Question

explain that the second ionization energy of Li us higher than that of F.​

Answer 1

Answer:

Lithium - atomic number - 3(electronic configuration of lithium- 1s^2 2s^1)

Boron - atomic number - 5(electronic configuration of boron- 1s^2 2s^2 2p^1)

Now,The ionization energy or ionisation enthalpy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation..

The 1st ionization energy corresponds to the energy required to remove the 1st loosely bound electron.

The 2nd ionization energy corresponds to the energy required to remove the 2nd loosely bound electron and so on…..

When electron is removed from lithium that is when electron is removed from 2s^1 lithium forms a 1s^2 configuration cation(which is extremely stable) on the other hand when electron is removed from boron that is when electron is removed from 2p^1 boron forms a 1s^2 2s^2 configuration cation(which is relatively less stable only in comparison to lithium’s 1s^2)

Now when it comes to second ionization enthalpy ,it becomes difficult to remove electron from 1s^2 of lithium as it is tightly held by the nucleus of lithium and also because of it’s small size in comparison to boron.

As a result second ionization energy of lithium is greater than boron.

Explanation: