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

1 Answer
Nov 4, 2015

Answer:

+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+"#