Question

Consider The Following Transformation:

Possible Transformations Are:

• A. 1 and 2
• B. 1 and 3
• C. 2 and 3
• D. 1,2, and 3

Answer 1

The correct option is D

1,2, and 3

The explanation for the correct answer:

Option (D): 1, 2, and 3

• The Fluoride of alkali metals$\left(\mathrm{A}\right)$ reacts with ${\mathrm{XeF}}_6$ to form ${\mathrm{A}}^{+}{\mathrm{XeF}}_7^{-}$
• Example: $\mathrm{NaF}+{\mathrm{XeF}}_6\to {\mathrm{Na}}^{+}{\mathrm{XeF}}_7^{-}$
• In solid-state, the compound ${\mathrm{PCl}}_5$ exist as ${\left[{\mathrm{PCl}}_4\right]}^{+}{\left[{\mathrm{PCl}}_6\right]}^{-}$.
• Because of the high charge in ${\mathrm{Al}}^{+3}$, it attracts the lone pair of $\mathrm{O}$ from ${\mathrm{H}}_2\mathrm{O}$ molecule, thus breaking ${\mathrm{H}}_2\mathrm{O}$ into ${\mathrm{H}}^{+}$ and ${\mathrm{OH}}^{-}$.
• The ${\mathrm{H}}^{+}$ ions in the reaction combines with ${\mathrm{H}}_2\mathrm{O}$ molecule to form ${\mathrm{H}}_3{\mathrm{O}}^{+}$.
• .

Thus, all the following transformations are possible. So, Option (D) is correct.