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

Last updated date: 14th Jul 2024
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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.

We are been asked to find the oxidation number of phosphorus in $N{a_3}P{O_4}$
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
$3 \times Na + P + 4 \times O = 0 \\ \Rightarrow 3 \times 1 + x + 4 \times ( - 2) = 0 \\ \Rightarrow 3 + x - 8 = 0 \\ \Rightarrow x = + 5 \\$
Therefore, the oxidation number of phosphorus in $N{a_3}P{O_4}$ is $+ 5$.