Question

The set with the correct order of acidity is:
A. $HClO < HCl{O_2} < HCl{O_3} < HCl{O_4}$
B. $HCl{0_4} < HCl{0_3} < HCl{0_2} < HCl0$
C. $HClO < HCl{0_4} < HCl{0_3} < HCI{O_2}$
D. $HCl{O_4} < HCl{O_2} < HCl{O_3} < HClO$

Answer 1

Oxyacid is an acid that contains oxygen. Specifically, it is a compound that contains hydrogen, oxygen, and at least one other element. The acidity of the oxy acid increases with the increase in the oxidation number.it also depends upon the stability of the conjugate base. Higher the stability of the conjugate base higher will be the acidity of the oxyacid.

Complete step by step answer:
The strength of an oxyacid is defined by the extent to which it dissociates in water (i.e. its ability to form ${H^ + }$ions). In general, the relative strength of oxy acid can be predicted based on the electronegativity and oxidation number of the central metal atom.
Oxidation number or oxidation state is the total number of electrons that an atom either gains or losses to form a chemical bond with another atom.
Therefore, the Oxidation number of Cl HClO is $+ 1$ $HCl{O_2}$ is $+ 3$ , $HCl{O_3}$ is $+ 5$ , and $HCl{O_4}$ is $+ 7$ .
We know that the acidity of oxyacids of halogen increases with an increase in the oxidation number of halogens. This is because the corresponding conjugate base of the oxoacids of chlorine gets stabilization.
Therefore, $HCl{O_4}$ should be the most acidic among the given choices.
Then according to the oxidation number of the chlorine, the correct order of acidity is,
$HClO < HCl{O_2} < HCl{O_3} < HCl{O_4}$

Hence, the correct option is A.

Additional information:
The term $p{K_a}$ equals to $- \log {K_a}$ , where ${K_a}$ is the dissociation constant of acid. Now from this formula of $p{K_a}$ it is clear that with increasing the value of the dissociation constant the value of $p{K_a}$ decreases and vice-versa. And with increasing the value of the dissociation constant the acidity increases. Therefore, it can be said that with increases $p{K_a}$ value acidity decreases and vice-versa.

Note:
Now to be acidic hydrogen that hydrogen should be attached with a high electronegative group or electron-withdrawing group. The higher the electronegativity of the group, the higher will be the electron deficiency of the hydrogen, attached to that group as well as the acidity. Now in the case or carbon, the electronegativity varies with hybridization. With increasing the s character in the hybridization electronegativity of the carbon increases and vice-versa. Now the order of the electronegativity of the different hybridization of carbon is $s{p^3} < s{p^2} < sp$ . Therefore, the order of the acidity of the hydrogen attached to these hybridized carbons is ${C_{s{p^3}}} - H < {C_{s{p^2}}} - H < {C_{sp}} - H$ .