Question
Write the condensed electron configuration and partial orbital diagram for Ga3+
Answers
GPT-4o mini
Gallium (Ga) has an atomic number of 31, and its ground state electron configuration can be written as:
\[
\text{Ga:} \, [Ar] \, 3d^{10} \, 4s^{2} \, 4p^{1}
\]
When gallium forms a \( Ga^{3+} \) ion, it loses three electrons. The electrons are removed from the outermost orbitals first, which in this case are the 4s and 4p orbitals. Thus, the ionization process will remove the two electrons from the 4s orbital and the one electron from the 4p orbital.
The electron configuration for \( Ga^{3+} \) becomes:
\[
\text{Ga}^{3+:} \, [Ar] \, 3d^{10}
\]
### Partial Orbital Diagram for \( Ga^{3+} \)
The partial orbital diagram for \( Ga^{3+} \) would look like this:
```
3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s:
4p:
```
In summary, the condensed electron configuration for \( Ga^{3+} \) is \([Ar] \, 3d^{10}\), and the partial orbital diagram shows that the 3d subshell is fully filled with 10 electrons. The 4s and 4p orbitals are empty.
\[
\text{Ga:} \, [Ar] \, 3d^{10} \, 4s^{2} \, 4p^{1}
\]
When gallium forms a \( Ga^{3+} \) ion, it loses three electrons. The electrons are removed from the outermost orbitals first, which in this case are the 4s and 4p orbitals. Thus, the ionization process will remove the two electrons from the 4s orbital and the one electron from the 4p orbital.
The electron configuration for \( Ga^{3+} \) becomes:
\[
\text{Ga}^{3+:} \, [Ar] \, 3d^{10}
\]
### Partial Orbital Diagram for \( Ga^{3+} \)
The partial orbital diagram for \( Ga^{3+} \) would look like this:
```
3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s:
4p:
```
In summary, the condensed electron configuration for \( Ga^{3+} \) is \([Ar] \, 3d^{10}\), and the partial orbital diagram shows that the 3d subshell is fully filled with 10 electrons. The 4s and 4p orbitals are empty.