Question

Oxidation number of sulphur in ${{S}_{8}}$ ​,${{S}_{2}}{{F}_{2}}$ and ${{H}_{2}}S$ are:
(A) +2, 0, +2
(B) 0, +1, −2
(C) −2, 0, +2
(D) 0, +1, +2

Answer 1

Oxidation state shows the gain or loss of total number of electrons of an atom to form a chemical bond with another atom. Oxidation state can also be phrased as oxidation number. The atom having higher electronegativity will gain electrons and the other one will lose electrons in order to form a bond.

Complete step by step answer:
- The oxidation number is the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element. So, the transfer of electrons is described by the oxidation state of the molecule. Now, we should know about some rules that will help us in determining oxidation number of ${{S}_{8}}$ , ${{S}_{2}}{{F}_{2}}$ and ${{H}_{2}}S$ . These rules are as follows:
- The oxidation number of an atom is zero in a neutral compound that contains atoms of only one element. So, from this we can say that the atoms in ${{O}_{2}},{{P}_{4}},{{O}_{3}}$ and aluminium metal all have an oxidation number of 0.
- To find the oxidation number of any atom in a compound, we can simply make an equation in which the overall charge on the compound is equal to the sum of oxidation states of all atoms in the compound. Thus, by this equation, we can find the unknown value of oxidation number of an atom if other values are known.
- The oxidation state of an atom depends upon which atom it is sharing the bond with. We can say that if an atom is more electronegative, then it will gain electrons and we can assign its oxidation number accordingly and vice versa.

So, let’s find the oxidation number of sulphur in given compounds.
i) ${{S}_{8}}$ :
- As we can see that ${{S}_{8}}$ contains only one type of atom that is of sulphur. So, as per rule, its oxidation state will be zero because the bond contains the same two atoms, hence there is no electronegativity difference. So, we can finalize that in this compound, the oxidation state of sulphur is zero.
ii) ${{S}_{2}}{{F}_{2}}$
Now, we will find the oxidation state of sulphur in ${{S}_{2}}{{F}_{2}}$. Note that the oxidation number of the halogen family is -1. So, this fluorine atom also has -1 oxidation state. The overall charge on this compound is zero. So, the calculation will be as follows:
Overall charge on ${{S}_{2}}{{F}_{2}}$ = 2(Oxidation number of S) + 2(Oxidation number of F)
0 = 2(Oxidation number of S) + 2(-1)
0 = 2(Oxidation number of S) – 2
So, 2 = 2(Oxidation number of S)
Oxidation number of S = $\frac{2}{2}$ = +1
So, from the above calculation, the oxidation state of sulphur in ${{S}_{2}}{{F}_{2}}$ is +1.
iii) ${{H}_{2}}S$
Now, we will calculate the oxidation state of sulphur in ${{H}_{2}}S$ . Here, we should note that the oxidation state of hydrogen is +1 because it is less electronegative. Overall charge on ${{H}_{2}}S$ is zero. So, we can use the formula of finding oxidation number here as:
Overall charge on ${{H}_{2}}S$ = 2(Oxidation number of H) + Oxidation number of S
0 = 2(+1) + Oxidation number of S
0 = 2 + Oxidation number of S
Oxidation number of S = -2
So, from above calculation oxidation state of sulphur in ${{H}_{2}}S~$ is -2.

So, from above discussion and calculation we now know our correct option.
So, the correct answer is “Option B”.

Note: We should understand this important concept about hydrogen oxidation number. The oxidation number of hydrogen is +1 when it is combined with a non-metal as in $C{{H}_{4}},\text{ }N{{H}_{3}},\text{ }{{H}_{2}}O$ , and HCl. And we should note that the oxidation number of hydrogen is -1 when it is combined with a metal as in LiH, NaH, $Ca{{H}_{2}}\text{ }and\text{ }LiAl{{H}_{4}}.$This happens because hydrogen is more electronegative than metals and less electronegative than nonmetals.