# B has a smaller first ionization enthalpy than Be.

Question:

B has a smaller first ionization enthalpy than Be. Consider the following statements:

(I) it is easier to remove $2 p$ electron than $2 s$ electron

(II) $2 p$ electron of B is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $\mathrm{Be}$

(III) $2 s$ electron has more penetration power than $2 p$ electron

(IV) atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$

(atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )

The correct statements are:

1. (I), (II) and (IV)

2. (II), (III) and (IV)

3. (I), (II) and (III)

4. (I), (III) and (IV)

Correct Option: , 3

Solution:

${ }_{5} \mathrm{~B}: 1 s^{2} 2 s^{2} 2 p^{1}$

${ }_{4} \mathrm{Be}: 1 s^{2} 2 s^{2}$

First ionisation enthalpy of $\mathrm{B}$ is lower than $\mathrm{Be}$ because Be has a stable electronic configuration. It required more energy to remove the first electron from $2 s$ (in Be) than $2 p$ (in B) because $2 s e^{-}$has more penetration power than $2 p$. Therefore options (I), (II) and (III) are correct. Atomic radius of B is less than Be.