Chapter 8 - Basic Concepts of Chemical Bonding

This is Chapter 8: Basic Concepts of Chemical Bonding. Whenever two atoms or ions are strongly held together, we say that there is a chemical bond between them. There are three general types of chemical bonds: ionic, covalent, and metallic. We can get a simple understanding of these three types of bonds by thinking about the simple act of using a stainless steel spoon to add table salt to a glass of water. Table salt is sodium chloride, which is held together by ionic bonds due to the electrostatic attraction between oppositely charged ions. The water in the glass mainly consists of H2O molecules, with the hydrogen and oxygen atoms bonded to each other through covalent bonds, formed by the sharing of electrons between atoms. On the other hand, the spoon consists mainly of iron metal, in which iron atoms are connected to each other via metallic bonds, formed by electrons that are relatively free to move within the metal. These different substances, sodium chloride, water, and iron metal, behave as they do because of the way in which their constituent atoms are connected to each other. For example, sodium chloride readily dissolves in water, but iron metal does not. What determines the types of bonding in any substance? How do the characteristics of these bonds give rise to different physical and chemical properties? The keys to answering these questions are found in the electronic structure of the atoms involved, which we discussed in Chapters 6 and 7. In this chapter and the next, we examine the relationship between the electronic structure of atoms and the ionic and covalent chemical bonds that they form. The learning objectives for this chapter are as follows: 1. 2. 3. 4. 5. 6. 7. To be able to draw structures. To learn the octet rule. To be able to distinguish between ionic and covalent bonding. To use electronegativity values to describe bond polarity. To describe localized bonding and use resonance structures to represent it. To determine the formal charge on atoms in a covalent structure. To understand relative bond order and bond length. Let's start by talking about the Lewis symbol. The electrons involved in chemical bonding are the valence electrons, which for most atoms are those in the outermost occupied shell. The table below shows the different elements from Groups 1 through 8, their electron configurations, and their Lewis symbols. | Element | Electron Configuration | Lewis Symbol ||----------|------------------------|---------------|| Lithium | 1s² 2s¹ | Li || Sodium | 1s² 2s² 2p⁶ 3s¹ | Na || Beryllium | 1s² 2s² | Be || Magnesium | 1s² 2s² 2p⁶ 3s² | Mg || Aluminum | 1s² 2s² 2p⁶ 3s² 3p¹ | Al || Carbon | 1s² 2s² 2p² | C || Silicon | 1s² 2s² 2p⁶ 3s² 3p² | Si || Nitrogen | 1s² 2s² 2p³ | N || Phosphorus | 1s² 2s² 2p⁶ 3s² 3p³ | P || Oxygen | 1s² 2s² 2p⁴ | O || Sulfur | 1s² 2s² 2p⁶ 3s² 3p⁴ | S || Fluorine | 1s² 2s² 2p⁵ | F || Chlorine | 1s² 2s² 2p⁶ 3s² 3p⁵ | Cl || Neon | 1s² 2s² 2p⁶ | Ne || Argon | 1s² 2s² 2p⁶ 3s² 3p⁶ | Ar | When forming compounds, atoms tend to gain, lose, or share electrons until they are surrounded by eight valence electrons. This is known as the octet rule. All atoms want to have eight electrons around them, as this gives them a lot of stability. This is why the noble gases, which have eight valence electrons and a complete octet, do not react with other elements. For example, fluorine has seven valence electrons, so it tends to gain one more electron to achieve a complete octet. On the other hand, elements such as sodium find it easier to lose one electron and achieve the electron configuration of neon, which also has a complete octet. Family 1 elements, like sodium, have a positive 1 charge as ions because it is easy for them to remove that electron and achieve a complete octet, similar to the noble gas configuration. Family 2 elements, like magnesium and beryllium, also find it easier to lose electrons and achieve stability. Electrons, that's why tablets I always get a plus two. Because when they do that, now they have the configuration of a noble gas. So let's talk now about the ionic formation either. Successivehood is resolved from interaction of metals that are on the left side of the periodic table with nonmetals that we can find them at the right side. Okay, so most of the time the metals are the ones that like to lose in the electrons and the nonmetals are the ones that gain electrons. And in that way, basically they can eventually create that noble configuration or configuration of noble gas. For example, we have sodium and we have chlorine. These are the new structures for sodium and for chlorine. Sodium is from family 1, so that's why we write Na and one electron. Chlorine is from family 7, so that's why we write 3 pairs of