Bonding and Chemical Interactions - Review - MCAT General Chemistry Review

MCAT General Chemistry Review

Part I Review

Chapter 3: Bonding and Chemical Interactions

The Maillard reaction is one of the more important chemical processes that occur in the process of cooking and baking. (Yes, here we go with food preparation, again. This is, in fact, relevant to your MCAT preparation.) The reaction mechanism itself is one with which you are (or will be) closely familiar from your studies of organic chemistry: a nucleophilic reaction between the amino terminus of the peptide chain of a protein and the carbonyl functionality of a sugar to form an N-substituted glycosylamine. This compound then undergoes a complex series of rearrangements and other reactions to produce a set of compounds that gives cooked food its pleasing brown color and delectable flavor. For no food preparation is this reaction more important than for meat (or meat substitute for all you vegetarians out there).

When the surface of the meat (or meat substitute) comes into contact with the hot surface of a pan or grill, the proteins and sugars on the meat exterior begin interacting through the Maillard reaction. The pan must be sufficiently hot to bring the exterior of the meat to a temperature of around 155°C (310°F), which is the optimal temperature for the reaction to occur. Of course, those willing to shell out $50 or more for a steak at a fine steakhouse are expecting perfection and good taste through and through, not just on the surface. So how does a grill master achieve the impossible: generating very high heat for the exterior but not overcooking the interior? The answer, or at least part of it, is this: drying the meat. When meat that has a lot of water on its exterior surface hits the hot pan, the first process that takes place is the boiling of the water. Boiling is a phase change from liquid to gas and occurs at a constant temperature; water’s boiling point is 100°C (212°F). Because this temperature is considerably lower than that necessary for the Maillard reaction, no browning will occur and the flavor compounds so sought after will not form. Rather, the meat will essentially steam, and the end product will be a lifeless, overcooked hunk of toughened proteins unworthy of its hefty price tag. The lesson here is, if you want a tasty steak, always dry your meat!

Of course, the real lesson is the topic of discussion for this chapter: bonding and chemical interactions. We will not actually address complex chemical bonding, such as that which takes place in the Maillard reaction, in this chapter. (We cannot stress enough, however, that the nucleophilic mechanism by which many reactions, like the Maillard reaction, proceed will be tested in the MCAT’s Biological Sciences section.) Rather, this chapter will address the basics of chemical bonding and interactions. Here, we will investigate the nature and behavior of covalent and ionic bonds. We will also review a system by which bonding electrons are accounted for, Lewis structures, and go over the main principles of valence shell electron pair repulsion (VSEPR) theory. Finally, we will recount the various modes of interaction between molecules, the intermolecular forces.



The atoms of most elements, except for a few noble gases, can combine to form molecules. The atoms in most molecules are held together by strong attractive forces called chemical bonds, which are formed via the interaction of the valence electrons of the combining atoms. The chemical and physical properties of the resulting compound are usually very different from those of the constituent elements. For example, elemental sodium, an alkali metal, is so reactive that it can actually produce fire when reacting with water (the reaction is highly exothermic), and elemental chlorine gas is so toxic that it was used for chemical warfare during World War I. However, when an atom of sodium and an atom of chlorine react, the produced ionic compound, sodium chloride, is safe for us to eat. You may know it better as common table salt!


Electronegativity (which we learned about in the last chapter) is a property that addresses how an individual atom acts within a bond and will help us understand the quality of the molecules formed from atoms with different electronegativities.

Why do atoms join together to form compounds? Why do the sodium atom and the chlorine atom form sodium chloride? For many molecules, the constituent atoms have bonded according to the octet rule, which states that an atom tends to bond with other atoms until it has eight electrons in its outermost shell, thereby forming a stable electron configuration similar to that of the noble gases. However, this is not a hard and fast rule, more a “rule of thumb,” and as we suggested in the first chapter, there are more elements that can be exceptions to the rule than there are elements that follow the rule without exception. These “exceptional” elements include hydrogen, which can only have two valence electrons (achieving the configuration of helium); lithium and beryllium, which bond to attain two and four valence electrons, respectively; boron, which bonds to attain six valence electrons; and all elements in period 3 and below, which can expand the valence shell to include more than eight electrons by incorporating d-subshell orbitals. For example, in certain compounds, chlorine can form seven covalent bonds, thereby holding 14 electrons in its valence shell.


The octet rule is kind of like the English language, with its eight parts of speech. Nobles generally speak properly, using all eight parts as they are intended. Most people aren’t nobles, however. Civilians might desire to be so sophisticated, but unless they find someone to share some wisdom with, they’ll speak in slang, perhaps neglecting some prepositions or adverbs.

A simple way to remember all the exceptions is as follows:

• Hydrogen is excused from the octet rule because it doesn’t have enough “space” for eight electrons, it only has the one s-subshell (which can hold a maximum of two electrons).

• Lithium, beryllium, and boron are just lazy—they have enough room because they have both s- and p-orbitals to hold a total of eight electrons, but they’d rather not put in all the hard work to get all eight.

• All the elements in period 3 and below have extra storage space in their attics, so they can hold more than eight electrons if they want to.

Another way to remember the exceptions (and one that’s even easier) is to remember the common elements that almost always abide by the octet rule: carbon, nitrogen, oxygen, fluorine, sodium, and magnesium. (We include the last two even though they lose—rather than gain—sufficient electrons to end up with a completed octet.)

We classify chemical bonds into two distinct types: ionic and covalent. In ionic bonding, one or more electrons from an atom with lower ionization energy, typically a metal, are transferred to an atom with greater electron affinity, typically a nonmetal, and the electrostatic force of attraction between opposite charges holds the resulting ions together. This is the nature of the bond in sodium chloride, where the positively charged sodium cation is electrostatically attracted to the negatively charged chloride anion. In covalent bonding, an electron pair is shared between the two atoms, typically two nonmetals, which have relatively similar values of electronegativity. The degree to which the pair of electrons is shared equally or unequally between the two atoms determines the degree of polarity in the covalent bond. If the electron pair is shared equally, the covalent bond isnonpolar; if the pair is shared unequally, the bond is polar. If both electrons being shared were contributed by only one of the two atoms, the bond is called coordinate covalent.

Ionic Bonds


Ionic bonds form between atoms that have significantly different electronegativities. The atom that loses the electron(s) becomes a cation, and the atom that gains the electron(s) becomes an anion. The resulting ionic bond is the electrostatic force of attraction between the opposite charges of the ions. There is no sharing of electrons in an ionic bond. For this electron transfer to occur, the difference in electronegativity must be greater than 1.7 on the 4.0-Pauling scale. In general, you will recognize ionic bonds forming between the alkali metals and the alkaline earth metals of Groups IA and IIA (Groups 1 and 2) and the halogens of Group VIIA (Group 17). The atoms of the active metals have one or two electrons, which they hold onto only loosely. The atoms of the halogens are strongly “interested in” gaining one more electron to complete their valence shells. These tendencies explain the formation of the ionic bond in, say, sodium chloride, which we’ve already discussed.


The t in cation looks like a plus sign: ca + ion.

