Electronic configurations of four elements $A$,$B$,$C$,$D$ are given below:
$A$) $1s^2 2s^2 2p^6 3s^1$
$B$) $1s^2 2s^2 2p^6 3s^2 3p^1$
$C$) $1s^2 2s^2 2p^6 3s^2$
$D$) $1s^2 2s^2 2p^6 3s^2 3p^2$
The correct order of first ionization enthalpy of these elements is:

  • A
    $D > B > C > A$
  • B
    $C > D > B > A$
  • C
    $C > A > B > D$
  • D
    $D > C > B > A$

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$(I)$ It is easier to remove a $2p$ electron than a $2s$ electron.
$(II)$ The $2p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of $Be$.
$(III)$ The $2s$ electron has more penetration power than the $2p$ electron.
$(IV)$ The atomic radius of $B$ is more than $Be$.
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