The decreasing order of the ionization potential of the following elements is:

N e > C l > P > S > A l > M g

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N e > C l > P > S > M g > A l

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N e > C l > S > P > M g > A l

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N e > C l > S > P > A l > M g

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Solution

The correct option is B $N e > C l > P > S > M g > A l$
The decreasing order of the ionization potential of the following elements is $N e > C l > P > S > M g > A l$ .
In a period, on moving from left to right, ionisation enthalpy increases. In a group, on moving from top to bottom, the ionisation energy decreases.
Closed shell $(N e)$ , half-filled ( $P$ ) and completely filled configuration ( $M g$ ) are the cause of the higher value of I.E.
Since octet is complete in $N e$ , a huge amount of energy is required to remove an electron as it will lead to disruption of stable electronic configuration. P has higher ionisation energy than S as P has a half-filled 3p subshell. Removal of an electron from P atom will break the stability of half-filled subshell.

Mg has higher ionisation energy than Al as Mg has completely filled 3s subshell. Removal of an electron from the Mg atom will break the stability of completely filled subshell. Also, less amount of energy is required to remove an electron from p-subshell than from s-subshell.

The outer electronic configurations of Mg and Al are $3 s^{2}$ and $3 s^{2} 3 p^{1}$ respectively.

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