Ionic compounds have characteristic physical properties, which you should recognize for Test Day. Because of the strength of the electrostatic force between the ionic constituents of the compound, ionic compounds have very high melting and boiling points. The melting point of sodium chloride is greater than 800°C. Many ionic compounds dissolve readily in aqueous and other polar solvent solutions and, in the molten or aqueous state, are good conductors of electricity. In the solid state, the ionic constituents of the compound form a crystalline lattice consisting of repeating positive and negative ions in which the attractive forces between oppositely charged ions maximize, while the repulsive forces between ions of like charge minimize.

For example, if you were to analyze the atomic structure of the salt you just spread over the meat that you’re going to grill up for dinner, you would see that each sodium ion is surrounded by six chloride ions and each chloride is surrounded by six sodium ions. This is a lattice formation known as 6:6 coordinated. (The optimal time to salt meat is about 30 minutes prior to cooking so that the salt has sufficient time to draw fluid to the surface by osmosis, creating a salt solution, which then gets drawn back into the muscle fiber once again. This results in well-seasoned meat through the entire thickness, not just at the surface.)

Covalent Bonds


When two or more atoms with similar electronegativities interact, the energy required to form ions through the complete transfer of one or more electrons is greater than the energy that would be released upon the formation of an ionic bond. That is to say, when two atoms of similar tendency form a compound to attract electrons in a bond, it is energetically unfavorable to form ions. So, rather than struggling to form ions, the atoms simply opt to share the electrons as a compromise, which allows them both to fill their valence shells. The binding force between the atoms is not ionic; rather, it is the attraction that each electron in the shared pair has for the two positive nuclei of the bonded atoms.

MCAT Expertise

Think of bonds as a tug-of-war between two atoms. When the difference in electronegativity is great (more than 1.7) then the “stronger” molecule wins all of the electrons and becomes the anion. When the electronegative values are relatively similar, then we have a stalemate, or a covalent bond with mostly equal sharing of electrons.

Covalent compounds contain discrete molecular units with relatively weak intermolecular interactions. Consequently, these compounds tend to have lower melting and boiling points, and because they do not break down into constituent ions, they are poor conductors of electricity in the liquid state or in aqueous solutions.


The formation of one covalent bond may not be sufficient to fill the valence shell for a given atom. Thus, many atoms can form bonds with more than one other atom, and each atom can form multiple bonds between itself and another atom, with few exceptions. Two atoms sharing one, two, or three pairs of electrons are said to be joined by a single, double, or triple covalent bond, respectively. The number of shared electron pairs between two atoms is called the bond order; hence, a single bond has a bond order of one, a double bond has a bond order of two, and a triple bond has a bond order of three. Three features characterize a covalent bond: bond length, bond energy, and polarity.

Bond Length

If the arrangement of atoms in covalent bonds can be likened to two wooden spools connected by a dowel in a very sophisticated children’s construction set, then the length of the dowel is analogous to the length of the bond between the atoms sharing the electron pair(s). Bond length is the average distance between the two nuclei of the atoms involved in the bond. As the number of shared electron pairs increases, the two atoms are pulled closer together, leading to a decrease in bond lengths. Thus, for a given pair of atoms, a triple bond is shorter than a double bond, which is shorter than a single bond.


We will see a great example of covalent bonds in organic chemistry, and we can see here the inverse proportionality between bond length and strength.

Bond Length

Bond Strength










Bond Energy

Bond energy is the energy required to break a bond by separating its components into their isolated, gaseous atomic states. Just as it becomes increasingly difficult to snap a tree branch of increasing thickness, the greater the number of pairs of electrons shared between the atomic nuclei, the more energy is required to “break” the bond(s) holding the atoms together. Thus, triple bonds have the greatest bond energy, and single bonds have the lowest bond energy. We will discuss bond energy and calculations involving actual bond energy values (called bond enthalpy) in Chapter 6, Thermochemistry. By convention, the greater the bond energy is, the “stronger” the bond.


As in the case of two people who decide to share some commodity for the purpose of achieving a larger goal and who must then decide the degree to which the common commodity will be shared equally or unequally between them (that commodity might be money, land, pizza, or an apartment), atoms that come together in covalent bonds also “negotiate” the degree to which their sharing of electron pair(s) will be equal or unequal. The nature and degree of sharing between the nuclei of two atoms in a covalent bond is determined by the relative difference in their respective electronegativities, with the atom of higher electronegativity getting the larger “share” of the electron pair(s). A polar bond is a dipole, with the positive end of the dipole at the less electronegative atom and the negative end at the more electronegative atom.

Nonpolar Covalent Bond

When atoms that have identical or nearly identical electronegativities share electron pair(s), they do so with equal distribution of the electrons. This is called a nonpolar covalent bond, and there is no separation of charge across the bond. Of course, only bonds between atoms of the same element will have exactly the same electronegativity and, therefore, share with perfectly equal distribution the pair(s) of electrons in the covalent bond. Examples of diatomic molecules include H2, Cl2, O2, and N2. At the same time, many bonds can be said to be approximately nonpolar. For example, the electronegativity difference between carbon and hydrogen is so sufficiently small that we can usually consider the C–H bond to be effectively nonpolar.

Polar Covalent Bond

Of course nobody really likes to share anything. We only say we do because we know that’s what expected of us. When nobody’s looking, a child will grab a few extra candies for herself, and even adults are known to fight over property lines, the last slice of pizza, or whether someone paid his fair share of the dinner tab. Atoms of elements that differ moderately in their electronegativities will share their electrons unevenly, resulting in polar covalent bonds. While the difference in their electronegativities (typically between 0.4 and 1.7 Pauling units) is not enough to result in the formation of an ionic bond, it is sufficient to cause a separation of charge across the bond, with the more electronegative element acquiring a greater portion of the electron pair(s) and taking on a partial negative charge, image, and the less electronegative element acquiring a smaller portion of the electron pair(s) and taking on a partial positive charge, image. For instance, the covalent bond in HCl is polar because the two atoms have a moderate difference in electronegativity (approximately 0.9). The chlorine atom gains a partial negative charge, and the hydrogen atom gains a partial positive charge. The difference in charge between the atoms is indicated by an arrow crossed at its tail end (giving the appearance of a “plus” sign) pointing toward the negative end, as shown in Figure 3.1.


Figure 3.1

A molecule that has such a separation of positive and negative charges is called a polar molecule. The dipole moment of the polar bond or polar molecule is a vector quantity, µ, defined as the product of the charge magnitude (q) and the distance between the two partial charges (r):

µ = qr

The dipole moment vector, represented by an arrow pointing from the positive to the negative charge, is measured in Debye units (coulomb-meter). Please note that the convention used by chemists for designating the direction of the dipole moment from positive to negative is the opposite of the convention used by physicists, who designate the direction of a dipole moment from negative to positive.


In a coordinate covalent bond, the shared electron pair comes from the lone pair of one of the atoms in the molecule, while the other atom involved in the bond contributes nothing to the relationship. (This works great for atoms but might not be the best way to form a lasting marriage!) Once such a bond forms, however, it is indistinguishable from any other covalent bond. The distinction is only helpful for keeping track of the valence electrons and formal charges (see Figure 3.2). Coordinate covalent bonds are typically found in Lewis acid-base compounds (see Chapter 10, Acids and Bases). A Lewis acid is any compound that will accept a lone pair of electrons, while a Lewis base is any compound that will donate a pair of electrons to form a covalent bond; for example, as in the reaction between borontrifluoride (BF3) and ammonia (NH3) shown in Figure 3.2.


