Question

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 :

• A. (I), (III) and (IV)
• B. (II), (III) and (IV)
• C. (I), (II) and (IV)
• D. (I), (II) and (III)

Hint:

Check each statement: B's 2p electron is higher in energy, better shielded and less penetrating than Be's 2s; but B is smaller than Be because of higher effective nuclear charge.

Answer 1

The Correct Option is D

(I) True: the 2p electron is higher in energy than 2s, so it is removed more easily.
(II) True: the 2p electron of B is shielded by the \(1s^2\) core and also by the 2s electrons (and penetrates less), so it feels a smaller effective nuclear charge than Be's 2s electron.
(III) True: penetration order is \(ns \gt np\), so 2s penetrates more than 2p.
(IV) False: atomic radius decreases across a period, so B (Z = 5) is smaller than Be (Z = 4).
Correct statements: (I), (II) and (III).
Answer: (4) (I), (II) and (III)