Question

Write the oxidation number of phosphorus in $ P{O_4}^{3 - } $ .

Answer 1

Hint :Here when the chemical formula of any compound is noted, we see that depending on the valence electrons’ count for every element, they take a positive or negative value; if atom gets more electrons to form this bond then it gets negative sign and if they leave some electrons then the number they take is positive. That number which represents the count of electrons (gain or loss) can be called oxidation number. Here remember that the total charge on an ion is the sum of each atom’s oxidation number.

Complete Step By Step Answer:
Let us understand that; if an atom gets more electrons to form this bond then it gets a negative sign and if they leave some electrons then the number they take is positive. This electron count in turn gives oxidation number. Then the total charge on an ion is the sum of each atom’s oxidation number.
Given compound is; $ P{O_4}^{3 - } $
We know that the oxidation state of oxygen here may be $ - 2 $ as commonly considered.
While the charge on the ions is $ - 3 $ .
Then we can also write the equation to get to the oxidation number of $ P $ in this compound;
 $ \Rightarrow 1 \times {P_{ON}} + 4 \times {O_{ON}} = - 3 $
Then we can easily substitute $ - 2 $ as $ {O_{ON}} $ , then ;
 $ \Rightarrow 1 \times {P_{ON}} + 4 \times - 2 = - 3 $
On solving we can see $ {P_{ON}} $ remains to be;
 $ \Rightarrow {P_{ON}} = + 5 $
 $ \therefore $ The oxidation state here in $ P{O_4}^{3 - } $ for $ P $ $ \Rightarrow {P_{ON}} = + 5 $ .

Note :
This should be taken care;
- Neutral atoms or molecules have no net charge. As a result, their total oxidation state will be zero.
- Keep in mind the oxidation state of some elemental atoms like sodium or iron will be zero. Similarly, a neutral molecule’s total oxidation state will be zero; like that of water, oxygen, methane, and ammonia.
- Atoms in homo-polar molecules have no oxidation state. For oxygen molecules, an atom's oxidation number shall be taken as zero.