Question

The coordination number and oxidation number of the central metal ion in $\left[ {Co{{(Br)}_2}{{(en)}_2}} \right]B{r_2}$
A) 6, 6
B) 2, 4
C) 6, 4
D) 4, 4

Answer 1

To answer this question, you should recall the concept of oxidation numbers. The oxidation state of an atom is defined as the number of electrons lost and, therefore, describes the extent of oxidation of the atom. For example, the oxidation state of carbon in${\text{C}}{{\text{O}}_{\text{2}}}$ would be $+ 4$ since the hypothetical charge held by the carbon atom if both of the carbon-oxygen double bonds were completely ionic would be equal $+ 4$.

Complete step by step solution:
The oxidation number of the central atom can be calculated by finding the charge associated with it when all the electron pairs that are donated by the ligands are removed from it. Consider the oxidation state of ${\text{Co}}$as $x$
$x + 2\left( { - 1} \right) + 0 = + 2$.
We will get the value of
$\Rightarrow x = 4$.
Hence, the oxidation state of central metal i.e. ${\text{Co}}$ is 4
The coordination number of the central atom is defined as the total number of sigma bonds through which the ligands are bound to the coordination center. In the given question the central metal ion has a total of 6 sigma bonds as $en$ is a bidentate ligand and ${\text{Br}}$ is a monodentate ligand.

Hence, the correct answer to this question is option C.

Note:
In most of the compounds, the oxidation number of oxygens is $- 2$. There are some exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $- 1$. Example, $N{a_2}{O_2}$
Superoxide- Every oxygen atom is allocated an oxidation number of $- \dfrac{{{\text{ }}1}}{2}$. Example, $K{O_2}$
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $+ 1$.