transition metals - What is the reason for the anomalous electronic configuration of palladium?

The p block elements generally have the general valence shell configuration as $$n\mathrm{p}^{1-6}\,n\mathrm{s}^{1-2}.$$ Expected electronic configuration for palladium is $$\ce{^46Pd} = \mathrm{1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^8}.$$ But I find that it's valence shell instead is $$\mathrm{4d^{10}\, 5s^0}$$ I first thought, for achieving fully filled d-subshell, 2 electrons go to the d-subshell from s-subshell. This electron transfer should happen for all members of its group, yet only palladium demonstrates this. Why is this so?