1. Write the electronic configuration of Ga

Gallium has atomic number $31$.

So, the electronic configuration of $\mathrm{Ga}$ is:

2. Find the configuration of $\mathrm{Ga}^+$

To form $\mathrm{Ga}^+$, one electron is removed from the outermost shell.

The outermost electron is the $4p^1$ electron, so:

3. Identify the valence electrons

Valence electrons are the electrons in the outermost shell, i.e. the shell with highest principal quantum number $n$.

For $\mathrm{Ga}^+$, the highest value of $n$ is $4$, and the electrons present there are:

So, the valence electrons are in the $s$-subshell.

4. Azimuthal quantum number for an $s$-subshell

The azimuthal quantum number $l$ is:

• $s \rightarrow l=0$
• $p \rightarrow l=1$
• $d \rightarrow l=2$
• $f \rightarrow l=3$

Since the valence electrons are in the $4s$ subshell,

5. Final answer

The azimuthal quantum number for the valence electrons of $\mathrm{Ga}^+$ is:

6. Comparison with stored correct answer

Stored correct answer = $0$.

This matches the derived answer.