Figure 3.2

NH3 donates a pair of electrons to form a coordinate covalent bond; thus, it acts as a Lewis base. BF3 accepts this pair of electrons to form the coordinate covalent bond; thus, it acts as a Lewis acid. Lewis acids, incidentally, are commonly encountered in some often-tested organic chemistry reactions, such as anti addition across double bonds, and as catalysts in electrophilic aromatic substitutions (EAS).


The electrons that are involved in a covalent bond are in the valence shell and are called bonding electrons, while those electrons in the valence shell that are not involved in the covalent bond are called nonbonding electrons. The unshared electron pairs can also be called lone electron pairs, because they are associated only with one atomic nucleus. Because atoms can bond with other atoms in many different possible combinations, the Lewis structure system of notation has been developed to keep track of the bonded and nonbonded electron pairs. You may think of Lewis structures as a kind of bookkeeping. The number of valence electrons attributed to a particular atom in the Lewis structure of a molecule is not necessarily the same as the number of valence electrons the atom would have as an isolated atom, and the difference accounts for what is referred to as the formal charge of that atom in a particular Lewis structure. Often, more than one Lewis structure can be drawn for a molecule. If the possible Lewis structures differ in the way the atoms are connected (that is to say, they differ in their bond connectivity or arrangement), then the Lewis structures represent different possible compounds. If the Lewis structures show the same bond connectivity and differ only in the arrangement of the electron pairs, then these structures represent different resonance forms for a single compound. Lewis structures do not represent the actual or even theoretical geometry of a real compound. Their usefulness lies in showing you the different possible ways in which atoms may be combined to form different compounds or resonances of a single compound. When more than one arrangement can be made, you can assess the likelihood of each arrangement by checking the formal charges on the atoms in each arrangement. The arrangement that minimizes the number of formal charges, or minimizes the value of formal charges, is the arrangement that most likely represents the actual compound.

Key Concept

When dealing with Lewis dot structures, we only deal with the eight valence electrons (s- and p-orbitals of the outer shell) on each atom. Remember that some atoms can expand their octets by utilizing the d-orbitals in this outer shell, but this will only take place with atoms in period 3 or greater.

Lewis Structures

A Lewis structure, or Lewis dot symbol, is the chemical symbol of an element surrounded by dots, each representing one of the s and/or p valence electrons of the atom. The Lewis symbols of the elements in the second period of the periodic table are shown in Table 3.1.

Table 3.1


MCAT Expertise

All of the Kaplan strategies are based on methodical approaches to problems and concepts, which save you time on Test Day.

Just as a Lewis symbol is used to represent the distribution of valence electrons in an atom, it can also be used to represent the distribution of valence electrons in a molecule. For example, the Lewis symbol of a fluoride ion, 047 the Lewis structure of the diatomic molecule 047 Certain steps must be followed in assigning a Lewis structure to a molecule. The steps are outlined here, using HCN as an example. (Hydrogen cyanide, an extremely poisonous compound, is found in small amounts in the pits of fruits such as cherries and apples.)

• Write the skeletal structure of the compound (i.e., the arrangement of atoms). In general, the least electronegative atom is the central atom. Hydrogen (always) and the halogens F, Cl, Br, and I (usually) occupy the end position.

In HCN, H must occupy an end position. Of the remaining two atoms, C is the least electronegative and, therefore, occupies the central position. The skeletal structure is as follows:


• Count all the valence electrons of the atoms. The number of valence electrons of the molecule is the sum of the valence electrons of all atoms present:

H has 1 valence electron;
C has 4 valence electrons;
N has 5 valence electrons; therefore,
HCN has a total of 10 valence electrons.

• Draw single bonds between the central atom and the atoms surrounding it. Place an electron pair in each bond (bonding electron pair).

H : C : N

Each bond has two electrons, so 10 - 4 = 6 valence electrons remain.

• Complete the octets (eight valence electrons) of all atoms bonded to the central atom, using the remaining valence electrons still to be assigned. (Recall that H is an exception to the octet rule because it can have only two valence electrons.) In this example, H already has two valence electrons in its bond with C.


• Place any extra electrons on the central atom. If the central atom has less than an octet, try to write double or triple bonds between the central and surrounding atoms using the nonbonding, unshared lone electron pairs.

The HCN structure above does not satisfy the octet rule for C because C possesses only four valence electrons. Therefore, two lone electron pairs from the N atom must be moved to form two more bonds with C, creating a triple bond between C and N. Finally, bonds are drawn as lines rather than pairs of dots.

H-CimageN :

Now, the octet rule is satisfied for all three atoms, because C and N have eight valence electrons and H has two valence electrons.

Formal Charge

In evaluating a Lewis structure to determine whether or not it may likely represent the actual arrangement of atoms in a compound, you will calculate the formal charge on each atom in the proposed Lewis structure. In doing so, you must be aware that you are assuming a perfectly equal sharing of all bonded electron pairs, regardless of actual differences in electronegativity, such that each electron pair is split evenly between the two atomic nuclei that share it. When you compare the number of electrons assigned to an atom in a Lewis structure (assigning one electron of each bonded pair to each of the atoms involved in the bond) to the number of electrons normally found in that atom’s valence shell, the difference between the two numbers is the formal charge. A fairly simple equation you can use to calculate formal charge is

Formal Charge = V - Nnonbonding - ½Nbonding

where V is the normal number of electrons in the atom’s valence shell, Nnonbonding is the number of nonbonding electrons, and Nbonding is the number of bonding

MCAT Expertise

Practice with many molecules and remembering the number of bonds that common central atoms usually form will make this formula easier to use on Test Day. For example, the nitrogen atom below normally has three bonds and one lone pair. Here, it is feeling generous and sharing more than usual; therefore, it will have a positive charge. If a molecule is selfish and shares less than usual, it will be negative (as we often see with oxygen atoms).

electrons. The charge of an ion or compound is equal to the sum of the formal charges of the individual atoms comprising the ion or compound.

Example: Calculate the formal charge on the central N atom of [NH4]+.

Solution: The Lewis structure of [NH4]+ is


Nitrogen is in group VA; thus it has five valence electrons. In [NH4]+,

N has 4 bonds (i.e., eight bonding electrons and no nonbonding electrons).

So V = 5; Nbonding = 8; Nnonbonding = 0.

Formal charge = 5 - 050 (8) - 0 = + 1

Thus, the formal charge on the N atom in [NH4]+ is +1.

Let us offer a brief note of explanation on the difference between formal charge and oxidation number, as we are sure that you lay awake at night pondering such questions. It’s quite simple, really: Formal charge underestimates (ignores, actually) the effect of electronegativity differences, while oxidation numbers overestimate the effect of electronegativity, assuming that the more electronegative atom will in fact have a 100 percent share of the bonding electron pair. For example, in a molecule of CO2 (carbon dioxide), the formal charge on each of the atoms would be 0, but the oxidation number of each of the oxygen atoms would be -2 and the carbon would have an oxidation number of +4. In reality, the distribution of electron density between the carbon and oxygen atoms lies somewhere between the extremes predicted by the formal charges and the oxidation states.


