Bonds and Interactions for the MCAT: Everything You Need to Know
A complete guide to MCAT bonds and interactions, covering key concepts, test strategies, and practice questions to improve your chemistry score.
(Note: This guide is part of our MCAT General Chemistry series.)
Table of Contents
Part 1: Introduction to bonds and interactions
Part 2: Interatomic forces
Part 2: Interatomic Forces
a) Ionic bonds
b) Covalent bonds
c) Coordinate covalent bonds
d) Single and double bonds
e) VSEPR theory
Part 3: Intermolecular Forces
a) Hydrogen bonds
b) Dipole-dipole interactions
c) London dispersion forces
Part 4: High-Yield Terms
Part 5: Passage-Based Questions and Answers
Part 6: Standalone Questions and Answers
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Part 1: Introduction to bonds and interactions
What holds everything in the universe together? Why don't atoms and molecules randomly split apart and float away? The answer lies in bonding and molecular interactions. Understanding how atoms and molecules interact with each other is crucial for much of the chemistry section of the MCAT. Atomic bonds and intermolecular attractions are the foundation for many additional concepts in organic chemistry and biochemistry.Chemical bonds are chemical interactions that create an attractive force between atoms or molecules. There are a variety of bonds, or attractive forces, that exist. Some exist between individual atoms of a molecule, while others exist between molecules.
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Part 2: Interatomic forces
a) Ionic bonds
The simplest form of interatomic bond is known as an ionic bond. Ionic bonds are formed through the interaction of two atoms with vastly different electronegativities. Because both atoms have different electronegativities, one atom donates electrons to another. This results in the creation of a positively charged ion (cation) and a negatively charged ion (anion). The negatively and positively charged atoms attract, creating an ionic bond.
Because ionic bonds take advantage of different electronegativities, the constituent elements usually have quite different characteristics. In fact, ionic bonds are typically composed of metal and nonmetal. Sodium chloride (chemical formula: NaCl) is a molecule that is held together by an ionic bond. Sodium is a highly reactive metal, and chlorine is a highly reactive gas. Together, they form table salt, which we consume every day!
Figure 1 A sodium atom and a chloride ion form an ionic bond to create table salt.
b) Covalent bonds
The second form of interatomic bond is the covalent bond. A covalent bond is usually formed between two atoms with similar electronegativities. Thus, covalent bonds are usually formed between nonmetals or metalloids.
Because the atoms have similar electronegativities, the atoms do not donate electrons to each other. Instead, they share electrons. This sharing between the two atoms can either be equal or unequal. Whether atoms in a covalent bond equally or unequally shared electrons depends on their difference in electronegativity. The greater the difference in electronegativity between two elements is, the more partial ionic character (e.g., the more similarity to an ionic bond) the bond between them has.
The polarity of a covalent bond can be described through a dipole moment. The dipole moment is a quantity that can be calculated using the following equation:
p=qd
where p = dipole moment,
q = net charge,
d = distance between the partial charges
This equation may look similar to another law: Coulomb’s law, which describes the attractive force generated between two charged particles. Here, the dipole moment describes an analogous quantity. A large dipole moment indicates a larger polarity, or difference in electron distribution, between partial charges.
Lewis structures are a common method of representing covalent bonds. In Lewis dot structures, electrons in the valence orbitals are represented by a dot on each side of an atom’s elemental symbol. (To learn more about valence electrons, be sure to refer to our guide on atoms and periodic trends.)
To draw a Lewis dot structure, begin by writing the elemental symbol for the atom of interest. Then, calculate the number of valence electrons in the s and p orbitals. Once calculated, place one dot to the top of the elemental symbol. Continue placing dots in a clockwise manner around the symbol until the number of dots equals the number of valence electrons. This process is repeated for each element in a covalent bond.
After each atom participating in a covalent bond is illustrated, electrons participating in sigma bonds can be connected with a line.
Figure 2 Constructing a Lewis structure.
