Question

In which of the following arrangements, the given sequence is not strictly according to the property indicated against it?
(A) ${{H}_{2}}O$ < ${{H}_{2}}S$ < ${{H}_{2}}Se$ < ${{H}_{2}}Te$; increasing $p{{K}_{a}}$ values
(B) $N{{H}_{3}}$ < $P{{H}_{3}}$ < $As{{H}_{3}}$ < $Sb{{H}_{3}}$; increasing acidic character
(C) $C{{O}_{2}}$ < $Si{{O}_{2}}$ < $Sn{{O}_{2}}$ < $Pb{{O}_{2}}$; increasing oxidizing power
(D) $HF$ < $HCl$ < $HBr$ < $HI$; increasing acidic strength

Answer 1

Hint: Solve the option one by one by taking all the factors into consideration. Acidic character for hydrides of group 16 and 17 are determined by the extent of hydrogen binding. For group 15, the acidic character is determined by the electron charge density over the central atom and size of the atom as well. For oxides of group 14, the extent of overlapping will determine its oxidizing power.

Complete step by step solution:
The elements of group 16 form compounds with hydrogen in which 2 atoms of hydrogen are bonded to one atom of the element of group 16. It is observed that the acidic character increases as we move down the group. So, the ${{K}_{a}}$ value increases down the group and thus the $p{{K}_{a}}$ value decreases down the group in the same order. So, the increasing order of $p{{K}_{a}}$ values will be:
${{H}_{2}}Te$ < ${{H}_{2}}Se$ < ${{H}_{2}}S$ < ${{H}_{2}}O$

Therefore, we can conclude that the option in which the given sequence is not strictly according to the property indicated against it is option (A). Thus, it is the correct answer.

Note: It is important to understand the difference between ${{K}_{a}}$ and $p{{K}_{a}}$ as they can sound confusing at times. ${{K}_{a}}$ stands for the dissociation constant for the dissociation reaction of an acid. On the other hand, $p{{K}_{a}}$ stands for the negative log of ${{K}_{a}}$. It is shown below.
$p{{K}_{a}}=-\log ({{K}_{a}})$
This is the reason why $p{{K}_{a}}$ decreases down the group mentioned in the solution.