Concept: Transition Metals: Atomic Size9m
Hey guys. In this new video we're going to take a look at the trend of atomic size and how it relates to transition metals. So, recall in general as we move from left to right Across a row or period our atomic size decreases. So, a decrease is going from left to right across the period and now remember going down to any group or column it's going to increase, but now when it comes to transition metals, we're going to say, when moving across a period the size for the most part remains constant. Now, why exactly is that? if we take a look here at this chart, what this chart here is it's looking at the atomic number of the different types of transition metals. So, we have these transition metals are from the 3d row, here this would be 4d and this one would be 5d, right? So, what we see here is that there's at first a sharp drop in our atomic size as we move from the first element in each row to the second but then after that the difference in size gets smaller and smaller over time, the reason for this is we're adding electrons to one of the inner orbital shells and not the outer shells. So, for example, if you think of manganese, manganese is argon 4s2 3d5. So, if you think about it, here's our nucleus here, this is the first shell, the second, shell, the third shell and this is our fourth shell. So, what's happening here is that 4s and 3d shells have very, very similar energies but the way that this is written out, we have electrons, these 3d electrons are being added after the 4s has been added. So, what's going on here is that we're actually injecting these 3d electrons into the third shell. So, what's happening here is the outer shell isn't getting any bigger, we're adding them to inner shell and by adding them to the inner shell there, we're actually basically protecting these outer shell electrons, the two that are out there from the effects of the nucleus. Remember, that there's an attraction for the nucleus to the electrons, this is called effective nuclear charge, as we're adding more and more electrons to the inner shells there's going to be basically a decrease in our effective nuclear charge, we're going to be less attracted to that nucleus, which is going to cause our atomic size not to decrease as much as it should. Now here we're going to say, when moving from the 3d transition row to the 4d, we're going to see a sharp increase in atomic radius, because remember, as we head to, as we move down any group there's an increase in our size, let me take a look at a periodic table, we're going to say, we're going to say here that this is 3d here, this is 4d, this is 5d. So, as I moved from this row, the 3d row, and we move down to the 4d row, there's going to be a sharp increase in our atomic size, which is the trend we are used to see, as we go down any group there should be an increase in size, but what we're going to notice here is, as we go from 4d to 5d there's not an increase in size, in fact, their sizes are very similar to each other, very, very similar to each other. Now, we're going to say what's causing this to happen, we say, we call this the lanthanide contraction. Now, if we take a look, as we move to 5d, row 5d. Remember, what's in between 6s and 5d. Notice how the numbers go 56, 57 and then they skip and go to 72, that's because here we have our lanthanides, our lanthanides are involved. So, we're talking about this row here, the reason that there's not a big increase in size as there should be is, this is what's going on, as we move from 4d to 5d, what has to happen here is, the difference is that the 4d row doesn't deal with f orbitals at all, when we get to 5d we start to deal with f orbitals. So, what starts to happen here is, the f orbitals means that we're going to have an additional 14 electrons, an additional 14 electrons being crammed into the fourth shell. Now, if we're adding 14 additional electrons, what does that really mean? Well, that means that not only are we adding 14 additional electrons, we're adding 14 additional protons to the nucleus. So, your proton has just become even more positive, we've added so many more protons to that nucleus. Now, remember protons are positive, electrons are negative, because we have so many more protons within the nucleus they're going to be more attracted to the electrons around them. So, there's going to be basically an increase in the attraction that electrons have towards the nucleus. So, they're going to move closer and closer to the nucleus because of this effective nuclear charge. Remember, effective nuclear charge means an attraction for nucleus. So, what's happening here is we have two things competing, as we go down a group our size should increase as we add more electrons but as a result of effective nuclear charge, instead of increasing there's a decrease, and basically they just even out. So, there's a decrease in size because of effective nuclear charge, there should be an increase because we're adding more electrons, they both kind of cancel each other out. So, that's why, when we're going from 4d to 5d there's kind of a same size scenario here, both relatively are the same size 4d's and 5d's, and again it's because there should be an increase in size but effective nuclear charge is basically decreasing that increase, it's causing the size to remain relatively constant. Now, because of these similar sizes, if we take a look, here we have zirconium, you might be familiar with the term cubic zirconium, like fake diamonds. So, we have zirconium here and then we have hafnium here, look at their atomic sizes, they're almost right on top of each other, their sizes are incredibly similar, because their sizes are incredibly similar they're going to behave chemically similar to one another and if we keep moving you can see that these, two their sizes are not as close to each other but pretty close. So, they're very similar together as well chemically because their atomic sizes are about the same, but notice as you move from left to right Across the period. So, as we move this way in the 4d row and this way in the 5d row these things, these elements are going to become chemically more different from each other, okay? Just realize that initially zirconium and hafnium they're going to be very similar to each other in terms of size because of the lanthanide contraction but as we move from left to right their differences are going to become more and more apparent because as you move across the period atomic size does change even for transition metal it just doesn't change as greatly as it would for the main group elements. So, remember the general idea of atomic size or atomic radius, it decreases the going across a period left to right increases going down a group, when it comes to the transition metals, we have to take into account similar sizes because of the lanthanide contraction periods, as we move from left to right size does decrease but much more slowly in the transition metals. So, there's going to be a change in size but because it's much slower these transition metals will behave chemically similar to one another.