As we’ve suggested, you may be able to draw two or more nearly identical Lewis structures that demonstrate the same arrangement between the atoms but differ in the specific placement of some pairs of electrons. These are called resonance structures. The actual electronic distribution in the real compound is a hybrid, or composite, of all the possible resonances. For example, SO2 has three resonance structures, two of which are minor: O=S–O and O–S=O. The third is the major structure: O=S=O. The nature of the bonds within the actual compound is a hybrid of these three structures; indeed, spectral data indicate that the two S–O bonds are identical and equivalent. This phenomenon is known as resonance, and the actual structure of the compound is called the resonance hybrid. Resonance structures are expressed with a double-headed arrow between them.


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]-.


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]-.


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


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:


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.


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.


Figure 3.5

In describing the shape of a molecule, only the arrangement of atoms (not electrons) is considered. Even though the electron pairs are arranged tetrahedrally, the shape of NH3 is pyramidal. It is not trigonal planar because the lone pair repels the three bonding electron pairs, causing them to move as far away as possible.

Example: Predict the geometry of CO2.

Solution: The Lewis structure of CO2 is 061

The double bond behaves just like a single bond for purposes of predicting molecular shape. This compound has two groups of electrons around the carbon. According to the VSEPR theory, the two sets of electrons will orient themselves 180° apart, on opposite sides of the carbon atom, minimizing electron repulsion. Therefore, the molecular structure of CO2 is linear: 062 .

Key Concept

The shapes from Table 3.2 refer to electronic geometry, which is different from molecular geometry. In Figure 3.5, we see an ammonia molecule that has a tetrahedral electronic structure but is considered to have a molecular structure that is trigonal pyramidal.

One subtlety that students sometimes miss is the difference between electronic geometry and molecular geometry. Electronic geometry describes the spatial arrangement of all pairs of electrons around the central atom, including the bonding and the lone pairs. Molecular geometry describes the spatial arrangement of only the bonding pairs of electrons, in much the same way that the geometry of an object made from our sophisticated construction set would be determined by the position of the wooden spools attached to the dowels. For example, consider the fact that CH4 (methane), NH3(ammonia), and H2O all have the same electronic geometry: In each compound, four pairs of electrons surround the central atom. This is tetrahedral electronic geometry. However, because each has a different number of lone pairs, each has a different molecular geometry. Methane has tetrahedral geometry, ammonia has trigonal pyramidal, and water has angular or bent geometry. The distinction is important, and the MCAT will primarily focus on molecular geometry, but there is one important implication of electronic geometry: the determination of the ideal bond angle. Tetrahedral electronic geometry, for example, is associated with an ideal bond angle of 109.5°. Thus, molecular geometries that deviate from tetrahedral electronic geometry, such as those of ammonia and water, have bond angles that are deviations from 109.5°. You may have been tempted to say, for example, that water’s bond angle is a deviation from linear geometry with its ideal bond angle of 180°, but this is not the case. The actual bond angle in water is around 104.5°, a deviation from 109.5°.

Polarity of Molecules

When two atoms of different electronegativities bond covalently by sharing one or more pairs of electrons, the resulting bond is polar, with the more electronegative atom possessing the greater share of the electron density. However, the mere presence of bond dipoles does not necessarily result in a molecular dipole; that is, an overall separation of charge across the molecule. We must first consider the molecular geometry and the vector addition of the bond dipoles based upon that molecular geometry. A compound with nonpolar bonds is always nonpolar; a compound with polar bonds may be polar or nonpolar, depending upon the spatial orientation of the polar bonds within the given molecular geometry.

A compound composed of two atoms bound by a polar bond must have a net dipole moment and is therefore polar. The two equal and opposite partial charges are localized on the two atoms at the ends of the compound. HCl (hydrogen chloride) is a good example of this, because the bond between the hydrogen and chlorine atom is polar (with the hydrogen atom assuming a partial positive charge and the chlorine atom assuming a partial negative charge); the compound must also be polar, with a molecular dipole moment in the same direction and same magnitude as the bond dipole moment. A compound consisting of more than two atoms bound with polar bonds may be either polar or nonpolar, because the overall dipole moment of a molecule is the vector sum of the individual bond dipole moments. If the compound has a particular molecular geometry such that the bond dipole moments cancel each other (i.e., if the vector sum is zero), then the result is a nonpolar compound. For instance, CCl4 has four polar C–Cl bonds, but because the molecular geometry of carbon tetrachloride is tetrahedral, the four bond dipoles point to the vertices of the tetrahedron and, therefore, cancel each other out, resulting in a nonpolar compound (see Figure 3.6).

MCAT Expertise

Back to that tug-of-war from earlier, sometimes we can see the winner before the final flag. A polar covalent bond will have one atom that carries more electron density than the other does (and therefore a partial negative charge) but hasn’t won the match yet.


Figure 3.6

However, when the molecular geometry is such that the bond dipoles do not cancel each other, the molecule will have a net dipole moment and, therefore, will be polar. For instance, the O–H bonds in H2O are polar, with each hydrogen molecule assuming a partial positive charge and the oxygen assuming a partial negative charge. Because water’s molecular geometry is angular (bent), the vector summation of the bond dipoles results in a molecular dipole moment from the partially positive hydrogen end to the partially negative oxygen end, as illustrated in Figure 3.7.

Key Concept

A molecule with polar bonds need not be polar: The bond dipole moments may cancel each other out, resulting in a nonpolar molecule. Although a molecule with polar bonds need not be polar overall, a polar molecule must have polar bonds.


Figure 3.7


To finish our discussion of covalent bonds, we need to address the issue of atomic and molecular orbitals. If you remember back to the first chapter, we described the modern understanding of the atom as a dense, positively charged nucleus surrounded by clouds of electrons organized into orbitals (regions in space surrounding the nucleus within which there are certain probabilities of finding an electron). The four quantum numbers completely describe the energy and position of any electron of an atom. While the principal quantum number indicates the average energy level of the orbitals, the azimuthal quantum number, l, describes the subshells within each principal energy level, n. When l = 0, this indicates the s-subshell, which has one orbital that is spherical in shape. The 1s-orbital (n = 1, l = 0) is plotted in Figure 3.8.


Quantum Numbers (Chapter 1) revisited:

• For any value of n, there are n values of l(0imagen- 1).

l= 0images
= 1 imagep
= 2imaged

• For any value of l, there are 2l + 1 values of ml (number of orbitals); values themselves will range from -l to l.


Figure 3.8

When l = 1, this indicates the p-subshell, which has three orbitals shaped like barbells along the x, y, and z axes at right angles to each other. The p-orbitals are plotted in Figure 3.9.


Figure 3.9

Although well beyond the scope of the MCAT, mathematical analyses of the wave function of the orbitals are used to determine and assign plus and minus signs to each lobe of the p-orbitals. The shapes of the five d-orbitals and the seven f-orbitals are more complex and need not be memorized for the MCAT. When two atoms bond to form a compound, the atomic orbitals interact to form a molecular orbital that describes the probability of finding the bonding electrons. Molecular orbitals are obtained by combining the wave functions of the atomic orbitals. Qualitatively, the overlap of two atomic orbitals describes this. If the signs of the two atomic orbitals are the same, a bonding orbital forms. If the signs are different, an antibonding orbital forms.

