The hybridization of $$Xe{O_3}{F_2}$$   is $$s{p^3}d$$  and its structure is trigonal bipyramidal in which oxygen atoms are situated on the plane and the fluoride atoms are on the top and bottom.

Pure $$Cl{O_2}$$  is obtained by passing dry $$C{l_2}$$ over $$AgCl{O_3}$$  at $${90^ \circ }C.$$
\[2AgCl{{O}_{3}}+C{{l}_{2}}\left( \text{dry} \right)\xrightarrow{{{90}^{\circ }}C}\]       \[2AgCl+2Cl{{O}_{2}}+{{O}_{2}}\]

Hypophosphorous acid, $${H_3}P{O_2},$$  has the following structure.

As it contains only one replaceable $$H$$-atom ( that is attached with $$O,$$  not with $$P$$  directly ) so it is a monoprotic acid.
All other given statements are true.

No explanation is given for this question. Let's discuss the answer together.

$$Xe{O_2}{F_2}$$   has trigonal bipyramidal geometry, due to presence of lone pair of electrons on equitorial position, its shape is see-saw.

As the size increases, bond dissociation enthalpy becomes lower. Also, as the size of atoms get smaller, ion pairs on the two atoms get close enough together to experience repulsion. In case of $${F_2},$$  this repulsion is bigger and bond becomes weaker.
Hence, the correct order is $$C{l_2} > B{r_2} > {F_2} > {I_2}.$$

\[\begin{align} & 2Al+6HCl\xrightarrow{\Delta ,air}2AlC{{l}_{3}}+3{{H}_{2}} \\ & 2Al+3C{{l}_{2}}\to 2AlC{{l}_{3}} \\ \end{align}\]

\[AlC{{l}_{3}}.6{{H}_{2}}O\xrightarrow{\Delta }Al{{\left( OH \right)}_{3}}+3HCl+3{{H}_{2}}O\]
Thus \[AlC{{l}_{3}}\]  cannot be obtained by this method.

\[NO\left( g \right)+N{{O}_{2}}\left( g \right)\xrightarrow{-{{30}^{\circ }}C}\underset{\left( \text{blue}\,\,\text{liquid} \right)}{\mathop{{{N}_{2}}{{O}_{3}}}}\,\]

$$S{O_2} + 2{H_2}O \to {H_2}S{O_4} + 2\left[ H \right]$$

\[2Al+3C{{l}_{2}}\xrightarrow{\Delta }2AlC{{l}_{3}}\,\text{(anhydrous)}\]