Concept: Transition Metals: Ionization Energy7m
Hey guys. In this new video we're going to take a look at the trend of ionization energy and how it relates to the transition metals. So, remember in general recall when it comes to ionization energy we're talking about the amount of energy required to remove an electron. Remember, as we move from left to right across a period or ionization energy is supposed to increase because if you look at a periodic table, as we move from left to right What's happening, we're here as metals, all of these are metals for the most part and as we move this way will becoming more like the nonmetals. Remember, metals want to lose electrons to become like noble gases. So, they want to give away their electrons more easily, but as we move from the left to the right side we're becoming more like the non metals, non-metals on the other hand want to gain electrons, meaning they don't want to give them away, which means more energy has to be invested in order to rip them off. So, as you move from left to right will become less metallic and more like the nonmetals, that's why ionization energy increases, also remember, as we move down any group, we become more metallic, you become larger and larger in size, the larger in size you become the farther away your electrons are from the nucleus therefore the easier it is for us to remove the electrons. So, as we move down any group, we're going to say our ionization energy decreases. Now this is the common trend when it comes to main group elements, but remember, with transition metals we're dealing with d orbitals and f orbitals. So, things are not exactly the way they should be in general terms, we're going to say here, however, when moving down a group, we think that the ionization energy of the third row is higher than the ionization energies of the first and the second rows. So, if we take a look, when we say the first row of the transition metals, we're talking about this row the 3 d row and as we talked about the second row, we're talking about 4d and this would be 5d. So, these are the rows that we're talking about. Now, normally as we move down a group your ionization energy should be decreasing because of their coming more like metals, which easily give away their electrons but here, 5d instead of having the lowest first ionization energy has the highest. Now, this opposite trend is the result of the elements atomic size. So, think of it like this atomic size, when it comes to atomic size we have different things playing a factor in determining the overall size. So, when we talk about overall size, we're talking about here's our nucleus, which is positive, right? With its protons then we have these electrons spinning around it. Remember, there's an attraction between the electrons and a nucleus, this is referred to as our effective nuclear charge, okay? So, the effective nuclear charge is basically the attraction the electrons have towards the positive nucleus, and remember, when we're moving from the 4d row to the 5d row there's a phenomenon called your like lanthanide contraction. So, remember lanthanide contraction, when we start talking about the 5d row. Remember, before we get to 5d we're talking about before f row. So, here it'd be 6s 2, then we talk about 4f 14 and then we start doing 5d, here when we include the four f orbitals, that's an addition of 14 additional electrons but more importantly, that's an addition of 14 additional protons in the nucleus.
