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Q.

${A^{-}}_{(g)} \to {A^{+}}_{(g)}, ΔH = + 600 KJ$ occurs in two steps. If I.P of A is 500KJ then electron gain enthalpy of “A” will be

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a

+100KJ

b

-100KJ

c

+1100KJ

d

-1100KJ

answer is C.

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Detailed Solution

$\begin{array}{l} A^{-} \to A : {Ip}_{1} = xKJ \\ A \to A^{+} : {Ip}_{2} = 500 KJ \end{array}$

$x + 500 = 600 : x = {Ip}_{1}$ of $A^{-}$ = +100KJ

So, EA of A =-100KJ

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