Question

How can we find the charge of a particular compound like ${\text{N}}{{\text{H}}_{\text{3}}}$…. as I want to take out the oxidation state of ${\left( {{\text{Ni}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{6}}}} \right)^{{\text{ + 2}}}}$?

Answer 1

Hint For finding the particular charge of ammonia (${\text{N}}{{\text{H}}_{\text{3}}}$), first we have to calculate the formal charge of whole molecule by adding the formal charge of each atom present in the ammonia.

Complete step by step solution:
As we know that structure of ammonia is shown as-

And formal charge of any compound will be calculated as follow:
Formal Charge (F.C. ) = No. of valence electrons – No. of non - bonded electrons – No. of bonds
First we calculate formal charge of each atom present in the ammonia molecule.
-F.C. on Nitrogen (${\text{N}}$) = $5 - 2 - 3 = 0$
-F.C. on each hydrogen atom (${\text{H}}$) = $1 - 0 - 1 = 0$
-From above two data it is clear that formal charge on ammonia molecules is zero and it is a neutral molecule.
Now we are known about the charge of particular compound like ammonia by which we calculate the oxidation state of given compound i.e. ${\left( {{\text{Ni}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{6}}}} \right)^{{\text{ + 2}}}}$. And oxidation state of metal Nickel $\left( {{\text{Ni}}} \right)$ is calculated as follow:
-Let us take the oxidation state of nickel is $x$, and charge of ammonia is zero, then we get
$x - 0 = + 2$
Or $x = + 2$

Hence oxidation state of metal Nickel in the compound ${\left( {{\text{Ni}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{6}}}} \right)^{{\text{ + 2}}}}$ is $+ 2$.

Note: In this question some of you may do wrong calculation if you are not familiar with the atomic number as well as with the electronic configuration of the atoms present in the ammonia molecule.