Formation of Ions – Quantum Numbers

by Adam Le Gresley, PhD

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    00:02 Atoms of other elements who do not have completed outer shells would like to achieve this in order to achieve a lower energetic state. And they can do this either by swapping or by sharing electrons. And this is the origins for ionic or covalent bonding, as we will see in subsequent lectures. For the first 18 elements, a filled valence shell usually consists of eight electrons. That's obviously because you can only complete—you can only have—eight electrons in the outer shell of the second principle quantum number. It's only when you go up to the third shell—the third principle quantum number—it's possible to put 18 electrons in. Elements with higher atomic numbers require 18. Elements in groups 1, 2, and 3 on the periodic table—so that from the extract I showed you at the beginning—will tend to lose electrons. Why? Well, because they usually only have 1, 2, or 3 electrons in the outer shell. It's easier to lose 1, 2, or 3 than to gain—in this case—7, 6, or 5. Any elements which find themselves in group 7 or group 6 will tend to want to gain electrons in order to complete their outer shell. Again, the reason for this is simple: because it's easier for them to acquire two electrons than it is for them to lose, in the case of fluorine, 7, and in the case of oxygen, 6. Now you may say, "Well, what do you mean it is easier, or it's simpler?" But the reality is, as you start stripping away electrons, you, of course, are increasing the size of your positive charge. So if, for example, you move a single electron from an outer shell of, let's say, lithium, and you form an Li+, that's relatively straightforward. But if you were to then… Let's say, for the case of aluminum, you remove one electron; you've created a disproportionality between the charges. You now have fewer electrons than you have protons. This increased nuclear charge density technically pulls electron density inwards and creates an ion, which is small. Moving an electron away from that shell again—so in other words, a second electron—becomes more challenging, because now you have a distribution of force, a distribution of electrostatic attraction, which is greater than before. So subsequent losses become more and more difficult to achieve. In the case of acquiring electrons—in the case of group 1 or group 2—of course, this means adding greater electron negativity, a greater electron density, onto those atoms, and electrons will repel each other. This, of course, creates a energetically unstable system. So let's consider the formation of ions in their simplest form. Let's look at lithium. That's a group 1 or alkali metal on the periodic table, and we can see, or we should note from its electron configuration, possessing three electrons, the first two go in the 1s—give us 1s2— and the third one goes into the 2s, which gives us 2s1. The easiest way that lithium can form a full outer shell is to lose a single electron. And indeed, that's how it exists. You tend to find that lithium itself is very reactive, because it tends to lose its electrons very quickly. In fact, it is one of the most electropositive elements in the periodic table. And so, in the case of our ion, lithium, which has been formed, it will now have the formal electron configuration of 1s2, because now that electron from the 2s2… 2s1 orbital has been lost and now has the same electron configuration as helium, which, as we discussed earlier, is stable by virtue of the fact it has a full outer shell.

    04:10 So if we look at group 17 elements or group 7 elements, such as fluorine, chlorine, bromine, or iodine—halogens—we can see that they have an electron configuration which is quite the opposite to the lithium. They, indeed, need to gain electrons in order to complete their outer shell. If we take fluorine as an example, can see we have two electrons in the 2s2 subshell and five electrons in the 2p subshell. By gaining a electron, it's possible to complete the outer shell—the second shell—by ensuring it now has eight electrons in it. This, in the case of fluorine, will generate the fluoride anion. Note the difference: It's an anion, not a cation, because it has gained an electron rather than lost an electron. In this case, F– (the anion) has the same electronic arrangement as neon.

    05:08 That is to say it has a full second shell (configuration being 1s2, 2s2, 2p6) and therefore mimics that stable noble gas. So fundamentally, it's important to understand that when it comes to ionic chemistry, electrons are not found on their own, and they must be attached to an atom. And ions are also not found on their own. It is not possible, for example, to buy fluoride an ion. They must be countered with a cation of some sort. Energetically, this would be impossible. So therefore, if we look at our simple example of lithium and fluorine in the formation of an ion, the lithium wishes to lose an electron in order to achieve its low-energy state. The fluorine wishes to gain an electron in order to achieve a relatively low-energy state. And so, therefore, an electron in the case of a reaction of lithium and fluorine (not that I recommend you do this; it would be very exothermic) would be the loss of an electron from the lithium outer shell and the gain by the fluorine of that electron into the fluorine outer shell. And this formal movement of charge of electrons from one atom to the other is the basis of ionic bond formation, where there is the formal transfer of an electron from one atom to the other. And the results of this, as you can see here in this simple example, is the formation of the lithium fluoride salt. Salts typically exist in solid form in crystal lattices. This is because (as we may catch up with a little later on) the nondirectional nature of this permanent electrostatic charge.

    07:01 So in other words, the Li+ doesn't just attract a single fluoride anion; it attracts any other fluoride anions that happen to be within the vicinity in a nondirectional fashion. Salts are usually water-soluble, and this is by virtue of the possibility of the ions to interact by dipole bonding with water molecules, which, as we'll see from a biological perspective in some of the later lectures, is particularly important when trying to come up with the best agonist or antagonist for a given receptor or enzyme.

