For the sodium cation (Na^+), how many electrons are present in its valence shell, and how many unpaired electron spins does it have?

• A. 1 valence electron; 1 unpaired electron spin
• B. 2 valence electrons; 2 unpaired electron spins
• C. 0 valence electrons; 0 unpaired electron spins
• D. 8 valence electrons; 0 unpaired electron spins

Answer: (C)

Verified step by step guidance

1. Identify the atomic number of sodium (Na), which is 11. This means a neutral sodium atom has 11 electrons.
2. Determine the electron configuration of neutral sodium: \$1s^2\ 2s^2\ 2p^6\ 3s^1$.
3. Since the problem asks about the sodium cation \(\text{Na}^+\), recognize that it has lost one electron compared to the neutral atom, so it has 10 electrons.
4. Write the electron configuration for \(\text{Na}^+\) by removing one electron from the outermost shell: \$1s^2\ 2s^2\ 2p^6$.
5. From this configuration, note that the valence shell is now the second shell (\(n=2\)) with \$2s^2\ 2p^6$, which is fully filled, so there are 8 valence electrons and no unpaired electron spins.