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In the formation of $A l C l_{3}$ Aluminium atom loses

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$2 e^{-}$

$3 e^{-}$

$1 e^{-}$

$4 e^{-}$

answer is C.

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Detailed Solution

$\begin{array}{l} A l \to A l^{+ 3} + e^{-} \\ 3 C l + 3 e^{-} \to 3 C l^{-} \end{array}} \to A l^{3 +} + 3 C l^{-} \to A l C l_{3}$

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