyl ligands. Thus, making it a homoleptic complex compound. The carbonyl ligands present are neutral ligands, so they do not have any charge.
Then, the oxidation state of the chromium atom in the complex will be given as follows:
Let the charge of the metal be X.
Then, $X+\text{charge}\,\text{on}\,\text{the}\,\text{ligand=charge on the complex}$.
But since the charge on the complex and of the neutral ligand is zero. Thus, making the chromium metal a zerovalent, that is, the oxidation state is zero and has the IUPAC name as Hexacarbonyl chromium (0).

So, the oxidation state of the chromium in $Cr{{(CO)}_{6}}$ is option (A)- $0$.

Additional information:
It has an octahedral geometry and is diamagnetic in nature. This is because of the strong-field ligand, which causes the valence electrons in the metal to pair up. Such that three $3d$-orbital has a total of five electrons and thus have ${{d}^{2}}s{{p}^{3}}$ hybridisation.
Since the inner $3d$-orbital is involved in the hybridisation, it is an inner orbital complex.

Note: The metal belonging to the $d$-block elements. Thus, showing varied oxidation states, but considering the charge of the ligands attached to it, having a fixed charge. The metal acquires a specific oxidation state in the complex.