Two different patterns of overlap are observed in the formation of molecular bonds. When orbitals overlap head-to-head, the resulting bond is called a sigma ( image) bond. Sigma bonds allow for relatively free rotation, because the electron density of the bonding orbital is a single linear accumulation between the atomic nuclei. When the orbitals overlap in such a way that there are two parallel electron cloud densities, a pi (image) bond is formed. Pi bonds do not allow for free rotation because the electron densities of the orbital are parallel. To picture this difference, imagine pushing your fists together and rotating them as they touch—this is a sigma bond. Now imagine putting your arms out in front of you, holding them parallel to each other with your elbows bent at about right angles while another person slips two rubber bands around your forearms, one near the elbows and the second closer to the wrists. With the rubber bands on, try bringing one forearm up toward you while you extend your other forearm out away from you. The parallel tension provided by the rubber bands makes it quite difficult to rotate your forearms in opposite directions. This is a pi bond.


The pi bonds of alkenes, alkynes, aromatic compounds, and carboxylic acid derivates are what lend the ever-important functionalities in organic chemistry.

The Intermolecular Forces


Like guests at a cocktail party who mingle but ultimately have little to say to each other, atoms and compounds participate in weak electrostatic interactions. The strength of these intermolecular interactions can impact certain physical properties of the compounds, such as melting and boiling point. The weakest of the intermolecular interactions are the dispersion forces, also known as London forces. Next, are the dipole–dipole interactions, which are of intermediate strength. Finally, we have the strongest type of interaction, the hydrogen bond, which is a misnomer because there is no sharing or transfer of electrons, and consequently, it is not a true bond. We must keep in mind, however, that even hydrogen bonds, the strongest of these interactions, have only about 10 percent the strength of a covalent bond, so these electrostatic interactions can be overcome with additions of small or moderate amounts of energy.


These intermolecular forces are the binding forces that keep a substance together in its solid or liquid state (see Chapter 8). These same forces determine whether two substances are miscible or immiscible in the solution phase (see Chapter 9).


The bonding electrons in nonpolar covalent bonds may appear, on paper, to be shared equally between two atoms, but at any point in time, they will be located randomly throughout the orbital. For these instantaneous moments, then, the electron density may be unequally distributed between the two atoms. This results in rapid polarization and counterpolarization of the electron cloud and the formation of short-lived dipole moments. These dipoles interact with the electron clouds of neighboring compounds, inducing the formation of more dipoles. The momentarily negative end of one molecule will cause the closest region in any neighboring molecule to become temporarily positive itself, thus causing the other end of this neighboring molecule to become temporarily negative, which in turn induces other molecules to become temporarily polarized, and the entire process begins all over again. The attractive interactions of these short-lived and rapidly shifting dipoles are called dispersion forces or London forces.

Real World

While London forces (a type of van der Waals force) are the weakest of the intermolecular attractions, when there are millions of these interactions, as there are on the bottom of a gecko’s foot, there is an amazing power of adhesion, which is demonstrated by the animal’s ability to climb smooth vertical, even inverted, surfaces.

Dispersion forces are the weakest of all the intermolecular interactions because they are the result of induced dipoles that change and shift moment by moment. They do not extend over long distances and are, therefore, significant only when molecules are close together. The strength of the London force also depends on the degree to which and the ease by which the molecules can be polarized (that is to say, how easily the electrons can be shifted around). Large molecules in which the electrons are far from the nucleus are relatively easy to polarize and, thus, possess greater dispersion forces. Think about two plates—a big dinner plate and a small teacup saucer—with nearly equal numbers of marbles on each. The marbles will roll around more, and may be distributed more unequally, on the bigger plate than on the smaller saucer. Nevertheless, don’t underestimate the importance of the dispersion forces. If it weren’t for them, the noble gases would not liquefy at any temperature because no other intermolecular forces exist between the noble gas atoms. The low temperatures at which the noble gases liquefy is to some extent indicative of the magnitude of the dispersion forces between the atoms.


Polar molecules tend to orient themselves in such a way that the opposite ends of the respective molecular dipoles are closest to each other: The positive region of one molecule is close to the negative region of another molecule. This arrangement is energetically favorable because an attractive electrostatic force is formed between the two molecules.

Dipole–dipole interactions are present in the solid and liquid phases but become negligible in the gas phase because of the significantly increased distance between gas particles. Polar species tend to have higher melting and boiling points than those of nonpolar species of comparable molecular weight. You should realize that London forces and dipole–dipole interactions are different not in kind but in degree. Both are electrostatic forces between opposite partial charges; the difference is only in the strength and in the permanence of the molecular dipole.


Hydrogen bonds are a favorite of the MCAT! So understand them well, and you will be able to answer every question pertaining to their nature, behavior, and effects. A hydrogen bond is actually only a specific, unusually strong form of dipole–dipole interaction, which may be intra- or intermolecular. Please understand that hydrogen bonds are not actually bonds—there is no sharing or transfer of electrons between two atoms. When hydrogen is bound to a highly electronegative atom, such as fluorine, oxygen, or nitrogen, the hydrogen atom carries little of the electron density of the covalent bond. The hydrogen atom acts essentially as a naked proton (it is a little scandalous!). The positively charged hydrogen atom interacts with the partial negative charge of oxygen, nitrogen, or fluorine on nearby molecules. Substances that display hydrogen bonding tend to have unusually high boiling points compared with compounds of similar molecular formula that do not hydrogen bond. The difference derives from the energy required to break the hydrogen bonds. A very good example of this difference, which you absolutely need to understand for the MCAT, is the difference between a carboxylic acid, which can form hydrogen bonds between molecules of itself, and the acyl halide, a derivative of the carboxylic acid, which has substituted a halogen for the hydroxyl group and which cannot form hydrogen bonds with molecules of itself. The boiling point of propanoic acid (MW = 74 g/mol) is 141°C, but the boiling point of propanoyl chloride (MW = 92.5 g/mol) is 80°C. Hydrogen bonding is particularly important in the behavior of water, alcohols, amines, and carboxylic acids. It is no overstatement to say that if it weren’t for water’s ability to form hydrogen bonds and exist in the liquid state at room temperature, we would not exist (at least not in the form we recognize as “human”).


