Analysis of the Statements

Statement A: The central atom in both $HCl{O_4}$ and $HClO$ is $s{p^3}$ hybridized.

1. For $HCl{O_4}$ (Perchloric acid):

   • The central atom is Chlorine (Cl). The structure is such that the hydrogen atom is bonded to an oxygen atom, which is then bonded to the chlorine atom ($H-O-ClO_3$).
   • The chlorine atom is bonded to four oxygen atoms. One is a single bond to the -OH group, and the other three are double bonds to the other oxygen atoms.
   • To find the hybridization, we calculate the steric number (SN) for the central Cl atom:
     • SN = (number of sigma bonds) + (number of lone pairs)
   • The Cl atom forms 4 sigma bonds (one with each oxygen atom) and has no lone pairs of electrons. (Valence electrons of Cl = 7; electrons used in bonds = 1 (in Cl-O) + 3×2 (in Cl=O) = 7; Lone pairs = 0).
   • SN = 4 + 0 = 4.
   • A steric number of 4 corresponds to $sp^3$ hybridization.

2. For $HClO$ (Hypochlorous acid):

   • The structure is $H-O-Cl$. The central atom in oxoacids is considered to be the non-metal, which is Chlorine (Cl).
   • The chlorine atom is bonded to one oxygen atom via a single bond.
   • To find the hybridization of Cl:
     • Number of sigma bonds = 1 (with O).
     • Number of lone pairs on Cl = (Total valence electrons - electrons in bonds) / 2 = (7 - 1) / 2 = 3.
   • SN = 1 (sigma bond) + 3 (lone pairs) = 4.
   • A steric number of 4 corresponds to $sp^3$ hybridization.

• Since the central atom (Cl) is $sp^3$ hybridized in both molecules, Statement A is correct.

Statement B: $HCl{O_4}$ is more acidic than $HClO$ because of the resonance stabilization of its anion.

1. Acidity Trend: The acidity of an acid is determined by the stability of its conjugate base. A more stable conjugate base corresponds to a stronger acid.

2. Conjugate base of $HCl{O_4}$: The conjugate base is the perchlorate ion, $ClO_4^-$. The negative charge is delocalized over all four oxygen atoms through resonance. This creates four equivalent resonance structures, spreading the negative charge evenly. This high degree of delocalization makes the $ClO_4^-$ ion extremely stable.

   $[O=Cl(=O)(=O)-O]^- \leftrightarrow [O-Cl(=O)(=O)=O]^- \leftrightarrow [O=Cl(=O)(O^-)=O] \leftrightarrow [O=Cl(O^-)(=O)=O]$

3. Conjugate base of $HClO$: The conjugate base is the hypochlorite ion, $ClO^-$. The negative charge is localized on the single oxygen atom, and there is no possibility of resonance stabilization.

4. Conclusion: Because the perchlorate ion ($ClO_4^-$) is much more stable than the hypochlorite ion ($ClO^-$) due to resonance, $HCl{O_4}$ is a much stronger acid than $HClO$. The reasoning provided in the statement is accurate.

• Therefore, Statement B is correct.

Statement C: $HCl{O_4}$ is formed in the reaction between $C{l_2}$ and ${H_2}O$.

1. The reaction of chlorine gas with water is a disproportionation reaction:

   $C{l_2}(g) + {H_2}O(l) \rightleftharpoons HCl(aq) + HClO(aq)$

2. This reaction produces hydrochloric acid ($HCl$) and hypochlorous acid ($HClO$), not perchloric acid ($HClO_4$). Perchloric acid is typically prepared by reacting a perchlorate salt (like $KClO_4$) with a strong, non-volatile acid like sulfuric acid.

• Therefore, Statement C is incorrect.

Statement D: The conjugate base of $HCl{O_4}$ is weaker base than ${H_2}O$.

1. Identify species: The conjugate base of $HClO_4$ is the perchlorate ion, $ClO_4^-$. We need to compare the basicity of $ClO_4^-$ and $H_2O$.

2. Brønsted-Lowry Acid-Base Theory: The strength of a conjugate base is inversely proportional to the strength of its corresponding acid. A stronger acid has a weaker conjugate base.

3. Consider the equilibrium: When perchloric acid dissolves in water, the following reaction occurs:

   $HClO_4 (Acid_1) + H_2O (Base_2) \rightleftharpoons H_3O^+ (Acid_2) + ClO_4^- (Base_1)$

4. Compare Acid Strengths: $HClO_4$ is a very strong acid, significantly stronger than the hydronium ion, $H_3O^+$. The equilibrium lies almost completely to the right.

5. Compare Base Strengths: In an acid-base equilibrium, the reaction favors the formation of the weaker acid and the weaker base. Since the equilibrium favors the products, $H_3O^+$ must be the weaker acid (compared to $HClO_4$), and $ClO_4^-$ must be the weaker base (compared to $H_2O$).

• Therefore, the conjugate base of $HCl{O_4}$ ($ClO_4^-$) is a weaker base than ${H_2}O$. Statement D is correct.

Final Conclusion

The correct statements are A, B, and D.