Question

Beryllium has higher ionisation enthalpy than boron. This can be explained as:
A. Beryllium has higher size than boron hence its ionisation enthalpy is higher
B. Penetration of $2p$ electrons to the nucleus is more than $2s$ electrons.
C. It is easier to remove an electron from $2p$ orbital as compared to $2s$ orbital due to more penetration of $s$ electrons.
D. Ionisation energy increases in a period.

Answer 1

The reason why ionisation enthalpy of beryllium is more than that of boron is that it is easier to remove an electron from the valence orbital of boron than from beryllium. We know that beryllium has $4{e^ - }$ and boron has $5{e^ - }$ . electrons in beryllium occupy the $s$ orbital only where as the electrons of boron occupy the $p$ orbital as well.

Complete step by step answer:
As mentioned in the hint, this answer can be solved using the information known about the number of valence electrons that are present in the orbitals of beryllium and boron.
The electronic configuration of beryllium is $1{s^2}2{s^2}$ . This configuration is due to the fact that there are $4{e^ - }$ in the outermost orbital of beryllium.
There are $5{e^ - }$ in the valence orbital of boron. Therefore, the electronic configuration will be:
$1{s^2}2{s^2}2{p^1}$ .
Now we shall understand the concept of penetration effect in the orbitals of an element. This effect can be defined as the measure of attraction of an electron to the nucleus based on the distance of the electron from the nucleus. This order of penetration or attraction can be represented as shown below:
$s > p > d > f$ . that is, the penetration effect of the $s$ orbital is more than the others. This implies that the electrons in this orbital are more attracted to the nucleus than the other orbitals. Therefore, it proves more difficult to remove an electron from these orbitals.
Therefore, in the case of beryllium the last electron which is usually removed will be removed from the $s$ orbital. Therefore, since we know that these electrons are strongly attracted to the nucleus it will be difficult to remove. But in the case of Boron it is easier to remove the electron from the valence orbital because we will be doing so from the $p$ orbital. Therefore, the ionisation enthalpy is more in beryllium.
Option a will be wrong as Beryllium does not have a larger radius than boron.
Option b is not applicable because penetration of $2p$ electrons to the nucleus is less than $2s$ electrons.
The fourth option is false as Ionisation energy decreases in a period due to the increase in the radius down the group.

So, the correct answer is Option C .

Note: Penetration effect can be used to determine the attraction of electrons in the orbitals by the nucleus. The order is: $s > p > d > f$ .
The closer the electron is to the nucleus more is the attraction between the electron and nucleus.
The more the penetration effect that is observed in an orbital more is the ionisation enthalpy of the electron.