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 cations 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 pass before we get to titanium is argon. So we'd start out with argon. Then we'd say 4s1, 4s2. So there's 2 electrons. And remember, we show those 2 electrons in this orbital. And according to the Pauli exclusion principle, they have to have opposite spins. One spins up, one spins down. Then we'd say that titanium is in our 3 d row, and it'd be 1 2. Hund's rule says that electrons that are in the same type of orbitals have same energy or degenerate, So we half fill first. So we go up, up. So if we wanted to translate this into simpler term, we just say it's argon 4 s 2 3 d 2. This would be the condensed electron configuration of titanium atoms. Now here we're looking at titanium 3. 3 plus means it's lost 3 electrons. Remember we take electrons from the highest n value. 4 s, 4 means that these electrons are in the 4th shell or n equals 4. Here, this means n equals 3. We have to lose 3 electrons total because the charge is 3 plus. So the first two electrons are lost from our 4 s orbitals. So we lose these 2 electrons, we have 0 left here. And then we still need to lose one additional electron, and that'll come from our 3 d, leaving us with one left. So if we wanted to fill out this electron orbital diagram, we'd still have argon. 4 s is empty, and then 3 d only has one electron left. So here we could say it's argon 4s03d1, or you can write it as argon3d1 because there's no electrons left in the 4 s orbit. So you can just ignore it. Right? So just remember, this is how we'd approach writing the condensed electron configuration of a transition metal atom and its respective cation.
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example
Electron Configurations Of Transition Metals Example
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53s
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Provide a condensed electron configuration for vanadium 3 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 argon4s23d3. Now with vanadium 3 ion, we've lost 3 electrons. Remember the first electrons come from the highest shell number, which would be electrons from the 4 s orbitals because they're in the 4th shell. So we'd lose 2 from there, so those are gone now, and we'd have to lose one additional electron because we need to lose 3. It would have to come from 3 d. So what we have left at the end is argon 3d2. This would be the condensed electron configuration of the vanadium 3 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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