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

We know that 

The electronic configuration of B and Be are:

$$\begin{gathered} \mathrm{B}\Rightarrow 1{ \mathrm{s}}^{2}2{ \mathrm{s}}^{2}2{\mathrm{p}}^1 \\ \mathrm{Be}\Rightarrow 1{ \mathrm{s}}^{2}2{ \mathrm{s}}^2 \end{gathered}$$

Now, We know that 2 s - orbital experienced more $Z_{\text{eff }}$ than 2 p therefore 2 p electron could be easily removed than $2 \mathrm{s}$.

Therefore, Statement 1 is correct.

Now,In case of B there are total 4 inner core electrons but in Be only $2{\mathrm{e}}^{-}$(in $1{ \mathrm{s}}^2$ ) therefore 2 p more shielded in B than $2 \mathrm{s}$ of Be.Hence, Statement (II) is also correct.

Since,s-orbital has spherical shape through that $Z_{\text{eff }}$ work 10 % but p-orbital is dumbbell shape therefore its $Z_{\text{eff }}$ less and penetrating power less than $2 \mathrm{s}$ that is why Statement (III) is also correct.

Now,Atomic radius of B less than Be due more $Z_{\text{eff }}$.

Hence, statement (IV) is incorrect.

Hence, option(a) is correct.