Which of the following ions has the same electronic configuration as the noble gas neon (Ne)?

• A. O^{2-}
• B. Mg^{2+}
• C. F^-
• D. Na^+

Answer: (D)

Verified step by step guidance

1. Recall that the noble gas neon (Ne) has an atomic number of 10, which means it has 10 electrons. Its electronic configuration is \$1s^2 2s^2 2p^6$.
2. For each ion, determine the number of electrons by adjusting the atomic number of the neutral atom according to the charge of the ion:
3. - For \(O^{2-}\): Oxygen has 8 protons, so the neutral atom has 8 electrons. Adding 2 electrons for the \$2-$ charge gives 10 electrons.
4. - For \(Mg^{2+}\): Magnesium has 12 protons, so the neutral atom has 12 electrons. Removing 2 electrons for the \$2+$ charge gives 10 electrons.
5. - For \(F^-\): Fluorine has 9 protons, so the neutral atom has 9 electrons. Adding 1 electron for the \(-\) charge gives 10 electrons.
6. - For \(Na^+\): Sodium has 11 protons, so the neutral atom has 11 electrons. Removing 1 electron for the \(+\) charge gives 10 electrons.
7. Since all these ions have 10 electrons, they all have the same electronic configuration as neon, which is \$1s^2 2s^2 2p^6$.