Figure 3 Comparing two resonance structures of ozone
One advantage to using a Lewis dot diagram is that resonance structures can be shown. A resonance structure is an alternative model of covalent bonding. A molecule that displays resonance may have electrons in any of these resonance configurations at any given time. In general, the bonds within such a molecule are assumed to be a combination of all of these resonance structures. The electrons within the molecule are said to be delocalized and can be found with equal probability within any of these resonance structures. The resulting combined configuration may be shown with a dashed line.
The most stable resonance form is one that provides the lowest formal charge of the atoms. Formal charge refers to the amount of “functional” charge an atom would have, assuming all electrons are shared equally. Thus, it can be calculated using the following equation:
formal charge = # valence electrons - nonbonding electrons - bonding electrons ÷ 2
Recall our previous discussion of ozone. In the ozone molecule, each oxygen atom has a different formal charge. One atom has a charge of 0, while the other two have charges of -1 and +1. This is the relatively stable form of ozone, as the formal charges of each species are fairly low. In contrast, creating only single bonds in ozone would create formal charges of -1, -1, and +2, which are much higher formal charges. This resonance form is highly unlikely to occur!
Molecules and chemical compounds can often be drawn in a variety of ways. For more information on this, be sure to refer to our guide on isomers.
c) Coordinate covalent bonds
Recall that covalent bonds are simply bonds formed between atoms in which electrons are unevenly shared. A coordinate covalent bond, sometimes referred to as a dative bond, is a special subtype of covalent bond that often forms between transition metals. In this type of covalent bond, both electrons in the bond are donated by a single atom.
Due to their unique orbital structure, transition metals can easily accept electrons from other atoms. For this reason, transition metals can form two, three, or four coordinate covalent bonds with other atoms! Zinc, iron, and magnesium are examples of such metals that usually bond with enzymes inside our bodies. Since they are integrated into enzymes and are often critical for the enzyme’s function, these transition metals are referred to as cofactors. (For more information on enzymes and cofactors, be sure to refer to our guide on enzymes.)
d) Single and double bonds
In chemistry, the octet rule states that each atom is most stable when it contains a full valence shell of 8 electrons. This means that atoms with 1 electron in their valence shell are ready to give up their electrons while those with a nearly full shell aren’t. Thus, each atom in a molecule must be exposed to eight electrons: either through electron transfer (e.g. ionic bonding) or sharing (e.g., covalent bonding).
Figure 4 Atoms seek to obtain a full octet of valence electrons.
There are many instances where atoms in a molecule have more than one bond. Single, double, and triple bonds vary in both bond length and bond energy. Bond length is the distance between the nuclei of two bonded atoms. Single bonds have the largest bond length, followed by double and triple bonds, respectively.
Bond energy quantifies the amount of energy needed to break the bond. Since triple bonds have more bonds to break, it follows that they would have the greatest bond energy.
Bond length and energy have important implications on the rigidity of molecules. Bonds that are shorter and have greater energy (for example, triple bonds) increase the rigidity of molecules. These bonded atoms are less prone to rotation and vibration. In comparison, a single bond allows much more mobility since the bond is longer and of lower energy.
Let’s take a closer look at sigma and pi bonds. Sigma bonds are bonds that form when two s orbitals overlap. Pi bonds are bonds that form when two p orbitals in the same plane overlap. (To learn more about s and p orbitals, be sure to refer to our guide on atoms and periodic trends.)
Figure 5 Formation of sigma (σ) and pi (π) bonds.
Note that multiple sigma and pi bonds can be formed between two atoms, resulting in the formation of single, double, or triple bonds. A single bond consists of solely a sigma bond. A double bond consists of a sigma and pi bond. A triple bond consists of one sigma and two pi bonds.
Figure 6 A single bond consists of a single sigma bond, while a double bond requires both a sigma and a pi bond.
