Write the ground-state electron configuration for the followin...

Question

Write the ground-state electron configuration for the following element:

Lead (Z = 82)

Accepted Answer

All right, so we have a really long one here. We have to write the ground state electron configuration for the following element lead, which has an atomic number of 82. Remember, when it's neutral, the number of protons and electrons are the same, so this would be equivalent to 82 electrons. So look at the periodic table. We start off with one s two to us too. Two p six, Then we go three s to three p six for us to now. We're going into the D block. Drops down by one three d. Counting that whole row is 10 four p six. Keep going. You still have a long ways to go five s to. Then we go for D 10. Got to keep going. Five p six six s to now. When we get the success to, we're gonna come into a situation where we're gonna both deal with the five D and also the four f orbital's. Alright, So after six s two, we can just go down to the four f row and just say four f 14. Then we could come back around to say that we have five Diaz, the whole room. Then we can say, here we have six p to. So this one is quite a long electron configuration. And with the heavier or higher numbered atomic number of elements, it takes a long time to write the whole thing out. But don't worry. We're gonna soon see how to write the electron configuration in, um, or condensed and faster way.

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