Follow the directions of Question 29 for (a) \(\mathrm{Mo}^{3+}\) (b) \(\mathrm{Pd}^{4+}\)

Molybdenum (Mo) is atomic number 42, while Palladium (Pd) is atomic number 46. This information can be found in the periodic table.

Using the periodic table or electron configuration rules, write the electron configurations for the neutral atoms of Mo and Pd. For Mo, with atomic number 42, the electron configuration is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^4\]For Pd, with atomic number 46, the electron configuration is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8\]

Now, we will remove the required number of electrons from the neutral atom configurations to get the ion configurations. Keep in mind that electrons are removed from the most recently added quantum level shells. (a) For Mo³⁺, we will remove three electrons from the neutral configuration:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^4 \Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^1\]So, the electron configuration for Mo³⁺ is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^1\] (b) For Pd⁴⁺, we will remove four electrons from the neutral configuration:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8 \Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^4\] So, the electron configuration for Pd⁴⁺ is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\]