Question

The oxidation state of Iodine in $IP{O_4}$ is-
a.) +1
b.) +3
c.) +5
d.) +7

Answer 1

Hint: We will first determine the oxidation state of each element in the compound given by the question i.e. $IP{O_4}$. We will let the oxidation state of iodine be zero and then solve it further. As we know that the cumulative number of electrons that the atom either absorbs or lacks to establish a chemical bond with another atom is the oxidizing number, also known as the oxidizing state.

Complete step-by-step answer:
An oxidation number indicative of the potential of each atom to obtain, give or exchange electrons is allocated to each atom involved in an oxidation reduction reaction. For examples, the iron $F{e^{3 + }}$ has a rating of +3 as it can gain three electrons to create a chemical bond, whereas the ion ${O^{2 - }}$ has a rating of −2, since the ion can give two electrons. The aggregate of oxidation numbers for the electrically neutral product is zero; for instance, the oxidation number for the two iron atoms (+6 in total) in hematite ($F{e_2}{O_3}$) is equal to the oxidation number for the three oxygen atoms (-6).
In the given compound $IP{O_4}$, oxygen is carrying the oxidation number of -2.
Phosphorus then has the oxidation state of +5.
Now, let the oxidation state of iodine be x.
Now, the oxidation state will be-
$\Rightarrow x + 5 + \left( {4 \times - 2} \right) = 0 \\\ \\\ \Rightarrow x + 5 - 8 = 0 \\\ \\\ \Rightarrow x - 3 = 0 \\\ \\\ \Rightarrow x = + 3 \\\$
Hence, the oxidation state of iodine is +3.
Thus, option B is the correct option.

Note: Many elements have the same number of oxidations in different compounds; for example, in both of its compounds, fluorine has the amount of oxidization − 1. Others may presume a number of oxidation numbers , especially of non-metallic elements and transitional elements; for example, nitrogen can have a number of oxidations between −3 (as in ammonia, $N{H_3}$) and +5 (as in nitric acid, $HN{O_3}$).