Answer :
To determine which ion has a noble gas electron configuration, we should start by understanding what a noble gas electron configuration is. Noble gases have completely filled energy levels, which makes them particularly stable. The electron configuration of a noble gas is fully occupied up to a certain principal quantum number.
Now, let’s analyze each of the ions provided:
1. \( Be^{2-} \) ion:
Beryllium normally has an electron configuration of \( 1s^2 2s^2 \). When it gains 2 electrons (forming \( Be^{2-} \)), its configuration becomes:
[tex]\[ 1s^2 2s^2 2p^2 \][/tex]
This does not match any noble gas configuration because it has more electrons than helium (1s^2) and fewer than neon (1s^2 2s^2 2p^6).
2. \( Na \) (sodium) atom:
Sodium normally has an electron configuration of \( 1s^2 2s^2 2p^6 3s^1 \).
3. \( Na^+ \) (sodium ion):
When sodium loses one electron to form \( Na^+ \), the electron configuration becomes:
[tex]\[ 1s^2 2s^2 2p^6 \][/tex]
This configuration matches the electron configuration of neon (Ne), a noble gas.
4. \( Na^{2+} \) ion:
This is a hypothetical ion, as sodium prefers to lose only one electron to achieve a stable configuration. However, its electron configuration would be:
[tex]\[ 1s^2 2s^2 2p^6 3s^{-1} \][/tex]
The negative exponent on the \( 3s \) orbital denotes something that is not physically possible in a stable, neutral atom or ion, indicating an incorrect configuration.
5. \( Be^+ \) ion:
Beryllium with one less electron would have the configuration:
[tex]\[ 1s^2 2s^1 \][/tex]
This is not a noble gas configuration.
Therefore, out of the ions given, \( Na^+ \) (Option 3) is the ion that has a noble gas electron configuration. The noble gas configuration referred to is that of neon (Ne), which is:
[tex]\[ 1s^2 2s^2 2p^6 \][/tex]
So, the correct answer is:
[tex]\( Na^+ \)[/tex] (Option 3)
Now, let’s analyze each of the ions provided:
1. \( Be^{2-} \) ion:
Beryllium normally has an electron configuration of \( 1s^2 2s^2 \). When it gains 2 electrons (forming \( Be^{2-} \)), its configuration becomes:
[tex]\[ 1s^2 2s^2 2p^2 \][/tex]
This does not match any noble gas configuration because it has more electrons than helium (1s^2) and fewer than neon (1s^2 2s^2 2p^6).
2. \( Na \) (sodium) atom:
Sodium normally has an electron configuration of \( 1s^2 2s^2 2p^6 3s^1 \).
3. \( Na^+ \) (sodium ion):
When sodium loses one electron to form \( Na^+ \), the electron configuration becomes:
[tex]\[ 1s^2 2s^2 2p^6 \][/tex]
This configuration matches the electron configuration of neon (Ne), a noble gas.
4. \( Na^{2+} \) ion:
This is a hypothetical ion, as sodium prefers to lose only one electron to achieve a stable configuration. However, its electron configuration would be:
[tex]\[ 1s^2 2s^2 2p^6 3s^{-1} \][/tex]
The negative exponent on the \( 3s \) orbital denotes something that is not physically possible in a stable, neutral atom or ion, indicating an incorrect configuration.
5. \( Be^+ \) ion:
Beryllium with one less electron would have the configuration:
[tex]\[ 1s^2 2s^1 \][/tex]
This is not a noble gas configuration.
Therefore, out of the ions given, \( Na^+ \) (Option 3) is the ion that has a noble gas electron configuration. The noble gas configuration referred to is that of neon (Ne), which is:
[tex]\[ 1s^2 2s^2 2p^6 \][/tex]
So, the correct answer is:
[tex]\( Na^+ \)[/tex] (Option 3)
