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Ch.9 - Periodic Properties of the Elements

Chapter 9, Problem 48

Use the periodic table to determine each quantity.

b. the number of 3d electrons in Cr

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Welcome back, everyone looking at its position in the periodic table, determine the number of five D electrons present in gold. Now, in this problem, what we're going to do is understand that gold is located in period six of the periodic table and belongs to the ninth column of the D block, right? So essentially, if we start writing the configuration of gold, we are going to look at the previous noble gas in the periodic table, which is xenon, right, we can use the noble gas configuration and then we are just going to fill the S electrons since we are going to have six S and the S shell can hold up to two electrons. So we're just going with six S two based on period number six followed by the F electrons. Let's recall that in period number six, we're dealing with the four F sub shell. So we're going to use four F and this F is completely filled, which has a total of 14 electrons followed by the five D. Right? Let's recall that in period number six, we have ad orbital that corresponds to five D. In other words, one unit lower relative to the S orbital or the period we are in. So that'd be five D. And based on the fact that it is in column number nine, we have a total of five D nine electrons. So we would expect to have a total of 90 electrons. However, we have to recall the stability trend of the D orbitals. We have a total of five D orbitals. And let's suppose that we have nine electrons. So if we draw these electrons based on the hunts rule, we have 123456789, let's notice that there is only one electron left to completely fill the, the orbital and the stability of an element highly increases if we manage to get that orbital completely filled. So what happens is that we're essentially taking one electron from the S orbital and we completely fill our five D orbital to increase stability. So the correct configuration would actually be zenon six S two 4 F-14. Now we are decreasing our s from 2 to 1 because we are essentially, let's say borrowing the electron to fill our D orbital, right? And then we end up with five D 10, which highly increases the stability based on the fact that the D orbital is now completely filled. And therefore the number of the five D electrons present in gold would be some. That's our final answer. Thank you for watching.