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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:

1. (1) (i), (iii) and (iv)
2. (2) (i), (ii) and (iii)
3. (3) (i), (ii) and (iv)
4. (4) (ii), (iii) and (iv)

Correct answer: (4) (ii), (iii) and (iv)

Solution

The correct option is right because the 2p electron in boron is indeed more shielded by inner electrons compared to the 2s electron in beryllium, which makes it easier to remove. Additionally, 2s electrons have greater penetration power than 2p electrons, allowing them to be closer to the nucleus and thus more tightly bound. Lastly, the atomic radius of boron is larger than that of beryllium, contributing to the lower ionization enthalpy.

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