Question

In which of the following acid peroxy linkage is present?
(A) ${{H}_{2}}S{{O}_{3}}$
(B) ${{H}_{2}}S{{O}_{4}}$
(C) ${{H}_{2}}C{{O}_{4}}$
(D) ${{H}_{2}}{{S}_{2}}{{O}_{8}}$

Answer 1

As we know that a peroxy bond is a bond between two oxygen atoms (generally of the type A-O-O-H or A-O-O-A). This peroxy bond between oxygen atoms is very weak and can homolyze easily to form radicals which are very reactive in nature. So here we have to tell that the acid in which peroxy linkage is present.

Complete answer:
Let us first begin with the concept of peroxy linkage and then try it on every acid compound given in the question as follows:-
-Peroxy linkage: It is a bond between two oxygen atoms (generally of the type A-O-O-H or A-O-O-A) and is very weak in nature. To know whether peroxy linkage is present or not, compare the oxidation state of the central atom in the compound to its maximum oxidation state.
-If the oxidation state of the central atom calculated by X-Method is greater than maximum oxidation state of the same atom, then that acid contains peroxy linkage.
-Now let us check the presence of peroxy linkage in each compound as follows:-
(A) ${{H}_{2}}S{{O}_{3}}$ :-
Maximum oxidation state of S: +6
Calculated oxidation state of S: 2(+1) + x + 3(-2) = 0
x = +4
As maximum oxidation state of ‘S’ is greater than its calculated oxidation state, so ${{H}_{2}}S{{O}_{3}}$do not have any peroxy linkage.
(B) ${{H}_{2}}S{{O}_{4}}$ :-
Maximum oxidation state of S: +6
Calculated oxidation state of S: 2(+1) + x + 4(-2) = 0
x = +6
As maximum oxidation state of ‘S’ is equal to its calculated oxidation state, so ${{H}_{2}}S{{O}_{4}}$do not have any peroxy linkage.
(C) ${{H}_{2}}C{{O}_{4}}$ :-
Maximum oxidation state of C: +4
Calculated oxidation state of C: 2(+1) + x + 4(-2) = 0
x = +6
Since the maximum oxidation state of ‘C’ is less than its calculated oxidation state, therefore ${{H}_{2}}C{{O}_{4}}$ have peroxy linkage.
(D) ${{H}_{2}}{{S}_{2}}{{O}_{8}}$ :-
Maximum oxidation state of S: +6
Calculated oxidation state of S: 2(+1) + 2x + 8(-2) = 0
x = +7
Since the maximum oxidation state of ‘S’ is less than its calculated oxidation state, therefore ${{H}_{2}}{{S}_{2}}{{O}_{8}}$ also have peroxy linkage.
-Conclusion from the above data is that peroxy linkage is present in (C) ${{H}_{2}}C{{O}_{4}}$and (D) ${{H}_{2}}{{S}_{2}}{{O}_{8}}$.

Note:
-The number of peroxy linkages in a compound can be calculated as follows:-
(A) If number of central atom =1, then number of peroxy linkage = $\dfrac{\text{Calculated oxidation state by X-Method - Maximum oxidation state}}{2}$
(B) If number of central atom =2, then number of peroxy linkage = $\dfrac{\text{Calculated oxidation state by X-Method - Maximum oxidation state}}{1}$