Arrange the elements in order of decreasing ionization energy. In no particular order, the elements are P, Si, S, Ca, and Ga.?

Firstly, consider the first three elements which belong to the Period 3 of the Periodic Table. Across the period (i.e. in order Si, P and S), there is a general trend of increasing ionization energy. This is because the increase in nuclear charge and the decrease in atomic size make the attractive forces the nucleus and the electrons in the outermost shell increase. However, P ([Ne] 3s² 3p³) has a higher ionization energy than S ([Ne] 3s² 3p⁴), because 3p subshell of P atom is half-filled and exhibits extra stability and thus more energy is needed to remove one electron from it. Besides, less energy is needed to remove one electron from S atom.

Hence, ionization energy: P > S > Si

Secondly, consider the last two element which belong to the Period 4 of the Periodic Table. Ca ([Ar] 4s²) has a higher ionization energy than Ga [Ar] 4s² 4p¹), because less energy is needed to remove an electron from the 4p subshell of B than from the 4s subshell of Be atom, as 4p subshell is in a higher energy level than 4s subshell. Besides, 4s subshell of Be atom is completely filled and exhibits extra stability, and thus more energy is needed to remove one electron from it.

Hence, ionization energy: Ca > Ga

Thirdly, consider Si and Ca. The effective nuclear charge of Si is 4+, while that of Ca is 2+. Si has 4 filled electronic shell, while Ca has only 3. Si has higher ionization energy that Ca because Si has higher effective nuclear charge and smaller atomic size than Ca, and thus the attractive forces between the nucleus and the electrons in the outermost shell in Si atom is stronger than that Ca atom.

Hence, ionization energy: Si > Ca

Conclusively, ionization energy (kJ/mol) : P (1012) > S (1000) > Si (787) > Ca (590) > Ga (579)