Question

Find the oxidation number of P in $Na{H_2}P{O_4}$.

Answer 1

Hint: $Na{H_2}P{O_4}$ is called sodium dihydrogen phosphate or monosodium phosphate. This compound contains Sodium (Na), Hydrogen (H), Phosphorus (P) and Oxygen (O). Phosphorus belongs to the nitrogen group and its atomic number is 15. The general oxidation number of Sodium is +1, Oxygen is -2 and Hydrogen is +1. So using these atoms’ oxidation numbers, we are going to find the oxidation number of Phosphorus.

Complete answer:We are given to find the oxidation number of P in $Na{H_2}P{O_4}$.
Phosphorus is a non-metal and it belongs to the 17^th group. Its atomic number is 15.
Oxidation number is also known as oxidation state, is the total number of electrons that an atom either loses or gains to form a bond with another atom.
Oxidation number of a complete compound is always zero which means a compound must always be neutral. So here also, the oxidation number of $Na{H_2}P{O_4}$ is zero, which means the sum of all the oxidation numbers of the atoms involved in forming $Na{H_2}P{O_4}$ must sum up to zero.
We know that the oxidation number of Sodium is +1, Hydrogen is +1 and Oxygen is -2 and let the oxidation number of Phosphorus in $Na{H_2}P{O_4}$ be x.
Then $1\left( { + 1} \right) + 2\left( { + 1} \right) + x + 4\left( { - 2} \right) = 0$, as there are 2 hydrogen atoms and 4 oxygen atoms
$ \Rightarrow 1 + 2 + x - 8 = 0$
$ \Rightarrow x - 5 = 0$
$\therefore x = + 5$

Therefore, the oxidation number of Phosphorus in $Na{H_2}P{O_4}$ is +5.

Note:Phosphorus exhibits variable valency (+5, +3) that is why it is called a multivalent nonmetal. Do not confuse oxidation number with valency. Valency and oxidation numbers can be the same sometimes and sometimes they can be different. Valency is the no. of electrons present in the outermost shell of a particular element whereas oxidation number is the no. of electrons that an element has lost or gained in forming a particular compound.