electrons and one single electron. And as we mentioned, in this case, we have a metal and we have a nonmetal. So, then metals tend to lose that electron, so they release that electron. And chlorine is from family 7, they love to gain electrons. Okay, so in that way, now when we release that electron, the sodium has a positive one charge. It was neutral before, but because we removed one negative particle, now the sodium is going to have a positive charge, a cation. And chlorine will receive one extra electron. It was neutral before, but when you receive that negative particle, now you have eight electrons, you complete your octet. So, we can put the octet rule. You're going to be negative one because you gained that electron. So, sodium, also because they lose that electron, they will gain that electronic configuration from the noble gases. And that's why we can create those ions and we can have that ionic compound as sodium chloride. So here we have, for example, the sodium as solid and we have the gaseous chlorine. We combine them in this chamber and at this point they start transferring the electrons. The sodium starts to release those electrons and they are transferred from the sodium to the chlorine gas. And because they transfer that, they are losing those electrons, we start to produce the sodium ions. And also, as well, when chlorine receives those electrons, it starts to produce the chloride anion. And eventually, they will form the sodium chloride. Okay, and this basically is known as an exothermic reaction. So let's talk about the energetics of ionic bonding. Simply, the Born-Haber cycle. Many factors affect the energy of ionic compounds or rewriting bonding. Start with the metal and the nonmetal element that you saw before. We have the sodium as a solid and the chlorine as a gas. Then we need to make them gaseous atoms. Okay, so the sodium gas will produce the sodium solid. We'll turn it to gas. And then we need to create those ions. And then those ions will combine to produce the sodium chloride, the molecule of sodium chloride or any compound. So this is the Born-Haber cycle for this type of reaction where we have initially the sodium as a solid and chlorine as a gas. We transform this gas to be solid. To the gas. And at this point, we have the ions produced. This process, all this process, absorbs energy. So that's why the energy is increasing. But then when you start to combine the ions, you will produce sodium chloride. And sodium chloride has lower energy than the initial reactants. So that means that this process, the whole process, is an exothermic process. You start at this point in energy and you finish at this point in energy. And the total energy of this process, for the production of sodium chloride, is 788 kilojoules. And this is also associated with the lattice energy for sodium chloride. So for sodium chloride, basically this process to produce will release a total of 788 kilojoules per mole. And this is also associated with the lattice energy because an exothermic, if we're talking about the energy that is released, you see that it's negative 788 kilojoules. But the positive of this value, that is 788 kilojoules, is the lattice energy. The energy that you will need to add to the sodium chloride to separate those ions. So the formation will release 788 kilojoules. That's why when released, you're going to have a negative value. But if you want to separate those ions, if you start, for example, with sodium chloride and you want to separate those ions, you will need to add 788 kilojoules to separate those ions. Now, let's discuss the ionization energies and electron affinity. Remember that the ionization energy is how easy it is to create an ion. And in this case, how easy it is to release the electrons. So that's why the cations basically have those ionization energies, while the nonmetals, we talk about the electron affinity because they would like to gain those electrons. It takes energy to convert the atoms to ions in an exothermic process. It takes energy to create a cation in an endothermic process. As we saw in the cycle in the past slide, but energy is released when making the anion. And also, the formation of the solid releases a huge amount of energy, being an exothermic reaction. So, this makes the formation of the salts from the elements exothermic, as we saw before. We saw that the energy of sodium chloride is lower than the energy of the reactants. So, that's why this process to produce sodium chloride is an exothermic kind of reaction. The lattice energy is a measurement and before, in this case, that huge exothermic transition is reversed. At the reverse of the lattice energy, the energy required to completely separate a mole of solid ionic compound into its gaseous ions. So, we