Question

What is the oxidation number of $Co$ in the compound?
${\left[ {Co\left( {{C_2}{O_4}} \right){{\left( {N{H_3}} \right)}_2}} \right]^ - }$

Answer 1

We have to know that the oxidation number, which is also called oxidation state, is the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom. The numerical value of the oxidation state is equal to the number of electrons lost or gained.

Complete answer:
We need to remember that any free element has an oxidation number equal to zero. For monatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion. The hydrogen atom (H) exhibits an oxidation state of $+ 1$. Oxygen has an oxidation of $- 2$ in most of its compounds. The oxidation number of an atom is the charge that atom would have if the compound was composed of ions. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Since the compound carries negative charge on it thus when we find the oxidation number we will add the sum of charges to the negative one. Oxalate is a bidentate ligand thus its contribution is four while ammonia does not carry any charge since it is a neutral ligand.
$x + \left( { - 4} \right) + 0 = - 1$
$x - 4 = - 1$
$x = - 1 + 4$
$x = 3$
The compounds contain one neutral ligand and one negative ligand. The neutral ligand is $N{H_3}$ and its charge is 0. The negative ligand is ${C_2}{O_4}$ and the charge of one molecule is $- 2$. As there are two molecules its charge is $- 4$.

Note:
We have to know that the electropositive metal atoms of group $1,2and3$ lose a specific number of electrons and have constant positive oxidation numbers. In molecules, more electronegative atoms gain electrons from a less electronegative atom and have negative oxidation states.