The reaction which is consistent with the fact that ${Cl}_{2} O_{6} (s)$ exists as $[{ClO}_{2}^{+}] [{ClO}_{4}^{-}]$ , would be:

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$\begin{array}{r} 2 {HClO}_{4} + P_{2} O_{5} ⟶ 2 {HPO}_{3} + {Cl}_{2} O_{7} \end{array}$

$\begin{aligned} {Cl}_{2} O_{6} + NaOH ⟶ {NaClO}_{3} + {NaClO}_{4} + H_{2} O \end{aligned}$

$\begin{array}{r} {Cl}_{2} O_{6} + HF ⟶ {ClO}_{2} F + {HClO}_{4} \end{array}$

$2 {ClO}_{2} + 2 O_{3} \overset{0^{\circ} C}{⟶} {Cl}_{2} O_{6} + 2 O_{2}$

answer is B.

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Detailed Solution

Here reaction is,

$\begin{array}{r} {Cl}_{2} O_{6} + HF ⟶ {ClO}_{2} F + {HClO}_{4} \end{array}$

The ions for the ${FClO}_{2}$ are,

$F - and Cl 2 +$

So, this ie reaction which is consistent with the fact that ${Cl}_{2} O_{6} (s)$ exists as $[{ClO}_{2}^{+}] [{ClO}_{4}^{-}]$ .

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