Which electron configuration is correct for a sodium ion, ${\mathrm{Na}}^{+}$?

• A. $1s^{2}2s^{2}2p^{6}3s^2$
• B. $1s^{2}2s^{2}2p^6$
• C. $1s^{2}2s^{2}2p^{6}3s^1$
• D. $1s^{2}2s^{2}2p^5$

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the electron configuration of a neutral sodium (Na) atom. Sodium has an atomic number of 11, so it has 11 electrons. The electron configuration is written by filling orbitals in order of increasing energy: \$1s^2 2s^2 2p^6 3s^1$.
2. Step 2: Recognize that a sodium ion, \(\mathrm{Na}^+\), is formed when sodium loses one electron. This means the ion has one less electron than the neutral atom, so it has 10 electrons.
3. Step 3: Remove one electron from the highest energy orbital of the neutral sodium atom. Since the last electron is in the \$3s\( orbital, removing it leaves the electron configuration as \)1s^2 2s^2 2p^6$.
4. Step 4: Write the electron configuration for \(\mathrm{Na}^+\) using the remaining 10 electrons, which corresponds to the electron configuration of neon, a noble gas: \$1s^2 2s^2 2p^6$.
5. Step 5: Confirm that this configuration is stable and corresponds to a full octet in the second shell, which is typical for a positively charged sodium ion.