Question

The first ionisation enthalpies of \[\ce{Na, Mg, Al}\] and \[\ce{Si}\] are in the order:

पर्याय

• \[\ce{Na < Mg > Al < Si}\]

• \[\ce{Na > Mg > Al > Si}\]

• \[\ce{Na < Mg < Al < Si}\]

• \[\ce{Na > Mg > Al < Si}\]

Answer 1

\[\ce{Na < Mg > Al < Si}\]

Explanation:

The electronic configurations of \[\ce{Na}\] and \[\ce{Mg}\] are:

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

The electronic configurations of Mg and Al are:

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

In \[\ce{Mg}\], the electron is to be removed from 3s-orbital while in \[\ce{Al}\], 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 \[\ce{Mg}\] is higher than \[\ce{Al (Mg > Al)}\].

The electronic configurations of \[\ce{Al}\] and \[\ce{Si}\] are:

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

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

Thus, ionisation enthalpy of \[\ce{Al}\] is less than \[\ce{Si}\].