Question

Among the second period elements the actual ionization enthalpies are in the order $Li < B < Be < C < O < N < F < Ne$.
With the help of information given above, explain the following:
(i) Be has higher ${\Delta _i}H$ than B.
(ii) O has lower ${\Delta _i}H$ than N and F.

Answer 1

Firstly, you should know the electronic configurations of B, Be, N and F. Element whose electronic configuration contains completely filled or half-filled orbitals, are highly stable and have high ionization enthalpy (${\Delta _i}H$). Also, attraction of nucleus for 2s electrons is always more than for 2p electrons.

Complete step by step answer:
The ionization enthalpy can be defined as the amount of energy required to remove an electron from an isolated gaseous atom in its ground state. If an element is more stable, then it has higher ionization enthalpy than the element which is less stable.

(i) The electronic configuration of Beryllium (Be) is: $1{s^2}2{s^2}$.
The electronic configuration of Boron (B) is: $1{s^2}2{s^2}2{p^1}$.
Be has higher ionization enthalpy (${\Delta _i}H$) than B due to the following reasons:
- The electronic configuration of Be has completely filled $2s$ orbital. Thus, Be has higher stability than B. When the element has higher stability, its ionization enthalpy is higher. Therefore, Be has higher ${\Delta _i}H$ than B.

- During ionization reaction of Be, electron in 2s orbital has to be removed while during ionization of B, electron in 2p orbital has to be removed. We know that the 2s electron penetrates to the nucleus to a greater extent than 2p electron. Therefore, 2p electrons are more shielded than 2s electrons. The attraction of the nucleus for the 2s electron will be more than that for the 2p electron. Hence, removal of 2s electrons requires a higher amount of energy than the removal of 2p electrons. Therefore, Be has higher ${\Delta _i}H$ than B.

(ii) The electronic configuration of O is: $1{s^2}2{s^2}2{p^4}$
Electronic configuration of N is: $1{s^2}2{s^2}2{p^3}$
Electronic configuration of F is: $1{s^2}2{s^2}2{p^5}$
The 2p orbital in oxygen (O) contains 4 electrons out of which 2 are present in the same 2p-orbital. Due to this, there is repulsion in the 2p electrons of O. Hence, valence electrons of O can be easily removed. N has stable half-filled p-orbital configuration. Therefore, its ${\Delta _i}H$ will be higher than O. F has greater nuclear charge due to which its ${\Delta _i}H$ will be higher than O. Hence, ionization enthalpy, ${\Delta _i}H$ of O is lower than that of N and F.
So, the correct answer is “Option C”.

Note: N, O and F belongs to the same period i.e., second period of the periodic table. When we move across a period, successive electrons are added to the same principal quantum level and the shielding of nuclear charge by the inner electrons is not much. Consequently, the outermost electrons are held more and more tightly and thus the ionization enthalpy increases across a period.