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For N^(3-) gt O^(2-) gt F^(-) and Na^(+)...

For `N^(3-) gt O^(2-) gt F^(-)` and `Na^(+)`, the order in which their ionic radii varies is

A

`N^(3-) gt O^(2-) gt F^(-) gt Na^(+)`

B

`N^(2) gt Na^(+) gt O^(2) gt F^(-)`

C

`Na^(+) gt O^(2)gt N^(3) gt F^(-)`

D

`O^(2) gt F^(-) gt Na^(+) gt N^(3-)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the order of ionic radii for the ions \(N^{3-}\), \(O^{2-}\), \(F^{-}\), and \(Na^{+}\), we need to consider the number of electrons and protons in each ion, as well as the effective nuclear charge experienced by the electrons. ### Step-by-Step Solution: 1. **Identify the Ions and Their Electron Configurations:** - \(N^{3-}\): Nitrogen has an atomic number of 7. When it gains 3 electrons, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). - \(O^{2-}\): Oxygen has an atomic number of 8. When it gains 2 electrons, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). - \(F^{-}\): Fluorine has an atomic number of 9. When it gains 1 electron, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). - \(Na^{+}\): Sodium has an atomic number of 11. When it loses 1 electron, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). 2. **Determine the Number of Protons:** - \(N^{3-}\) has 7 protons. - \(O^{2-}\) has 8 protons. - \(F^{-}\) has 9 protons. - \(Na^{+}\) has 11 protons. 3. **Analyze the Ionic Sizes:** - All ions have the same electron configuration (10 electrons), but they differ in the number of protons. - The more protons an ion has, the greater the effective nuclear charge acting on the electrons, which pulls the electrons closer to the nucleus and results in a smaller ionic radius. 4. **Order the Ions by Ionic Radius:** - \(N^{3-}\) (7 protons) has the largest radius because it has the least effective nuclear charge. - \(O^{2-}\) (8 protons) is smaller than \(N^{3-}\) but larger than \(F^{-}\). - \(F^{-}\) (9 protons) is smaller than both \(N^{3-}\) and \(O^{2-}\). - \(Na^{+}\) (11 protons) has the smallest radius due to the highest effective nuclear charge. 5. **Final Order of Ionic Radii:** - The order from largest to smallest ionic radius is: \[ N^{3-} > O^{2-} > F^{-} > Na^{+} \]

To determine the order of ionic radii for the ions \(N^{3-}\), \(O^{2-}\), \(F^{-}\), and \(Na^{+}\), we need to consider the number of electrons and protons in each ion, as well as the effective nuclear charge experienced by the electrons. ### Step-by-Step Solution: 1. **Identify the Ions and Their Electron Configurations:** - \(N^{3-}\): Nitrogen has an atomic number of 7. When it gains 3 electrons, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). - \(O^{2-}\): Oxygen has an atomic number of 8. When it gains 2 electrons, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). - \(F^{-}\): Fluorine has an atomic number of 9. When it gains 1 electron, its electron configuration becomes \(1s^2 2s^2 2p^6\) (total 10 electrons). ...
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