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Statement I: The ionization energy of Na...

Statement I: The ionization energy of `Na^(2+)` is less than that of `Na^(+)` .
Statement II: `Na^(+)` has an inert gas configuration.

A

If both Statement I and Statement II are true and Statement II is the correct explanation of Statement I.

B

If both Statement I and Statement II are true but Statement II is not the correct explanation for Statement I.

C

If Statement I is true but Statement II is false.

D

If Statement I is false but Statement II is true.

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the statements provided in the question, we will evaluate each one step by step. ### Step 1: Analyze Statement I **Statement I:** The ionization energy of `Na^(2+)` is less than that of `Na^(+)`. 1. **Understanding Ionization Energy:** Ionization energy is the energy required to remove an electron from an atom or ion. 2. **Electronic Configuration of `Na^(+)`:** Sodium (Na) has an atomic number of 11. When it loses one electron to form `Na^(+)`, its electronic configuration becomes: - `Na`: 1s² 2s² 2p⁶ 3s¹ - `Na^(+)`: 1s² 2s² 2p⁶ (which is the same as the configuration of Neon, a noble gas). 3. **Electronic Configuration of `Na^(2+)`:** When `Na^(+)` loses another electron to form `Na^(2+)`, its electronic configuration becomes: - `Na^(2+)`: 1s² 2s² 2p⁵. 4. **Comparison of Ionization Energies:** Since `Na^(+)` has a noble gas configuration (which is stable), removing an electron from it (to form `Na^(2+)`) requires more energy than removing an electron from `Na` (which has a less stable configuration). Therefore, the ionization energy of `Na^(2+)` is actually greater than that of `Na^(+)`. **Conclusion for Statement I:** This statement is **false**. ### Step 2: Analyze Statement II **Statement II:** `Na^(+)` has an inert gas configuration. 1. **Inert Gas Configuration:** An inert gas configuration refers to a completely filled electron shell, which is very stable. 2. **Electronic Configuration of `Na^(+)`:** As established earlier, the electronic configuration of `Na^(+)` is: - `Na^(+)`: 1s² 2s² 2p⁶. 3. **Comparison with Noble Gas Configuration:** This configuration is identical to that of Neon (Ne), which is indeed an inert gas. **Conclusion for Statement II:** This statement is **true**. ### Final Conclusion - **Statement I:** False - **Statement II:** True Thus, the correct option is that Statement I is false and Statement II is true. ---
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