Question

What is the unabbreviated electron configuration of gold?

Answer 1

Predicted: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^9$

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

Explanation:

When. The atomic number of gold $\left(Au\right)$ is $79$. Therefore, its ground-state electron configuration should have a total number of $79$ electrons.

$Au$ is one of those elements that does not follow the predicted order of the electron configuration (the other elements are $Cu,Ag,Pd,Cr$, and $Mo$). This happens simply because completing the d orbital brings lower energy (lower risk of electron repulsion) compared to filling up the s orbital first.