Question

The first ionisation enthalpies of $$\mathrm{Na},\mathrm{Mg},\mathrm{Al}$$ and $$\mathrm{Si}$$ are in the order:

Options

• $$\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}$$

• $$\mathrm{Na}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}$$

• $$\mathrm{Na}<\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}$$

• $$\mathrm{Na}>\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}$$

Answer 1

$$\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}$$

Explanation:

The electronic configurations of $$\mathrm{Na}$$ and $$\mathrm{Mg}$$ are:

$$\mathrm{Na}$$ (11): $$[\mathrm{Ne}]$$ 3s¹ and $$\mathrm{Mg}$$ (12): $$[\mathrm{Ne}]$$ 3s² In both the atoms, the electron is to be removed from 3s-orbital but nuclear charge in $$\mathrm{Na}$$ is less than $$\mathrm{Mg}$$. Thus, ionisation energy of $$\mathrm{Na}$$ is less than $$\mathrm{Mg}(\mathrm{Na}<\mathrm{Mg})$$.

The electronic configurations of Mg and Al are:

$$\mathrm{Mg}:[\mathrm{Ne}]$$ 3s²;  $$\mathrm{Al}:[\mathrm{Ne}]$$ 3s² 3p¹

In $$\mathrm{Mg}$$, the electron is to be removed from 3s-orbital while in $$\mathrm{Al}$$, it is to be removed from 3p-orbital. Since it is easier to remove an electron from 3p-orbital in comparison to 3s-orbital, the ionization enthalpy of $$\mathrm{Mg}$$ is higher than $$\mathrm{Al}(\mathrm{Mg}>\mathrm{Al})$$.

The electronic configurations of $$\mathrm{Al}$$ and $$\mathrm{Si}$$ are:

$$\mathrm{Al}$$ (13): $$[\mathrm{Ne}]$$ 3s² 3p¹ $$\mathrm{Si}$$ (14): $$[\mathrm{Ne}]$$ 3s² 3p²

In both the atoms, the electron is to be removed from 3p-orbital but nuclear charge in $$\mathrm{Si}$$ is more than $$\mathrm{Al}$$.

Thus, ionisation enthalpy of $$\mathrm{Al}$$ is less than $$\mathrm{Si}$$.