B has a smaller first ionization enthalpy than Be. Consider the following statement:
(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 first write the ground-state electronic configurations. For beryllium we have $$\mathrm{Be}(Z = 4):\;1s^{2}\;2s^{2}.$$ For boron we have $$\mathrm{B}(Z = 5):\;1s^{2}\;2s^{2}\;2p^{1}.$$

The first-ionisation enthalpy is the energy required to remove the outer-most (valence) electron from a gaseous atom. According to the general periodic trend, ionisation enthalpy increases from left to right in a period because nuclear charge increases while the principal quantum number remains the same. However, boron shows an exception: its first-ionisation enthalpy is lower than that of beryllium. To understand why, we analyse the nature of the electrons that are removed.

Be loses a $$2s$$ electron ($$2s^{2}\rightarrow2s^{1}$$), whereas B loses a $$2p$$ electron ($$2p^{1}\rightarrow2p^{0}$$). Now we discuss the four given statements one by one in the light of shielding, penetration and atomic size.

Statement (I) says “it is easier to remove a $$2p$$ electron than a $$2s$$ electron.” A $$2p$$ electron is farther from the nucleus on average and is shielded more effectively by the inner $$1s^{2}$$ and $$2s^{2}$$ electrons than a $$2s$$ electron. Hence the attraction between the nucleus and a $$2p$$ electron is weaker, so less energy is needed to remove it. Therefore statement (I) is true.

Statement (II) says “the $$2p$$ electron of B is more shielded from the nucleus by the inner core of electrons than the $$2s$$ electrons of Be.” Inside boron the $$2p$$ electron experiences the repelling effect of the already present $$2s^{2}$$ electrons in the same shell as well as the $$1s^{2}$$ core. By contrast, each $$2s$$ electron in beryllium is only shielded by the $$1s^{2}$$ core. Thus the effective nuclear charge felt by the $$2p$$ electron in B is lower than that felt by a $$2s$$ electron in Be. So the shielding of the $$2p$$ electron in B is indeed greater. Hence statement (II) is true.

Statement (III) says “$$2s$$ electron has more penetration power than $$2p$$ electron.” Penetration refers to the probability of finding an electron close to the nucleus. For a given principal quantum number $$n=2$$, the order of penetration is $$2s > 2p.$$ Because an $$s$$ orbital possesses a finite probability density at the nucleus whereas a $$p$$ orbital has a node there, the $$2s$$ electron penetrates closer to the nucleus, feels a higher effective nuclear charge, and is held more tightly. Hence statement (III) is true.

Statement (IV) says “atomic radius of B is more than Be.” Moving from Be to B across the same period, the effective nuclear charge increases (because of the extra proton) and the principal quantum number remains $$n=2$$. Greater nuclear attraction pulls the electron cloud inward, so the atomic radius actually decreases. Therefore $$r_{\text{B}} < r_{\text{Be}}.$$ Statement (IV) is false.

Summarising, statements (I), (II) and (III) are correct, while statement (IV) is incorrect.

Hence, the correct answer is Option C.