$B$ has a smaller first ionization enthalpy than Be. Consider the following statements
(I) It is easier to remove $2 p$ electron than $2 s$ electron.
(II) $2 p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of Be.
(Ill) $2 s$ electron has more penetration power than $2 p$ electron
(IV) Atomic radius of $B$ is more than Be. (Atomic number $B = 5 . B e = 4$ )
The correct statements are

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(I). (II) and (III)

(II). (Ill) and (IV)

(]). (Ill) and (IV)

(]). (II) and (IV)

answer is A.

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Detailed Solution

$\begin{aligned} B e \Rightarrow 1 s^{2} 2 s^{2} \\ B \Rightarrow 1 s^{2} 2 s^{2} 2 p^{1} \end{aligned}$

It is easier to remove $2 p$ electron than $2 s$ electron due lo less penetration effect of $2 p$ than $2 s$ . $2 p$ electron of Boron is more shielded from the nucleus by the inner core of electron than the $2 s$ electron of Be $B$ has a smalier size than $B e$ .