Question

Oxidation number of cobalt in $\left[Co\right(N{H}_3{)}_6]C{l}_{2}Br$ is -

• A. 6
• B. zero
• C. 3
• D. 2

Answer 1

The correct option is C 3

Let the O.N. of $Co$ be $x$
Where O.N. represnts the oxidation number.

here also,
O. N. of $N{H}_3$ is zero
O. N. of $Cl$ is - 1
O. N. of $Br$ is -1

$\therefore O.N.of\left(Co\right)+6\times O.N.of\left(N{H}_3\right)+2\times O.N.of\left(Cl\right)+O.N.of\left(Br\right)=0$

putting the values,
$\Rightarrow x+6\left(0\right)-1\times 2-1=0$
$\therefore x=+3$
So, the oxidation number of cobalt in the given complex compount is $+3$.