Among the second period elements the actual ionisation enthalpies are in the order 
Li<B<Be<C<O<N<F<Ne.
Explain why
(i) Be has higher  than B
(ii) O has lower  than N and F?

i) In case of Be ( 1s² 2s²) the outermost electron is present in 2s- orbital  while in B(1s²2s²2p¹). IT is present in 29-orbital. Since 2s- electrons, due to greater penetration, are more strongly attracted by the nucleus than 2p-electrons, therefore, more amount of energy is required to knowck out a 2s-electron than a 2p-electron. Consequently,  of Be is Higher than that of  of B.

ii) The electronic configuration of N(1s² 2s² 2p¹_x 2p¹_y 2¹_z) in which 2p-orbital are exactly half filled is more stable than the electronic configuration of O( 1s² 2s² 2p²_x2p¹_y 2¹_z) in which the 12p-orbital are neither half filled nor completely filled. Therefore, it is  difficult to remove an electron from N than from O. As a result,  of O is less than that of N. Because of higher nuclear charge(+9) and smaller atomic size, the first ionization enthalpy of F is higher than that of O. Therefore, ionization enthalpy of O is less than that N as well as of F.