This chapter built upon our knowledge of the atom and the trends demonstrated by the elements in the periodic table to understand the different ways by which atoms partner together to form compounds, either (a) exchanging electrons to form ions, which are then held together by the electrostatic force of attraction that exists between opposite charges, or (b) sharing electrons to form covalent bonds. We discussed the nature and characteristics of covalent bonds, noting their relative lengths and energies as well as polarities. The review of Lewis structures and VSEPR theory will prepare you for predicting likely bond arrangements, resonance structures, and molecular geometry. Finally, we compared the relative strengths of the most important intermolecular electrostatic interactions, noting that even the strongest of these, the hydrogen bond, is still much weaker than an actual covalent bond. The next time you’re searing meat in a skillet or browning toast in a toaster, take a moment to consider what’s actually going on at the atomic and molecular level. It’s not just cooking; it’s science!



image Atoms come together to form compounds through bonds. Many atoms interact with other atoms in order to achieve the octet electron configuration in their valence shell. There are two types of bonds: ionic and covalent.

image Ionic bonds form when a very electronegative atom gains one or more electrons from a much less electronegative atom; the strong electrostatic force of attraction between opposite charges holds the resulting anion and cation together. Covalent bonds form when two atoms share one or more pairs of electrons. The force of the bond is due to the electrostatic attraction between the electrons in the bond and each of the positively charged nuclei of the bonding atoms.

image Covalent bonds are characterized by their length, energy, and polarity. A single covalent bond exists when two atoms share one pair of electrons, a double bond exists when two atoms share two pairs of electrons, and a triple bond exists when two atoms share three pairs of electrons. Triple bonds are the shortest and have the highest bond energy. Single bonds are the longest and have the lowest bond energy.

image Although all covalent bonds involve a sharing of one or more electron pairs, the electron density of the bonding electrons may not be distributed equally between the bonding atoms. This gives rise to a polar bond, with the more electronegative atom receiving the larger share of the bonding electron density.

image Lewis structures are bookkeeping devices used to keep track of the valence electrons of atoms and their various possible arrangements in a compound. Evaluation of formal charge on each atom in the arrangement can help determine which arrangement is most likely to be representative of the actual bond connectivity. Lewis structures that demonstrate the same atomic arrangement but differ in the distribution of electron pairs are called resonance structures. The actual molecular positioning of electrons is a “weighted” hybrid of all the possible resonance structures called a resonance hybrid.

image VSEPR theory helps us predict the actual three-dimensional arrangement of bonded and nonbonded electron pairs around the central atom in a compound. The theory states that nonbonded electrons (lone pairs) exert strong electrostatic repulsive forces against the bonded pairs of electrons and, as a result, the electron pairs arrange themselves as far apart as possible in order to minimize the repulsive forces.

image A molecular dipole is the resultant of all the bond dipole vectors in the molecule. A compound consisting of only nonpolar bonds will by definition be nonpolar. A compound consisting of one polar bond will by definition be polar (although perhaps insignificantly). A compound consisting of two or more polar bonds may be polar or nonpolar, depending on the molecular geometry and the vector addition of the bond dipoles.

image Atomic orbitals overlap their + and - lobes to result in bonding or antibonding molecular orbitals. When the signs of the overlapping atomic orbitals are the same, the result is a bonding molecular orbital. When the signs are opposite, the result is an antibonding orbital.

image Sigma bonds involve head-to-head (or end-to-end) overlap of electron cloud density and thus allow for relatively low-energy rotation. Pi bonds involve parallel overlap of electron cloud density and thus do not allow for low-energy rotation.

image The intermolecular interactions are electrostatic interactions that weakly hold molecules together. They generally are significant only over short distances and in the solid or liquid state. The weakest intermolecular interaction is the dispersion force (London force), which results from the moment-by-moment changing unequal distribution of the electron cloud density across molecules. A more moderate-strength intermolecular interaction is the dipole–dipole interaction, which exists between the opposite ends of permanent molecular dipoles. The strongest of these intermolecular interactions is the hydrogen bond, which is a special case of the dipole–dipole interaction between the hydrogen atom attached to either oxygen, nitrogen, or fluorine and another oxygen, nitrogen, or fluorine on another molecule—or sometimes, if the molecule is large enough, even the same molecule. One of the most important occurrences of hydrogen bonding is that between water molecules.


imageµ = qr

image Formal charge = V - NnonbondingNbonding

Practice Questions

1. What is the character of the bond in carbon monoxide?

A. Ionic

B. Polar covalent

C. Nonpolar covalent

D. Coordinate covalent

2. Which of the following molecules has the oxygen atom with the most negative formal charge?

A. H2O

B. CO32-

C. O3


3. Which of the following structure(s) contribute most to NO2’s resonance hybrid?


A. I only

B. III only

C. I and II

D. I, II, and III

4. Order the following compounds shown from lowest to highest boiling point:





IV. Isopropyl alcohol

A. I < II < IV < III

B. III < IV < I < II

C. II < IV < I < III

D. III < I < IV < II

5. Both CO32– and ClF3 have three atoms bonded to a central atom. What is the best explanation for why CO3 has trigonal planar geometry, while ClF3 is trigonal bipyramidal?

A. CO3 has multiple resonance structures, while ClF3 does not.

B. CO3 has a charge of -2, while ClF3 has no charge.

C. ClF3 has lone pairs on its central atom, while CO3 has none.

D. CO3 has lone pairs on its central atom, while ClF3 has none.

6. Which of the following has the largest dipole moment?


B. H2O

C. CCl4

D. SO2

7. Despite the fact that both C2H2 and NCH contain triple bonds, the lengths of these triple bonds are not equal. Which of the following is the best explanation for this finding?

A. In C2H2, since the triple bond is between similar atoms, it is shorter in length.

B. The two molecules have different resonance structures.

C. Carbon is more electronegative than hydrogen.

D. Nitrogen is more electronegative than carbon.

8. Which of the following is the best explanation of the phenomenon of hydrogen bonding?

A. Hydrogen has a strong affinity for holding onto valence electrons.

B. Hydrogen can only hold two valence electrons.

C. Electronegative atoms disproportionately carry shared pairs when bonded to hydrogen.

D. Hydrogen bonds have ionic character.

9. Which of the following best describes the character of the bonds in a molecule of ammonium?

A. Three polar covalent bonds

B. Four polar covalent bonds

C. Two polar covalent bonds, two coordinate covalent bonds

D. Three polar covalent bonds, one coordinate covalent bond

10. Although the octet rule dictates much of molecular structure, some atoms can exceed the octet rule and be surrounded by more than eight electrons. Which of the following is the best explanation for why some atoms can exceed the octet rule?

A. Atoms that exceed the octet rule already have eight electrons in their outermost electron shell.

B. Atoms that exceed the octet rule only do so when bonding with transition metals.

C. Atoms that exceed the octet rule can do so because they have d-orbitals in which extra electrons can reside.

D. Some atoms can exceed the octet rule because they are highly electronegative.

11. Which of the following types of intermolecular forces is the most accurate explanation for why noble gases can liquefy?

A. Van der Waals forces

B. Ion-dipole interactions

C. Dispersion forces

D. Dipole-dipole interactions

12. What is correct electron configuration for elemental chromium?

A. [Ar] 3p6

B. [Ar] 3d54s1

C. [Ar] 3d6

D. [Ar] 3d44s2

13. In the structure shown, which atom(s) has/have the most positive charge?


A. The phosphorus atom has the most positive charge.

B. All atoms share the charge equally.

C. The four oxygens share the highest charge.

D. The oxygen at the peak of the trigonal pyramidal geometry has the most positive charge.

14. Which of the following is the best name for the new bond formed in the reaction shown?


A. Polar covalent bond

B. Ionic bond

C. Coordinate covalent bond

D. Hydrogen bond

15. Both BF3 and NH3 have three atoms bonded to the central atom. Which of the following is the best explanation of why the geometry of these two molecules is different?