saw how from the gaseous ions create that solid, and that process was -788 kilojoules. But now, if you have this solid ionic compound and you want to return it to gaseous ions, you need to add 788 kilojoules per mole for sodium chloride. The energy associated with the electrostatic interaction is governed by Coulomb's law. The electrostatic interaction here is equal to Kappa times Q1 and Q2 divided by D. In this equation, Q1 and Q2 are the charges on the particles in Coulombs, with their sign, and D is the distance between their centers in meters. And Kappa is a constant. So basically, this equation indicates that the attractive interaction between two oppositely charged ions increases as the magnitudes of their charges increase, and as the distance between their centers decreases. Thus, for a given arrangement of ions, the lattice energy increases as the charges on the ions increase, and as their radii decrease. So that's why it's important to understand Coulomb's law. This law basically gave us that relationship between the charges and the distance that they are really important for that lattice energy. The lattice energy increases with an increase in charge on the ions and a decrease in size of the ions. So here we have the lattice energy. As we can see here, lithium, when you combine it with fluoride, chlorine, iodine, fluorine is smaller than chlorine, chlorine is smaller than iodine. So as we can see, the lattice energy will increase when the size of the ion is small. The lithium is constant, so it's not making a difference. And it differs here, the one that is changing is the anion. As you increase the size of the anion, the energy will decrease. So that's why the energy will increase when you decrease the size and also increase in charge on the ions, as we can see here. We have lithium and we have magnesium. Lithium has a positive 1 charge and magnesium has a positive 2 charge. So that means that we are increasing the charge, and what happens to the lattice energy? It also increases. So the lattice energy increases with an increase in charge of the ions and by decreasing the size of the ions. Now let's talk about covalent bonds. In covalent bonds, atoms share electrons. In ionic bonds, electrons are transferred from the metals to the nonmetals, creating ions. In covalent bonding, the electrons are shared between atoms. There are several electrostatic interactions, including attractions between the electrons and the nuclei. Remember that electrons have a negative charge, and the nucleus has a positive charge. So there is an attraction between the nuclei and the electrons. There is also repulsion between the electrons and repulsion between the nuclei due to the high concentration of positive charge. For a bond to form, the attraction must be greater than the repulsion. In a covalent bond, the attraction must be greater than the repulsion. Now let's talk about Lewis structures and how they can help us understand covalent bonding. Lewis structures demonstrate the sharing of electrons to form covalent bonds. We start by trying to give each atom the same number of electrons as the nearest noble gas. By sharing electrons, we fulfill the octet rule. The simplest examples are hydrogen (H2) and chlorine (Cl2). In the case of hydrogen, each hydrogen atom shares one electron, resulting in a bond with two electrons. In the case of chlorine, each chlorine atom shares one electron, resulting in a bond with two electrons. The electrons in the bonds are shared between the atoms. This allows both hydrogen and chlorine to fulfill the octet rule. In Lewis structures, lone pair electrons are located on only one atom, while bonded pair electrons are shared between atoms. Lone pair electrons can be represented by dots, while shared electrons can be represented by lines. In many molecules, atoms attain complete octets by sharing more than one pair of electrons. When two electron pairs are shared between two atoms, a double bond is formed. When three electron pairs are shared, a triple bond is formed. For example, in carbon dioxide (CO2), the carbon atom shares two pairs of electrons with each oxygen atom, creating double bonds. This allows carbon to have a total of eight electrons and oxygen to have a total of eight electrons, fulfilling the octet rule for both atoms. Lewis structures help us understand the distribution of electrons in covalent bonds. Polar covalent bonds occur when electrons are not shared equally between atoms. Fluorine pulls harder on the electrons it shares with hydrogen, creating a higher electron density around fluorine compared to hydrogen. This creates a polarity in the bond, with fluorine being more negatively charged and hydrogen being more positively charged. In contrast, when two identical atoms