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In the reaction $A^{- n_{2}} + x e^{-} ⟶ A^{- n_{1}}$ . Here, X will be :

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$n_{2} - n_{1}$

$n_{1} - n_{2}$

$n_{1} + n_{2}$

$n_{1} \times n_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} A^{- n_{2}} + {xe}^{-} \to A^{- n_{1}} \\ n_{2} + x = n_{1} \\ \Rightarrow x = n_{1} - n_{2} \end{array}$

So, In the reaction $A^{- n_{2}} + x e^{-} \to A^{- n_{1}}$ Here. X will be : $n_{1} - n_{2}$

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