    07:40 So if we look at magnesium, for example, which I alluded to earlier, it has two electrons in its outer shell. When it loses these two electrons, in order to give the complete octet (the 2s2, 2p6), it affords us the cation Mg2+. This is stable by virtue of the neon being the resulting electron configuration, which is a stable configuration. The same applies with aluminum. Here, we're losing three electrons, the 3s2 and 3p1, to afford us the aluminum cation, yeah? So this also is stable as a consequence of having a complete outer shell in the form of the neon electron configuration. However, if we can pair these fixed possible ionizations to the transition metal or the d-block region of the periodic table, we see that there is a discrepancy. So as I said, magnesium and aluminum either exist as their elements or as 2+ and 3+ cations respectively. However, in the case of the transition metal or d-block elements, it is possible for them to have multiple oxidation states or multiple ionizations. If we look, for example, at copper in its elemental form, it exists as Cu. In its ionized form, it can either exist as Cu+ or Cu2+. Therefore, unlike magnesium, it can either shed one of its outer shell electrons or both of them. The same applies with tin.

    09:20 It can either shed two of its outer shell electrons to give Sn2+ or four to give Sn4+.

    09:28 And it's this diversity which affords transition metals a wide diversity and variety of ionic reactions, which we will touch upon in later lectures.

    09:39 So salts made up from monoatomic cations and anions are named in the following way (so this is the nomenclature part): NaCl, sodium chloride; LiF, lithium fluoride; and MgBr2, magnesium bromide. So let's just touch upon these in isolation. If we look at sodium chloride, what we're saying here is that the cation actually retains its original elemental name, and it's only the group 7 (in this case) element which has become an anion, which adopts the suffix -ide, okay? So the anion takes i-d-e at the end of its name. However, the cation remains the same. Note: In the case of sodium chloride, it is a one-to-one formula unit.

    10:30 There is one positive charge on the sodium and one negative charge on the chlorine. So the ratio between the two is 1:1, otherwise known as the stoichiometry. The same applies with lithium fluoride. However, if we look at magnesium bromide, we can see that there are actually two negatively charged cat… anions, Br–, in order to offset the positive 2+ charge of the magnesium. But we do not call it magnesium dibromide. It is assumed that you understand that magnesium can only bind to two things which are monovalent, like bromine. And remember, at all times, salts must be charge-neutral, which is the reason where you have the two anions, Br–; there must be two of those to offset the 2+ charge of the magnesium. And if we were to, for example, look at aluminum chloride, it would have to be AlCl3, where we have a aluminum ion with a formal charge of 3+ and three chlorine anions, each with a charge of 1–.

    About the Lecture

    The lecture Formation of Ions – Quantum Numbers by Adam Le Gresley, PhD is from the course Chemistry: Introduction.

    Included Quiz Questions

    1. It achieves a noble gas like configuration by swapping or sharing the electrons
    2. It achieves a halogen like configuration by swapping or sharing the electrons
    3. It achieves an alkali metal like configuration by swapping or sharing the electrons
    4. It achieves an alkali earth metal like configuration by swapping or sharing the electrons
    5. It achieves a transition metal like configuration by swapping or sharing the electrons
    1. The addition of an electron in the outermost shell of fluorine converts it permanently into neon gas
    2. The elements in groups 1, 2 and 3 of the periodic table like to lose the electrons in a chemical reaction
    3. The halogens readily pick up the electrons during a chemical reaction to achieve noble gas like configuration
    4. The noble gases are highly reluctant to lose or gain electrons during a chemical reaction
    5. The free halogen elements do not exist in nature due to their highly reactive nature
    1. The increase in effective nuclear charge pulls the electron density inwards more strongly in the sodium cation
    2. The decrease in effective nuclear charge pulls the electron density inwards more strongly in the sodium cation
    3. The repulsive forces between the core and outer shell electrons push the core electrons more towards the nucleus
    4. The shield effect of the core electrons vanishes due to certain unknown reasons
    5. The electrons shrink in their size, hence leading to an overall reduction in the ionic radius
    1. Due to fully filled electron configuration and high second ionization energy values
    2. Due to half filled electron configuration and high value of second ionization energies
    3. Due to partially filled electron configuration and low value of second ionization energies
    4. Due to alkali earth metal like configuration and low ionization energy values
    5. Due to halogen elements like electron configurations and high ionization energy values
    1. Because of the tendency of ions to interact with the water molecules via dipole bonding
    2. Due to the tendency of anions to lose electrons to water molecules
    3. Due to the tendency of ions to form a covalent bond with water molecules
    4. Due to the tendency of ionic compounds to donate their protons to the lone pair of electrons of the oxygen atom on the water molecule
    5. Due to the tendency of cations to pick up the lone pair of oxygen atoms attached to two hydrogen atoms in a water molecule
    1. Due to empty d-orbitals, unpaired electrons and the small energy difference between 4s and 3d orbitals
    2. Due to fully filled 4s orbital and high energy difference between 4s and 3d orbitals
    3. Due to fully filled 3d orbitals
    4. Due to paired electrons in 4s and 3d orbitals
    5. Due to larger energy gaps between 4s and 3d orbitals

    Author of lecture Formation of Ions – Quantum Numbers

     Adam Le Gresley, PhD

    Adam Le Gresley, PhD

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