STD 12 - 4. d and f- block elements — JEE STD 11 Science — Question

$[A]$ $HClO_4$ is more acidic than $\mathrm{HClO}$ because of the resonance stabilization of its anion

$[B]$ $HClO_4$ is formed in the reaction between $Cl_2$ and $H_2 O$

$[C]$ The central atom in both $HClO_4$ and $HClO$ is $s p^3$ hybridized

$[D]$ The conjugate base of $HClO_4$ is weaker base than $H_2 O$

$\begin{array}{*{20}{c}}
  {\,\,\,\,\,O} \\ 
  {\,\,\,\,\,||} \\ 
  {HOC{H_2}C{H_2} - C - OC{H_2}C{H_3}} 
\end{array}$ $\xrightarrow{{PCC}}(A)\,\xrightarrow[{(1\,\,molar\,\,equivalent)}]{{{H_2}C \equiv CHMgBr}}(B)$ $\xrightarrow{{N{H_4}Cl\,/\,{H_2}O}}\,(C)\,\,\xrightarrow[{{H_2}O}]{{KOH}}\,$ $\xrightarrow{{{H_3}{O^ \oplus }}}\,\,\xrightarrow[{pyridine}]{{\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,O} \\ 
  {\,\,\,\,\,\,||} \\ 
  {{{(C{H_3} - C)}_2}O} 
\end{array}}}(D)$