Question

Find the oxidation number of P in $M{g_2}{P_2}{O_7}$.

Answer 1

The oxidation number of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom.

Complete step by step answer:
Let X be the oxidation number of P in
$M{g_2}{P_2}{O_7}$
The Mg atoms have oxidation number of +2.
The O atoms have an oxidation number of -2.
2(+2)+2X+7(−2)=0
2X−10=0
2X=10
X=+5
The oxidation number of P in
$M{g_2}{P_2}{O_7}$ is +5.
So +5 is the answer.

Additional Information:
Oxidation state and oxidation number are quantities that commonly equal the same value for atoms in a molecule and are often used interchangeably. Most of the time, it doesn't matter if the term oxidation state or oxidation number is used. Oxidation state refers to the degree of oxidation of an atom in a molecule.
Oxidation states are typically represented by integers, which can be positive, negative, or zero. In some cases, the average oxidation state of an element is a fraction, such as 8/3 for iron in magnetite

Note:
Phosphorus belongs to group 15. It has 5 electrons in its valence shell. So it can either lose 5 e - to attain +5 oxidation state or gain 3 e- to attain −3 oxidation state. Hence its oxidation state varies from −3 in $P{H_3}$ to +5 in ${H_3}P{O_4}$