What do \(\mathrm{F}^{-}, \mathrm{O}^{2-}, \mathrm{Na}^{+}\), and \(\mathrm{Mg}^{2+}\) all have in common?

Short Answer

Expert verified
All the ions, \(\mathrm{F^{-}}\), \(\mathrm{O^{2-}}\), \(\mathrm{Na^{+}}\), and \(\mathrm{Mg^{2+}}\), have a complete octet in their outer electron shell, making them more stable than their neutral counterparts.

Step by step solution

01

Write the electron configurations for the neutral atoms

Using the periodic table, we can write the electron configurations for the neutral atoms: fluorine (F), oxygen (O), sodium (Na), and magnesium (Mg). - Fluorine (F): \( 1s^2 2s^2 2p^5 \) - Oxygen (O): \( 1s^2 2s^2 2p^4 \) - Sodium (Na): \( 1s^2 2s^2 2p^6 3s^1 \) - Magnesium (Mg): \( 1s^2 2s^2 2p^6 3s^2 \)
02

Determine the electron configurations of the ions

Now, we need to find the electron configurations for the given ions by adding or removing electrons from the neutral atoms, based on their charges. - \(\mathrm{F^{-}}\): Add 1 electron to complete the outer shell: \( 1sg^2 2s^2 2p^6 \) - \(\mathrm{O^{2-}}\): Add 2 electrons to complete the outer shell: \( 1s^2 2s^2 2p^6 \) - \(\mathrm{Na^{+}}\): Remove 1 electron from the outer shell: \( 1s^2 2s^2 2p^6 \) - \(\mathrm{Mg^{2+}}\): Remove 2 electrons from the outer shell: \( 1s^2 2s^2 2p^6 \)
03

Identify the common property among the ions

As we can see from their electron configurations, all these ions now have a complete octet in their outer shell (2s and 2p orbitals are fully occupied).
04

Conclusion

Therefore, the property that \(\mathrm{F^{-}}\), \(\mathrm{O^{2-}}\), \(\mathrm{Na^{+}}\), and \(\mathrm{Mg^{2+}}\) have in common is that they all have a complete octet in their outer electron shell, which makes them more stable than their neutral counterparts.

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