Question #ed336

1 Answer
Nov 3, 2015

Answer:

#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.