Question

What are the oxidation states of phosphorus in the following?
$N{a_3}P{O_4}$

Answer 1

Hint: $N{a_3}P{O_4}$ is called sodium phosphate. The compound contains sodium, phosphorus and oxygen. Phosphorus belongs to the nitrogen group and its atomic number is fifteen. The general oxidation number of oxygen and sodium are $ - 2,+ 1$ respectively.

Complete answer:
We are been asked to find the oxidation number of phosphorus in $N{a_3}P{O_4}$
Now to calculate oxidation number of an element in a compound first let’s study the elements
Phosphorus is a nonmetal of the seventeenth group. Its atomic=c number is fifteen. Oxidation number can also be defined as the oxidation state, that is the number of electrons an element can gain or lose to form a bond with another atom.
Oxidation number of a whole compound is always zero which means that a compound should and has to have a neutral oxidation state. So here also $N{a_3}P{O_4}$ is zero, which means the sum of the oxidation numbers of all the elements present in $N{a_3}P{O_4}$ is equal to zero.
We already know that oxidation number of oxygen and sodium are $ - 2, + 1$ respectively and the whole charge on the compound is zero therefore now let the oxidation number of phosphorous be $x$, then
$
  3 \times Na + P + 4 \times O = 0 \\
   \Rightarrow 3 \times 1 + x + 4 \times ( - 2) = 0 \\
   \Rightarrow 3 + x - 8 = 0 \\
   \Rightarrow x = + 5 \\
 $
As there are three sodium units for every unit of the compound and four units of oxygen for every unit of the compound.
Therefore, the oxidation number of phosphorus in $N{a_3}P{O_4}$ is $ + 5$.

Note:
Phosphorus exhibits a lot of varied oxidation states and that is why it is called a multivalent nonmetal. The oxidation number should never be confused with the valency. Valency is the number of electrons present in the outermost shell of a particular atom.