bond, such as chloride or hydrogen, the electron pairs are shared equally, creating a nonpolar covalent bond. In an ionic compound like lithium fluoride (LiF), there is a high electron density on the fluoride side, creating a negative charge, and a high positive charge on the lithium side. In a covalent bond, like hydrogen fluoride (HF), there is a higher electron density on the fluorine side compared to the hydrogen side, but it is not as extreme as in an ionic compound. The electrons are still being shared between the atoms. It is important to note that while fluorine is pulling more on the shared electrons in a covalent bond with hydrogen, it is not completely taking the electrons away from hydrogen. The electrons are still being shared between the two atoms. Electrons bond with hydrogen, but they pull those electrons a little bit more than the hydrogen does. That's why we create polarity in those covalent bonds. So, you can have nonpolar covalent bonds and polar covalent bonds, but you will never have a polar ionic bond. This is because in an ionic bond, you transfer all the electrons from one atom to another. When we talk about polar or nonpolar, we are referring to covalent bonds. We use a quantity called electronegativity to estimate whether a given bond is nonpolar covalent, polar covalent, or ionic. Electronegativity is defined as the ability of an atom in a molecule to attract electrons to itself. The greater an atom's electronegativity, the greater its ability to attract electrons. Electronegativity generally increases as you go from left to right across a period in the periodic table, and as you go from bottom to top in a group. Fluorine has the highest electronegativity, with a value of 4.0. In a polar covalent bond, two atoms share electrons unequally, resulting in a partial negative charge (represented by Δ-) and a partial positive charge (represented by Δ+). This creates polarity in the bond. Another way to represent polarity is by using an arrow, with the head of the arrow pointing towards the atom with higher electron density (Δ-) and the cross of the arrow indicating the more positive end (Δ+). To determine the type of bond (ionic or covalent), we can consider a few hints. If there is a metal with a nonmetal, it is likely ionic. If there are two nonmetals, it is likely covalent. We can also look at the difference in electronegativity between the two atoms involved in the bond. Additionally, the properties of the compound, such as melting point, can provide clues about the type of bond. Lewis structures are used to represent covalent molecules. To write a Lewis structure, we need to sum the valence electrons from all atoms, taking into account the overall charge. For an anion, we add one electron for each negative charge, and for a cation, we subtract one electron for each positive charge. Once we have the total number of valence electrons, we distribute them around the molecule, following the octet rule. We write the symbols for the atoms, showing which atoms are attached to each other, and connect them with single bonds. In the case of phosphorus and chlorine, phosphorus is in the middle because it can form more bonds. Let a pair of electrons, a lone pair of electrons, and three unpaired electrons. So, these three and the lone pair can form bonds. You have three possibilities to create bonds with chlorine because it has seven electrons. You will have three lone pair electrons in one individual electron. So, that means you can only create one bond with chlorine. So, you're going to put the atom that can create more bonds in the middle, which is sulfur, and put one bond between them. Each bond represents two electrons, as we mentioned before. Okay, so we need to keep track of the electrons. We start with 26 valence electrons and subtract six that we used to create the three single bonds. Now, we have 20 electrons left to distribute around the molecule. Apply the octet rule to all the atoms bonded to the central atom. Here we have 2, 4, 6, and 2. Here we have 8, 2, 4, 6. Here we also have 8, and here we have 8. So, we have a total of 9 pairs, which equals 18 electrons. Therefore, we have 20 - 18 = 2 electrons remaining. We can add them to the molecule. We can see here that phosphorus has only 6 electrons around it, so we can place those 2 electrons on the central atom to complete its octet. So, we just need to look for 40 valence electrons in a way that we can draw the Lewis structures. Now, 2 - 2 is equal to zero, so we've used all the possible valence electrons to draw the structure. Now, if there are not enough electrons to give the central atom enough for the octet, we need to use multiple bonds. In this case, we have HCN as the molecule and this is the initial structure, but we can see here