Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct statements(s) about the oxoacids, ${\text{HClO}}_4$ and $\text{HClO}$, is (are) -

Welcome to an exciting exploration of oxoacids! In this problem, we are tasked with evaluating four different statements regarding two well-known chlorine oxoacids: perchloric acid (${\text{HClO}}_4$) and hypochlorous acid ($\text{HClO}$). This question beautifully weaves together concepts of acidity, resonance, chemical reactions, and hybridization. Let's break it down step by step.

The first statement claims that ${\text{HClO}}_4$ is more acidic than $\text{HClO}$ due to the resonance stabilization of its anion. To verify this, we must look at what happens when these acids donate a proton (${\text{H}}^{+}$).

When ${\text{HClO}}_4$ loses a proton, it forms the perchlorate ion (${\text{ClO}}_4^{-}$). In this ion, the central chlorine atom is bonded to four oxygen atoms. The single negative charge is not localized on just one oxygen; instead, it is delocalized across all four oxygen atoms through resonance. This extensive delocalization makes the perchlorate ion incredibly stable.

On the other hand, when $\text{HClO}$ loses a proton, it forms the hypochlorite ion (${\text{ClO}}^{-}$). Here, the negative charge is localized entirely on the single oxygen atom, with no possibility for resonance stabilization. Because a more stable conjugate base corresponds to a stronger acid, ${\text{HClO}}_4$ is significantly more acidic than $\text{HClO}$. Thus, Statement (A) is correct.

The second statement suggests that ${\text{HClO}}_4$ is formed by the reaction of chlorine gas (${\text{Cl}}_2$) with water (${\text{H}}_2\text{O}$). Let's recall the chemistry of halogens.

When chlorine gas dissolves in water, it undergoes a classic disproportionation reaction:

In this reaction, chlorine (which starts at an oxidation state of $0$) is simultaneously oxidized to $+1$ in hypochlorous acid ($\text{HClO}$) and reduced to $-1$ in hydrochloric acid ($\text{HCl}$). It does not form perchloric acid (${\text{HClO}}_4$). Therefore, Statement (B) is incorrect.

Next, we examine the hybridization of the central atoms in both molecules. The steric number (number of sigma bonds plus the number of lone pairs) determines the hybridization.

In ${\text{HClO}}_4$, the central chlorine atom forms four sigma bonds with the surrounding oxygen atoms and has zero lone pairs. A steric number of $4$ corresponds to ${\text{sp}}^3$ hybridization.

In $\text{HClO}$, the connectivity is $\text{H}-\text{O}-\text{Cl}$. The central atom here is oxygen, which forms two sigma bonds (one with hydrogen, one with chlorine) and possesses two lone pairs. This also gives a steric number of $4$, meaning the oxygen is ${\text{sp}}^3$ hybridized. Even if one were to consider chlorine as the atom of interest in $\text{HClO}$, it has one sigma bond and three lone pairs, which again yields a steric number of $4$ and ${\text{sp}}^3$ hybridization. Consequently, Statement (C) is correct.

The final statement compares the basic strength of the conjugate base of ${\text{HClO}}_4$ to that of water.

Consider the dissociation of perchloric acid in water:

Because ${\text{HClO}}_4$ is one of the strongest known acids, this reaction proceeds almost entirely to the right. This complete dissociation implies that the water molecule (${\text{H}}_2\text{O}$) is much more effective at accepting and holding onto the proton than the perchlorate ion (${\text{ClO}}_4^{-}$) is. In acid-base terms, this means that water is a stronger base than the perchlorate ion. Therefore, Statement (D) is correct.

After a thorough analysis, we have determined that statements (A), (C), and (D) are correct. This problem serves as a fantastic reminder of how fundamental principles like resonance and steric numbers govern the macroscopic properties of molecules.