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 $\mathrm{Be}$
(III) $2 s$ electron has more penetration power than $2 p$ electron
(IV) atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$
(atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )
The correct statements are:
Correct Option: , 3
${ }_{5} \mathrm{~B}: 1 s^{2} 2 s^{2} 2 p^{1}$
${ }_{4} \mathrm{Be}: 1 s^{2} 2 s^{2}$
First ionisation enthalpy of $\mathrm{B}$ is lower than $\mathrm{Be}$ because Be has a stable electronic configuration. It required more energy to remove the first electron from $2 s$ (in Be) than $2 p$ (in B) because $2 s e^{-}$has more penetration power than $2 p$. Therefore options (I), (II) and (III) are correct. Atomic radius of B is less than Be.
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