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 statement are :
Thoery Based.
Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. Beryllium (Be, atomic number 4) has a higher first ionization enthalpy than Boron (B, atomic number 5). Let's analyze the statements step by step to understand why.
Be:
B:
Be has a fully filled 2s orbital, which is stable. B has one electron in the 2p orbital.
(I) It is easier to remove 2p electron than 2s electron
This is correct. 2p orbitals are higher in energy and more diffuse than 2s orbitals. The 2s electron is closer to the nucleus on average and experiences a greater effective nuclear charge, making it harder to remove. Therefore, the 2p electron in B is easier to remove than a 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
This is correct. The 2p orbital has poorer penetration towards the nucleus compared to the 2s orbital. In B, the 2p electron is shielded by the 1s² core and the 2s² electrons. In Be, the 2s electrons are also shielded by the 1s² core, but the 2s orbital penetrates better, so they experience a higher effective nuclear charge and are less shielded.
(III) 2s electron has more penetration power than 2p electron
This is correct. Due to its shape and probability distribution, the 2s orbital has greater penetration towards the nucleus compared to the 2p orbital. This means a 2s electron spends more time closer to the nucleus and is harder to remove.
(IV) Atomic radius of B is more than Be
This is incorrect. Atomic radius generally decreases across a period from left to right due to increasing effective nuclear charge. Be has a larger atomic radius than B. The increased nuclear charge in B (Z=5) compared to Be (Z=4) pulls the electrons closer, making B smaller.
Statements (I), (II), and (III) are correct. Statement (IV) is false.
The correct option is: (I), (II) and (III)
Effective Nuclear Charge (Zeff):
Where Z is the atomic number and σ is the shielding constant. Higher Zeff means electrons are held more tightly.
Penetration Power Order: ns > np > nd > nf
Ionization Enthalpy Trend: Generally increases across a period, but exceptions occur due to stability of half-filled and fully filled orbitals.