Writing Electron Configurations for Cations

In this example problem, we write electron configurations for three different cations. Our problem reads: "Write the electron configurations for lead 4+, iron 3+, and silver+." We'll start with our periodic table. We can assign our different blocks as follows: - The S-block is our first two columns, including helium. - The P-block is the last six columns. - The D-block is our transition metals. - The F-block is our inner transition metals. According to our rules for quantum numbers, the S-block begins when N equals 1, and we can continue numbering down for the rows. The P-block begins when N equals 2, and we can continue numbering down. The D-block begins when N equals 3, and we continue numbering. The F-block begins when N equals 4. Now we can find the electron configuration for lead and adjust it to create lead 4+. We'll write the shorthand electron configuration. The noble gas that precedes lead is xenon. After that, we fill our S-block with cesium and barium, which is 6s2. Then, we move through our F-block so that it's 4F14. To go to mercury, we transition to the D-block, which is 5D10. Lastly, we have two additional electrons in the 6P subshell. Now we have the electron configuration for lead, and we need to adjust it for lead 4+. When creating the electron configuration for a cation from the electron configuration for the neutral atom, we remove the number of electrons according to the charge from the highest principle quantum number. The highest principle quantum number with electrons is N equals 6. So, we first remove the two electrons from the 6P subshell. Next, we have to remove the electrons from the 6S subshell since 6 is our highest principle quantum number. Notice that we do not remove electrons from the 5D subshell, which would be the reverse of the order in which they're filled. We have to remove them from the 6S subshell. What remains is the electron configuration for lead 4+ which is xenon 4F14 5D10. Next, let's write the electron configuration for iron. The noble gas that precedes iron is argon, and then we begin filling the 4A subshell with potassium to calcium, so it's 4S2. We then have six electrons to move over to iron in the 3D subshell. Thus, iron's electron configuration is argon 4S2 3D6. To adjust it for iron 3+, we need to remove three electrons, and we remove them from the highest principal quantum number. So, we remove two electrons first from the 4S subshell, and we still need to remove one more electron, which we remove from the 3D subshell. What we're left with for iron 3+ is argon 3D5. Finally, let's write the electron configuration for silver. The noble gas that precedes silver is krypton. The next electrons go into the 5S subshell, and then there are nine additional electrons to get over to silver. We may think that this is the correct electron configuration for silver. However, one of the electrons from the 5S subshell actually makes it into the 4D subshell to fully fill our 4D. This is one of our exceptions when filling our transition metals. Silver has one electron in the 5S subshell and ten electrons in the 4D subshell because this is a lower energy orientation for our electrons in our orbitals. Now that we have the correct electron configuration for silver, we can create the electron configuration for silver+. We need to remove one electron, and it gets removed from the highest principal quantum number, which is the 5S subshell. What remains is krypton 4D10.