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Representing ionic solids using particulate models

The cations and anions in an ionic solid are arranged in a lattice structure that maximizes the attractive forces between opposite charges and minimizes the repulsive forces between like charges. The structure of an ionic solid can be represented using a particulate model. Created by Sal Khan.

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  • blobby green style avatar for user Michael Rugaard
    Hi - I don't understand, why the rubidium with its greater nucleus will end up being smaller than the rubidium?

    I'm thinking of what Sal says here:
    "So both of these have the same number of electrons, but rubidium has two more protons than bromide does. And so the rubidium is going to attract that outer shell of electrons, that fourth shell of electrons, more than the bromide nucleus is going to. And so, the rubidium in this example is going to be smaller than the bromide."

    Thanks for any help.
    (4 votes)
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    • leaf red style avatar for user Richard
      “why the rubidium with its greater nucleus will end up being smaller than the rubidium?”
      Sounds like one of those rubidiums should be a bromide. Also, what exactly do you mean by greater nucleus?

      Sal’s reasoning is correct. If bromine gains an electron to become an anion, the bromide ion, and if rubidium loses an electron to become a cation, they would both have the same electron configuration. This means they would have the same number of electrons occupying the same electron orbitals in each atom. The only difference then would be the number of protons in the nuclei of the atoms. Bromine has less protons than rubidium, so bromine’s nucleus exerts less of a pull on its valence electrons than rubidium’s nucleus does to its valence electrons. If there’s less force holding the nucleus and the valence electrons together, the valence electrons can exist at a greater distance from the nucleus. This results in a larger atomic radius and a larger ion for the bromide compared to the rubidium. The opposite effect occurs in rubidium which has more protons in its nucleus to pull the valence electrons closer into the nucleus resulting in a smaller atomic radius and smaller ion.

      Looking at the actual data, bromide has an ionic radius of 182 pm, while rubidium has an ionic radius of 166 pm. Bromide is larger than rubidium as expected.

      Hope that helps.
      (22 votes)
  • stelly green style avatar for user schnj0468
    Hi! In one of the "Structure of Ionic Solids" practice problems, they show a particulate model of SrO, where the Sr cations appear to be much smaller than the O anions. Why might this be the case when Sr appears to be so much farther down the periodic table?
    (6 votes)
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    • leafers tree style avatar for user jaredthefuturephysican
      "SrO" is just showing you the atoms that bond together, not the bond formation (negative and positive charges).

      In this case, the bond form would be Sr positive 2 and O minus 2, in which Sr loses 2 electrons, which reduces the size of the atom through the nuclear charge being not as divided between as many electrons and in this case, takes off the third shell of the Sr atom. Which causes a smaller radius.

      The S, on the other hand, has a negative charge of two, which means the negative charge from more electrons is split which causes a smaller radius.
      (5 votes)
  • hopper cool style avatar for user Iron Programming
    Howdy Richard,

    It's been a while but I'm still here studying. :)

    I have a question about the central atom in a molecule. I've been studying hybridization and molecular geometry in more detail and so I wanted to make a specificiation.

    People say that the hybridization of the central atom in a molecule determine its molecular geometry, but what about a molecule that doesn't have just one central atom.

    For example there are many molecules that don't really have any central molecule.

    Now I believe that the way you would do it is basically determine the hybridization for each individual "central" atom, that is, every time that is not on the outside.

    Then, we can determine the molecular geometry by basically arranging the molecule in a certain way according to the rule we've already learned.

    Does this seem right?

    Hope to hear from you soon!

    NOTE: I asked this question here because KA is have technical issues with their notifications working on threads.
    (3 votes)
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    • leaf red style avatar for user Richard
      Hey Iron, in case KA didn't give you the notification for my last answer, I'm just going to copy/paste it here.

      So yes you are correct in that the central atom's hybridization determines the overall molecule's shape. These shapes includes things like tetrahedral, trigonal planar, octahedral and a few other shapes covered in VSEPR theory. https://en.wikipedia.org/wiki/VSEPR_theory

      But most molecules do not possess easy to describe shapes since most molecules do not have a single central atom. Particularly in organic chemistry where we can have large atoms with no real center. Instead yes we determine the bond hybridizations around each atom and the shape it produces(excluding hydrogen since it only has a single s orbital and do not engage in hybridization).

