To determine which atom's ionization energy is greater than that of phosphorus (P), we need to consider various elements and their positions in the periodic table. Ionization energy is the energy required to remove an electron from an atom. This property typically varies as follows:
1. Across a Period: Ionization energy generally increases as we move from left to right across a period because the nuclear charge increases while the atomic radius decreases, making it harder to remove an electron.
2. Down a Group: Ionization energy generally decreases as we move down a group because the atomic radius increases, making it easier to remove an electron.
Let us review the given elements:
- Ba (Barium):
- Group: 2
- Period: 6
- Barium is far down in group 2, where elements tend to have lower ionization energies.
- K (Potassium):
- Group: 1
- Period: 4
- Potassium is in group 1, where elements typically have very low ionization energies compared to other groups.
- As (Arsenic):
- Group: 15
- Period: 4
- Arsenic is located in the same group as phosphorus, but a period below it. Typically, its ionization energy will be lower than phosphorus but not drastically lower.
- Cl (Chlorine):
- Group: 17
- Period: 3
- Chlorine is close to phosphorus in period 3 and belongs to group 17, known for very high ionization energies due to their high nuclear charge and smaller atomic radius.
Based on this information:
- Ba (Barium) and K (Potassium) have significantly lower ionization energies than phosphorus due to their positions in the periodic table.
- As (Arsenic) has a lower but comparable ionization energy to phosphorus, as they are in the same group but arsenic is one period lower.
- Cl (Chlorine), being in group 17 and period 3, has a higher ionization energy than phosphorus due to its position in the periodic table.
Thus, the correct answer, indicating the atom with ionization energy greater than that of phosphorus (P), is:
D. Cl (Chlorine)
