Question

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

• A. $H^{-}>He>L{i}^{+}>B{e}^{2+}$
• B. $L{i}^{+}>B{e}^{2+}>He>H^{-}$
• C. $B{e}^{2+}>L{i}^{+}>He>H^{-}$
• D. $H^{-}>L{i}^{+}>He>B{e}^{2+}$

Answer 1

The correct option is A $H^{-}>He>L{i}^{+}>B{e}^{2+}$

The energy of an electron in the $1s$ orbital of $H^{-},He,L{i}^{+}$ and $B{e}^{2+}$ follows the order $H^{-}>He>L{i}^{+}>B{e}^{2+}$.
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. $H^{-},He,L{i}^{+}$ and $B{e}^{2+}$ are isoelectronic species and each one has two electrons.