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# The first ionization potential of Na, Mg, Al and Si are in the order:A. $Na > Mg > Al > Si$B.$Na > Mg > Al < Si$C. $Na < Al < Mg < Si$D. $Na < Mg < Al < Si$ Verified
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Hint: The first ionisation potential is an atomic property that represents the outermost electronic state. It is an estimate of the energy needed to remove one valence electron from a neutral atom.

$N{a_{11}} = 1{s^2},2{s^2},2{p^6},3{s^1}$
$M{g_{12}} = 1{s^2},2{s^2},2{p^6},3{s^2}$
$A{l_{13}} = 1{s^2},2{s^2},2{p^6},3{s^2},3{p^1}$
$S{i_{14}} = 1{s^2},2{s^2},2{p^6},3{s^2},3{p^1}$
That is, in the case of Al$\left( {3{s^2}3{p^1}} \right)$, the electron must be removed from the outer 3 p orbital, while in the case of Mg$\left( {3{s^2}} \right)$, the electron must be removed from the stable fully filled orbital. As a result, the ionisation energy of Na is lower than that of Mg (Na < Mg).But, the electron must be separated from the 3p-orbital of both Al and Si, but Si has a higher nuclear charge than Al. As a result, Al's ionisation enthalpy is lower than Si's. A p electron requires less energy to remove than a filled s electron from the same principal quantum shell.
Thus, correct order of first ionisation potential is option C ,$Na < Al < Mg < Si$