Arrange the following elements in order of decreasing first ionization energy: Bi, Ar, Sb, Br. Rank elements from largest to smallest.
The Correct Answer and Explanation is:
Answer (from largest to smallest first ionization energy):
Ar > Br > Sb > Bi
Explanation:
First ionization energy refers to the energy required to remove the outermost electron from a neutral atom in its gaseous state. This property is influenced by several factors, including nuclear charge, atomic radius, and electron shielding.
In general, ionization energy increases across a period (from left to right on the periodic table) and decreases down a group (from top to bottom). This trend occurs because, across a period, atoms have more protons in their nuclei (increased nuclear charge) and the added electrons enter the same principal energy level, so the attraction between the nucleus and electrons becomes stronger. Conversely, moving down a group adds more electron shells, increasing atomic size and electron shielding, which reduces the nucleus’s effective pull on the valence electrons.
Now, consider the given elements:
- Argon (Ar) is a noble gas located in Period 3 and Group 18. It has a very high first ionization energy because its outer electron shell is full and very stable. Removing an electron from such a stable configuration requires a significant amount of energy.
- Bromine (Br) is a halogen in Period 4 and Group 17. It also has a relatively high ionization energy due to its high effective nuclear charge, but it is lower than that of argon because bromine has a larger atomic radius and more electron shielding.
- Antimony (Sb) and Bismuth (Bi) are both group 15 elements, but Sb is in Period 5 and Bi is in Period 6. As we move down the group, ionization energy decreases due to increasing atomic size and shielding. Therefore, Sb has a higher ionization energy than Bi.
By applying these periodic trends, the correct order of decreasing first ionization energy is:
Ar > Br > Sb > Bi.
