Lewis structures

Lewis structures - A Theory of Bonding – Sharing Pairs of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A Theory of Bonding – Sharing Pairs of Electrons Electron Pair Representations, Lewis Structures and the Octet Rule In 1916 G. N Lewis proposed a theory of bonding based on a large amount of empirical chemical information. Lewis noted that inert gases were especially stable to chemical reactions, and that the chemical formulas of many compounds, particularly those in the first three rows of the periodic table (H through Ar), could be understood through a few simple rules. According to these rules, atoms formed stable compounds by attaining intert gas configurations. This simply means that all atoms want to contain an outer shell that is full of electrons. In order to achieve this, atoms either transfer electrons (as in ionic compounds) or share pairs of electrons (as in covalent compounds). In other words, for covalently bonded compounds, a chemical bond would be formed by two atoms sharing 2, 4, or 6 electrons. These rules, seemingly naïve at the time, have proved very useful, particularly to organic chemists. Indeed, you will make extensive use of “Lewis Dot Structures” when you take organic chemistry. To understand this scheme, lets look at some examples. We will begin with the simplest of molecules, H2. Each hydrogen atom has one electron, and the bond between the two atoms in the hydrogen molecule, H2, is a result of the two atoms sharing the two electrons between them. If we represent each electron by a dot, then the hydrogen atom is represented as: And the formulation of the hydrogen molecule as: where each atom can be seen as having two electrons: In the hydrogen molecule, the electrons in the bond are shared between both H atoms. Each atom thus has 2 electrons. This is the same number of electrons as the inert gas helium. Now, lets consider a slightly more complicated molecule. Lets consider one of the halogens. Recall that all the halogens, fluorine, chlorine, bromine and iodine each form diatomic molecules. The simplest of these is fluorine (because it has only 2 energy levels ‐ shells). In examining fluorine we will only consider the electrons that have been added in addition to helium’s very stable pair (i.e. those electrons in the outer energy level ‐ shell). These electrons are called valence electrons. From the periodic table we can see that fluorine has 7 valence electrons1. 1 Helium has the atomic number 2 while fluorine has the atomic number 9. Thus an uncharged fluorine atom has seven more electrons than does an uncharged helium atom. 1 Thus the fluorine atoms can be represented as: Note that only seven electron are shown surrounding the fluorine atom, although we know that a fluorine atoms has nine electrons. We know from experience that the two most “buried” electrons (those in the inner shell), the helium occupancy, are not involved in the bonding of fluorine. Thus they are not included in the representation. Using Lewis’ scheme we can think of two fluorine atoms bonding by sharing a pair of electrons to become a diatomic fluorine molecule: On the left of the reaction equation each fluorine atom is shown to have seven valence electrons for a total of 14 electrons. On the right the molecule, F2, also has 14 electrons (7 pairs), but now the pair of electrons between the atoms is shared. Therefore each fluorine atom in the F2 molecule is surrounded by eight valence electrons, the same number of valence electrons that the inert gas neon has. Now think about a hydrogen atom and a fluorine atom bonding together to become a hydrogen fluoride molecule: As was true before, the number of valence electrons shown on the left is the same as those shown on the right – in this case, eight. On the left of the equation neither atoms has a completed valence shell. Hydrogen has only 1 electron instead of 2, and fluorine has only 7 electrons instead of 8. Here, the sharing of the two electrons to form a bond results in the electron occupancy of helium for the hydrogen atom (2 electrons) and the electron occupancy of neon for the fluorine atom (8 electrons). Recall that through the sharing of electrons by bonding the atoms are seen as obtaining the electron occupancy of the nearest heavier inert gas. For hydrogen the nearest heavier inert gas in helium, and for fluorine the nearest heavier inert gas is neon. Hopefully in general the idea is becoming clearer. Atoms seek to become more stable by adopting the electron occupancy of inert gases. They do this through sharing electrons to form bonds. The shared electrons are considered to “count” towards the number of electrons for both atoms involved in forming the bond. To summarize, hydrogen will strive to have two electrons in its valence shell, but the other atoms will form bonding arrangements, which give them eight electrons in their valence shell. 2 Let us now use this idea to consider the electron dot configuration of water, H2O: To be able to draw the Lewis structures yourself you will first have to be able to determine the number of valence electrons there are for each atom in the compound, and then you will have to find some arrangement for sharing pairs of electrons to form bonds that will allow all the atoms in the bonded compound to have a full valence shell – in other works, for each atom other than hydrogen to be surrounded by eight electrons. The total number of valence electrons available can be found by totaling the number of valence electrons in each atom. For the example above: one from each of the hydrogen atoms and six from the oxygen atom – for a total of eight electrons. The simple model for bonding that Lewis developed says that some of these eight valence electrons will be shared between two atoms, thereby forming a bond. The idea is that the most stable bonding arrangement is one where each atom has a full valence shell. To have a full valence