Question

Oxidation number of Sulphur in $\mathop {Na}\nolimits_2 \mathop S\nolimits_2 \mathop O\nolimits_3 $ would be
A. +4
B. +2
C. -2
D. 0

Answer 1

Hint: It is known to us that oxidation number is the number of electrons that an atom either gains or loses in order to form a chemical bond with another atom and oxidation number is also known as oxidation state.

Complete step by step solution:
There are some rules to assign an oxidation number to a given element or compound.
The rules are as follows:
Rule 1: The oxidation number of any free element is equal to 0. For example: the oxidation number of $Na$ is 0.
Rule 2: If any monatomic ion is given, then the oxidation number of that ion is its net charge.
Rule 3: In most compounds, the oxidation number of oxygen atom is -2 except in case of peroxides where it is -1.
Rule 4: All alkali metals i.e. group 1 elements exhibit oxidation number of +1 in their compounds.
Rule 5: Oxidation number of all alkaline earth metals i.e. group 2 elements are taken +2 in their compounds.
Rule 6: Oxidation number of all halogens are taken as -1 except some cases. But the oxidation number of fluorine is always taken as -1 in their compounds
Rule 7: If the given compound is neutral then, the sum of all the oxidation numbers of the constituent atoms equals to 0.
Coming to the question, now we can easily calculate the oxidation number of $S$ in $\mathop {Na}\nolimits_2 \mathop S\nolimits_2 \mathop O\nolimits_3$ by using the above stated rules:
Calculations:
In given compound,
Oxidation number of oxygen atom, $O = - 2$
Oxidation number of sodium atom, $Na = + 1$
Let the oxidation number of the Sulphur atom in the given compound be x.
So, by following Rule 7 we can now calculate the oxidation number of Sulphur:

$\,2 \times ( + 1) + 2 \times x + 3 \times ( - 2) = 0$
$2 + 2x - 6 = 0$
$2x = 4$
$x = + 2$

Therefore, we can now easily say that option B is the correct option.

Note: It should be noted that if polyatomic ions would have been given then in that case, the sum of all oxidation numbers of all atoms that constitute them equals the net charge of the polyatomic ion.