Question

Find the oxidation number of S in\[{{S}_{2}}C{{l}_{2}}\].

Answer 1

Hint: For finding the oxidation number we have to find valence electrons of every atom in the molecule and if any charge is given on the molecules and then we have to balance the overall charge.

Step by step answer:
 We know that the oxidation number (also called an oxidation state) is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom. Each atom that participates in an oxidation-reduction is assigned an oxidation number that tells about its ability to acquire, donate, or share electrons.
\[\overset{-1}{\mathop Cl}\,-\overset{+1}{\mathop S}\,-\overset{+1}{\mathop S}\,-\overset{-1}{\mathop Cl}\,\]
Each chloride ion in \[{{S}_{2}}C{{l}_{2}}\] has a charge of -1 since chlorine is in Group 7 and its valency is one, it requires one electron to complete its octet. Let ‘x’ is the oxidation number of sulfur in \[{{S}_{2}}C{{l}_{2}}\] then by balancing the total charge on\[{{S}_{2}}C{{l}_{2}}\], it is a neutral atom it means the sum of the total charge is equal to zero:

2(x) + 2(-1) = 0
 x = 1.

So, the oxidation number or oxidation state is ‘+1’.

Note: We have to note that the sum of the oxidation numbers is zero since \[{{S}_{2}}C{{l}_{2}}\] is neutral. The oxidation state of S is not the same in every compound, it is different in different compounds and it depends on the atoms with which it is bonded.