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
Select the correct option:
A
(I), (II) and (IV)
B
(I), (III) and (IV)
C
(I), (II) and (III)
D
(II), (III) and (IV)
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
1st I.E. of Be > B
In case of Be, electron is removed from 2s orbital which has more penetration power, while in case of $B$ electron is removed from $2 p$ orbital which has less penetration power.
2 p electron of B is more shielded from nucleus by the inner electrons than 2 s electrons of Be
∴ It is easier to remove $2 p$ electron than $2 s$ electron.
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