A. BF3 has three bonded atoms and no lone pairs, which makes its geometry trigonal pyramidal.

B. NH3 is sp3 hybridized, while BF3 is sp2 hybridized.

C. NH3 has one lone pair.

D. BF3 is nonpolar, while NH3 is polar.

16. Which of the following is a proper Lewis structure for BeCl2?


17. Which of the following best describes an important property of bond energy?

A. Bond energy increases with increasing bond length.

B. The more shared electron pairs comprising a bond, the higher the energy of that bond.

C. Single bonds are more difficult to break than double bonds.

D. Bond energy and bond length are unrelated.

18. Which of the following is true about the polarity of molecules?

A. Polarity is dependent on the vector sum of dipole moments.

B. Polarity does not depend upon molecular geometry.

C. If a molecule is comprised of one or more polar bonds, the molecule is polar.

D. If a molecule is comprised of one or more nonpolar bonds, the molecule is nonpolar.

Small Group Questions

1. Why does bond length decrease with increasing bond strength?

2. Predict two elements whose hydrides (H-) would contain incomplete octets.

Explanations to Practice Questions

1. B

Carbon monoxide, CO, has a double bond between carbon and oxygen, with the carbon retaining one lone pair and oxygen retaining two lone pairs. In polar covalent bonds, the difference in electronegativity between the bonded atoms is great enough to cause electrons to move disproportionately toward the more electronegative atom but not great enough to transfer electrons completely (i.e., to form an ionic bond). This is the case for CO. Oxygen is more electro-negative than carbon, so electrons will be disproportionately carried on the oxygen, leaving the carbon atom with a slight positive charge. Nonpolar covalent bonds occur between atoms that have the same electronegativity, for example, the carbon–carbon bond in CH3–CH3. Coordinate covalent bonding occurs through donation of a lone pair from a Lewis base to a Lewis acid.

2. B

To answer this question, you must understand the contribution of resonance structures to average formal charge. In CO32-, there are three possible resonance structures. Each of the three oxygen atoms carries a formal charge of -1 in two out of the three structures. This averages to approximately -2/3charge on each oxygen atom, which is more than the other answer choices. To prove this, estimate the formal charges on each oxygen atom in the other answer choices. There are no formal charges in H2O, which has no resonance structures. In ozone, O3, there are two possible resonance structures. The central oxygen carries a positive charge in both structures, and each outer oxygen carries a negative charge in one of the two possible resonance structures. This leaves an average charge of -½ on the two outer oxygens, which is less than the -2/3 in CO32-. Finally, CH2O has no resonance structures and, therefore, no formal charge on oxygen.

3. C

The two greatest contributors are structures I and II. Resonance structures are representations of how charges are shared across a molecule. In reality, the charge distribution is a weighted average of contributing resonance structures. The most stable resonance structures are those that minimize charge on the atoms in the molecule; the more stable the structure, the more that it will contribute to the overall charge distribution in the molecule. Structure I has a -1 charge on one oxygen and a single positive charge on the central nitrogen atom. Structure II has the exact same distribution, except the negative charge is on the other oxygen. It is equally as stable as structure I, so they both contribute equally to the hybrid. Structure III involves two negative charges (one on each oxygen) and a +2 charge on the central nitrogen. With so many charges, it’s less stable than structures I and II and, thus, is not an important resonance structure for NO2.

4. D

The key to answering this question is to understand the types of intermolecular forces that exist in each of these molecules. Kr (III) is a noble gas with a full octet, so the only intermolecular forces present are dispersion forces (also called London forces), the weakest type of intermolecular forces. This means that these molecules are held together extremely loosely and will be the easiest to transition from the organized liquid phase to the disorganized gaseous phase. In fact, because Kr is a noble gas, assume that it’s usually in the gaseous state and therefore has a low boiling point. Next, decide between two polar molecules: acetone (I) and isopropyl alcohol (IV). Both have the benefit of dipole–dipole forces, which are stronger than dispersion forces. Dipole–dipole forces arrange polar molecules such that the positive ends associate with their neighbors’ negative ends, which keeps these molecules in close proximity to one another. So which boils at a higher temperature? Alcohols are known for their ability to hydrogen bond, which significantly raises boiling point. Hydrogen bonding occurs when an electronegative atom (fluorine, oxygen, or nitrogen) binds to hydrogen, stripping it of electron density and leaving it with a partial positive charge. That partial positive charge can then associate with nearby partial negative charges. These forces are even stronger than dipole–dipole interactions, so isopropyl alcohol will boil at a higher temperature than acetone. Finally, the strongest interactions are ionic, and these exist in compounds such as potassium chloride (II).

5. C

The central atom in CO3, carbon, has no lone pairs. It has three resonance structures, each of which involves a double bond between carbon and one of the three oxygens. Having made four bonds, carbon has no further orbitals for bonding or to carry lone pairs. This makes CO3’s geometry trigonal planar. Alternatively, ClF3 also has three bonds, one to each of three fluoride atoms. However, chloride still maintains two extra lone pairs (without which the formal charge on the central chloride atom is +4; with the two lone pairs it is zero, a more stable configuration). These lone pairs each inhabit one orbital, meaning that the central chloride must organize five items about itself: three bonds to fluorides and two lone pairs. The best configuration for maximizing the distance between all of these groups is trigonal bipyramidal. (A) is true but does not account for the difference in geometry. Similarly, although (B) is true, the charge does not explain a difference in geometry. (D) is incorrect; CO3 has no lone pairs on its central atom, while ClF3 has two lone pairs.

6. A

The best way to approach this problem is to draw the structure of each of these molecules, then consider the electronegativity of each bond as it might contribute to an overall dipole moment. HCN is the correct answer because of large differences in electronegativity, aligned in a linear fashion. There is a strong dipole moment in the direction of nitrogen, without any other moments canceling it out. H2O has two dipole moments, one from each hydrogen pointing in the direction of oxygen. The molecule is bent, and the dipole moments partially cancel out. There is a molecular dipole, but it is not as strong as HCN’s. Sulfur dioxide has a similar bent configuration, so its dipole will be smaller than that of HCN. Another consideration for sulfur dioxide is that oxygen and sulfur do not have a large difference in electronegativity, so even the individual bond dipoles are smaller than those we have seen so far. CCl4 has a tetrahedral geometry. Although each of the individual C–Cl bonds is highly polar, the orientation of these bonds causes the dipoles to cancel each other out fully, yielding no overall dipole moment.

7. D

Bond lengths decrease as the bond order increases, and they also decrease in a trend moving up the periodic table’s columns or to the right across to the periodic table’s rows. In this case, because both C2H2 and NCH have triple bonds, we cannot compare the bond lengths based upon bond order. We must then rely on other periodic trends. The bond length decreases when moving to the right along the periodic table’s rows because more electronegative atoms have shorter atomic radii. The nitrogen in NCH is likely to hold its electrons closer, or in a shorter radius, than the carbon in C2H2. (B) is incorrect because there are no significant resonance structures contributing to the character of either triple bond. (C) expresses a true statement but one that is irrelevant to the length of the carbon–carbon triple bond.

