Question

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

• A. Ne > Cl > P > S > Al > Mg
• B. Ne > Cl > P > S > Mg > Al
• C. Ne > Cl > S > P > Mg > Al
• D. Ne > Cl > S > P > Al > Mg

Answer 1

Answer: (B)

The decreasing order of the ionization potential of the following elements is $Ne>C1>P>S>Mg>Al.$
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 $(Ne)$, half-filled $(P)$ and completely filled configuration $(Mg)$ are the cause of the higher value of I.E.
Since octet is complete in $Ne$, 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 subshell.
The outer electronic configurations of $Mg$ and $Al$ are $3s^2$ and $3s^{2}3p^1$ respectively.