shell as atom in a bonded compound must be surrounded by eight valence electrons (with the exception of hydrogen, which requires only two electrons). The game, when you are looking for the correct Lewis dot structure for a compound, is to find some way to situate the shared electron pairs so that the need for eight valence electrons is satisfied. This is sometimes called the Octet Rule. Be sure to remember that while the octet rule arses from a useful model of bonding, it only applies to elements in the first three rows of the periodic table, and not even always to those2. Before you can figure out the proper bonding scheme, you first need to know the number of valence electrons. Finding the number of valence electrons for a given unbonded and uncharged atom is relatively easy. Just follow the following rules: 1) For hydrogen it is always one 2) For any other uncharged and unbonded atom you simply take the atomic number and subtract from that the atomic number of the closest inert element with a lower atomic number. For example, the atomic number of oxygen is 8, and helium (atomic number 2) is the closest (and only) inert gas with a lower atomic number. So oxygen has a total of 8 – 2 = 6 valence electrons. Similarly, nitrogen (atomic number 7) has a total of 7 – 2 = 5 valence electrons and carbon (atomic number 6) has 4. 2 Boron is a commonly cited exception to the octet rule. Boron often forms compounds that result in only six surrounding valence electrons. Boron trihydride. BH3, is an example of a compound where the boron atom is electron deficient according to the Lewis scheme of bonding. 3 How many valence electrons do you think Silicon has? 3 What about sulfur? 4 Now that you know how to determine the number of valence electrons, why don’t you try to write electron dot (Lewis) structures for some simple molecules like NH3, CH4, C2H6 and H2O2. Finding a structure that will give the desired number of electrons around each of the bonded atoms is mostly a matter of trial and error. However, as you gain experience in this skill you will find the task less daunting, and become much more efficient. NH3 CH4 C2H6 H2O2 Good, the last two compounds, C2H6 and H2O2 are a little bit more difficult and you should not feel despondent if you found them so. The trick to solving these two structures was to realize that the carbons in C2H6 and the oxygens in H2O2 must be bonded together as follows: H3C‐‐CH3 and H‐‐O‐‐O‐‐H To simplify our illustrations we can represent the bonds formed in these compounds (or any Lewis structures) with solid lines, where the single solid line represents a bond formed from the sharing of the electron pair. A single solid line is 3 Silicon has four valence electrons. 4 Sulfur has six valence electrons. 4 also used to represent a pair of electrons that are not involved in bonding (this will be discussed again later). Thus, we can represent the structures you just drew as follows: NH3 CH4 C2H6 H2O2 Note that the carbon atoms, because they have 4 valence electrons of their own, usually need to form four bonds to reach a total of 8 surrounding electrons. So when you are trying out possible bonding arrangements, carbon is often the central atom surrounded by others (as was the case for C2H6 as shown above). Nitrogen on the other hand, because it has 5 valence electrons (i.e., it needs only 3 more electrons to reach a total of 8) often forms three bonds, while oxygen, because it has 6 valence electrons, usually forms two bonds. You should use this information (tabulated below) as a guide to help you solve the correct Lewis structure for a given compound. Atom Hydrogen Carbon Nitrogen Oxygen # Valence electrons 1 4 5 6 # Bonds usually formed 1 4 3 2 Now, lets look at some slightly more complicated structures. In all the compounds we have drawn Lewis structures for so far, there have only been single bonds. There are however, compounds that involved double, or even triple bonds. Carbon dioxide, CO2, is one such example. For this compound it is not possible to draw a Lewis structure that has eight electrons around each of the 3 atoms using only single bonds. When I speak of a single bond, I simply mean that the atoms forming the bond are sharing only one pair of electrons. It is however, possible for atoms to share two, or even three, pairs of electrons. By doing this the pair of atoms form a double bond (sharing of 2 electron pairs) or a triple bond (sharing of 3 electron 5 pairs). To help illustrate this lets look at the Lewis structure for CO2. We can draw the bonding pattern for CO2 as follows: The complete Lewis structure for CO2 is: Observe that in CO2 the carbon atom is the central atom and that it forms 4 bonds (yes, each double bond counts as two bonds) and each oxygen atom in the molecule has two bonds. We can once again represent all the electron pairs in CO2 with short solid lines, as we have done for the bonding electrons: In the above illustration, the solid lines between the atoms are seen to represent bonds, and the other solid lines represent non‐bonding pair of electrons (often referred to as lone pairs of electrons). For convenience sake, you may find it a little easier to use these solid lines (dashes) when playing with possible Lewis structures instead of drawing all the dots (simply because the dashes are quicker to draw). Now that we have covered the basic principles, try and draw Lewis structures for the compounds shown below. Note that the way HCO2H and CH3CO2H are written. The order of the atoms sort of suggests the proper way to arrange them in the Lewis structure. This is common for the way chemists often write out organic compounds. It isn’t perfect, of course, but it is useful to pay attention to the molecular structure suggested by the way the compound is written. C2H4 O2 N2 C2H2 6 HCO2H CH3CO2H If you had difficulty doing these Lewis structures, be sure you remembered the number of bonds that each of