Jonathan Fong

**U.C.Berkeley**

M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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So for Titanium, it would be pretty easy. So you would write 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, and then 3d2 or if we did the short hand notation, it would be [Ar] then you'd have 4s²3d², because the Ar basically covers stuff through 3p6.

Now if you take Ti2+, the ion, basically you're taking away two electrons. Now where does it take away from? Well, it takes it away from the outermost shell. Where is the outermost shell? Well, it's in 4s4. It's the outermost energy level. So basically it would take away these 4s electrons first. So Transition metals lose s-electrons before d-electrons. So that would be good to know especially for these ions.

If we wrote this out, it would be 1s2, 2s2, 2p6, 3s2, 3p6. And since we're taking away two electrons, because it's 2+, we'll omit the 4s2 electrons, and we'll just write 3d2. So basically the 4s-electrons are taken away. So if you write it out short hand, it'll be Argon [Ar]3d2 showing that 4s2-electrons were taken away.

Now if it's Ti4+, now we've taken away two additional electrons compared to the Ti2+. So we would write out 1s2, 2s2, 2p6, 3s2, 3p6, and then the 3d2-electrons would be gone. So compared to the original Titanium atom, we gave away four electrons. So both the 4s2, and the 3d2 electrons would be gone. Then you can write [Ar] or you can write it as instead of [Ar], we do Neon, because that would be 2p6, and then 3s2, 3p6. So that would be for Titanium.

So we have three other Transition metals that we can take a look at. So for each of these I'll do the short-cut, but just to show. For Copper, remember Copper is one of those exceptions to the Electron Configurations, because normally you would write it 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, and 3d9.

What happens is 4s, and 3d are pretty close in energy to each other. So what happens is one of the 4s-electrons goes in, helps fill up the d orbitals. So 3d actually completes, and has 10 because that helps complete the third energy level. So we're only left with one valence electron or one electron in the fourth energy level for Copper. So if you wrote it out, it would be Argon [Ar]4s1, 3d10. So not 4s13d9 because this is an exception.

So then if we have Cu+, then basically Transition metals lose their s-electrons, so we would end up with [Ar]. Then of course one electron would go away, because we lose one electron, and it be 3d10.

Then for Cu2+ we basically would take away one more electron. So we would have 3d9, or one less electron. So it would take away one of the electrons for 4s, and one of the electrons from 3d, from the original atom. So that's for Copper.

For Chromium if you write it out 1s2, 2s2, 2p6, 3s2, 3p6. Then instead of 4s23d4, this ends up being 4s13d5, because is you notice in 3d, you have five orbitals. So let's draw the boxes out; 1, 2, 3, 4, 5 and then so we fill this up, now we have five orbitals instead of four, because we took one away from 4s-electrons. So basically 4s instead of having 2, we'll only have 1 because 3d borrowed one of them.

So if you write it out, then you're going to write out the shorthand Argon [Ar]4s13d5. Now you can do the other longhand ones yourself. I'll show you the shorthand ones just in case. Remember follow the rule, so Transition metals lose their s electrons before the d electrons, so it will take one from the 4s, and one from the 3d, because it's Cr2+. So basically, it loses two electrons. So that would be [Ar]3d4. Then this would be [Ar]3d³.

Then the last one is Vanadium. If you do Vanadium, then we end up with [Ar]4s23d3. Then the 2+ takes away 2 from the 4s, so we have [Ar]3d3. Then for Vanadium 3+, we have [Ar]3d2. So these last two I did just so you can double check your answer so you can have some practice just in case if you have some trouble. So hopefully this helps you in understanding Electronic Configurations for Transition metals. Have a good one.