$\mathrm{AlI}_{3}$, when reacts with $\mathrm{CCl}_{4}$, gives
$\mathrm{AlI}_{3}$, when reacts with $\mathrm{CCl}_{4}$, gives
- $\mathrm{AlCl}_{3}$
- $\mathrm{CI}_{4}$
- $\mathrm{Al}_{4} \mathrm{C}_{3}$
- both (a) and (b)
Solution
$\mathrm{AlI}_{3}$, on reaction with $\mathrm{CCl}_{4}$, gives the $\mathrm{AlCl}_{3}$ $4 \mathrm{AlI}_{3}+3 \mathrm{CCl}_{4} \longrightarrow 4 \mathrm{AlCl}_{3}+3 \mathrm{CI}_{4}$
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Asked in: JEE-TOPICTESTS-CHEMISTRY
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