the atoms usually forms. Remember that carbon is usually the central atom surrounded by other atoms, and that the way the compound is written often gives you a hint about how it is bonded. Another way to approach these is to use the following formula to calculate the number of shared electrons according to: S = N – A Where S is the number of shared electrons (i.e., twice the number of bonds formed), N is the number of valence electrons needed (8 for each of the atoms except hydrogen, which required only 2), and A is the number of valence electrons available for bonding. Lets reconsider the compound CO2. As you may recall, the Lewis dot structure for CO2 is: or The compound has two double bonds and since each bond arises from the sharing of two electrons, that means that there are 8 total shared electrons. Now lets try and use the formula S = N – A: N: There are a total of three atoms in the compound, so the total number of electrons needed for bonding, N = 3 x 8 = 24. A: The total number of electrons available for bonding is simply the sum of the valence electron for each atom, 4 from carbon and 6 from each of the two oxygen atoms for a total of, A = 4 + 6(2) = 16. Thus, the total number of shared electrons, S = N – A = 24 – 16 = 8. The structure we drew above has four bonds, so there are indeed 8 shared electrons in the Lewis structure. IT WORKS! Lets do another example. How many shared electrons would you expect to find in PCl3? 7 We have a total of 4 atoms here. Each P has 5 valence electrons and each Cl has 7 valence electrons. N = 4(8) = 32 A = 3(7) + 5 = 26 S = 32 – 26 = 6. Six shared electrons means 3 bonds, so a reasonable Lewis dot structure would be: Okay, now you try and draw the Lewis structures for the following compounds: CS2 SiCl4 H2C2 CH3CH2COH Do you think you go them right? Check the number of bonds of each of the compounds to make sure it is sensible. How many bonds do you think silicon should have? HINT: Compare the number of valence electrons silicon has with the number of valence electrons carbon has, and then think about the number of bonds it might form. If there are any structures you couldn’t draw Lewis structures for before, go back to them now and try again. OK. Now for a new wrinkle to consider on this same topic. What happens when you have a charged compound? For example, like NH4+, the ammonium cation. The +1 charge means that it has one fewer electron than the uncharged species. The number of valence electrons for an uncharged species would be 5 + 4(1) = 9. But since it has a +1 charge the number of available valence electrons must be 9 – 1 = 8. 8 The number of valence electrons needed is 8 + 4(2) = 16. So the number of shared electrons, S = N – A = 16 – 8 = 8. So the compound has 8 shared electrons (forming 4 bonds), and the Lewis dot structure is: Now consider an anion like cyanide, CN‐. A good start is to figure the number of shared electrons. The number of valence electrons needed is 16, and the number of available electrons is 10 (including an extra one because of the one negative charge on the anion). So the number of shared electrons is 6, indicating that there is a triple bond between the carbon atom and the nitrogen atom. The Lewis dot structure for this compound is: And finally – one closing caveat to consider: RESONANCE. Some compounds can have more than one valid Lewis structure that can be drawn. In cases like these all the Lewis structures satisfy the “rules of the game” and each structure is considered to be equivalent to the next. As an example let’s consider nitrogen dioxide, NO2. To obtain an octet of 8 electrons around each atom, you could draw either of the following Lewis structures for NO2: or The two structures shown both satisfy the rules for Lewis dot structures and are equivalent. So, which one is correct? According to the structures we have drawn there should be two types of bonds in an NO2 molecule, one shorter bond (the double bond) and one longer bond (the single bond). Experimentally however, it has been shown that the two N‐O bonds in NO2 are of the same length (and strength). This implies that neither structure is correct. Both are incorrect. The true structure of NO2 is actually a hybrid of the two equivalent Lewis structures: Instead of having one nitrogen‐oxygen single bond and one nitrogen‐oxygen double bond, there are two equivalent bonds that are hybrids (an average of the proposed 9 inequivalent bonds). Each bond is sort of a 1.5 bond, somewhere between a single bond and a double bond.5 In reality the molecule continuously resonates between the two structures shown above and what we see is an average bond length. (This is known as the delocalized electron model and is a model for bonding that we will cover in more detail when we consider covalent bonding next semester). Now take a look at the formate anion, HCO2‐. Try and draw two possible Lewis structures for the formate anion below: (Don’t forget when you are counting the number of available valence electrons to include the extra electron that comes from the negative charge). If you have drawn the Lewis structures correctly, this simple theory of bonding (as proposed by Lewis) predicts the molecule to have one carbon‐oxygen double bond and one carbon‐oxygen single bond. Do you think this is correct? Or do you think both bonds will be somewhere in between the single and double bonds? Considering what you now know about resonance (and the NO2 example above), draw the true Lewis structure for the formate anion: 5 You might ask what experimental measure there is for determining what kind of bond is formed. An easy measure is bond length. Single bonds are longer than double bonds, which in turn are longer than triple bonds. 10 11 ...
View Full Document

This note was uploaded on 02/13/2011 for the course CHEM 111 taught by Professor Bastos during the Fall '08 term at CUNY Hunter.

Ask a homework question - tutors are online