that for hydrogen, it only needs two electrons around it. So, with hydrogen, we only need two electrons around it, which completes the octet rule exception. With nitrogen, we have eight electrons around it, but with carbon, we only have four. So, we need to create double bonds between carbon and nitrogen to fulfill the octet rule. We can use one of the lone pairs of nitrogen to create a double bond. From these two electrons, we can create one bond between these two atoms. Also, with these two electrons, we can create one bond between those two atoms. Now, we have 1, 2, 4, 6, 8 electrons around carbon, all of them shared between the three atoms. Here, nitrogen has six and a triple bond because we have two electrons for each bond plus two more. Now, we have the octet around nitrogen and carbon as well. Sometimes, we need to assign formal charges because we have ions. The formal charge is the charge an atom would have if all the electrons in a covalent bond were shared equally. The formal charge is equal to the valence electrons minus half of the bonding electrons minus all nonbonding electrons. This formula gives us the formal charge. In the case of carbon dioxide, we have carbon double bonding to oxygen and oxygen. Here, we can see that we have the octet for this oxygen. We have two, four, and four more, giving us eight electrons around this oxygen. Similarly, for carbon, we have four bonds around it, so each bond is represented by two electrons. Therefore, we have eight electrons. The valence electrons for oxygen is 6 and for carbon is 4. Now, we need to subtract the assigned electrons from atoms. For carbon, we have four bonding electrons, so half of them is 2. For this oxygen, we have 2, 4, 6 bonding electrons, plus 1 nonbonding electron, which equals 7. Therefore, we have 6 - 7 = -1 for this oxygen. For carbon, we have 4 bonding electrons, so half of them is 2. For this oxygen, we have half of the 6 electrons, which is 3, plus 2 nonbonding electrons, which gives us 5. So, the formal charge for carbon is 4 - 4 = 0, and for oxygen it is 6 - 5 = 1. In another structure of carbon dioxide, we have a single bond between carbon and one oxygen, and a triple bond between the other oxygen and carbon. Even though there is a different arrangement of electrons, we still have eight electrons around each oxygen and carbon. Now, we need to look at the number of electrons assigned to each atom. For this oxygen, we have 1/2 of the bonding electrons, which is 1, plus all non-bonding electrons. We have 6 bonding electrons and 1 non-bonding electron, giving us a total of 7 electrons assigned for this oxygen. For carbon, we have only bonding electrons, which is 8. So, 4 - 4 = 0 for carbon, and for this oxygen, 6 - 7 = -1. This structure demonstrates that both oxygens have a formal charge of 0, while the carbon has a formal charge of -1. So, we see that we have two different Lewis structures for carbon disulfide. Which one is the dominant one? The one in which the atoms have formal charges closest to 0. In the first structure on the left, the negative formal charge is on the most electronegative atom, sulfur. In the second structure, we have formal charges of 0 and -1, and in the third structure, we have formal charges of 0 and -1 as well. So, they have the same distribution of formal charges, but we need the negative charge on the most electronegative atom, which in this case is sulfur. Therefore, the most dominant structure for the Lewis structure is the first one on the left. These are correct Lewis structures. Be also the dominant Lewis structure for nCS. So, one Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures known as resonance structures to describe the molecule. This is a zone here, and we can see that sometimes we have a multiple bond (double bond) between these two oxygens, and sometimes we have the multiple bond between these two oxygens. That means that the electrons have been moving around this area. Okay, we have the double bond here, and then we have the total here. The double bond is a resonance structure. It is called resonance structures. So, this is a resonance structure of this structure. That is the distribution of the electrons across the molecule. And here we have one example of benzene. This is a benzene ring. We have the double bond between these two carbons, between these two carbons, and these two carbons. And we can see that we can move those double bonds in a way that instead of it being here, now they are going to be in other places. These do a resonance structure. This is also the same representation but without showing the atoms. Okay, but this represents the carbons. Okay, and we don't write hydrogens in this type of representation, but we still have the double bond between these two carbon