8. C

(C) correctly describes the underlying forces of hydrogen bonding. Electronegative atoms bound to hydrogen disproportionately pull covalently bonded electrons toward themselves, which leaves hydrogen with a partial positive character. That partial positive charge is attracted to nearby negative or partial-negative charges, such as those on other electronegative atoms. (A) is not true; hydrogen has little electronegativity and does not hold its valence electrons closely. (B) is a true statement but not descriptive enough to explain hydrogen bonding. (D) is not correct; although these bonds are highly polarized, they are not ionic.

9. D

First recall that ammonium is NH4+, while ammonia (commonly confused) is NH3. It helps to associate the suffix -ium with a charged form of the molecule. Once you remember that ammonium is NH4+, eliminate any answer choice that only accounts for three bonds [answer choice (A)]v. Next, it helps to recall that ammonium is formed by the association of NH3 (uncharged, with a lone pair on the nitrogen) with a positively charge hydrogen cation (no lone pairs). In other words, NH3 is a Lewis base, while H+ is a Lewis acid. This type of bonding, between Lewis acid and base, is a coordinate covalent bond. Thus, you know that there is one coordinate covalent bond in this molecule, making (D) the correct answer.

10. C

This question addresses the issue of when the octet rule can be violated. All atoms that are in the third or higher period have d-orbitals, each of which can hold 10 electrons. The typical eight “octet” electrons reside in s- and p-orbitals. (A) and (B) do not explain why more than eight electrons can be held. Electronegativity is irrelevant to whether or not an atom can exceed the octet rule, making (D) incorrect.

11. C

All of the listed types of forces dictate interactions among different types of molecules. However, noble gases are entirely uncharged and do not have polar covalent bonds, ionic bonds, or dipole moments. Recognize that the only types of forces listed that could relate to noble gases could be van der Waals forces (A) or dispersion forces (C). Of these two, van der Waals forces is an umbrella term that includes both dispersion and dipole–dipole interactions (which you’ve already eliminated). Because not all van der Waals forces apply, rule that out as the correct answer and stick with dispersion forces. Dispersion forces are a specific type of interaction that occurs among all bonded atoms due to the unequal sharing of electrons at any given moment in the electron’s orbit. This unequal sharing allows for instantaneous partial positive and partial negative charges within the molecule. Though these interactions are small, they are necessary for liquefaction.

12. B

The key to this question is understanding the pattern of filling for s- and d-orbitals among the transition metals. There are 24 electrons in chromium, 6 more than are present in argon. Where will these six electrons lie? It is advantageous for d-orbitals to be half filled. Therefore, one electron will fill each of the five 3d-orbitals. It will take less energy to put the sixth electron into an s-orbital than it would to add it to one of the d-orbitals that already has an electron (and force it to be no longer half filled).

13. A

In this Lewis diagram, the PO43- molecule has an overall formal charge of -3. The four oxygens each would be assigned a formal change of -1, based on the following formula: Formal charge = V (valence electrons in the free atom) -½ Nbonding (electrons shared in bonds) -Nnonbonding (lone pairs/free electrons). For each oxygen, we calculate: FC = 6 - ½ (2) - 6 = -1. For the central phosphorus, assume then that with a total formal charge of -3 and four oxygens with a change of -1 each, the phosphorus must have a formal charge of +1. Alternatively, calculate its formal charge as FC = 5 - ½ (8) - 0 = +1. Considering this molecule’s other resonance structures, you’d come to the same conclusion—that phosphorus is the most positive atom.

14. C

The reaction in this question shows a water molecule, which has two lone pairs of electrons on the central oxygen, combining with a free hydrogen ion. The resulting molecule, H3O+, has formed a new bond between H+ and H2O. This bond is created through the sharing one of oxygen’s lone pairs with the free H+ ion. This is essentially a donation of a shared pair of electrons from a Lewis base (H2O) to a Lewis acid (H+, electron acceptor). The charge in the resulting molecule is +1, and it is mostly present on the central oxygen, which now only has one lone pair. This type of bond, formed from a Lewis acid and Lewis base, is called a coordinate covalent bond.

15. B

NH3 has three hydrogen atoms bonded to the central nitrogen and one lone pair on the central nitrogen. These four groups—three atoms, one lone pair—lead NH3 to be sp3 hybridized. By hybridizing all three p-orbitals and the one s-orbital, four groups are arranged about the central atom, maximizing the distances between the groups to minimize the energy of the configuration. NH3’s hybridization leads to its tetrahedral electronic geometry yet trigonal pyramidal molecular geometry. In contrast, BF3 has three atoms but no lone pairs, resulting in sp2 hybridization. Its shape is called trigonal planar. (A) is incorrect; although BF3 does has three bonded atoms and no lone pairs, its geometry is not trigonal pyramidal. (C) is tempting because NH3 has a lone pair, but this answer choice is not complete enough to explain the differences in molecular geometry. Finally, although (D) is also true, the polarity of the molecules does not explain their geometry; rather, the molecules’ different geometries contribute to the overall polarity of the molecules.

16. B

Most atoms require eight valence electrons to follow the octet rule. However, some atoms, like beryllium, can have fewer than eight valence electrons (suboctet). As a result, when bonding with chloride, beryllium is likely to form only two bonds, using its own two outer valence electrons and one from each chloride to form BeCl2, as drawn in (B). (A) is incorrect because Be cannot both bond to chloride and have lone pairs. (C) assumes that all three atoms need their octets filled, and (D) places 10 electrons on each chlorine.

17. B

This answer requires an understanding of the trends that cause higher or lower bond energies. Bonds of high energy are those that are difficult to break. These bonds tend to have more shared pairs of electrons and, thus, cause a stronger attraction between the two atoms in the bonds. This stronger attraction also means that the bond length of a high-energy, high-order bond (i.e., a triple bond) is shorter than that of its lower-energy counterparts (i.e., single or double bonds). Thus, (A) is incorrect; bond energy increases with decreasing bond length. Single bonds are longer than double bonds. Thus, (C) is incorrect; single bonds are easier to break than double bonds. Finally, as previously discussed, bond energy is inversely related to bond length, making (D) incorrect.

18. A

Polarity is precisely described by the first answer choice. Dipole moments describe the relationship of shared electrons between two bonded atoms. If there is a dipole moment between two atoms, then the electrons in their shared bond are preferentially centered on one atom, typically the more electronegative of the two. The polarity of a molecule is then defined as the vector sum of these dipole moments in the molecule’s three-dimensional configuration. For example, although individual carbon–chloride bonds have dipole moments with a partial negative charge on chloride, the molecule CCl4 has a tetrahedral configuration. Thus, the four polarized bonds cancel each other out for no net dipole moment, creating a nonpolar molecule. Based on this description, it is clear that (B) is incorrect, because molecular geometry is an essential component of determining molecular polarity. For the same reasons described previously, it is possible for a nonpolar molecule to contain one or more polar bonds, making (C) incorrect. Finally, it is possible for a molecule to contain at least one nonpolar bond and yet be a polar molecule. For example, CH3CH2Cl has a nonpolar bond (C–C), yet a molecular dipole exists. In its tetrahedral arrangement, the one polar bond between C–Cl will have a dipole moment, creating a net polarity of the molecule in the direction of the chloride atom.