How would you explain the fact that the first ionization enthalpy of Lithium is lesser than that of Beryllium but its second ionization enthalpy is greater than that of Beryllium?

Answer

The electronic configurations of `Li` and `Be` are:
`Li: 1s^(2) 2s^(1) Be: 1s^(2) 1s^(2)`
Thus, the first electron in both the cases has to be removed from the `2s-`orbital. But the nuclear charge of `LI(+3)` is lower than that of `Be(+4)`. Therefore, the firsy ionization energy of Lithium is lower than that of Beryllium.
After the loss of first electron. the electronic configuration of `Li^(+)`is `1s^(2)` Here, the electron is to be removed from inert (helium) gas configuration, which is very stabel and hence removal of second electron from Lithium is very difficult. However, is caase of Beryllium, after the ioss of first electron, the electronic configuration of `Be^(+)` is `1s^(2) 2s^(1)`. Hence, the electron is to be removed from a `2s` orbital which is much easier than to remove an electron from inert gas configuration. Therefore, the second ionization enthalpy of Lithium is higher than that of Beryllium.