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5.3: Electron-Dot Model of Bonding - Lewis Structures

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    156241
  • Using Lewis Dot Symbols to Describe Covalent BondingEdit section

    This sharing of electrons allowing atoms to "stick" together is the basis of covalent bonding. There is some intermediate distant, generally a bit longer than 0.1 nm, or if you prefer 100 pm, at which the attractive forces significantly outweigh the repulsive forces and a bond will be formed if both atoms can achieve a completen s2np6 configuration. It is this behavior that Lewis captured in his octet rule. The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:

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    99cbb09f87af5a4b09dd56499e843e08.jpg

    Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair ; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond.Edit section

    We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:

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    254015a5c4a159aa4e955da701127624.jpg

    The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:

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    67f67edbdb9eef2f11217fe1ff08c85b.jpg

    The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:

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    1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32−, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.Edit section

    Note

    The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.

    Edit section
    1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall from Chapter 2 that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32−, for example, we add two electrons to the total because of the −2 charge.
    2. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In \(H_2O\), for example, there is a bonding pair of electrons between oxygen and each hydrogen.
    3. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
    4. If any electrons are left over, place them on the central atom. Some atoms are able to accommodate more than eight electrons.
    5. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.

    Edit section

    Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.Edit section

    H2OEdit section

    1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.Edit section

    2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.Edit section

    3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.Edit section

    4. Each H atom has a full valence shell of 2 electrons.Edit section

    5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:

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    f6d9a24622184ca561822f70eb85e58b.jpg

    This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.

    OClEdit section

    1. With only two atoms in the molecule, there is no central atom.Edit section

    2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.

    3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.

    4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:

    7129216d30dabf7573f148e54f8dcc51.jpg

    Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.

    CH2OEdit section

    1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:

    ef5c576c0c84ded384007a33c963a2e1.jpg

    2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.

    3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:

    c30b4f42355f76f67e81f90a19a8f0bd.jpg

    Six electrons are used, and 6 are left over.

    4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:

    c8ddcf2f0df54fc20f18d0e60408951c.jpg

    Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.

    5. There are no electrons left to place on the central atom.

    6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:

    2f734e95e4e4513e5106a1b61e604cc8.jpg

    Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.

    An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.Edit section

    Example

    Write the Lewis electron structure for each species.

    1. NCl3
    2. S22−
    3. NOCl

    Given: chemical species

    Asked for: Lewis electron structures

    Strategy:

    Use the six-step procedure to write the Lewis electron structure for each species.

    Solution:

    Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:

    579121db32b7b48c36ae54dcf17ee6d8.jpg

    Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.

    1. e9847c1f1728e285bb9dee914470d140.jpg
    2. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:

      6e69300f5796dbb06798c387bd7bcc89.jpg
    3. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:

      dc6a5041bcdabb91cb727522339fe44b.jpg

      Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:

      312e28bcf7eb8ae248c0e7fa83ea0f22.jpg
    4. Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:

      56192089dcd6a3ab3cfc307d5c85f9b0.jpg

      All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.

      5880b3fce5aac0e0cc466c9d707c9ee6.jpg

    Exercise

    Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.

    Answer:

    1. 03bae19e91ea497c755e3fb932ed38d6.jpg83b695097f238c87d66b81488d2c724a.jpg
    2. 5a3bdca4a8ddf4937091e34428109a33.jpg
      1f1706c3f1e84e24caca141d0bcf5843.jpg

    Formal ChargesEdit section

    It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.

    To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:

    • Nonbonding electrons are assigned to the atom on which they are located.
    • Bonding electrons are divided equally between the bonded atoms.

    For each atom, we then compute a formal charge:

    To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH3) whose Lewis electron structure is as follows:

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    768e2d063688200c5482e54c7905ab81.jpg

    A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e + (6 bonding e/2)]. Substituting into Equation 5.3.1, we obtain

    A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e + (2 bonding e/2)]. Using Equation 4.4.1 to calculate the formal charge on hydrogen, we obtain

    The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH3 molecule.

    Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.

    Example

    Calculate the formal charges on each atom in the NH4+ ion.

    Given: chemical species

    Asked for: formal charges

    Strategy:

    Identify the number of valence electrons in each atom in the NH4+ ion. Use the Lewis electron structure of NH4+ to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation 4.4.1 to calculate the formal charge on each atom.

    Solution:

    The Lewis electron structure for the NH4+ ion is as follows:

    67c2f8a9e1f13d7325f739b98bcc30e9.jpg

    The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation 4.4.1, the formal charge on the nitrogen atom is therefore

    formalcharge(N)=5(0+82)=0

    Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore

    formalcharge(H)=1(0+22)=0

    The formal charges on the atoms in the NH4+ ion are thus

    5e0f7f069aee8f92ef91dd65654d5a9c.jpg

    Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.

    Exercise

    Write the formal charges on all atoms in BH4.

    Answer:

    c36560409246c6322bb1d1a06fed7586.jpg

    If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero.

    Using Formal Charges to Distinguish between Lewis Structures

    As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO2. Both structures conform to the rules for Lewis electron structures.

    CO2

    1. C is less electronegative than O, so it is the central atom.

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    2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.

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    3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.

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    4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:

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    2e184f83e0554f7763706e62c4cac493.jpg

    This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.

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    5. No electrons are left for the central atom.

