What is the electron configuration of Au+?

Jan 3, 2016

$\left[X e\right] 4 {f}^{14} 5 {d}^{10}$

Explanation:

The atomic number of Au is 79.

Therefore, its configuration is:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{6} 4 {f}^{14} 5 {d}^{10} 6 {s}^{1}$

or, $\left[X e\right] 4 {f}^{14} 5 {d}^{10} 6 {s}^{1}$

For $A {u}^{+}$, one electron is removed from the outermost $6 s$ orbital, making the configuration,

$\left[X e\right] 4 {f}^{14} 5 {d}^{10}$