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27. Transition Metals

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The 18 and 16 Electron Rule

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In this video, we're gonna take a look at the 18 and 16 electron role. Now, remember main group elements want to follow the octet rule because that gives them the ideal number of electrons just like a noble gas. Now, main group elements remember our groups one A 28 A and if we look there are eight slots, so 12345678. That's what that's what explains why eight is the ideal number. In terms of the octet rule, exceptions to this are hydrogen and helium. Helium only has two valence electrons involved with it. Hydrogen only needs to gain one more to have the configuration like helium. Now we're gonna say when it comes to transition metals though transition metal chemistry we use the 18 and 16 electron rule as an indicator for the reactivity of a transition metal. So we're gonna say the most stable transition metal complexes in several cases have electron counts of 18. Okay, so 18 electrons. This trend because we're dealing with 18 electrons is called the 18 electron rule. How do we come up with this number of 18? Well, the most stable number of electrons represent the number of total SP and D electrons together. Remember your ass have to d have 10 electrons max they can hold and P has six. So when we add those up together, two plus 10 plus six, that gives us 18. So that's where this 18 electron rule comes from. And remember when it comes to these transition metals they have number of valence electrons based on the number of S and D orbital electrons. So from three All the way up to 12. Now there are exceptions. Uh 18 for transition metal is great for it to get to a noble gas configuration by incorporating its d orbital's. But there are exceptions to this. We're gonna say exceptions often happen with transition metals from 8 to 11 valence electron groups. So we're talking about the transition metals within this sector and below. Now we're gonna say the tendency of these medals to be happy with 16 electrons is called the 16 electron rule. Now for a lot of the transition metal catalyzed reactions that we're going to see in organic chemistry, the two most common transition metals that do this. 16 electron rule are palladium and nickel. So they could follow the 16 electron rule of course all transition metals would love to get 18 electrons because that would be the best configuration to become more like a noble gas. But palladium and nickel especially are also okay with 16. So we take this into consideration when thinking about transition metals. Whereas main group elements are trying to get eight for the octet rule, transition metals in terms of stability are always aiming for 18 and a few cases where 16 is okay Now that now that we've talked about the 18 and 16 electron role for transition metals will move on to examples in terms of electron counts. So just remember transition metals ideally want 18 electrons. Whereas main group metals, we're used to seeing them ideally wanting eight electrons.

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The 18 and 16 Electron Rule Example 1

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So here the example states, what is the electron count of the neutral transition metal complex of B R two? P D C 02. Alright, so first we have palladium as our transition metal, we have to give its electron configuration to determine the number of valence electrons it possesses. So remember, palladium has a unique electron configuration. It's just krypton four D 10. It has no s orbital electrons. Those two are promoted in order to completely fill up our d orbital's. So we have 10 d orbital electrons. So the valence electron count for palladium is 10 minus the charge of our complex. Here. It has no charge, we're told it's neutral, so it's zero plus the number of X Liggins, which in this case would be the two bro means Plus two times are elegans. Here we have carbon neal as our two elegans. So that's two times two. So here this comes out to 10 plus two plus four. So in this case, palladium is following a 16 electron rule. Now that we've seen, this one will move on in the next video to the second example. This one may not be what you think it is? Think of what kind of Ligon is? R E N. What is that abbreviation for? That's gonna play a role in terms of calculating our electron count for this particular complex. So come back and see how I approach. Example too

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The 18 and 16 Electron Rule Example 2

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So this one is not exactly what you may think it is. So here it says, what is the electron account of the neutral transition metal complex of palladium? E N two. So remember E N is an abbreviation for ethylene dia mean, remember ethylene di amine is NH two CH two CH 2 and then another NH two. Here we have two locations for a lone pair. This is a by Dent ated Ligon. So we're gonna figure out what our number of valence electrons are for palladium. So it's still 10. That's not gonna change minus the charge, which is zero. Ethylene Damian is not an X type ligand. It's an L type Ligon. So X. Here is zero plus two times the number of L type Liggins. Now for the formula two times L type Liggins is based on the number of of lone pairs or electron donating sites on the L type Ligon. But this is a by dented Ligon. It has two locations that can donate electrons. So basically we have to buy dented Liggins. So that's equivalent to four L type Liggins Because there are four four sites total which we can have the donation of electrons through lone pairs. So this is actually two times four. So that's 18 for the electron count. So, again, going back to the original formula for electron count in the part of the equation where it's two times l type Ligon. That's based on the number of electron donating sites. So for Mono, dent ated, it would just uh still stay as one but for by dented, it be double double that. So that's why we have 18 as our electron count here. So just keep that in mind when figuring out the electronic count for any transition metal complex.

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Problem

How many NH_{3} ligands would need to be placed onto a nickel atom if we assume the resulting complex follows the 18-electron rule?

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Problem

Using the 18-electron rule, explain why V(CO)_{6} can be easily reduced to [V(CO)_{6}]^{–}.

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