Question

What is the effective atomic number of dichlorobis (ethylenediamine) cobalt(III)?

Answer 1

Hint: The effective atomic number of a compound is given by taking into account all the electrons present on the central metal atom and the valence electrons donated by the surrounding ligands. The electrons should be counted properly to find the exact Effective atomic number (EAN).

Complete answer:
Firstly, we will find the charge on the compound dichlorobis (ethylenediamine) cobalt(III) - ${\left[Co{\left(en\right)}_{2}C{l}_2\right]}^{+}$
The single oxidation state of the atom will be considered its common oxidation number. The addition of charges on the atoms will result in the overall charge on the complex.
In the given complex,
Cobalt has an oxidation number of +3. So, its charge will be +3.
The oxidation number of ethylenediamine is 0. So, its charge will be 0.
The oxidation number of Chloride is -1. But there are two chlorine atoms. So, its charge will be -2.
$Overall\text{ charge on }{\left[Co{\left(en\right)}_{2}C{l}_2\right]}^{+}\text{ = (+3) + (2}\times 0)\text{ + (2}\times -1)\text{ = +1}$
The atomic number of Cobalt is 27. The electronic configuration is $$[Ar]3d^{7}4s^2$$. However the Cobalt in ${\left[Co{\left(en\right)}_{2}C{l}_2\right]}^{+}$ is Cobalt (III). It has a charge of +3 on it which means it loses 3 electrons to become Co (III).
Thus the atomic number of Co(III) = 27 – 3 = 24.
Now, Ethylenediamine is a bidentate ligand. So each nitrogen atom the Ethylene diamine will donate two electrons each i.e. 4 by one (en) group. There are two Ethylene diamine groups. So, the total valence electrons by Ethylene diamine ligand will be equal to 8.
For chlorine atoms, each atom donates 2 electrons. There are 2 chlorine atoms in the compound. So, the total valence electrons by $C{l}^{-}$ is 4.
Thus, $$Effectiveatomicnumber=\text{ 24 + 8 + 4 = 36}$$.
Final answer: The effective atomic number of dichlorobis (ethylenediamine) cobalt(III) = 36.

Note:
Ethylenediamine is a neutral $\sigma$donor and the chlorine ion $$C{l}^{-}$$ is a $\pi$ donor. For any central metal remember to subtract the charge present on the atom from its atomic number. The overall charge does not affect the effective atomic number of the compound.