Using a four-block periodic table as a guide, determine the ground-state electron configuration of palladium (Pd).

First, we need to find the position of palladium (Pd) within the periodic table. Pd is located in the transition metal group in period 5 and group 10.

To build the electron configuration for Pd, we will use the four-block periodic table and follow the order of the blocks: s, d (if applicable), p, f (if applicable). 1. The first two electrons go into the 1s orbital: \(1s^2\) 2. The next two electrons go into the 2s orbital: \(1s^2,2s^2\) 3. The next six electrons go into the 2p orbitals: \(1s^2,2s^2,2p^6\) 4. The next two electrons go into the 3s orbital: \(1s^2,2s^2,2p^6,3s^2\) 5. The next six electrons go into the 3p orbitals: \(1s^2,2s^2,2p^6,3s^2,3p^6\) 6. The next two electrons go into the 4s orbital: \(1s^2,2s^2,2p^6,3s^2,3p^6,4s^2\) 7. The next ten electrons go into the 3d orbitals: \(1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10}\) 8. The next six electrons go into the 4p orbitals: \(1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6\) 9. The next two electrons go into the 5s orbital: \(1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2\) We have now placed a total of 36 electrons. Palladium, with an atomic number of 46, has 10 more electrons. These remaining 10 electrons should be placed into the 4d orbitals. So, the electron configuration becomes: \(1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10}\)

The ground-state electron configuration for palladium (Pd) is: \(1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10}\)