So, now you have an injection of a lot more protons into the nucleus, meaning a nucleus becomes even more positive, meaning that there's a greater attraction for the electrons towards the nucleus, this attraction caused the electrons to move in closer to the nucleus. So, basically the size of the atom gets smaller. Remember, ionization energy is the energy to remove an electron, if my electrons are more attracted to my positive nucleus then it's going to become really hard for me to come in and rip one of them off and that's what's going on here, that's why 5-d actually has a higher first ionization energy than 3-d or 4-d because of the lanthanide contraction, the electrons are more attractive to the more tart on positive nucleus. So, if we take a look at this trend you can see that the third row for the most part much higher first ionization energies, here we don't see it happening with scandium but as we move up, well, not scandium, you see that not happening with that 5-d. So, here we're talking about our 5-d, we're talking about hafnium actually, okay? So, hafnium starts up here, but then look at how much the energy increases. So, 5-d higher first ionization energy and then there's a competition between our 3-d row and our 4-d row, just remember here, these two kind of follow the trend, for them the atomic size does increase, when you move down the group, right? So, that means that 4-d is larger than 3-d and since it's larger its electrons are farther away from the nucleus, they're not as attracted to the nucleus. So, it's easier for me to come in and rip off an electron, okay? So, that's why 4-d has a lower first ionization energy for the most part in 3-d. Remember, with the lanthanide contraction that's when we start talking about 5-d in terms of the row and that phenomenon causes it to have a higher first ionization energy, this is an important term that you need to remember in terms of transition metals, that's why they have variable charges, that's why they're chemically different from the other metals in the periodic table.
Concept: Transition Metals: Oxidation States8m
Hey guys. In this new video we're going to take a look at the different oxidative states that are allowed to transition metals. So, remember that when we talk about oxidation, oxidation is just simply removing electrons and we're removing electrons or creating different types of positive ions or cations. Now, one of most common features of transition metals is that they possess multiple oxidation states, meaning they have more than one positive charge. Remember, certain types of transition metals though only have one charge, for example, silver is always plus 1, cadmium is always plus 2, zinc is always plus 2. So, these are type 1 transition metals, meaning they have only one charge. Now, if we take a look here and there's a lot to look at. So, let me take myself out of the image, here we see the different types of oxidative States that are allowed for just the first row or 3d row for our transition metals. So, we're starting out with scandium all the way to zinc, here we have basically them in their natural state. So, when they're in their natural state they have no charge. So, their oxidation number is 0. So, all of them have a natural States all of them 0. Notice, here we have plus 1 charges for some of them, we have, what we have plus 3 charges plus 2 charges and notice here that, when it comes to these different types of charges, manganese by far has the most oxidation states allowed, meaning it can have any charge from 0 to plus 7. Now, what are some of things we should see, what are some of the trends that we should notice from here, what we should. Notice here is that the highest oxidation state of elements from groups 3b to 7b the highest one that you can possibly have is +7, okay? So, that's the highest charge you can have, going from 3B to 7B, okay? You should also. Notice that these types of oxidation states they're basically seen when we combine them with very electronegative elements, so the highest oxidation states that can have is all the way at the +7 and that's when they're connected to oxygen or fluorine these are the two most electronegative elements on the periodic table. Now, also notice here we have our 8v elements. So, remember 8b, okay? So, we have these three here, they're all considered 8b sometimes, we'll refer to this as 8B1 8B2 and 8B3 okay.
