Question: Difference in the oxidation number of sulphur atom in \[{{N}}{{{a}}_{{2}}}{{{S}}_{{4}}}{{{O}}_{{6}}}...

Question

Difference in the oxidation number of sulphur atom in ${{N}}{{{a}}_{{2}}}{{{S}}_{{4}}}{{{O}}_{{6}}}$ is x, that of ${{{H}}_{{2}}}{{{S}}_{{2}}}{{{O}}_{{5}}}$ is y. Find value of x+y is:

Answer 1

Solution

The oxidation number gives the valency of an atom in a compound. Most of the atoms form a compound by the transfer of electrons available in them and that changes their oxidation state which may be positive or negative.

Complete step by step answer:
We can simply find out the oxidation state of an atom if we know the valencies or oxidation states of other elements or elements present in it.
First, we are taking the case of ${{N}}{{{a}}_{{2}}}{{{S}}_{{4}}}{{{O}}_{{6}}}$ .
We know that sodium has the common oxidation state of alkali metals, that is, ${{ + 1}}$ and oxygen has the oxidation state as ${{ - 2}}$ and as the compound doesn’t show any charge, the total charge will be zero.
We can write it in the form of an equation;
${{2 \times ( + 1)}}\;{{ + 4x + 6 \times ( - 2) = 0}}$
On solving, we get
${{2 + 4x - 12 = 0}} \Leftrightarrow {{4x - 10 = 0}} \Leftrightarrow {{x = 2}}{{.5}}$
Next, we consider ${{{H}}_{{2}}}{{{S}}_{{2}}}{{{O}}_{{5}}}$ . As in the above case we have hydrogen having oxidation state as +1 and oxygen with oxidation state -2. We are writing it in an equation form;
${{2 \times ( + 1)}}\;{{ + 2y + 5 \times ( - 2) = 0}}$
On solving, we get
${{2 + 2y - 10 = 02y - 8 = 0}} \Leftrightarrow {{y = 4}}$
Now we want to find out the value of x+y and we can write it as;
${{x + y = 2}}{{.5 + 4 = 6}}{{.5}}$ .
Now we got the value as $6.5$

Note: Several elements like the d-block elements whose d- subshells are getting filled can have variable valency. Among them, the element manganese has the highest range of variable valencies. This enables them to form a wide range of compounds.
The oxidation state of an atom in a compound helps us to know whether it has gained electrons or lose it. Also, when we consider a full reaction, by finding out the oxidation state of an element, we can say whether it is oxidised or reduced in the reaction.