What is the electronic configuration of Mercury, and how many electrons does it have for $n = 3$, in the $4d$ atomic orbitals, in the $4p_z$ atomic orbital, and in the valence shell with a spin of $s = -1/2$?

Okay, let's break down the electronic configuration of Mercury (Hg, atomic number 80). The full electronic configuration for Mercury is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ Now, let's answer your questions: i. **Electrons with n = 3:** This includes all electrons in the 3s, 3p, and 3d subshells. 3s² + 3p⁶ + 3d¹⁰ = 2 + 6 + 10 = **18 electrons**. ii. **Electrons in 4d atomic orbitals:** From the configuration, the 4d subshell has **10 electrons**. iii. **Electrons in 4pz atomic orbital:** The 4p subshell (4p⁶) is completely filled. It consists of three orbitals: 4px, 4py, and 4pz. Since it's full, each of these orbitals contains **2 electrons**. iv. **Electrons in the valence shell with spin "up" (s = -1/2):** The valence shell for Mercury is the 6s subshell (6s²). It contains two electrons with opposite spins. One electron will have a spin of +1/2, and the other will have a spin of -1/2. Therefore, **1 electron** will have a spin of -1/2.