Notice that these here, they have fewer oxidation states than some of the others and the highest oxidation state is less common and it never equals the group number, okay? So, here the highest possible oxidation state plus 7 is not existent with these 8B groups and what we should notice here is what's the common trend that you should have noticed, look at 3b, 3b, what's its highest oxidation state plus 3, 4b what's it oxidation state plus 4, 5B is plus 5, 6B is plus 6 and 7B is plus 7. So, basically the trend is from groups 3b to 7b their up there group number equals their highest oxidation state, but that's not seen with 8B 1B or 2b, well, actually it is seen with 2B, 2B as well but it's not seen with the 8B's, okay? Also, what we should realize here is that what's the most common oxidation state from what we see in terms of this chart, the most common oxidation state is plus 2. So, plus 2 is the most common oxidation state for these transition metals but why is that because if we take a look, we'd say here we have argon 4s2, here this is 3d and we don't know, which number of electrons we're talking about because we're talking about all of them, right? We're talking about all 3d, why is their most common oxidation state plus two? that's because those two electrons come from the outer shell in this case the fourth shell, that's where we lose electrons. So, losing those two electrons from the outer shell gives us a charge of plus 2, that's why for the first row transition metals the most common oxidation state is plus2, other things that we should take from this is that when it comes to ionic bonding versus covalent bonding, what's the trend? the trend here is for ionic bonding, it happens when we have lower oxidation states, so the lower the positive charge, and then we're going to say covalent bonding, it is more common when we have higher oxidation state. So, basically what I mean by this is this, let's say I gave us titanium with chlorine, here titanium is plus 2 because the 2 came from titanium and here titanium is plus 4 because that's where the 4 came from, here this would be titanium 2 positive, that's a lower oxidation number. So, this here, represents an ionic solid here, on the other hand this is titanium 4 positive since it's a higher oxidation state and when we say higher oxidation state we're talking about greater than plus 3, this will represent a molecular compound, okay? So, that's the trend you need to take from here, this goes more in depth than things we've talked about in the past when it comes to oxidation numbers and oxidation states or focusing strictly on transition metals now. So, just remember, if you have an oxidation number of plus 1 to plus 3, that's more ionic bond, greater than plus 3, we say that's more covalent bonding, okay? So, yes some of these metals can represent covalent compounds even though traditionally we'd say that an ionic compound is a metal with a nonmetal, here we see exceptions to that. So, there's a lot that goes into these oxidation states. So, remember the trends groups 3B to 7b, their group number equals the highest oxidation state possible for those elements, we don't see that trend with the other groups, and realize that manganese is just in a special spot where it can have the greatest number of oxidation states also notice that we don't get to plus eight ever in terms of our oxidation number, here we go up to plus 7 and that's where we cap off, knowing these trends are essential when it comes to identifying what's been oxidized and what's been reduced and from there who's the reducing agent and who's the oxidizing agent, make sure you refresh yourself in terms of the notes that we talked about, in terms of redox reactions and try to apply them to these transition metals.
Concept: Transition Metals: Oxidation States (Example 1)3m
Hey guys. In this new video we're going to take a look at how do we calculate the oxidation number of certain transition metals based on what they're connected to. So, we take a look at this first example, it says, determine the oxidation number of the underlined element. So, we're looking for the oxidation state of nickel. Now here this a little bit different from what we're used to seeing, here we actually have nickel connected to six waters inside of brackets and then chlorines to the right of it. Now, remember since we're looking for nickel here, nickel will be x, then we have to ask ourselves what's the oxidation number of water. Now, just think of it in simple terms, water is a covalent compound and it has no charge, therefore it's oxidation number will be equal to 0, next, we have halogens here, we have chlorine. Now, remember chlorine here is connected to the nickel not to the oxygen and water and remember the rules that we've talked about in the past, group 7a elements, halogens, they're equal to negative one unless they're connected to oxygen, here they're not connected to the oxygen, they're basically closely related to the nickel therefore their oxidation number here is minus one, then we're going to use a, treated like a math problem, we're going to say here we have one nickel, which is x plus six waters, each one is 0, plus two chlorine could little to their, each one is minus one. Remember, our equation equals the charge of the molecule, here it has no charge, there's no positive charge or negative charge that we can see. So, its charge is 0, this drops out because it's just 0. So, x minus 2 equals 0, x equals plus two. So, nickel here would be plus two in terms of its charge, we've done this example here, I want you guys to attempt to you to the next one, see if you can figure out what it would be. Remember, tell yourself, this is a covalent compound, does it have a charge or not, that determines its oxidation state, from there treat it like a math problem, here we're looking for Cobalt, I want you guys to attempt to do this on your own, come back and see how I approach the question.
