Question

Two sets of ionizations are shown in the tables below. Complete the tables by ordering each set of ionizations by increasing amount of energy required. In other words, for each set choose "1" next to the ionization that would require the least energy, "2" next to the ionization that would require the next least energy, and so on. ionization energy required ionization energy required Ne → Ne Fr — Fr te ? Al → Alt + é ? Li – Li + e K → K* té ? | Rb → Rb + é ? x 6 ?

Answer 1

Set 1

Ionization	Energy Required
Ne → Ne+ + e-	3
Al → Al+ + e-	2
K → K+ + e-	1

Set 2

Ionization	Energy Required
Fr → Fr+ + e-	1
Li → Li+ + e-	3
Rb → Rb+ + e-	2

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

a) Ionization energy generally decreases moving from top to bottom down an element group due to an increase in the principal quantum number of the outermost electron moving down a group. The nuclear charge increases going down the group as well as the distance of the outermost electron from the nucleus. But going down the group the effect of distance of the outermost electron from the nucleus becomes predominant over nuclear charge. Hence the decrease in ionization energy is observed.

b) Ionization energy generally increases moving from left to right across an element period due to the decrease in the atomic radius moving across a period, so there is a greater effective attraction between the negatively charged electrons and positively charged nucleus.

In Set 1 K requires the least energy to ionize as the principle quantum no of its outermost electron is highest among the three elements. Among Ne and Al, although the principle quantum no of its outermost electron in Al is higher than the Ne; their ionization energy follows the same trend. Furthermore, the outermost atomic orbital has been filled to its maximum capacity of 8 electrons in the case of Ne. Having 8 valence electrons is the most stable energy state (meaning lowest in energy) for an atom in its respective period that it can exist in. So, the removal of an electron from Ne requires far more energy than that of Al.

In Set 2 Fr requires the least energy to ionize as the principle quantum no of its outermost electron is highest among the three elements. And as in Li, the principle quantum no of its outermost electron is lowest among the three elements, the energy required to remove an electron is highest in Li.