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
1st I.E. of Be > B
<br><br>In case of Be, electron is removed from 2s
orbital which has more penetration power,
while in case of B electron is removed from 2p
orbital which has less penetration power.
<br><br>2p electron of B is more shielded from nucleus
by the inner electrons than 2s electrons of Be
<br><br>$\therefore$ It is easier to remove 2p electron than 2s
electron.
About this question
Subject: Chemistry · Chapter: Periodic Table and Periodicity · Topic: Periodic Trends
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