Question

The correct statement(s) about the oxoacids, $HCl{O_4}$and $HClO$, is(are):
This question has multiple correct options
A. The central atom in both $HCl{O_4}$ and $HClO$ is $s{p^3}$ hybridized
B. The conjugate base of $HCl{O_4}$ is weaker base than$\;{H_2}O$
C. $HCl{O_4}$ is formed in the reaction between$C{l_2}$ and $\;{H_2}O$
D. $HCl{O_4}$ is more stable than $HClO$ because of the resonance stabilization of its anion.

Answer 1

We know that oxoacids are the acids that have oxygen atoms in it. They are different from hydrochloric acid. Oxoacid may contain sulphur, carbon and nitrogen elements from p-block.

Complete answer:
Option (A) The central atom in both $HCl{O_4}$and $HClO$ is $s{p^3}$ hybridized. The central $Cl$ atom in $HCl{O_4}$ has $4$ bonding domains and zero lone pairs of electrons. The oxygen atom in $HClO$ has two bonding domains and two lone pairs of electrons.
Option (B) The conjugate base of $HCl{O_4}$ is weaker than water. This is because $HCl{O_4}$ is stronger acid than water. A stronger acid has weaker conjugate base and vice versa.
Option (D)$HCl{O_4}$ is more acidic than $HClO$ because of the resonance stabilization of its anion. Greater is the extent of delocalisation, greater is the stabilisation of negative charge and higher is the acidity. In $HCl{O_4}$, negative charge is delocalised over four oxygen atoms.
Hence, the above statements (A), (B) and (D) are correct.
Now, let’s talk about the statement (C).
(C) $HCl{O_4}$ is formed in the reaction between $C{l_2}$ and ${H_2}O$. This is not the correct statement. Instead, the following methods are used.
$NaCl{O_4} + HCl \to NaCl + HCl{O_4}$

**Hence, the answer is (A)(B) and (D).

Note: **
Here the halogen can be any halogen from the halogen group of p-block.

Here as shown in the figure the chlorine will have a different oxidation state. The change in oxidation state will be as per the bonding of the atoms with chlorine and hybridisation will be dependent on that.