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(B) Ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.$A$. $[Ne] 3s^2 3p^1$ (Aluminum):

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$[Ne] 3s^2 3p^3$

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(B) Ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.$A$. $[Ne] 3s^2 3p^1$ (Aluminum): Has one electron in the $p$-orbital,relatively easy to remove.$B$. $[Ne] 3s^2 3p^3$ (Phosphorus): Has a half-filled $p$-orbital $(p^3)$,which is highly stable,making it difficult to remove an electron.$C$. $[Ne] 3s^2 3p^2$ (Silicon): Has a partially filled $p$-orbital,easier to remove than $p^3$.$D$. $[Ar] 3d^{10} 4s^2 4p^3$ (Arsenic): Also has a half-filled $p$-orbital $(4p^3)$,but since it is in the $4^{th}$ shell,the electron is further from the nucleus compared to the $3p$ electron in Phosphorus.Comparing $B$ and $D$,the $3p$ electrons are more strongly attracted to the nucleus than $4p$ electrons due to smaller atomic size. Therefore,the ionization energy of $[Ne] 3s^2 3p^3$ is higher than that of $[Ar] 3d^{10} 4s^2 4p^3$.

$I.P.$ \text{ Trend}	Reason
$A$. $N > O$	Half-filled configuration
$B$. $Zr < Hf$	Lanthanide contraction
$C$. $Na > K$	$Z_{eff}$
$D$. $Al < Ga$	Transition contraction

Amongst the following elements with their electronic configurations given below,which one has the highest ionization energy?

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$I.P.$ \text{ Trend} Reason

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