Question

Give the oxidation state of sulfur in (a)${\mathbf{S}}_{\mathbf{8}}$ ; (b)${\mathbf{SF}}_{\mathbf{4}}$ ; (c)${\mathbf{SF}}_{\mathbf{6}}$ ; (d)${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ ; (e)${\mathbf{FeS}}_{\mathbf{2}}$; (f)${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ ; (g) ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{·}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ .

Short answer

(a) 0

(b) +4

(c) +6

(d) -2

(e) -1

(f) +6

(g) +2

Step-by-step solution

Oxidation state of an element in a compound

The extent of oxidation or reduction of an atom in a molecule is represented by its oxidation number.

Therefore, it expresses the number of electrons added or removed from it to form the compound.

The oxidation state of an atom depends on the number of valence electrons in it. Atoms remove or accept electrons to obtain a stable electronic configuration.

Finding the oxidation state of an atom

The first step to finding the oxidation state of an atom in a compound is to determine the charge of the species. If it is a neutral compound, the charge is zero; if it is an ion, the total charge will be a non-zero value.

The oxidation state of some atoms is always fixed. For example, the oxidation state of alkali metals is generally +1, and the oxidation state of halogens is -1. The oxidation state of all the other atoms in the compound other than the one to be calculated is determined.

Then the total oxidation state of all the other atoms are calculated by multiplying the oxidation state with number of atoms. This sum will be equal to the total charge of the compound.

The oxidation state of atoms in their elemental state will be zero. In molecules containing more than one atom of the element with unknown oxidation state, the calculated oxidation state will be its average value.

Finding the oxidation state of sulfur in given compounds

(a)

A single molecule of ${\mathrm{S}}_8$ contains eight sulfur atoms joined together by covalent bonds. These sulfur atoms do not lose or gain electrons to form ${\mathrm{S}}_8$ and, therefore, are in their elemental states. Therefore, the oxidation state of sulfur in ${\mathrm{S}}_8$ is zero.

(b)

Sulfur tetrafluoride ${\mathrm{SF}}_4$ is a neutral molecule and contains one sulfur atom and four fluoride ions. Fluorine is a halogen, and its oxidation state is -1.

Consider the oxidation state of sulfur in the compound to be x.

Then,

$\begin{array}{c}\left(1\times x\right)+\left[4\times \left(-1\right)\right]=0\\ x-4=0\end{array}$

Therefore, x can be calculated as:

x =+4

The oxidation state of sulfur in sulfur tetrafluoride is +4.

(c)

Sulfur hexafluoride ${\mathrm{SF}}_6$ is a neutral molecule and contains one sulfur atom and six fluoride ions. Fluorine is a halogen, and its oxidation state is -1.

Consider the oxidation state of sulfur in the compound to be x.

Then

$\begin{array}{c}\left(1\times x\right)+\left[6\times \left(-1\right)\right]=0\\ x-6=0\end{array}$

Therefore, x can be calculated as

x = +6

The oxidation state of sulfur in sulfur hexafluoride is +6.

(d)

Hydrogen sulfide ${\mathrm{H}}_2\mathrm{S}$ is a neutral molecule and contains one sulfur atom and two hydrogen ions. Hydrogen is an alkali metal, and its oxidation state is +1.

Consider the oxidation state of sulfur in the compound is x.

Then

$\begin{array}{c}\left[2\times \left(+1\right)\right]+\left[1\times x\right]=0\\ 2+x=0\end{array}$

Therefore, x can be calculated as

x = -2

The oxidation state of sulfur in hydrogen sulfide is -2.

(e)

Ferrous sulfide FeS is a neutral molecule containing two sulfur atoms and one ferrous ion. Iron is a transition metal, and its oxidation state is +2.

Consider the average oxidation state of sulfur in the compound is x.

Then

$\begin{array}{c}\left[1\times \left(+2\right)\right]+\left[2\times x\right]=0\\ 2+2x=0\end{array}$

Therefore, x can be calculated as

x = -1

The average oxidation state of sulfur in hydrogen sulfide is -2.

(f)

Sulfuric acid ${\mathrm{H}}_2{\mathrm{SO}}_4$is a neutral molecule that contains one sulfur atom, four oxygen atoms, and two hydrogen ions. The oxidation states of oxygen and hydrogen are -2 and +1, respectively.

Consider x to be the oxidation state of sulfur in the compound.

Then

$\begin{array}{c}\left[2\times \left(+1\right)\right]+\left(1\times x\right)+\left[4\times \left(-2\right)\right]=0\\ 2+x-8=0\end{array}$

Therefore, x can be calculated as

x = 8-2

= +6

The oxidation state of sulfur in sulfuric acid is +6.

(g)

Sodium thiosulfate ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ is a neutral molecule containing two sulfur atoms, three oxygen atoms, and two sodium ions. The oxidation states of oxygen and sodium ions are -2 and +1 respectively.

Consider x to be the average oxidation state of sulfur in the compound.

Then

$\begin{array}{c}\left[2\times \left(+1\right)\right]+\left(2\times x\right)+\left[3\times \left(-2\right)\right]=0\\ 2+2x-6=0\end{array}$

Therefore, x can be calculated as

$\begin{array}{c}x=\frac{6-2}{2}\\ =+2\end{array}$

The average oxidation state of sulfur in sodium thiosulfate is +2.