The decreasing order of the first ionization energy (in kJ `mol^(-1)`) of He, Mg and Na is `HegtMggtNa`. The increasing order of the `2^(nd)` ionization energy (in kJ `mol^(-1)`) of these elements will be

Options

A. `Na lt Mg lt He `
B. `Mg lt Na lt He`
C. `Mg lt He le Na`
D. `Na lt He lt Mg`

Answer

B

Explanation

`He : `1s^2 underset(-1e^(-)) overset(DeltaH_1) to 1s^1`
`Na : 1s^2 , 2s^2 2p^6 , 3s^1 underset(-1e^(-)) overset(DeltaH_1) to 1s^2 , 2s^2 2p^6`
Mg: `1s^2, 2s^2 2p^6 , 3s^2 underset(-1e^(-)) overset(DeltaH_1) to 1s^2, 2s^2 2p^6 3s^1`
He has highest value of II ionization energy due to smallest size. After removal of one electron from Na, the `Na^(+)` ion has inert gas configuration. To remove the electron from this configuration, a very high energy is required, hence a higher II ionization energy than Mg.
So, the order of II ionization energy is :
`Mg lt Na lt He`.