We just did scandium. Next let's move on to titanium. Thinking about titanium, so the next element in the periodic table if your question on the test was write the electron configuration for titanium, the easiest way to do it is just once again to think about argon. Put argon in brackets and then think to yourself, this would be 4s 1, this would be 4s 2, this would be 3d 1 and this would be 3d 2.
You could write 4s 2 and then 3d 2 or once again you could switch 3d 2 and 4s 2. Once again this is implying the d orbitals fill after the 4s orbital which isn't true but it does get you the right answer. It's useful to think about it both ways. It's useful to think about the energy levels properly but the same time if your goal is to get the answer the fastest way possible, looking at the periodic table and running through the electron configuration might be the best way to do it on test.
Let's look at some of these other elements here so we've just talked about scandium and titanium. All right, so let's go down here. Let's look at this little setup here. All right, so we just did scandium and titanium. All right, so scandium was argon 4s 2, 3d 1. We talked about two electrons in the 4s orbital, one electron in the 3d orbital.
We just did titanium 4s 2, 3d 2 or once again you could switch any of these. When you're doing orbital notation, adding that second electron to a d orbital. Here's the electron that we added so we didn't pair up our spins. We're following Hund's rule here. Next element is vanadium so we do the same thing.
One more electron, we add that electron to a d orbital but we add it to, we don't add it to one of the ones that we've already started the fill here, we add that electron to another d orbital, so once again following Hund's rule. Things get weird when you get to chromium. Let me use a different color here for chromium. If you're just thinking about what might happen for chromium, chromium one more electron to think about than vanadium.
You might think, let's just add that one electron to a 3d orbital like that and then be done with it. If you think about it, you might guess 4s 2, 3d 4. Let's go ahead and write that.
We get 4s 1, 3d 5. That electron, this electron here, let me go ahead and use red. So you could think about this electron. We expect it to be there, we expect it to be 4s 2, 3d 4. It's like that electron has moved over here to this empty orbital to give you this orbital notation. All right, so that's just an easy way of thinking about it and in reality that's not what's happening if you're building up the atom here because of the different energy levels.
But just to make things easier when you're writing electron configurations, you can think about moving an electron from the 4s orbital over to the last empty d orbital here. Some people say that this half filled d subshell, let me go and circle it here. This half filled d subshell is extra stable and that might be true for the chromium atom but it's not always true so it's not really the best explanation.
The real explanation is extremely complicated and actually just way too much to get into for a general chemistry course. Next element is manganese. All right, let me go ahead and stick with blue here. Manganese, one more electron than chromium here. Chromium we had six electrons here, and manganese we need to worry about seven electrons.
This is kind of what we expect, just going across the periodic table. Let me go ahead and do this for manganese. Let me use green here.
You might say okay, that's 4s 1, that's 4s 2 and then 3d 1, 3d 2, 3d 3, 3d 4, 3d 5. You might say to yourself 4s 2, 3d 5. This precedes how we would expect it to. All right, and the same thing with iron, so 4s 2, 3d 6. Electron configurations can be written directly from the periodic table, without having to memorize the aufbau scheme, using the following patterns:.
Half-filled and filled subshells are especially stable, leading to some anomalous electron configurations:. In the case of chromium, an electron from the 4 s orbital moves into a 3 d orbital, allowing each of the five 3 d orbitals to have one electron, making a half-filled set of orbitals.
In the case of copper, silver and gold, an electron from the highest-occupied s orbital moves into the d orbitals, thus filling the d subshell. Many anomalous electron configurations occur in the heavier transition metals and inner transition metals, where the energy differences between the s , d , and f subshells is very small. Key: s-block p-block d-block f-block. The 3d sublevel is being filled in the transition metals in period 4.
There are five d-orbitals in any d sublevel, so there are a total of 10 electrons possible in a d sublevel 2 electrons per orbital. Chromium, nickel, and copper, all in period 4, have unexpected electron configurations from what you would expect from the Aufbau principle.
However, experimental evidence supports the unexpected configurations. They still have electrons in the 3d sublevel. The filling of which orbital is represented by the transition metals in period 4?
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