In Platinum $\text{Pt}$, why do $\text{s}$ and $\text{p}$ orbitals contract when electrons start travelling at relativistic speeds $|v|\sim c$, while $\text{d}$ orbital expands?
The way I reasoned about this is that due to relativistic effects, relativistic mass of an electron increases:
$$|v|\sim c\rightarrow m_c = \frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}\uparrow$$
Which means that nucleus exerts higher attractive force on electrons, and so, they possess lower energy. However, due to $\text{d}$ orbital suffering from shielding effect, it expands while $\text{s}$ and $\text{p}$ orbitals contract in size.
