Question

What is the electron configuration of ${\mathrm{Au}}^{+}$?

Answer 1

Gold(Au):

• Gold$\left(\mathrm{Au}\right)$ is the d-block element.
• The atomic number of $\mathrm{Au}$ is $79$.

Electronic configuration of d-block

• In general, the electronic configuration of these elements is $(\text{n}-1){\mathrm{d}}^{1-10}\text{n}{\mathrm{s}}^{1-2}$.
• The (n–1) stands for the inner $d$ orbitals which may have one to ten electrons and the outermost $\text{n}s$ orbital may have one or two electrons.

Electronic configuration of $\mathrm{Au}$

• The electronic configuration of $\mathrm{Au}$ is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}6{\mathrm{s}}^1$.
• For ${\mathrm{Au}}^{+}$ one electron is removed from the outermost $6s$ orbital.
• Hence the electronic configuration of ${\mathrm{Au}}^{+}$ is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}4{\mathrm{s}}^{2}4{\mathrm{p}}^{6}4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^{6}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}$.