      For example, a molecule like adrenaline (https://en.wikipedia.org/wiki/Adrenaline) has no central atom and has atoms with different bond hybridizations. The carbons in the hexagon ring are all sp2 and therefore have trigonal planar electron and physical geometries. The rest of the atoms are sp3 hybridized and each have sp3 orbitals, however they have different shapes depending on how many lone pairs they have. The oxygens will have tetrahedral electron geometry but bent physical geometry around them. The nitrogen will also have tetrahedral electron geometry but trigonal pyramidal physical geometry around it. So we can describe the shape around each individual atom, but the whole molecule has no easily described shape so it just doesn't have a name. Trying to name the shape is made more difficult by the fact that the atoms are rotating around on the single bonds moving into different conformations constantly.

      If you want to get a sense of how weird the shapes of molecules can get, I use this one website for something called group theory in my own work. Group theory is a type of geometry used in chemistry basically obsesses over the shape of molecules and how they can be used to explain bonding. You can play around with a bunch of examples using the drop down menus on the left. https://symotter.org/gallery

      But in conclusion, small single central atom molecules, we have names for the shapes they make. But larger molecules with multiple central atoms becomes more difficult to do so and often do not have names. The best we can do is identify the individual atom's hybridizations and shapes around each atom.

      Hope that helps.
      (8 votes)
  • purple pi teal style avatar for user kyleea06
    Does the atomic radius play a part on the element's size? I know Sal says bromine is larger than rubidium, but just looking at the periodic table I would think rubidium is. I learned that the radius gets smaller the further to the right you move across the periods. It's in unit one-'atomic radii' I think.
    (2 votes)
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    • leaf red style avatar for user Richard
      Your atomic radii trends are correct; however, they only apply to neutral atoms. Ions will have different sizes compared to their neutral versions. This is because as you add or remove electrons you’re changing the amount of negative charge in the electron’s orbit. So now electrons will feel either less or greater attraction to the protons in the nucleus and be able to orbit the nucleus at a farther or closer distance compared to a neutral atom, thereby altering the atomic radius. Additionally, the changing of electrons in a neutral atom can change where the outermost valance shell is. Electrons occupying a completely new shell will have a large impact on the size of the atom since electron shells are different distances from the nucleus.

      Looking at rubidium chloride, RbBr, we can see it is composed of rubidium cations, Rb^(+), and bromide anions, Br^(-).

      Rubidium loses an electron to become the cation and has an electron configuration identical to neutral krypton. This means that there is one less electron in the electron cloud causing repulsions to the other remaining electrons which means they feel a stronger attraction to the nucleus and can orbit at a closer distance to the nucleus thereby reducing the atom’s size. Additionally, the valance shell goes from the 4th shell to the 3rd shell. The 3rd electron shell is closer to the nucleus than the 4th shell so that also contributes to the rubidium cation’s reduced size compared to its neutral version.

      Bromide gains an electron to become the anion and also has an electron configuration identical to neutral krypton. This time the extra electron causes additional repulsions between the electrons weakening the attraction to the nucleus and therefore increasing the size of the ion compared to the neutral atom.

      These size differences are confirmed when examining their ionic radii. Rubidium has an ionic radius of 166 pm, while bromide has an ionic radius of 182 pm.

      Hope that helps.
      (8 votes)
  • blobby green style avatar for user ChloeW
    What type of force keeps the particles in ionic compounds together?
    (3 votes)
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    • leaf red style avatar for user Richard
      It’s just electrostatic attractions. Attractions between charged particles of different signs, positive and negative. Positively charged ions, cations, are naturally attracted to negatively charged ions, anions. That attraction between ions, called an ionic bond, is what hold ionic solids together.
      (5 votes)
  • blobby green style avatar for user Michael
    How would the structure look like if the ratio of cations to anions wasn't 1:1? For example, something like magnesium chloride (MgCl₂).
    (2 votes)
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    • leaf red style avatar for user Richard
      A particulate model depicts all ionic solids similarly. The only difference being the size of the ions. So a particulate model of sodium chloride (NaCl) or rubidium bromide (RbBr) with a 1:1 ratio, would be almost identical to magnesium chloride with a 1:2 ratio. They would be a 2D repeating pattern of + and – circles. The only difference being the relative sizes of the circles.

