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 statement are :
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Boron (B) has a smaller first ionization energy than Beryllium (Be) due to its electron configuration. Be (4) is 1s²2s², so its 2s electron is stable. B (5) is 1s²2s²2p¹; its 2p electron is easier to remove.
Key reasons:
- (II) The 2p electron in B is more shielded from the nucleus by the inner 1s² core than the 2s electrons in Be.
- (III) A 2s electron has greater penetration power (closer to nucleus on average) than a 2p electron, making it harder to remove.
- (IV) Atomic radius of B is indeed greater than Be, reducing effective nuclear charge on the outer electron.
Statement (I) is incorrect; it is not inherently easier to remove a 2p than a 2s electron. The difference arises from the specific shielding and penetration in these atoms.
Final Answer: (II), (III) and (IV)