To determine which ion has a noble gas electron configuration, we need to look at the electron configurations of each and see if they match the electron configuration of a noble gas.
1. \( Na^{2-} \):
- Sodium (Na) has an atomic number of 11, so it normally has 11 electrons.
- The \( Na^{2-} \) ion would have 11 + 2 = 13 electrons.
- The electron configuration for 13 electrons is \( 1s^2 2s^2 2p^6 3s^2 3p^1 \), which is not a noble gas configuration.
2. \( Mg^{2+} \):
- Magnesium (Mg) has an atomic number of 12, so it normally has 12 electrons.
- The \( Mg^{2+} \) ion would have 12 - 2 = 10 electrons.
- The electron configuration for 10 electrons is \( 1s^2 2s^2 2p^6 \), which is the electron configuration of Neon (Ne), a noble gas.
3. \( Mg^{2} \):
- This notation seems to be incorrectly formatted. We typically see \( Mg^{2+} \) or \( Mg \). Assuming a typo, and considering the same logic applied to \( Mg^{2+} \) earlier, this would have the noble gas configuration.
4. \( B^{4+} \):
- Boron (B) has an atomic number of 5, so it normally has 5 electrons.
- The \( B^{4+} \) ion would have 5 - 4 = 1 electron.
- The electron configuration for 1 electron is \( 1s^1 \), which is not a noble gas configuration.
5. \( Na \):
- Sodium (Na) has an atomic number of 11, so it normally has 11 electrons.
- The electron configuration for 11 electrons is \( 1s^2 2s^2 2p^6 3s^1 \), which is not a noble gas configuration.
Based on this analysis, the ion \( Mg^{2+} \) has a noble gas electron configuration. Therefore, the correct answer is:
[tex]\( 2. \, Mg^{2+} \)[/tex]
