Question

Energy required for the ionisation reactions are shown:

$Na\left(g\right)\stackrel{+E_1}{\to }N{a}^{+}\left(g\right)$

$Mg\left(g\right)\stackrel{+E_2}{\to }M{g}^{+}\left(g\right)$ and

$N{a}^{+}\left(g\right)\stackrel{+E_3}{\to }N{a}^{2+}\left(g\right)$

Then the correct order of absorbed energy is :

• A. $E_1>E_2>E_3$
• B. $E_3>E_1>E_2$
• C. $E_2>E_3>E_1$
• D. $E_3>E_2>E_1$

Answer 1

The correct option is D $E_3>E_2>E_1$

The absorbed energy is directly proportional to the energy require to remove the outermost electron.

In the reaction, $N{a}^{+}\to N{a}^{2+}$ removing the electron from the $N{a}^{+}$ requires a high amount of energy because $N{a}^{+}$ has a stable configuration of fully filled p orbital and this energy is more than the energy required to remove an electron from $3s^2$ orbital in magnesium.

Among the given reaction the lowest energy is required to remove the last electron from sodium because removing an electron Na leads to $p^6$ configuration which is a stable configuration.

Therefore,

$E_3>E_2>E_1$

Thus, option D is correct.