Question

For the following questions state the oxidation number of the element in the given compound.

1. sul fur in Li2SO#3
2. chlorine in MgCl2
3.Silicon in SiO2
4.sul fur in H2SO4
5.Sulfur in SO4^2-
6. manganese in MnO4^-
7. Cr in Cr2O7^2-

Answer 1

in the periodic tables

elements in the same column have the same charge

in a compound oxidation numbers have to equal 0

1.

sulfur in Li2SO3

charges :

Li = +1 S = +4 O = -2

multiply the charge with the subscripted number next to the element

Li = +1

Li2 = +2

O = -2

O3 = -8

in a compound oxidation numbers have to equal 0

+2 S -6 = 0

S -4 = 0

S = +4

Sulfur = +4

2.

MgCl2

Mg = +2

+2 Cl2 = 0

Cl2 = -2

Cl by itself = -2 divided by 2 = -1

Chlorine = -1

3.

SiO2

Silicon = +4

4.

Sulfur in H2SO4

H: +1 S: +6 O: -2

H2SO4

H2 : +2 O4: -8

+2 +S -8 = 0

S - 6 = 0

S = +6

Sulfur = +6

5.

Sulfur in SO4^2-

Sulfur in SO4

Sulfur = +4

6.

Manganese in MnO4^-

Manganese in MnO4

Manganese = +4

7.

Cr2O7^2-

Dichromate

Cr in Cr2O7^2- or Cr2O7-2-

Cr: +6 O: -2