Which one of the following species has the electron configuration 1s^2 2s^2 2p^6?

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

Answer: (A)

Verified step by step guidance

1. Identify the electron configuration given: \$1s^2 2s^2 2p^6\(. This configuration corresponds to a total of 10 electrons because \)1s^2\( has 2 electrons, \)2s^2\( has 2 electrons, and \)2p^6$ has 6 electrons, summing to 10 electrons.
2. Recall that the neutral neon atom (Ne) has 10 electrons, so its electron configuration is exactly \$1s^2 2s^2 2p^6$.
3. Analyze each ion to determine how many electrons they have: Sodium ion (\(Na^{+}\)) loses one electron from neutral sodium (which has 11 electrons), so \(Na^{+}\) has 10 electrons.
4. Fluoride ion (\(F^{-}\)) gains one electron from neutral fluorine (which has 9 electrons), so \(F^{-}\) has 10 electrons.
5. Oxide ion (\(O^{2-}\)) gains two electrons from neutral oxygen (which has 8 electrons), so \(O^{2-}\) has 10 electrons.
6. Since all \(Na^{+}\), \(F^{-}\), and \(O^{2-}\) ions have 10 electrons, they all share the electron configuration \$1s^2 2s^2 2p^6$, which is the same as neon (Ne).