Answer
The electron configuration for $Au^{3+}$ is $$Au^{3+}: [Xe]4f^{14}5d^{8}$$
$Au^{3+}$ does not have noble-gas configuration.
Work Step by Step
*RULES TO WRITE ELECTRON CONFIGURATIONS FOR IONS:
- When electrons are removed to form cations, those of the orbitals that have the largest principal quantum number $n$ will be removed first.
- If there are more than one subshell for a given $n$, electrons of the subshell with the highest value of $l$ will be removed first.
- When electrons are added to form anions, they would be added to the empty or partially occupied orbital with the lowest value of $n$.
1) The electron configuration of a neutral $Au$ atom is $$Au: [Xe]4f^{14}5d^{10}6s^1$$
2) Here we need to find the electron configuration for $Au^{3+}$, an ion having 3 electrons removed from a neutral $Au$ atom.
Therefore, according to the rules, when electrons are removed, we would look for the orbital with the largest $n$ first. Subshell $6s$, with $n=6$, has the largest $n$.
There is 1 electron occupying subshell $6s$. Surely, this electron would be removed. However, still 2 more electrons need to be removed to make $Au^{3+}$.
3) So we move to find the next largest $n$. Subshell $5d$, with $n=5$, is the subshell with the second largest $n$.
10 electrons occupy this subshell. Only 2 more electrons need to be removed, so 2 out of these 10 electrons would be removed to make $Au^{3+}$.
That means the electron configuration for $Au^{3+}$ is $$Au^{3+}: [Xe]4f^{14}5d^{8}$$
The above configuration is not the configuration of any noble gases, so $Au^{3+}$ does not have noble-gas configuration.