      The biggest issue with a particulate model is that its limited in what it can represent. Looking at only a particulate model, we would conclude that rubidium bromide and magnesium chloride are identical in structure. But of course we know that they both have different ratios of cations to ions because we know their chemical formulae. However, these ratios can’t be represented in the particulate models. Particulate models are also limited in that they are 2D depictions, while ionic solids true structures are 3D lattices. We need to employ more advanced models using 3D unit cells to better understand the structure of ionic compounds.

      Hope that helps.
      (4 votes)
  • blobby green style avatar for user mr.sashashapkin
    What's the difference between a regular ion structure and crystal structure? Like for example table solt is a white powder but it can under certin conditions form crystals
    (2 votes)
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    • leaf red style avatar for user Richard
      Solid ionic compounds naturally organize themselves into ordered, crystalline patterns. The empirical formula for any ionic compound is the smallest unit which repeats in an ionic solid. Each ionic solid adopts a slightly different 3-D crystalline structure in order to minimize lattice energy.

      Sodium chloride, common table salt, is a repeating pattern of sodium cations and chloride anions. The actual crystalline pattern of sodium chloride is body-centered cubic. Each repeating unit cell is a cube where every sodium cation is surrounded by six chloride anions, and each chloride anion is surrounded by six sodium cations (still in a 1:1 ratio from the empirical formula NaCl). This means a sodium chloride crystal grows as a cube as more sodium and chloride ions are added. The mineral form of sodium chloride is halite (which is where the group name halogens comes form). If you have a halite crystal grow equally on all six sides, then it grows into a perfect cube of sodium chloride. In reality most times halite crystals grow at slightly different rates around the edges of the cube or merge when two cube crystals grow into each other, but you can still make out the smooth cube faces.

      When used in everyday life, sodium chloride doesn’t come as nice huge cubes, but rather have been broken up into small grains. Despite that, at the very small level it’s still a repeating pattern of sodium and chloride ions so it still retains that cubic crystalline pattern with even a few formula units. If you’re curious you can observe the cubic structure of sodium chloride, even the small grains, under a microscope. So regardless whether we observe ionic solids at the microscopic or macroscopic level, they maintain their crystalline structure.

      Hope that helps.
      (4 votes)
  • primosaur sapling style avatar for user reyansh_gupta
    What keeps the ions together?
    (2 votes)
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  • blobby green style avatar for user ebattallio
    The video describing atomic Radii Trends mentions that as you move down a group, say group 1, the radius increases. And as you move left to right in a period the radius decreases. So how is Br (group 17, period 4) larger than Rb (group 1, period 5)
    (2 votes)
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    • leaf red style avatar for user Richard
      Your atomic radii trends are correct; however, they only apply to neutral atoms. Ions will have different sizes compared to their neutral versions. This is because as you add or remove electrons you’re changing the amount of negative charge in the electron’s orbit. So now electrons will feel either less or greater attraction to the protons in the nucleus and be able to orbit the nucleus at a farther or closer distance compared to a neutral atom, thereby altering the atomic radius. Additionally, the changing of electrons in a neutral atom can change where the outermost valance shell is. Electrons occupying a completely new shell will have a large impact on the size of the atom since electron shells are different distances from the nucleus.

      Looking at rubidium chloride, RbBr, we can see it is composed of rubidium cations, Rb^(+), and bromide anions, Br^(-).

      Rubidium loses an electron to become the cation and has an electron configuration identical to neutral krypton. This means that there is one less electron in the electron cloud causing repulsions to the other remaining electrons which means they feel a stronger attraction to the nucleus and can orbit at a closer distance to the nucleus thereby reducing the atom’s size. Additionally, the valance shell goes from the 4th shell to the 3rd shell. The 3rd electron shell is closer to the nucleus than the 4th shell so that also contributes to the rubidium cation’s reduced size compared to its neutral version.

      Bromide gains an electron to become the anion and also has an electron configuration identical to neutral krypton. This time the extra electron causes additional repulsions between the electrons weakening the attraction to the nucleus and therefore increasing the size of the ion compared to the neutral atom.

      These size differences are confirmed when examining their ionic radii. Rubidium has an ionic radius of 166 pm, while bromide has an ionic radius of 182 pm.

