Question #ed336

Question

Question #ed336

1 Answer

Impact of this question

1981 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-03T00:53:03

$N$ $N$ = +5; $O$ $O$ = -2; and the whole ion $N O_{3}^{-}$ $NO_3^-$ 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 $H_{2} O$ $H_2O$ 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 $N O_{3}^{-1}$ $NO_3^"-1"$ ion is -1).

(3) All elements from Group 1A has an oxidation state of +1 (e.g. $N a^{+1}$ $Na^"+1"$ , $L i^{+1}$ $Li^"+1"$ ). All Group 2A and 3A elements have an oxidation state of +2 and +3, respectively. (e.g. $C a^{2+}$ $Ca^"2+"$ , $M g^{2+}$ $Mg^"2+"$ , $A l^{3+}$ $Al^"3+"$ )

(4) Oxygen always have a charge -2 except for peroxide ion ( $O_{2}^{2-}$ $O_2^"2-"$ ) 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 $H C l$ $HCl$ ) and always have a -1 charge if it is bonded with a metal (as in $A l H_{3}$ $AlH_3$ ).

Now let's solve your problem.

Based on rule 2, the oxidation state of the $N O_{3}^{-}$ $NO_3^-$ ion is -1.

$N O_{3}^{-}$ $NO_3^-$ = -1

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

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

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

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

Therefore your oxidation states are $N$ $N$ = +5; $O$ $O$ = -2 and the whole ion $N O_{3}^{-}$ $NO_3^-$ is -1.

Science

Math

Humanities

... and beyond

Question #ed336

1 Answer

Explanation:

Related questions

Impact of this question