Answer :
Let's solve this question by analyzing the periodic trends and comparing the elements in the given table.
Step 1: Understanding Periodic Trends
1. Atomic Radius:
- Within a group (vertical column) on the periodic table, the atomic radius increases as we move down the group.
- Within a period (horizontal row), the atomic radius decreases as we move from left to right across the period.
2. Ionization Energy:
- Within a group, the ionization energy decreases as we move down the group.
- Within a period, the ionization energy increases as we move from left to right across the period.
Step 2: Position of Elements
- Calcium (Ca) is in Group 2 and Period 4.
- Potassium (K) is in Group 1 and Period 4.
- Thus, Potassium is directly to the left of Calcium and is below Calcium within the same period.
Step 3: Applying Periodic Trends
- Since Potassium (K) is directly to the left of Calcium (Ca) in the periodic table, Potassium will have:
- A larger atomic radius than Calcium.
- A lower ionization energy than Calcium.
Given data for Calcium:
- Atomic radius of Calcium = 194 pm
- First ionization energy of Calcium = 590 kJ/mol
Step 4: Analyzing the Given Options
Let's analyze each option in the context of the periodic trends:
- Option A: 242 pm, 633 kJ/mol
- Atomic radius is larger than Calcium (correct).
- Ionization energy is higher than Calcium (incorrect, should be lower).
- Option B: 242 pm, 419 kJ/mol
- Atomic radius is larger than Calcium (correct).
- Ionization energy is lower than Calcium (correct).
- Option C: 120 pm, 633 kJ/mol
- Atomic radius is smaller than Calcium (incorrect).
- Ionization energy is higher than Calcium (incorrect, should be lower).
- Option D: 120 pm, 419 kJ/mol
- Atomic radius is smaller than Calcium (incorrect).
- Ionization energy is lower than Calcium (correct, but atomic radius is incorrect).
Conclusion:
Based on the periodic trends and the given data, the most probable values for the atomic radius and the first ionization energy for Potassium are given in Option B:
- Atomic radius = 242 pm
- Ionization energy = 419 kJ/mol
Thus, the answer is:
(B) 242 pm, 419 kJ/mol
Step 1: Understanding Periodic Trends
1. Atomic Radius:
- Within a group (vertical column) on the periodic table, the atomic radius increases as we move down the group.
- Within a period (horizontal row), the atomic radius decreases as we move from left to right across the period.
2. Ionization Energy:
- Within a group, the ionization energy decreases as we move down the group.
- Within a period, the ionization energy increases as we move from left to right across the period.
Step 2: Position of Elements
- Calcium (Ca) is in Group 2 and Period 4.
- Potassium (K) is in Group 1 and Period 4.
- Thus, Potassium is directly to the left of Calcium and is below Calcium within the same period.
Step 3: Applying Periodic Trends
- Since Potassium (K) is directly to the left of Calcium (Ca) in the periodic table, Potassium will have:
- A larger atomic radius than Calcium.
- A lower ionization energy than Calcium.
Given data for Calcium:
- Atomic radius of Calcium = 194 pm
- First ionization energy of Calcium = 590 kJ/mol
Step 4: Analyzing the Given Options
Let's analyze each option in the context of the periodic trends:
- Option A: 242 pm, 633 kJ/mol
- Atomic radius is larger than Calcium (correct).
- Ionization energy is higher than Calcium (incorrect, should be lower).
- Option B: 242 pm, 419 kJ/mol
- Atomic radius is larger than Calcium (correct).
- Ionization energy is lower than Calcium (correct).
- Option C: 120 pm, 633 kJ/mol
- Atomic radius is smaller than Calcium (incorrect).
- Ionization energy is higher than Calcium (incorrect, should be lower).
- Option D: 120 pm, 419 kJ/mol
- Atomic radius is smaller than Calcium (incorrect).
- Ionization energy is lower than Calcium (correct, but atomic radius is incorrect).
Conclusion:
Based on the periodic trends and the given data, the most probable values for the atomic radius and the first ionization energy for Potassium are given in Option B:
- Atomic radius = 242 pm
- Ionization energy = 419 kJ/mol
Thus, the answer is:
(B) 242 pm, 419 kJ/mol