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 followng:

(i) Be has higher ${\Delta }_i$ H than B.
(ii) O has lower ${\Delta }_i$ H than N and F.

Answer 1

(i) The electronic configuration of Be is $1s^{2}2s^2$.
The electronic configuration of B is $1s^{2}2s^{2}2p^1$.
Be has higher ${\Delta }_{i}H$ than B due to following reasons.
(a) The electronic configuration of Be has higher stability than the electronic configuration of B due to completely filled 2s orbital.
(b) During ionization of Be, s electron is removed.
During ionization of B, p electron is removed.
2s electron penetrates to the nucleus to greater extent than 2p electron. Thus, 2p electron is more shielded than 2s electron. The attraction of nucleus for 2s electron is higher than the attraction of nucleus for 2p electron. Thus, removal of 2s electron requires higher energy than the removal of 2p electron. Therefore, the ionization enthalpy of Be is higher than the ionization enthalpy of B.
(ii) The electronic configuration of oxygen is $1s^{2}2s^{2}2p^4$.
The 2p orbital contains 4 electrons out of which 2 are present in the same 2p-orbital. Due to this, the electron repulsion increases. N has stable half filled configuration. F has greater nuclear charge. Hence, the ionization enthalpy of O is lower than that of N and F.