Question

Find the oxidation number of nitrogen in $[Co(NH_3)_3Cl_3]$.

Answer 1

The oxidation number of an element is the number of electrons lost or gained by an atom of the element to form a chemical bond with another atom. To find the oxidation number of nitrogen in $[Co(NH_3)_3Cl_3]$, we shall first find the oxidation number of $NH_3$ in $[Co(NH_3)_3Cl_3]$.

Complete step by step solution:
The oxidation number of a compound is the sum of the oxidation numbers of all the species present in it.
Let the oxidation number of $NH_3$ be taken as $x$.
The oxidation number of $Co = +3$
The oxidation number of $Cl = -1$
Also, the oxidation number of each species is multiplied by the number of atoms of that species present in the compound.
$[Co(NH_3)_3Cl_3]$ is a coordination sphere with no charge on it. This means that the sum of the oxidation numbers of Co $NH_3$ and Cl in Co $NH_3$ and Cl is zero.
Therefore,
$+3 +3x + (-1)\times 3 = 0$
$\Rightarrow$ +3 +3x + (-3) = 0
$\Rightarrow$ +3 +3x -3 = 0
$\Rightarrow$ 3x = 0
$\Rightarrow x = \dfrac{0}{3}$
$\Rightarrow$ x = 0
Therefore, the oxidation number of $NH_3$ is $0$.
Now, from the oxidation number of $NH_3$, we shall find the oxidation number of $N$.
Again, the oxidation number of $NH_3$ is the sum of the oxidation numbers of nitrogen $(N)$ and hydrogen $(H)$ present in it.
Let the oxidation number of nitrogen in $NH_3$ be taken as $y$.
Oxidation number of $H = +1$
Therefore,
$y + (1\times 3) = 0$ [Since $3$ atoms of hydrogen are present]
$\Rightarrow$ y + 3 = 0
$\Rightarrow$ y = -3

Therefore, the oxidation number of nitrogen in $[Co(NH_3)_3Cl_3]$ is $-3$.

Note: If no charge is present on a coordination complex, it means the complex is neutral.
$NH_3$ is a ligand in the coordination compound $[Co(NH_3)_3Cl_3]$, surrounding the metal ion, $Co$. A ligand is either negatively charged or neutral.
If we remember that $NH_3$ is a neutral ligand, then finding the oxidation number of nitrogen would be much simpler. We can then avoid calculating the oxidation number of $NH_3$ first.