Question:
Observe the following statements:
Statement A: In general, the ionization potential value decreases on moving down in a group.
Statement B: The first ionization potential of sodium is greater than that of potassium.
Observe the following statements:
Statement A: In general, the ionization potential value decreases on moving down in a group.
Statement B: The first ionization potential of sodium is greater than that of potassium.
Statement A: In general, the ionization potential value decreases on moving down in a group.
Statement B: The first ionization potential of sodium is greater than that of potassium.
Show Hint
Ionization energy generally decreases down a group because atomic size and shielding effect increase. In Group 1:
\[
Li \gt Na \gt K \gt Rb \gt Cs
\]
with respect to first ionization energy.
Updated On: Jun 26, 2026
- Both A and B are wrong
- Both A and B are correct
- A is correct but B is wrong
- A is wrong but B is correct
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Examine Statement A.
Ionization potential (or ionization energy) is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom.
As we move down a group in the periodic table: Atomic radius increases and Shielding effect increases Due to the increased distance of the valence electron from the nucleus and greater shielding by inner electrons, the attraction between the nucleus and the outermost electron decreases.
Therefore, less energy is required to remove the electron.
Hence, ionization potential generally decreases down a group.
Thus, Statement A is correct.
Step 2: Examine Statement B.
Sodium \((Na)\) and Potassium \((K)\) belong to Group 1 of the periodic table.
Their electronic configurations are \[ Na : 1s^2\,2s^2\,2p^6\,3s^1 \] \[ K : 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1 \] Potassium has one additional shell compared to sodium.
Therefore, \[ \text{Atomic radius of K} \gt \text{Atomic radius of Na} \] and the outermost electron in potassium experiences a weaker nuclear attraction.
Hence, \[ IE_1(K) \lt IE_1(Na) \] Therefore, \[ IE_1(Na) \gt IE_1(K) \] Thus, Statement B is also correct.
Step 3: Final conclusion.
Both Statement A and Statement B are correct.
Therefore, \[ \boxed{\text{Both A and B are correct}} \] Hence, the correct option is \[ \boxed{(2)} \]
Ionization potential (or ionization energy) is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom.
As we move down a group in the periodic table: Atomic radius increases and Shielding effect increases Due to the increased distance of the valence electron from the nucleus and greater shielding by inner electrons, the attraction between the nucleus and the outermost electron decreases.
Therefore, less energy is required to remove the electron.
Hence, ionization potential generally decreases down a group.
Thus, Statement A is correct.
Step 2: Examine Statement B.
Sodium \((Na)\) and Potassium \((K)\) belong to Group 1 of the periodic table.
Their electronic configurations are \[ Na : 1s^2\,2s^2\,2p^6\,3s^1 \] \[ K : 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1 \] Potassium has one additional shell compared to sodium.
Therefore, \[ \text{Atomic radius of K} \gt \text{Atomic radius of Na} \] and the outermost electron in potassium experiences a weaker nuclear attraction.
Hence, \[ IE_1(K) \lt IE_1(Na) \] Therefore, \[ IE_1(Na) \gt IE_1(K) \] Thus, Statement B is also correct.
Step 3: Final conclusion.
Both Statement A and Statement B are correct.
Therefore, \[ \boxed{\text{Both A and B are correct}} \] Hence, the correct option is \[ \boxed{(2)} \]
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