The incorrect statement among the following is :

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the ${IE}_{1}$ of $Al$ is less than ${IE}_{1}$ of $Mg$

the ${IE}_{2}$ of $Mg$ is less than ${IE}_{2}$ of $Na$

the ${IE}_{1}$ of $Na$ is less than ${IE}_{1}$ of $Mg$

the ${IE}_{3}$ of $Mg$ is less than ${IE}_{3}$ of $Al$

answer is D.

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Detailed Solution

Aluminum has an outside electron in the p sub-level. Less energy is required to remove it because it has a greater energy than the outer electron in magnesium, which is in the s sub-level. Because the second electron in the noble gas core of $Na$ must be removed, whereas the second electron in the noble gas core of $Mg$ can be removed without a noble gas configuration, $IE (II)$ of $Na$ is higher than that of $Mg$ . $Al$ has three valence electrons and $Mg$ has fewer than three, thus instead of ionising another valence electron, they would have to remove an electron from their full octet or shift an already-ionized electron to a higher energy level. $Al$ thus has a lower third ionisation energy than magnesium.