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 :

Solution

Chemistry ma’am · 3 steps

Boron's single 2p electron is easier to pull off than Be's 2s, and the size statement is the false one.

1. The electrons removed. Be is 1s² 2s², so its first electron comes from 2s. B is 1s² 2s² 2p¹, so its first electron comes from 2p.
2. Penetration and shielding. A 2s electron spends more time close to the nucleus than a 2p electron, so it penetrates better. The 2p electron is therefore more shielded by the inner electrons and feels less pull, which makes it easier to remove.

   Statement	True or false	Why
   (I) 2p is easier to remove than 2s	True	2p feels a weaker effective nuclear charge
   (II) 2p of B is more shielded than 2s of Be	True	2p penetrates the 1s² core less
   (III) 2s penetrates more than 2p	True	Penetration order in a shell is s > p > d > f
   (IV) B is larger than Be	False	Radius falls across a period: B is about 88 pm, Be about 111 pm

3. So. Statements (I), (II) and (III) are correct, and (IV) is wrong. The answer is (d).

The idea to keep. When IE dips from Be to B or from N to O, explain it with the electron being removed, not with atomic size. Size shrinks across a period, so it cannot explain a dip.

Answer from the NTA final answer key. Solution by PrepWiser.