The energy of an electron in the $1 s$ orbital of $H^{-}, H e, L i^{+}$ and $B e^{2 +}$ follows the order.

H^{-} > H e > L i^{+} > B e^{2 +}

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L i^{+} > B e^{2 +} > H e > H^{-}

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B e^{2 +} > L i^{+} > H e > H^{-}

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H^{-} > L i^{+} > H e > B e^{2 +}

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Solution

The correct option is A $H^{-} > H e > L i^{+} > B e^{2 +}$
The energy of an electron in the $1 s$ orbital of $H^{-}, H e, L i^{+}$ and $B e^{2 +}$ follows the order $H^{-} > H e > L i^{+} > B e^{2 +}$ .
The energy of an electron $\propto Z^{2}$ and has negative sign. $Z$ is the atomic number.
As the atomic number increases, the energy of an electron becomes more negative. $H^{-}, H e, L i^{+}$ and $B e^{2 +}$ are isoelectronic species and each one has two electrons.

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