single bonds in these two carbon, double bond between these two carbons. So, it's been going to discover, you can see that you have double bond single bond double bond single bond double bond single bond. As well as here, now this double bond moves to this point, this one moves between this two carbons, and this one moves between this carbon. And have this structure. And this is happening really, really fast. So, we can represent this as this a ring with a circle inside, which represents resonance. It means that the electrons are moving all around the molecule. Okay, because we can see that we can move this to here to create this double bond, this double bond will move to here to create this one. So, if they are moving all around the molecule, so this is an organic compound, benzene (C6H6) that has these two resonance structures, and it's commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring. So, localized electrons are specifically on one atom or shared between two atoms. So, this one is also localized electrons; they are always between these two carbon, but this one of this is going to be always between these two carbon, but the other one is going to be moving all around the molecule. So, we're going to have the localized electrons that are shared by multiple atoms. Okay, so those that create those multiple bonds are delocalized because they are moving all around the molecule. Well, for example, so this one is localized, these two electrons are always between these two hydrogen and carbon, these two atoms. It is this these electrons are localized as well as this one, as well as one of these two double bonds. Okay, one of these bonds is going to be moving all the way around the molecule, but the other one is going to stay doing the bond between these two carbons. So, one of this is going to be localized, and the other electrons are going to be delocalized. Now, there are some exceptions to the octet rule. There are three types of ions or molecules that do not follow the octet rule. First, ions or molecules with an odd number of electrons, known as radicals. Ions or molecules with less than an octet. And ions or molecules with more than eight valence electrons, known as the expanded octet. Here we have an example of the number of electrons. The relatively rare and unstable and reactive ions and molecules with an odd number of electrons. Here we have nitrogen, which has one electron. That means that we're going to have five electrons around this nitrogen. And when we do a resonance structure of this one, we can create here also a radical. And we have around this oxygen just five electrons. And this is an exception to the octet rule. The other one is when we have fewer than eight electrons. Elements in the second period before carbon can make stable compounds with fewer than eight electrons. And we saw this with hydrogen. Okay, and also, for example, boron (BF3). BF3 gives boron a full octet. Places a negative charge on the boron and a positive charge on fluoride. This would not be an accurate picture for the distribution of the electrons in BF3. So, we have here the boron, and we can write this Lewis structure for BF3 where each fluorine has basically eight electrons, so it has an octet complete. While boron still happens to have just six. Okay, so we can rewrite to make boron have the octet complete. Okay, we can draw a double bond. We can use one of these unpaired electrons to create that double bond. But if we do this, each fluorine here with a double bond is going to be partially positive. And we know that fluorine is the element with the highest electronegativity. They love electrons. They want to have the electrons. They don't want to get rid of electrons. So, that's why these are less important, even though they are correctly drawn. But because of that difference in charge, we're going to have a positive charge here and a negative charge here. Well, here we don't have any charges. So, that's why this one is the dominant structure. The lesson is that if filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative atom, don't fill the octet. The central atom is not going to be favorable. Okay, so that's why, in this case, boron is really happy if we just have fewer than eight electrons, even though it's not an octet. We also have examples or exceptions with atoms that could have more than eight electrons. When an element is in period three or below in the periodic table, we have four or five, it can use the orbitals to make more than four bonds. The examples are PF5 and also phosphate. So, we can have here the PF5. We have the phosphorus in the middle, and around the phosphorus, we have 10 electrons. It has more than an octet. But because it has d orbitals available, they can use those orbitals to accept