How do you find the oxidation numbers for compounds?

1 Answer
Oct 30, 2015

Answer:

By remembering some important rules.

Explanation:

(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. You don't see it being written as #H_2O^+# or #H_2O^-# : these molecules simply does not exist.

(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the in 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"#)

(4) All Group 2A and 3B elements have an oxidation state of +2 and +3, respectively. (e.g. #Ca^"2+"#, #Mg^"2+"#,#Al^"3+"#)

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

(6) 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#).

If you can remember and apply these rules, assigning an oxidation state for a given element would be easy for you.

Let's try an example. Find the oxidation state of #S# in the formula #H_2SO_4#.

Following the rules above and converting the atoms according to their oxidation states,

H : (+1)(2 atoms) = 2

This based on Rule 6 wherein the #H# atom is bonded to nonmetallic #S# atom; and based on the subscript, there are a total of two #H# atoms in the formula.

S : #color (red) y# or unknown
O : (-2) (4 atoms) = -8

Again, based on the Rule 5. The oxidation state is multiplied by the number of atoms based on the subscript.

Getting the sum, we have

(+2) + #color (red) y# + (-8) = 0 (based on Rule 1)

Thus,

#color (red) y# + (-6) = 0

#color (red) y# = +6

Therefore, the oxidation state of #S# is +6 or #S^"6+"#.