Question

Oxidation number of $'S'$ in $N{a_2}{S_4}{O_6}$ is:
A.$ + 0.5$
B.$2.5$
C.$ + 4$
D.$ + 6$

Answer 1

Hint: At first think about oxidation number. The oxidation number is also called an oxidation state. The oxidation number is defined as the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

Complete step by step answer:
The structure of $N{a_2}{S_4}{O_6}$ is $ONa - S{(O)_2} - S - S - S{(O)_2} - ONa$.
From the above structure, we can say that the oxidation numbers of the first sulphur atom are six, the second sulphur atom is two, third sulphur is two, the fourth sulphur atom is six. The sum of the oxidation numbers is zero for an electrically neutral compound and equals the overall charge for anionic species.
Lets take the oxidation number of sulphur as $x$, sodium is $ + 1$ and oxygen is $ - 2$, then
The equation to find the unknown oxidation number according to the structure is
$2( + 1) + 4x + 6( - 2) = 0$
$
  2 + 4x - 12 = 0 \\
  4x = 10 \\
  x = + 2.5 \\
 $
So the oxidation number of S in $N{a_2}{S_4}{O_6}$is $ + 2.5$. The answer is B.

Additional Information:-
Sulphur is abundant, multivalent and nonmetallic. Under normal conditions, sulphur atoms form cyclic octatomic molecules with a chemical formula ${S_8}$. Elemental sulphur is a bright yellow crystalline solid at room temperature. It is tasteless and odourless. It is a poor conductor of electricity. It reacts with all metals except gold and platinum forming sulphides. It also forms compounds with several nonmetallic elements. Sulphur occurs in the uncombined state as well as in the combination with the other elements in rocks and minerals that are widely distributed. It is classified among the minor constituents of the earth crust.

Note:
Element with outer electronic configuration as $3d^1$ $4s^2$ shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as $3d^2$ $4s^2$ shows variable oxidation states of +2, +3 and +4. e.g. transition metals.