Question

Determine the Azimuthal quantum number of the last electron of $G{a}^{+}$ ion

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Step 1: Atomic number of Gallium $\left(Ga\right)$ is $31$.

Step 2: Electronic configuration of $Ga$ is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^1$ or $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^1$

Step 3: Determine the electronic configuration of the $G{a}^{+}$ ion:

1. Electronic configuration of $G{a}^{+}$ ion is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^0$ or $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^0$
2. Here the electron loses from the $\text{'}4\mathrm{p}\text{'}$ subshell as the outermost electron is present in that subshell.

Step 4: Determine the Azimuthal Quantum Number $\left(l\right)$ of the last electron of the $G{a}^{+}$ ion:

1. The last electron is present in the $\text{'}4\mathrm{s}\text{'}$ subshell.
2. Hence, the Principle Quantum number is $n=4$.
3. The Azimuthal Quantum Number for $\text{'}s\text{'}$ subshell is $l=0$.

Hence, option B is correct. The Azimuthal Quantum number for $\text{'}4\mathrm{s}\text{'}$ subshell is $l=0$.