Concept: Transition Metals: Oxidation States (Example 2)2m
Hopefully you guys attempted to do on your own. Now, let's take a look at it together, here we have cobalt, which is x again, and here we have ammonia NH3 and we have water again? Well, we know water is 0? Well, remember ammonia is NH3, NH3 is neutral. So, it's oxidation number is 0, if I had given you ammonium ion, ammonium since it's 1+, its oxidation number would have been 1+ just remember, this is ammonia and this here is ammonium, okay? So, remember the difference. So, this would also be 0, these halogens here, each would be 1- treat it like a math problem x1 cobalt plus 4 ammonia each one is 0, plus 1 water, which is 0, plus 1 bromine, which is 1-, plus 2 more bromines, each one is 1- equals a charge of 0. So, here this dropped out, this drops out x minus 2 minus one minus 2 equals 0, x minus 3 equals 0. So, here x equals plus 3. So, that'd be the oxidation state of cobalt. Now, for this practice 1, I want you guys to try to remember what is going on in terms of this question, it's asking us which one has greater metallic behavior, okay? And remember I kind of talked about this when we talked about oxidation states. Remember, the small the oxidation state, what does that mean in terms of the types of bonds we form and the greater the oxidation state of the transition metal, what kind of bonds do we form, this is key to answering this question, attempt to do it on your own and then come back and see how I approach the same question, good luck guys.
Concept: Transition Metals: Oxidation States (Practice 1)2m
Hey guys, let's take a look at the practice question, which was asking us about metallic behavior. So, here recall that we talked about, we said that if you had a lower oxidative number then you would represent an ionic bond but if you had a larger ionic number you'd represent a covalent bond. So, here if we're talking about an ionic bond that would mean that the transition metal will behave as a metal or have metallic behavior. So, if you calculated this correctly, titanium here would have been +2 and titanium here would have been +6, here since it has a lower oxidation number it will behave as a metal so that we can have ionic bonding, here since its oxidation number is much larger it represents covalent bonding because it behave more like a nonmetal. So, here the answer would be titanium 4 fluoride, this titanium will exhibit more metallic behavior, so that was the heat answering this question. Now, that we've done this I want you guys to try to attempt to see if you can get the right answer for this one here, here what we should realize is, I'll give you guys a hint on how to approach it, just realize when it comes to oxides they become less basic as the oxidation state increases. So, that's the clue here, they become less basic as the oxidative state increases, think of that and approach this question and see which one is the better answer, okay? So, good luck guys.
Concept: Transition Metals: Oxidation States (Practice 2)4m
Hey guys. Let's attempt to do the practice question was left at the bottom of a page. So, here I say, which oxide forms a more acidic aqueous solution. So, what did we say, we say that these oxides become less basic as their oxidation states increase and so that's what we're talking about here. So, here if we wanted to figure out their oxidation state. So, if we look at this first one, we have CrO3, we'd say here, chromium we don't know it's oxidation number is, it's x, oxygen here is minus 2 then we say, here we have x plus 3 oxygens, each one is minus 2, equals in charge of my molecule, which is 0, x minus 6 equals 0. So, x here is plus 6. So, this will be chromium 6, then we'd say what? we'd say here, this is x and then this is minus 2, since the charge overall is neutral then this x must of course be plus 2. So, what do we just say, we say oxides become less basic as their oxidation states increase, less basic means more acidic. So, this first compound has a greater oxidation number, chromium plus 6 therefore it's more acidic. So, CrO3 would have been your answer for this question. So, remember all that's entailed in terms of oxidation numbers and what factors that play in identifying the bonds being either ionic or covalent and then determining if it creates a basic oxide, an acidic oxide or an amphoteric oxide, amphoteric oxides are a little bit trickier to determine. Remember, amphoteric means it behaves as an acid or a base, here since the number it's so large +6, we're going to say that it can behave more as an acid than as a base or an amphoteric, good example of something that can be amphoteric is if we look at titanium 4 oxide TiO2, that is amphoteric plus four, is mid range in terms of oxidation numbers, by mid range we're meaning around plus 3, plus 4, but this varies, if you want more specific types of amphoteric oxide just make sure in lecture as your professor is discussing, this they'll point out which transition metals behaving more amphotericly than others, but right now just realize as your oxidation number increase so does your acidity.
The members of which pair are most similar in color?
a) chlorine, Cl2(g), and Bromine, Br2(l)
b) silver chloride, AgCl(s), and silver sulfide, Ag2S(s)
c) copper (II) chloride, CuCl2(s), and calcium chloride, CaCl2(s)
d) potassium permanganate, KMnO4(aq), and iodine, I2(g)