Chemical Properties and Bonding Characteristics of Chlorine Oxoacids

To determine the correct statements regarding the given oxoacids of chlorine—(i) $\text{HClO}$, (ii) $\text{HClO}_2$, (iii) $\text{HClO}_3$, and (iv) $\text{HClO}_4$—let us analyze their chemical structures, bonding characteristics, hybridization, and acidity.

1. Structural Analysis of the Chlorine Oxoacids

Chlorine has 7 valence electrons. In each oxoacid, one hydroxyl group ($\text{-OH}$) is attached to the central chlorine atom via a single $\sigma$-bond, while additional oxygen atoms form $\text{Cl}=\text{O}$ double bonds ($\sigma + \pi$).

• (i) $\text{HClO}$ (Hypochlorous acid):

  • Structure: $\text{H}-\text{O}-\text{Cl}$
  • Number of $\text{Cl}=\text{O}$ double bonds = $0$
  • Valence electrons used by $\text{Cl}$ = $1$
  • Number of lone pairs on $\text{Cl}$ = $\frac{7 - 1}{2} = 3$ lone pairs

• (ii) $\text{HClO}_2$ (Chlorous acid):

  • Structure: $\text{H}-\text{O}-\text{Cl}=\text{O}$
  • Number of $\text{Cl}=\text{O}$ double bonds = $1$
  • Valence electrons used by $\text{Cl}$ = $1 \, (\text{for } \text{Cl}-\text{OH}) + 2 \, (\text{for } \text{Cl}=\text{O}) = 3$
  • Number of lone pairs on $\text{Cl}$ = $\frac{7 - 3}{2} = 2$ lone pairs

• (iii) $\text{HClO}_3$ (Chloric acid):

  • Structure: $\text{H}-\text{O}-\text{Cl}(=\text{O})_2$
  • Number of $\text{Cl}=\text{O}$ double bonds = $2$
  • Valence electrons used by $\text{Cl}$ = $1 \, (\text{for } \text{Cl}-\text{OH}) + 4 \, (\text{for } 2\times\text{Cl}=\text{O}) = 5$
  • Number of lone pairs on $\text{Cl}$ = $\frac{7 - 5}{2} = 1$ lone pair

• (iv) $\text{HClO}_4$ (Perchloric acid):

  • Structure: $\text{H}-\text{O}-\text{Cl}(=\text{O})_3$
  • Number of $\text{Cl}=\text{O}$ double bonds = $3$
  • Valence electrons used by $\text{Cl}$ = $1 \, (\text{for } \text{Cl}-\text{OH}) + 6 \, (\text{for } 3\times\text{Cl}=\text{O}) = 7$
  • Number of lone pairs on $\text{Cl}$ = $\frac{7 - 7}{2} = 0$ lone pairs

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2. Evaluation of Options

• Option A: The number of $\text{Cl}=\text{O}$ double bonds in (ii) and (iii) combined is: $\text{Total } \text{Cl}=\text{O} \text{ bonds} = 1 + 2 = 3$ Therefore, Statement (A) is incorrect.

• Option B: The total count of non-bonding electron pairs (lone pairs) on $\text{Cl}$ in (ii) and (iii) together is: $\text{Total lone pairs} = 2 + 1 = 3$ Therefore, Statement (B) is correct.

• Option C: In (iv) $\text{HClO}_4$, the central chlorine atom forms 4 $\sigma$-bonds (1 to $\text{-OH}$ and 3 to oxygen atoms) and has $0$ lone pairs. $\text{Steric Number} = (\text{Number of } \sigma\text{-bonds}) + (\text{Number of lone pairs}) = 4 + 0 = 4$ A steric number of 4 corresponds to $sp^3$ hybridization. Therefore, Statement (C) is correct.

• Option D: As the oxidation state of chlorine increases from $+1$ in $\text{HClO}$ to $+7$ in $\text{HClO}_4$, the conjugate base formed ($\text{ClO}_4^-$) is progressively stabilized by resonance across multiple terminal oxygen atoms. Thus, the acidic strength order is: $\text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4$ Therefore, $\text{HClO}_4$ (iv) is the strongest acid, making Statement (D) incorrect.

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Conclusion

The correct statements are (B) and (C).