Question

Find the oxidation number of $\mathrm{N}$ in${\mathrm{NH}}_4$ ion.

Answer 1

Step-1 Calculating the overall charge on${\mathrm{NH}}_4$ion

• The ${{\mathrm{NH}}_4}^{+}$also known as ${{\mathrm{NH}}_4}^{+}$ ion has an overall charge on compound=$+1$

Step-2 Calculating the oxidation number of $\mathrm{N}$ in${\mathrm{NH}}_4$ ion.

1. For ${{\mathrm{NH}}_4}^{+}$, the oxidation number of Hydrogen= $+1$.

$\mathrm{Formula}\mathrm{to}\mathrm{calculate}\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{N}\mathrm{in}{{\mathrm{NH}}_4}^{+}$

$4\times \left(\mathrm{Oxidation}\mathrm{of}\mathrm{H}\right)+\left(\mathrm{Oxidation}\mathrm{of}\mathrm{N}\right)=\mathrm{Charge}\mathrm{on}\mathrm{the}{{\mathrm{NH}}_4}^{+}$

$+1=4\times \left(+1\right)+\left(\mathrm{Oxidation}\mathrm{of}\mathrm{N}\right)$

$+1-4=\left(\mathrm{Oxidation}\mathrm{of}\mathrm{N}\right)$

$\mathrm{Oxidation}\mathrm{of}\mathrm{N}=-3$

Hence, the oxidation number of $\mathrm{N}$ in${{\mathrm{NH}}_4}^{+}$ ion $=\left(-3\right)$