Sure! Let’s discuss how ionization energy trends work in the periodic table, particularly for the alkali metals, and then determine the order of their first ionization energies.
Ionization energy is the energy required to remove an electron from an atom in its gaseous state. When we look at the periodic table, several trends help us understand how ionization energy changes across different elements:
1. Within a group (vertical column): As you move down a group in the periodic table, ionization energy generally decreases. This is because each successive element has an additional electron shell, increasing the distance between the nucleus and the outermost electron and thereby reducing the effective nuclear charge experienced by the outermost electron. Consequently, it becomes easier to remove an electron from atoms lower in the group.
2. Within a period (horizontal row): As you move from left to right across a period, ionization energy generally increases. This is because the effective nuclear charge experienced by the valence electrons increases, but here we’re only looking at elements within the same group.
The elements in question—lithium (Li), sodium (Na), potassium (K), and rubidium (Rb)—all belong to Group 1, the alkali metals. The first ionization energy trend for alkali metals is as follows:
- Lithium (Li)
- Sodium (Na)
- Potassium (K)
- Rubidium (Rb)
According to the trend, lithium should have the highest first ionization energy because it’s at the top of the group, and rubidium should have the lowest because it’s at the bottom.
Therefore, the order of first ionization energies from highest to lowest is:
[tex]\[ \text{Li} > \text{Na} > \text{K} > \text{Rb} \][/tex]
This corresponds to the third option:
[tex]\[ \boxed{Li > Na > K > Rb} \][/tex]
