Consider the following statements regarding the first ionisation enthalpy of Boron is less than Berrylium.
I. It is easier to remove 2p electron than 2s electron.
II. 2p electron of Boron 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
The correct statements are

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I, II and III

I, III and IV

I, II and IV

II, III and II

answer is A.

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Detailed Solution

$\begin{array}{l} 5^{B} : 1 S^{2} 2 S^{2} 2 P^{1} . \\ 4^{B e} : 1 S^{2} 2 S^{2} \end{array}$

First ionization enthalphy of B is lower than Be because Be has stable electronic configuration. It require more energy to remove the first electron from 2S(in Be) than 2P(in B) because 2S electron has more penetration power than 2P. Therefore I, II, III are correct. Atomic radius of B is less than Be. Therefore IVth is incorrect.

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