The following diagrams represent the electron population of the composite s–d band for three metals—Ag, Mo, and Y:

Which diagram corresponds to which metal? (LO 12.7) (a) Ag = 3, Mo = 1, Y = 2 (b) Ag = 2, Mo = 1, Y = 3 (c) Ag = 2, Mo = 3, Y = 1 (d) Ag = 1, Mo = 2, Y = 3

Video transcript

Hello everyone today. We have the following problem. The image below shows the electron distribution in the composite SDI band of an unknown metal identified the most likely identity of the metal. So first thing we wanna do is we want to assume that the S. D. Band completely overlaps. And if it can the S. D. Band can accommodate According to this image, 12 valence electrons. This diagram however accommodates only 2468 10 valence electrons. So this diagram Displays only 10 of those valence electrons. And so we essentially have to find which of these metals has 10 valence electrons. So if we have zirconium here, if you look at the electron configuration of that, the ground or the noble gas configuration, we look at the noble gas that's directly before zirconium which is krypton. And then we have 4 D2, five S two. And we essentially just add up the exponents to give us the valence electrons. So in this case for zirconium we have four valence electrons. So a cannot bear an answer. If you look at B chromium, we look at the noble gas before that which is are gone. And then we fill out the rest of this electron configuration as present. So that's five D three D 54 S one And the exponents. We get six valence electrons still not 10. So B is incorrect. If you look at sea palladium, we look at the noble gas configuration which has crept on and then filling out the rest of this electron configuration, we have four D. 85 S. Two Counting up our valence electrons. We get 10 valence electrons, which is actually our answer. So our answer is going to be C palladium. And with that we've answered our question overall. I hope this helped, and until next time.