Question

The oxidation number of iodine in $I{F_5}$ is:

Answer 1

Hint: In order to find the oxidation number of iodine in $I{F_5}$, it is important to know what oxidation numbers are. The oxidation number of an atom is a number that represents the total number of electrons lost or gained by the atom.

Complete step by step solution:
There are some rules on which oxidation numbers can be assigned. Two rules which you should know to answer this question are:
-For monatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.
-In the compounds made of two elements, a halogen will always have an oxidation number of -1 assigned to them. When polyatomic ions are given, the sum of all the oxidation numbers of the atoms that constitute them equals the net charge of the polyatomic ion.
Therefore, following these two rules for the compound $I{F_5}$,
Let the oxidation number of iodine (I) in $I{F_5}$ is $x$. In the compound, $I{F_5}$, fluorine (F) belongs to the group 17 called halogens. Thus, according to the rules, fluorine will have oxidation number -1.
On the whole compound charge is zero. Therefore, on addition of oxidation numbers of iodine and fluorine, we get:
$x + 5( - 1) = 0$
$ \Rightarrow x = + 5$ .

Therefore, the oxidation number of iodine is +5 in the compound $I{F_5}$.

Note: Any element in the free or uncombined state (like Na, Mg) bears an oxidation number of zero. The oxidation number of oxygen in most compounds is -2. All alkali metals (group 1 elements) have an oxidation number of +1 in their compounds. All alkaline earth metals (group 2 elements) exhibit an oxidation number of +2 in their compounds.