Electron Configurations of Transition Metals: Exceptions
24. Transition Metals and Coordination Compounds
Electron Configurations of Transition Metals: Exceptions - Video Tutorials & Practice Problems
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concept
Exceptions (I)
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In this video, we're gonna take a look at the first type of exceptions that exist for certain transition metals. Here we're going to say starting from chromium, so chromium is right here, as the atomic number z increases, exceptions to electron configurations can be observed. So our memory tool here is that Chromium has an atomic number of 24, and there are 24. So we have these two elements that have this type of exception, exception 1, and then these 4 here. So to get the other column, remember, 2, skip, next 4. Alright. So here, we're gonna say 2. So we're starting on chromium, and then we skip the next 4. And then here goes our group of 4. Now all of these elements have our shell number n, and all of them have one electron in their s orbital. So this will be n s to the 1. If we take a look here, we're gonna say an s orbital electron can be promoted to create half filled orbitals with d 4 and d nine elements. So if we were to look at the electron configurations of these, these would be d 4 initially, and these would be d 9 initially before the promotion of an electron from the s orbital. So, for example, if we took a look at chromium, chromium has an atomic number of 24. We would think that its electron configuration is argon 4s23d4. But, again, since it's an exception 1 type transition metal, one of the electrons from the s orbital would get promoted over to our d. So now its electron orbital diagram would look like 1 in the 4s and 5 electrons in the 3 d orbitals, giving us a new electron configuration of argon for s 13d5. So just remember, when it comes to these elements in particular, they fit under exception 1. They are d 4 and d 9 elements, but because of the promotion of an electron from the s orbital, we change the d fours into d 5, as we see here, and d nines would become d tens. Alright. So just remember these particular elements under this first type of exception.
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example
Electron Configurations Of Transition Metals: Exceptions Example
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Here it says write the condensed electron configuration of the following ion. So here we have to do it for the silver ion. Its atomic number is 47. Neutral silver, if we look at the periodic table, would look like it's initially Krypton 5 s 24d9. But remember, it's one of the elements that exist under exception 1. We would take an electron from an d s orbital and promote it to the d. So the real electron configuration of neutral silver would be Krypton 5 s 14d10. So we're looking at this as the real electron configuration of neutral silver. Now plus 1 means we've lost 1 electron, which is gonna come from the highest shell number. Here, this one electron is in the 5th shell, and these 10 are on the 4th shell. The one electron we need to lose would come from the 5th shell. So here, its electron con its condensed electron configuration would be krypton 4 d 10. And the way it would look is it'd be Krypton. Our 5 s is gone now, so that orbital would not be filled in. These 5 will represent our 4 d orbitals and be up, up, up, up, up, and then come back around down, down, down, down, down. So here we have our condensed electron configuration and our electron orbital diagram.
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Problem
Problem
Provide a condensed electron configuration for Au atom and Au (III) ion.
A
Au: [Xe]6s24f145d9 Au3+: [Xe]6s24f145d6
B
Au: [Xe]6s24f145d10 Au3+: [Xe]4f145d9
C
Au: [Xe]6s14f145d10 Au3+: [Xe]4f145d8
D
Au: [Xe]6s14f145d10 Au3+: [Xe]6s14f145d7
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concept
Exceptions (II)
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2m
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In this video, we take a look at exception type twos that we have. We're gonna say additional exceptions to the electron configuration are observed in period 5 or row 5 of the periodic table. And a memory tool to help us recall which one they are is 58, the number 8, b. So if we take a look here, we're dealing with 5, so group 5 b, and 88, so 8 b. So we're looking at these 4 elements in particular. Now we're gonna say all are gonna have 1 electron in their 5 s orbital, but notice that palladium is in a different color. That's because palladium is different from the other 3. So it has its own exception on top of the others. For palladium, we're gonna say palladium has 0 electrons in its 5 s orbital. Now if we take a look here, we're gonna say that for these elements in particular, the 5 s orbital electron or electron is are promoted to a 4 d orbital. N b here is Niobium. Its atomic number is 41. Its initial electron configuration looks like it would be Krypton 5 s 243. But, again, it's one of the ones in blue, and they must have one electron in their s orbital. So that would mean that we promote this electron to our d orbital. So the new electron configuration would actually be 5 s 1, 4d4. Filling up our electron orbital diagram, we'd have 1 electron in the 5 s orbital. And then following Hunt's rule, we'd have Phil to put in our 4 electrons that are in our 4 d orbitals. So this would represent the real electron configuration of Niobium. So just remember, these are exceptions too for these particular transition metals.
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example
Electron Configurations Of Transition Metals: Exceptions Example
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Here it says, provide a condensed electron configuration for ruthenium atom. So atom means its neutral form. If we look at the periodic table, it would look like ruthenium is krypton 5 s 2 4d6. But remember, it's one of our exception twos. In it, we can only have 1 electron in our s orbital. So one of these electrons gets promoted to our 4 d orbital. So the real electron configuration of ruthenium is krypton 5 s 14d7. This will be our final answer for the condensed electron configuration of this particular transition metal atom.
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Problem
Problem
Given the following electron configuration, determine the identity of ion with +2 charge.
A
Pd2+
B
Ru2+
C
Pb2+
D
Cd2+
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