Question

What is the oxidation state of Co in the complex ${{\text{(CoC}{{\text{l}}_{2}}{{\text{(en)}}_{2}}\text{)}}^{+}}$?
(Note: Ethylenediamine (en) = $\text{N}{{\text{H}}_{2}}\text{ - C}{{\text{H}}_{2}}\text{ - C}{{\text{H}}_{2}}\text{ - N}{{\text{H}}_{2}}$)

Answer 1

For this problem, we have to consider the oxidation state of cobalt as x and then by adding the oxidation state of other molecules which will be equal to the net charge on the molecule, we can calculate the oxidation state of cobalt.

Complete Step-By-Step Answer:
- In the given question we have to calculate the oxidation state of Co the given complex.
- So, firstly as we know that the given complex is a coordination compound in which the molecules attached are an 'en' group which is also called ethylenediamine, a chlorine molecule.
- Ethylenediamine is a type of bidentate ligand that is the molecule which consists of two donor atoms, but it is a neutral atom which means that the oxidation state of ethylenediamine is zero.
- So, let the oxidation state of cobalt is x and we know that the halogens have an oxidation state of -1 as they need only one electron to fill their outermost shell.
- So, the value of x will be:
$\text{x + 0 + (2 }\times \text{-1) = +1}$
$\text{x - 2 = +1}$
$\text{x = +3}$
- Moreover, the name of the complex is Dichlorobis(ethylenediamine)cobalt II ion.
Therefore, $+3$ is the correct answer.

Note: In the given problem, chlorine is an unidentifiable molecule which consists of only one donor atom and makes only a single bond with the central molecule. When a ligand has multiple sites which act as donor atoms then it forms the ring-like structure called chelate and the ligand is known as a chelating ligand which forms the stable structure.