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Question
The first ionisation enthalpies of \[\ce{Na, Mg, Al}\] and \[\ce{Si}\] are in the order:
Options
\[\ce{Na < Mg > Al < Si}\]
\[\ce{Na > Mg > Al > Si}\]
\[\ce{Na < Mg < Al < Si}\]
\[\ce{Na > Mg > Al < Si}\]
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Solution
\[\ce{Na < Mg > Al < Si}\]
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 \[\ce{Si}\].
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