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    6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:

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    953bde827ee40a4f9b2f65dc8de4f90a.jpg

    Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO2 are as follows:

    6b2663b9111b53775052ac27c63f60db.jpg

    Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.Edit section

    Note

    The Lewis structure with the set of formal charges closest to zero is usually the most stable.

    Example

    The thiocyanate ion (SCN), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.

    Given: chemical species

    Asked for: Lewis electron structures, formal charges, and preferred arrangement

    Strategy:

    A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN.

    B Calculate the formal charge on each atom using Equation 4.4.1.

    C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.

    Solution:

    A Possible Lewis structures for the SCN ion are as follows:

    ab56216b8fdfa8b4be145f90f26bc72e.jpg

    B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore 6(6+22)=1.5(4+42)=1 In (c), nitrogen has a formal charge of −2.

    C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2.

    Exercise

    Salts containing the fulminate ion (CNO) are used in explosive detonators. Draw three Lewis electron structures for CNO and use formal charges to predict which is more stable. (Note: N is the central atom.)

    Answer:

    b8461925b8631574db7a41c2b7a2f0ac.jpg

    The second structure is predicted to be more stable.

    Edit sectionc36560409246c6322bb1d1a06fed7586.jpg

    Drawing isomers of Lewis Structures

    Contributors

    • Anonymous

    Layne Morsch (University of Illinois Springfield)

    Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions. The octet rule is violated in these three scenarios:

    1. When there are an odd number of valence electrons
    2. When there are too few valence electrons
    3. When there are too many valence electrons

    Exception 1: Species with Odd Numbers of Electrons

    The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide (NO ,refer to figure one). Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in. The formula to find a formal charge is:

    Formal Charge= [# of valence e- the atom would have on its own] - [# of lone pair electrons on that atom]

    - [# of bonds that atom participates in]

    No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitrogen monoxide. Nitrogen monoxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitrogen monoxide had ten valence electrons we would come up with the Lewis Structure (Figure 8.7.1):

    Figure 8.7.1. This is if Nitrogen monoxide has only ten valence electrons, which it does not.

    Let's look at the formal charges of Figure 8.7.2 based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure 8.7.1, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure 8.7.1, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitrogen monoxide (Figure 8.7.2):

    Figure 8.7.2. The proper Lewis structure for NO molecule

    Free Radicals

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    There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with OH, the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted Cl. Interestingly, odd Number of Valence Electrons will result in the molecule being paramagnetic.

    Exception 2: Incomplete Octets

    The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH3 (Borane).

    If one was to make a Lewis structure for BH3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure 8.7.3):

    Figure 8.7.3

    The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH3 with Lewis theory. One of the things that may account for BH3's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps.

    Let's take a look at another incomplete octet situation dealing with boron, BF3 (Boron trifluorine). Like with BH3, the initial drawing of a Lewis structure of BF3 will form a structure where boron has only six electrons around it (Figure 8.7.4).

    BF3.jpg

    Figure 8.7.4

    If you look Figure 8.7.4, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure 8.7.5):

    BF3 pt. 2.jpg

    Figure 8.7.5

    Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1.

    This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure 8.7.5, a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (χ=4.0). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible.

    However the large electronegativity difference here, as opposed to in BH3, signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure 8.7.6:

    BF3 pt. 3.jpg

    Figure 8.7.6

    None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure 8.7.4), the one with the double bond (Figure 8.7.5), and the one with the ionic bond (Figure 8.7.6). The most contributing structure is probably the incomplete octet structure (due to Figure 8.7.5 being basically impossible and Figure 8.7.6 not matching up with the behavior and properties of BF3). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure.

    As a side note, it is important to note that BF3 frequently bonds with a F- ion in order to form BF4- rather than staying as BF3. This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF3 .

    Example 8.7.1: NF3

    Draw the Lewis structure for boron trifluoride (BF3).

    SOLUTION

    1. Add electrons (3*7) + 3 = 24

    2. Draw connectivities:

    3. Add octets to outer atoms:

    4. Add extra electrons (24-24=0) to central atom:

    5. Does central electron have octet?

    • NO. It has 6 electrons
    • Add a multiple bond (double bond) to see if central atom can achieve an octet:

    6. The central Boron now has an octet (there would be three resonance Lewis structures)

    However...

    • In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
    • The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
    • Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure

    BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:

    Exception 3: Expanded Valence Shells

    Edit section

    More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below, which those terminal atoms bond to. For example, PCl5 is a legitimate compound (whereas NCl5) is not:

    Note

    Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond

    The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available (l=2). The orbital diagram for the valence shell of phosphorous is:

    Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration:

    • The larger the central atom, the larger the number of electrons which can surround it
    • Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.

    There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule.

    One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule.

    Example 8.7.2: The SO24 ion

    Such is the case for the sulfate ion, SO4-2. A strict adherence to the octet rule forms the following Lewis structure:

    Sulfate.jpg

    Figure 8.7.12

    If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2.

    If instead we made a structure for the sulfate ion with an expanded octet, it would look like this:

    Sulfate pt. 2.jpg

    Figure 8.7.13

    Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure 8.7.12, as opposed to +2 and -1 (difference of 3) in Figure 8.7.12) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case.

    Example 8.7.3: The ICl4 Ion

    Draw the Lewis structure for ICl4 ion.

    SOLUTION

    1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons

    2. Draw the connectivities:

    3. Add octet of electrons to outer atoms:

    4. Add extra electrons (36-32=4) to central atom:

    5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)

    Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies.