Question

Based solely on their location on the periodic table and the predicted periodic trends, rank the following atoms in IONIZATION ENERGY, from SMALLEST to LARGEST:

6C, 13Al, 49In, 9F

Answer 1

The correct order from smallest to largest ionization energy is Indium (In) < Aluminum (Al) < Carbon (C) < Fluorine (F), based on periodic trends of increasing ionization energy across a period and decreasing ionization energy down a group.

Step-by-step explanation:

To rank the atoms 6C (carbon), 13Al (aluminum), 49In (indium), and 9F (fluorine) in terms of ionization energy, we need to consider two main trends in the periodic table: ionization energy increases from left to right across a period and decreases down a group. With these trends in mind, we proceed to rank the ionization energies as follows:

• 49In (Indium): Located in Period 5, Group 13, it will have the smallest ionization energy due to a larger atomic radius and more electron shielding.
• 13Al (Aluminum): Positioned in Period 3, Group 13, aluminum will have a higher ionization energy than indium due to being higher up in the same group.
• 6C (Carbon): Being in Period 2, Group 14, carbon has a higher ionization energy than aluminum due to being further right (higher effective nuclear charge) and in an earlier period.
• 9F (Fluorine): Fluorine, located in Period 2, Group 17 has the highest ionization energy due to it being furthest to the right (highest effective nuclear charge) and higher up in the periodic table in the same period as carbon.

Thus, the correct order from smallest to largest ionization energy is Indium (In) < Aluminum (Al) < Carbon (C) < Fluorine (F).