Bridge
Resonance will be important when we discuss aromatic compounds and carboxylic acids in Organic Chemistry. It allows for great stability by spreading electrons and negative charges over a larger area.
Figure 3.3
The last two resonance structures of sulfur dioxide shown in Figure 3.3 have equivalent energy or stability. Often, nonequivalent resonance structures may be written for a compound. In these cases, the more stable the structure, the more it contributes to the character of the resonance hybrid. Conversely, the less stable the resonance structure is, the less that structure contributes to the resonance hybrid. It should be apparent by now that the minor resonances for SO2 are so because they induce a separation of charge such that the oxygen with the single bond and the extra lone-pair electron has a formal charge of -1 and the sulfur with the three bonds has a formal charge of +1. The major resonance structure is so because each atom has a formal charge of 0. Use formal charge to assess the stability of particular resonance structures qualitatively according to the following guidelines:
• A Lewis structure with small or no formal charges is preferred over a Lewis structure with large formal charges.
• A Lewis structure with less separation between opposite charges is preferred over a Lewis structure with a large separation of opposite charges.
• A Lewis structure in which negative formal charges are placed on more electronegative atoms is more stable than one in which the negative formal charges are placed on less electronegative atoms.
Key Concept
As noted before about atoms striving for nobility, here we see that molecules do the same. Charges that are spread over multiple atoms are more stable because they are essentially diluted.
Example: Write the resonance structures for [NCO]-.
Solution:
1. C is the least electronegative of the three given atoms, N, C, and O. Therefore the C atom occupies the central position in the skeletal structure of [NCO]-.
NCO
2. N has 5 valence electrons; C has 4 valence electrons; O has 6 valence electrons; and the species itself has one negative charge. Total valence electrons = 5 + 4 + 6 + 1 = 16.
3. Draw single bonds between the central C atom and the surrounding atoms, N and O. Place a pair of electrons in each bond.
N : C : O
4. Complete the octets of N and O with the remaining 16 - 4 = 12 electrons.
5. The C octet is incomplete. There are three ways in which double and triple bonds can be formed to complete the C octet: Two lone pairs from the O atom can be used to form a triple bond between the C and O atoms:
Or one lone electron pair can be taken from both the O and the N atoms to form two double bonds, one between N and C, the other between O and C:
Or two lone electron pairs can be taken from the N atom to form a triple bond between the C and N atoms:
These three are all resonance structures of [NCO]-.
6. Assign formal charges to each atom of each resonance structure.
The most stable structure is this:
because the negative formal charge is on the most electronegative atom, O.
Exceptions to the Octet Rule
We have stated repeatedly throughout this and earlier chapters: The octet rule has many exceptions. In addition to hydrogen, helium, lithium, beryllium, and boron, which are exceptions because they cannot or do not reach the octet, all elements in or beyond the third period may be exceptions because they can have more than eight electrons in their valence shells. These electrons can be placed into orbitals of the d-subshell, and as a result, atoms of these elements can form more than four bonds. On Test Day, don’t automatically discount a Lewis structure that shows an atom with more than four bonds—the test makers may be testing your ability to recognize the capability of many atoms to expand their valence shells beyond the octet. Always think critically about the information provided in the passage or question stem and synthesize it with the understanding that you are gaining here today and in your ongoing preparation.
Consider the sulfate ion, SO42-. When drawing the Lewis structure of the sulfate ion, giving the sulfur 12 valence electrons permits three of the five atoms to be assigned a formal charge of zero. The sulfate ion can be drawn in six resonance forms, each with the two double bonds attached to a different combination of oxygen atoms (see Figure 3.4).
MCAT Expertise
As with all rules, the octet “rule” has exceptions. It always applies to neutral atoms and anions of C, N, O, and F (which are common on the MCAT). It often (not always!) applies to the halogens and other representative elements. However, it never applies to H, He, Li, Be, or neutral B and Al. Other notable exceptions are elements beyond period 3 because these elements have d-orbitals. (Sulfur and phosphorus will be the most common examples seen on Test Day.)
Figure 3.4
GEOMETRY AND POLARITY OF COVALENT MOLECULES
Because Lewis structures do not in any way suggest or reflect the actual geometric arrangement of atoms in a compound, we need another system that provides that information. One such system is known as valence shell electron pair repulsion theory (VSEPR theory). This theory is actually dependent upon Lewis structure, so they go hand in hand.
Valence Shell Electron Pair Repulsion (VSEPR) Theory
VSEPR theory uses Lewis structure to predict the molecular geometry of covalently bonded molecules. It states that the three-dimensional arrangement of atoms surrounding a central atom is determined by the repulsions between the bonding and the nonbonding electron pairs in the valence shell of the central atom. These electron pairs arrange themselves as far apart as possible, thereby minimizing the repulsive forces (see Table 3.2). The following steps are used to predict the geometrical structure of a molecule using the VSEPR theory:
• Draw the Lewis structure of the molecule.
• Count the total number of bonding and nonbonding electron pairs in the valence shell of the central atom.
• Arrange the electron pairs around the central atom so that they are as far apart from each other as possible. For example, the compound AX2 has the Lewis structure X : A : X. The A atom has two bonding electron pairs in its valence shell. To position these electron pairs as far apart as possible, their geometric structure should be linear:
X-A-X
MCAT Expertise
Knowing the tetrahedral shape will be particularly useful because it is often present in carbon, nitrogen, and oxygen.
Table 3.2
Example: Predict the geometry of NH3.
Solution:
1. The Lewis structure of NH3 is
2. The central atom, N, has three bonding electron pairs and one nonbonding electron pair, for a total of four electron pairs.
3. The four electron pairs will be farthest apart when they occupy the corners of a tetrahedron. As one of the four electron pairs is a lone pair, the observed geometry is trigonal pyramidal, shown in Figure 3.5.