Why is the second ionization energy higher than the first ionization energy in each of the atoms in the alkali family?
In each of the atoms in the alkali family, the second ionization energy involves removing an electron from an orbital closer to the nucleus compared to the first electron removed. All alkaline metals have 1 s-electron on their outer shells (e.g. for Na it is 3s1 electron). This electron is easily removed when expending first ionization energy. The second electron to be removed is p-electron of lower electron shell (e.g. for Na it is any of six 2p electrons). Electrons in the closer orbitals experience greater forces of electrostatic attraction (according to Coulomb’s law, electrostatic force between charged particles is inversely proportional to the square of the distance between them) thus, their removal requires considerably more energy.