Question

How do you determine the oxidation number of sulfur in each of the following substances?
a) Barium sulfate,$BaS{O_4}$
b) Sulfurous acid,${H_2}S{O_3}$
c) Strontium sulfide,$SrS$
d) Hydrogen sulfide,${H_2}S$

Answer 1

While solving these types of questions, we will assume the oxidation number of sulfur to be x. We will then proceed by writing the oxidation numbers of all the remaining atoms and putting the overall sum of the oxidation numbers to be equal to the overall charge of the molecule/compound. As we can clearly see that the overall charge in all the compounds mentioned in question is 0 as these all are neutral compounds.

Complete answer:
We will follow these steps in all the parts.
Let the oxidation number of S is x.
Write the oxidation number of all the other atoms.
Calculate the sum of the oxidation number of all the atoms.
Put this sum equal to the overall charge of the molecule.
a) $BaS{O_4}$
Let the oxidation number of S in$BaS{O_4}$ is x
Write the oxidation number of all the other atoms
$B{a^{ + 2\,}}\,\,\,{S^x}\,\,\,{O^{ - 2}}_4$
Calculate the sum of the oxidation number of all the atoms
$+ 2 + x + 4\left( { - 2} \right) = 2 + x - 8$
Put this sum equal to the overall charge of the molecule
$x - 6 = 0 \\\ x = 6 \\\$
Oxidation number of sulfur in$BaS{O_4}$is +6.
b) ${H_2}S{O_3}$
Let the oxidation number of S is x.
Write the oxidation number of all the other atoms.
${H^{ + 1}}_2\,\,\,{S^x}\,\,\,{O^{ - 2}}_3$
Calculate the sum of the oxidation number of all the atoms.
$2( + 1) + x + 3( - 2) \\\ 2 + x - 6 \\\ x - 4 \\\$
Put this sum equal to the overall charge of the molecule
$x - 4 = 0 \\\ x = 4 \\\$
c) $SrS$
Let the oxidation number of S is x.
Write the oxidation number of all the other atoms.
$S{r^{ + 2}}\,\,\,{S^x}$
Calculate the sum of the oxidation number of all the atoms.
$2 + x$
Put this sum equal to the overall charge of the molecule
$2 + x = 0 \\\ x = - 2 \\\$
d) ${H_2}S$
Let the oxidation number of S is x.
Write the oxidation number of all the other atoms
${H^{ + 1}}_2S$
Calculate the sum of the oxidation number of all the atoms
$2 + x$
Put this sum equal to the overall charge of the molecule
$2 + x = 0 \\\ x = - 2 \\\$

Note: We need to write the oxidation number of all the other atoms carefully. The oxidation number of an element in this elementary form is zero. In a single monatomic atom, the oxidation number of the atom is the same as the charge on the ion.