those bonds from the fluorine. This is basically another kind of three-dimensional structure for PF5. And here we have the phosphate. The phosphate here has basically eight electrons around. And also has a complete octet, as well as the oxygen. But that, we can see here, we have five charges, a complete of five charges for negative one and a positive one. While if we create a double bond between an oxygen and phosphorus, then we're just going to have three charges. So, we are looking for the structure that has the least number of charges in each atom. So, that's why, even though this is correct and it completes all the octets, because phosphorus has d orbitals also available, we can put those electrons in those orbitals, and by that way, decrease the partial charges in the molecule. So, we can have just three instead of five. And even though we're still going to have the ion being minus three, in this case, we're going to have a lesser distribution of charges in the molecule. So, this is predominant over this one. Now, let's talk about covalent bond strength. Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. This is called the bond enthalpy. So, the bond enthalpy for chlorine bond (Cl-Cl) is measured to be 242 kilojoules per mole. We write out a reaction for breaking one mole of those bonds. So, we can break the bond by saying that we're going to distribute these two electrons. It's going to give one electron for each side. So, we're going to have two radicals of chlorine. And now we talk about the average bond enthalpy. These are positive because bond breaking is an endothermic process. You need to add energy. So, the bond enthalpy is going to be positive. So, one way to estimate the delta H for a reaction is to use the bond enthalpies of bonds broken and the new bonds formed- so energy is added to break bonds and release when making bonds. In other words, the delta H of the reaction is going to be equal to the sum of the bond enthalpies for all bonds broken minus the sum of the bond enthalpies of all bonds formed. Okay, so here for example, we have methane and we have chlorine. So here, we're also, you can see at this point, we broke one bond between a carbon and a hydrogen. So that's why we have a hydrogen here and the CH3. Now, it's before what's CH4, but now we remove one hydrogen, we have the CH3. To do that, we need to add energy. We need that energy to bring this bond. The same thing happened with the chloride, with the CO2. We create these two radicals here, we need to break this bond and we need to add energy. Now, when we create the bonds between this chlorine and this hydrogen, and this chlorine and this carbon, energy is released. And we can see here that actually the products have less energy than the reactants. So that means tat was an exothermic kind of reaction because it releases more energy than when you use to break the bonds to create these radicals. So from the figure on the last line, we have that it is methane and the chlorine producing the chlorine methane and the HCl. In this example, one carbon-hydrogen bond and one chlorine-ClC bond were broken here and one carbon-chlorine and one HCl bond are formed. We form with one of these chlorine bond to one of the carbon and the hydrogen was released one with the other chloride. So now we need to look for the enthalpy for each of the bonds. Okay, and that's in the table- here we have 413 kilojoules and here we had 242 kilojoules for the chlorine-chlorine bond. So this is the energy required to break those bonds. Then we look for the enthalpy for each of the chlorine that was formed. So it's going to be 328 kilojoules for carbon-chlorine and 431 kilojoules for hydrogen-chlorine. And that means that we have 655 kilojoules minus 759 kilojoules. And we see here that the reaction is an exothermic one and releases energy. This agrees with the figure that we saw before where we saw that the products have lower energy than the reactants. And in this way, we can calculate the enthalpy of a reaction. Remember that it will be the sum of the enthalpies of the broken bonds minus the enthalpy of the formed bonds. Also, we can relate the bond delta P and the bond length. We can also measure an average bond length for different bond types. As the number of bonds between the two atoms increases, the bond lengths decrease. Okay, so the single bond between carbon-carbon is going to be longer than the double bond between carbon and carbon. Here, this is the bond length. We can see that the bond length for the carbon-carbon single bond is going to be 1.54 angstroms, while the double bond carbon-carbon is 1.34 angstroms, and the triple bond is 1.25 angstroms. As the number of bonds that you can have between the atoms increases, you're going to have the length be