Question

Question #ed336

Answer 1

`N` = +5; `O` = -2; and the whole ion `NO_3^-` is -1.

Explanation:

Here are a few rules to remember when solving for oxidation state.

(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`).

Now let's solve your problem.

Based on rule 2, the oxidation state of the `NO_3^-` ion is -1.

`NO_3^-` = -1

based on rule 4, the oxidation state of `O` atom is -2.

`x` + [(3) (-2)] = -1 where `x` is the oxidation state of `N`.

Notice that since `O` atom has subscript, I needed to multiply the oxidation state by 3. Also, there are no rules for the oxidation state of `N` so we need to solve it long hand.

`x` + (-6) = -1
`x` = -1 + (+6)
`x` = +5

Therefore your oxidation states are `N` = +5; `O` = -2 and the whole ion `NO_3^-` is -1.