Question

Acidity of diprotic acids in aqueous solutions increase in the order:
(A) ${H_2}S < {H_2}Se < {H_2}Te$
(B) ${H_2}Se < {H_2}S < {H_2}Te$
(C) ${H_2}Te < {H_2}S < {H_2}Se$
(D) ${H_2}Se < {H_2}Te < {H_2}S$

Answer 1

Hint: The strength of the acid is dependent upon the stability of its conjugate bases. The increase in the atomic size the strength of the bond between hydrogen and chalcogen atom decreases

Complete answer:
We are given three diprotic acids of chalcogen elements. Chalcogen elements are the elements of the oxygen family. We know that S (Sulphur), Se (Selenium) and Te (Tellurium) are situated in the third, fourth and fifth period in the periodic table.
- We know that the strength of the acid is dependent upon the stability of its conjugate bases. So, as we go down in the periodic table, the size of the elements increases. So, we can say that the charge density decreases as we go down in the periodic table. So, S atom has maximum affinity for proton among the three and the Te atom will have the minimum affinity for proton.
- Alongside this, with the increase in the atomic size, the strength of the bond between hydrogen and chalcogen atoms decreases. This phenomenon facilitates the release of protons.
- Thus, from both of these above shown points, we can say that the acidity of the given diprotic acids can be expressed in the increasing order as below.
     \[{H_2}S < {H_2}Se < {H_2}Te\]

So, we can say that the correct answer of this question is (A).

Note:
Note that the acids are the compounds which can donate protons or accept electrons from other species. Diprotic acids are the acids which give two protons upon complete dissociation when they are dissolved in water.