The second ionization energy is the energy required to remove a second electron from a positively charged ion. In general, the second ionization energy is higher than the first ionization energy because it becomes more difficult to remove an electron from an already positively charged ion. The decreasing order of the second ionization energy for potassium (K), calcium (Ca), and barium (Ba) can be explained as follows:


Potassium has the atomic number 19, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1. In the first ionization, one electron is removed from the 4s orbital, resulting in the formation of K+ ion. The second ionization energy of potassium involves removing an electron from the 3p orbital.

The second ionization energy of potassium is higher than the first ionization energy, but it is relatively low compared to calcium and barium. This is because the 4s orbital is further from the nucleus compared to the 3p orbital, and the shielding effect of the inner electrons reduces the attractive force from the nucleus on the outermost electron. Therefore, it requires less energy to remove the second electron from potassium compared to calcium and barium.


Calcium has the atomic number 20, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2. In the first ionization, two electrons are removed from the 4s orbital, resulting in the formation of Ca2+ ion. The second ionization energy of calcium involves removing an electron from the 3p orbital, similar to potassium.

The second ionization energy of calcium is higher than that of potassium because the effective nuclear charge (the attractive force experienced by the outermost electron) is greater due to the removal of two electrons. Additionally, the 4s orbital is closer to the nucleus compared to the 3p orbital, increasing the attractive force on the outermost electron. Therefore, it requires more energy to remove the second electron from calcium compared to potassium.


Barium has the atomic number 56, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2. In the first ionization, two electrons are removed from the 6s orbital, resulting in the formation of Ba2+ ion. The second ionization energy of barium involves removing an electron from the 5p orbital.

The second ionization energy of barium is the highest among the three elements because the effective nuclear charge is the greatest due to the removal of two electrons. The 6s and 5p orbitals are further from the nucleus compared to the previous cases, but the significant increase in effective