Question

What is the oxidation number of nitrogen in the nitrate ion?

Answer 1

+5

Explanation:

Here are some of the rules for solving oxidation state of any element:

(1) The total charge of a stable compound is always equal to zero (meaning no charge).

For example, the `H_2O` molecule exists as a neutrally charged substance.

(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of `NO_3^"-1"` ion is -1).

(3) All elements from Group 1A has an oxidation state of +1 (e.g. `Na^"+1"`, `Li^"+1"`). All Group 2A and 3A elements have an oxidation state of +2 and +3, respectively. (e.g. `Ca^"2+"`, `Mg^"2+"`, `Al^"3+"`)

(4) Oxygen always have a charge -2 except for peroxide ion (`O_2^"2-"`) which has a charge of -1.

(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of `HCl`) and always have a -1 charge if it is bonded with a metal (as in `AlH_3`).

So for your problem, according to rule 2, the overall oxidation state of `NO_3^-` is -1.

`NO_3^-` = -1

According to rule 4, the `O` atom should have a -2 charge. But since the oxygen has a subscript in the formula, we need to multiply the charge by the subscript.

`color (red) x` + [(-2) (3)] = -1 where `color (red) x` is the oxidation state on `N` atom

solving for `color (red) x`,

`color (red) x` + (-6) = -1

`color (red) x` = -1 + (+6)

`color (red) x` = +5

Therefore, answer is `N^"5+"`