      Hope that helps.
      (3 votes)
  • primosaur ultimate style avatar for user William Crye
    Are the bonds always perfect like a checker-board pattern? what if they are not? what difference would it make?
    (3 votes)
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    • sneak peak green style avatar for user Equilibrium
      In most cases, the bonds do not look like a checkerboard pattern due to impurities disrupting the crystal structure. For example, high carbon steel does not pack the carbon in with the iron in such a pattern because there is a tighter way to put the material together. In geology, crystals are classified by their shape (i.e. hexagonal, cubic, etc). These differences in structure are caused by multiple elements trying to stack in the tightest possible form.
      (1 vote)

Video transcript

- [Instructor] In this video, we're gonna think about how ions will arrange themselves when they form solid crystals. When they form these lattice structures. So just in very broad brush terms, let's say that we have a bunch of this white cation, and we have a bunch of this green, or this blue-green anion. So let's say they're in a one-to-one ratio. How will that look? How will the solid look, if you were to take a two-dimensional slice of it? To imagine that, we can draw what we could call particulate models. We're just imagining a two-dimensional slice of the solid, and we're just drawing these ions as particles. Would it look something like this, where maybe the positive ion is all on one side, and then the negative ion is on the other side, is on the bottom if we were to take a slice? Would something like this make sense? Or maybe it's random. Maybe you have a positive there, and then you have some negatives right over there. And then, maybe you have a positive and a positive, and then a positive right over there. And then maybe you have some negatives right over there. Would this be a reasonable configuration, as they form these ionic bonds? Well, when we think about Coulomb forces, we know that like charges repel each other and unlike charges, or opposite charges, attract each other. And so, when these ionic solids form, they're unlikely to form in this way, or even in this way, because they're gonna form in a way that maximizes the attractive forces and minimizes the repulsive, the repelling forces. And so what would be an arrangement that does that? Pause this video and think about it. Well, all the positive charges are gonna try to get as close as possible to the negative charges and as far as possible from other positive charges. And the same thing is going to be true of negative charges. They're gonna try to get as far away from other negative charges as possible, and as close to other positive charges as possible. So the arrangement that you are likely to see is going to look something more like a checkerboard pattern. So it may be a positive there, a positive there, a positive there, a positive there, and a positive there. These are all the same ion, I'm not drawing it perfectly, they'd be the same size. And when you do these two-dimensional representations, these particulate models, it is important to get the size right, 'cause we're gonna think about that in a second. And then the negative charges would be in between. So notice. In this configuration, every negative is surrounded by positives, and every positive is surrounded by negatives. So it's maximizing the attractive forces and it's minimizing the repulsive forces. And if you were to think about it in three dimensions, you would have a lattice structure that looks something like that. And we have seen this in other videos. Now another interesting thing to think about is the size of the ions that form that ionic solid. Let's say we wanted to deal with rubidium bromide. Rubidium bromide. What would this look like if I were to draw it in a two-dimensional particulate model like this, and I wanted to make the size roughly comparable to what we would see between the rubidium and the bromide? Pause this video and think about that, and I'll give you a little bit of a hint. It might be useful to look at this periodic table of elements. All right, if we were to separate this out into its ions, it is a rubidium cation, and a bromide anion. Now a rubidium cation, it has lost an electron. So even though it still has 37 protons, its electron configuration now looks like that of krypton. Now, the bromide anion, even though it only has 35 protons, it's going to gain an electron to become a bromide anion, and it also has an electron configuration of krypton. So both of these have the same number of electrons, but rubidium has two more protons than bromide does. And so the rubidium is going to attract that outer shell of electrons, that fourth shell of electrons, more than the bromide nucleus is going to. And so, the rubidium in this example is going to be smaller than the bromide. And so if I were to draw one of these diagrams, it would look something like this. Let me draw the bromide first. So I have a bromide anion, I have another bromide anion, another bromide anion, maybe I have a bromide anion right over here, bromide anion over there, maybe a few more. Make 'em a little bit, if I was doing this with a computer, I would make them all the same size. So these are our bromide anions. And then your rubidium cations would be a bit smaller. And so, our particulate model right over here might look something like this. We wanna make it clear that the cation is a bit smaller than the anion. It would arrange, it would likely arrange in a pattern that looks like this. And notice, I am trying to make the sizes roughly accurate, to show that the cation is indeed smaller than the anion. Although it wouldn't be dramatically smaller. Remember, they have the same number of electrons. And they don't have that dramatically different number of protons. And this is just a very rough drawing. If they were dramatically different, you might show that in the sizes on this diagram.