Animated Solution for Chemistry - s and p-Block Elements: The correct statement(s) regarding, (i) $\text{HClO}$, (ii) ${\text{HClO}}_2$, (iii) ${\text{HClO}}_3$ and (iv) ${\text{HClO}}_4$, is(are)

When we dive into the world of p-block elements, the oxoacids of halogens stand out as a fascinating study of structure, bonding, and acidity. In this problem, we are tasked with analyzing four specific oxoacids of chlorine: hypochlorous acid ($\text{HClO}$), chlorous acid (${\text{HClO}}_2$), chloric acid (${\text{HClO}}_3$), and perchloric acid (${\text{HClO}}_4$). Let's break down their molecular anatomy step by step.

To understand the properties of these molecules, we must first visualize them. Chlorine, being a Group 17 element, possesses $7$ valence electrons. In all its oxoacids, one oxygen atom is bonded to a hydrogen atom, forming a hydroxyl ($-\text{OH}$) group. This group attaches to the central chlorine atom via a single $\sigma$-bond.

Any additional oxygen atoms present in the molecule will form double bonds ($\text{Cl}=\text{O}$) with the central chlorine. This happens because chlorine, being in the third period, has empty d-orbitals and can expand its octet to minimize formal charges.

Let's map out the structures: - $\text{HClO}$: Chlorine forms one single bond with the $-\text{OH}$ group. It uses $1$ valence electron, leaving $6$ electrons, which pair up to form $3$ lone pairs. - ${\text{HClO}}_2$: Chlorine forms one single bond with $-\text{OH}$ and one double bond with an oxygen atom. It uses $3$ valence electrons, leaving $4$ electrons, which form $2$ lone pairs. - ${\text{HClO}}_3$: Chlorine forms one single bond with $-\text{OH}$ and two double bonds with oxygen atoms. It uses $5$ valence electrons, leaving $2$ electrons, which form $1$ lone pair. - ${\text{HClO}}_4$: Chlorine forms one single bond with $-\text{OH}$ and three double bonds with oxygen atoms. It uses all $7$ valence electrons, leaving $0$ lone pairs.

Now, let's put the given statements to the test.

Statement (A) claims that the number of $\text{Cl}=\text{O}$ bonds in ${\text{HClO}}_2$ and ${\text{HClO}}_3$ together is two. Looking at our structures, ${\text{HClO}}_2$ has $1$ double bond, and ${\text{HClO}}_3$ has $2$ double bonds. The total is $1+2=3$. Therefore, Statement (A) is incorrect.

Statement (B) states that the number of lone pairs on Cl in ${\text{HClO}}_2$ and ${\text{HClO}}_3$ together is three. As we calculated earlier, ${\text{HClO}}_2$ has $2$ lone pairs, and ${\text{HClO}}_3$ has $1$ lone pair. The sum is indeed $2+1=3$. Thus, Statement (B) is correct.

Statement (C) focuses on the hybridization of chlorine in ${\text{HClO}}_4$. To find the hybridization, we calculate the steric number, which is the sum of the number of $\sigma$-bonds and the number of lone pairs on the central atom. In ${\text{HClO}}_4$, chlorine forms $4$ $\sigma$-bonds (one with each oxygen) and has $0$ lone pairs.

A steric number of $4$ corresponds to ${\text{sp}}^3$ hybridization. Therefore, Statement (C) is correct.

Statement (D) asserts that $\text{HClO}$ is the strongest acid among the four. The acidic strength of oxoacids containing the same central atom increases with the oxidation state of that central atom. A higher oxidation state means the central atom is more electron-withdrawing, which polarizes the $\text{O}-\text{H}$ bond more strongly and stabilizes the resulting conjugate base via inductive effects.

Let's calculate the oxidation states of chlorine: - $\text{HClO}$: $+1$ - ${\text{HClO}}_2$: $+3$ - ${\text{HClO}}_3$: $+5$ - ${\text{HClO}}_4$: $+7$

Since chlorine has the highest oxidation state ($+7$) in ${\text{HClO}}_4$, perchloric acid is the strongest acid, not hypochlorous acid. Hence, Statement (D) is incorrect.

By systematically analyzing the molecular structures, electron domains, and oxidation states, we have successfully navigated through the options. The correct statements are (B) and (C).