Answer :
To determine which set correctly orders the atoms from highest to lowest ionization energy, we need to understand how ionization energy trends on the periodic table.
Ionization energy is the energy required to remove an electron from an atom. The general trends for ionization energy are:
1. Ionization energy increases from left to right across a period (row) due to increasing nuclear charge.
2. Ionization energy decreases from top to bottom down a group (column) due to increasing atomic size, which means the outer electrons are farther from the nucleus and are less tightly bound.
Given the elements in question: Sulfur (S), Phosphorus (P), Aluminum (Al), and Argon (Ar), we need to arrange them in order of their ionization energies.
Let's analyze each element:
- Argon (Ar): This is a noble gas located at the far right of the periodic table in period 3, which means it has a very high ionization energy.
- Sulfur (S): This element is also in period 3, but it is to the left of Argon (closer to the middle), so it will have a lower ionization energy than Argon.
- Phosphorus (P): Phosphorus is also in period 3, but it is to the left of Sulfur, meaning it has a lower ionization energy than Sulfur.
- Aluminum (Al): This element is in period 3 but is located to the left of Phosphorus, so it will have the lowest ionization energy among these elements.
Based on the above analysis, the order from highest to lowest ionization energy is:
- Argon (Ar) has the highest ionization energy.
- Sulfur (S) comes next, as it is to the right of Phosphorus and Aluminum.
- Phosphorus (P) follows Sulfur, being to the left of Sulfur but to the right of Aluminum.
- Aluminum (Al) has the lowest ionization energy among the given elements.
Thus, the correct order is:
[tex]\[ \text{Ar} > \text{S} > \text{P} > \text{Al} \][/tex]
Therefore, the correct option is:
[tex]\[ \boxed{3} \][/tex]
Ionization energy is the energy required to remove an electron from an atom. The general trends for ionization energy are:
1. Ionization energy increases from left to right across a period (row) due to increasing nuclear charge.
2. Ionization energy decreases from top to bottom down a group (column) due to increasing atomic size, which means the outer electrons are farther from the nucleus and are less tightly bound.
Given the elements in question: Sulfur (S), Phosphorus (P), Aluminum (Al), and Argon (Ar), we need to arrange them in order of their ionization energies.
Let's analyze each element:
- Argon (Ar): This is a noble gas located at the far right of the periodic table in period 3, which means it has a very high ionization energy.
- Sulfur (S): This element is also in period 3, but it is to the left of Argon (closer to the middle), so it will have a lower ionization energy than Argon.
- Phosphorus (P): Phosphorus is also in period 3, but it is to the left of Sulfur, meaning it has a lower ionization energy than Sulfur.
- Aluminum (Al): This element is in period 3 but is located to the left of Phosphorus, so it will have the lowest ionization energy among these elements.
Based on the above analysis, the order from highest to lowest ionization energy is:
- Argon (Ar) has the highest ionization energy.
- Sulfur (S) comes next, as it is to the right of Phosphorus and Aluminum.
- Phosphorus (P) follows Sulfur, being to the left of Sulfur but to the right of Aluminum.
- Aluminum (Al) has the lowest ionization energy among the given elements.
Thus, the correct order is:
[tex]\[ \text{Ar} > \text{S} > \text{P} > \text{Al} \][/tex]
Therefore, the correct option is:
[tex]\[ \boxed{3} \][/tex]
