2. give the complete electron configuration for the two gallium ions. the oxidation numbers from many, but…

Explanation:

Step1: Recall gallium's neutral - state electron configuration

Gallium (Ga) has an atomic number of 31. Its neutral - state electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{1}$.

Step2: Determine the electron configuration of $Ga^{+}$

The oxidation state of $Ga^{+}$ means it has lost 1 electron. The electron is removed from the highest - energy orbital, which is the 4p orbital. So the electron configuration of $Ga^{+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{0}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}$.

Step3: Determine the electron configuration of $Ga^{3+}$

The oxidation state of $Ga^{3+}$ means it has lost 3 electrons. First, the 4p electron is removed, then the two 4s electrons. So the electron configuration of $Ga^{3+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{0}4p^{0}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$.

Answer:

$Ga^{+}$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}$; $Ga^{3+}$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$