Electron Configurations of Transition Metals - Video Tutorials & Practice Problems
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1
concept
Transition Metals
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3m
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When it comes to the transition metals. Remember that they occupy the D block of the periodic table or the group B elements of the periodic table for CS where we're losing electrons, electrons are lost from the highest shell number. First. Remember your shell number uses the principal quantum number. N. If we take a look here, remember we have our S block, we have our P block, we have our D block and down here we have our F block. If we're to take a look at Titanium, here goes titanium right here. And we've talked about condensed electron configurations before in earlier chapters. If you don't recall that, make sure you go take a look at our videos on condensed electron configurations. Right now, let's look at titanium. We would say that Titanium has an atomic number of 22 which means it has 22 electrons when it's neutral. The last noble gas we passed before we get to titanium is argon. So we start out with argon, then we'd say four S 14 S two. So there's two electrons. And remember we show those two electrons in this orbital and according to the poly exclusion principle, they have to have opposite spins, one spins up, one spins down. Then we'd say that titanium is in our 3d row and it would be 12 hans rule says that electrons that are in the same type of orbitals have same energy or are degenerate. So we have the first. So we go up, oh So if we wanted to translate this into simpler term, we just say it's argon for S 2 3d 2, this would be the condensed electronic configuration of titanium atoms. Now, here we're looking at titanium 33 plus means it's lost three electrons. Remember we take electrons from the highest end value 44 means that these electrons are in the fourth show or N equals four. Here, this means N equals three, we have to lose three electrons total because the charge is three plus. So the first two electrons are lost from our four S orbitals. So we lose these two electrons. So we have zero left here and then we still need to lose one additional electron and that will come from our 3d leaving us with one left. So if we wanted to fill out this electron orbital diagram, we'd still have Argon four S is empty and then 3D only has one electron left. So here we could say it's argon for 0 3d 1 or you can write it as argon 3d 1 because there's no electrons left in the four S orbit. So you can just ignore it, right. So just remember this is how we would approach writing the condensed electron configuration of a transition metal atom and its respective cion.
2
example
Electron Configurations Of Transition Metals Example
Video duration:
53s
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Provide a condensed electron configuration for vanadium three ion. So what we should do first is write the electron configuration of neutral vanadium. So vanadium atom, if we look at the periodic table would be argon four S 2, 3d 3. Now, with vanadium three ion, we've lost three electrons. Remember the first electrons come from the highest shell number, which would be electrons from the four S orbitals because they're in the fourth shell. So we'd lose two from there. So those are gone now and we'd have to lose one additional electron because we need to lose three. It would have to come from 3d. So we have left at the end is argon 3d 2. This would be the condensed electron configuration of the vanadium three ion.
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Problem
Problem
Write an electron configuration for ion of Tungsten (IV).
A
[Xe]6s24f145d4
B
[Xe]6s24f14
C
[Xe]6s24f145d8
D
[Xe]4f145d2
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Problem
Problem
Determine electron configuration for the ion of Cd in cadmium sulfide (CdS) compound.
A
[Kr]5s24d10
B
[Kr]4d10
C
[Kr]5s24d8
D
[Kr]5s24d105p2
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