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The correct increasing order of ionization enthalpy of $H e, L i^{+}$ and $B e^{+ 2}$ is

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$B e^{+ 2} < L i^{+} < H e$

$B e^{+ 2} < H e < L i^{+}$

$H e < L i^{+} < B e^{+ 2}$

$L i^{+} < B e^{+ 2} < H e$

answer is A.

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Detailed Solution

For ions with two electrons, more be the $\frac{Z_{(C h \arg e o f p r o t o n)}}{e_{(C h \arg e o f e l e c t r o n)}}$ ratio more is the ionization energy.

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