What is the full electron configuration for Na ^3+
The Correct Answer and Explanation is:
The electron configuration for the sodium ion (Na³⁺) can be derived by first considering the neutral sodium atom and then removing electrons, since Na³⁺ is a positively charged ion.
- Neutral Sodium (Na): Sodium has an atomic number of 11, which means a neutral sodium atom has 11 electrons. The electron configuration for a neutral sodium atom is: 1s2 2s2 2p6 3s11s^2 \, 2s^2 \, 2p^6 \, 3s^11s22s22p63s1 This represents the distribution of 11 electrons in different orbitals, with the first shell (1s) holding 2 electrons, the second shell (2s and 2p) holding 8 electrons, and the third shell (3s) holding 1 electron.
- Sodium Ion (Na³⁺): The Na³⁺ ion is formed by removing 3 electrons from the neutral sodium atom. Since the 3s orbital has the outermost electrons, these are the first to be removed. Therefore, we remove the 1 electron from the 3s orbital. After removing 3 electrons from the neutral sodium atom (2 from the 3s orbital and 1 more), the electron configuration becomes: 1s2 2s2 2p61s^2 \, 2s^2 \, 2p^61s22s22p6 This is the electron configuration for the Na³⁺ ion, which is the same as that of the noble gas neon (Ne), with a stable, filled shell configuration of 2 electrons in the first shell and 8 electrons in the second shell.
Explanation: When a sodium atom loses 3 electrons to form the Na³⁺ ion, it reaches the same electron configuration as neon (Ne), which is a stable configuration. This is because neon has a full second shell (2s² 2p⁶), and the loss of electrons from the 3rd shell allows sodium to achieve this stable state, resembling the noble gas configuration.
Thus, the full electron configuration for Na³⁺ is:1s2 2s2 2p61s^2 \, 2s^2 \, 2p^61s22s22p6