Recall that atoms arrange electrons in orbitals, which include s and p orbitals. When molecules form, s and p orbitals from atoms combine to create hybrid orbitals. A process known as hybridization allows two or more atomic orbitals to become a molecular orbital.
| Hybridization | Bond angle | Predicted geometry | Example |
|---|---|---|---|
One atom bonded to two others |
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One atom bonded to three others |
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One atom bonded to three other atoms and one lone pair |
SP Hybridize
SP2 Hybridize
SP3 Hybridize
SP2 Nitrogen
e) VSEPR Theory
Valence-shell electron pair repulsion theory, or VSEPR theory, is used to predict the shape of molecules. The core intuition for this theory is that electrons are negatively charged particles that repel each other. To minimize this repulsion, the bonds and lone pairs around a central atom will equally disperse themselves as far away from each other as possible in a three-dimensional space. Take the example of carbon dioxide, a molecule with 2 bonds and 0 lone pairs: the oxygens orient themselves on opposite sides of the carbon, exactly 180॰ away from each other, in linear geometry.
Figure 7 Carbon dioxide visualized in linear molecular geometry.
Methane, with 4 bonds and 0 lone pairs, reaches the lowest possible electron repulsion by orienting its hydrogens exactly 109.5॰ away from each other in tetrahedral geometry. The oxygen in water has 2 bonds and 2 lone pairs. The presence of the lone pairs causes the hydrogens to orient themselves in a bent version of the linear CO2 geometry.
Figure 8 Water and methane visualized in their respective bent and tetrahedral molecular geometries.
Therefore, we can understand molecular geometry in terms of the number of lone pairs and bonds around a central atom. The total of these groups is known as the steric number. For example, an atom with 1 lone pair and 2 bonds has a steric number of 3. The following table summarizes the most important combinations of bonds and lone pairs with their corresponding steric numbers, geometries, and angles. Though many more potential molecular geometries exist, they are typically beyond the scope of the MCAT.
Figure 9 A comprehensive list of the various molecular geometries you will see on the MCAT, characterized by steric number. Pay extra attention to the following as they are the most likely to appear on test day: linear, trigonal planar, tetrahedral, trigonal pyramidal, angular, and octahedral.
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Part 3: Intermolecular forces
Ionic and covalent bonds are examples of intramolecular forces, or forces that link a molecule together. In contrast, intermolecular forces are forces that attract molecules without physically linking them together.
a) Hydrogen bonds
Hydrogen bonds form as a result of a hydrogen atom binding to either oxygen, fluorine, or nitrogen atom (O, F, or N). The electronegativity of these atoms creates an unbalanced sharing of electrons. As a result, the hydrogen gains a partial positive charge, while the O, F, or N atom gains a partial negative charge.
The partial positive charge of the hydrogen atom on one molecule can interact with the partial negative charges of O, F, or N on a second molecule—thus creating a bond between the two molecules. Hydrogen bonds are particularly strong. For instance, the hydrogen bond attraction between water molecules is what causes the boiling point of water to be so high! Lots of energy in the form of heat is required to disrupt the hydrogen bond.
Hydrogen bonds are found in many different places, such as nucleotides, water, and organic solvent like alcohol.
Figure 10 Hydrogen bonding between functional groups on different molecules.
In depicting intermolecular interactions, it is convention to illustrate hydrogen bonds with dashed lines.
b) Dipole-dipole interactions
Dipole-dipole interactions are attractive forces that form as polar molecules align themselves in a liquid or solid. The partial positive character of a polar molecule interacts with the partial negative character of a neighboring molecule. Recall that this polarity is a result of uneven electron sharing between two atoms that have differing electronegativities.
Figure 11 An example of dipole-dipole interactions between hydrochloric acid. Note that the symbol indicates a "partial" positive or negative charge--rather than a completely positive or negative charge.
Hydrogen bonding is a special form of dipole-dipole interaction. Hydrogen bonds tend to be highly stable due to the large electronegativity difference between hydrogen and N, O, or F atoms. Dipole-dipole interactions may be less stable than hydrogen bonds due to a lower electronegativity difference and more equal sharing of electrons.
c) London dispersion forces
London dispersion forces, also known as van der Waals interactions, form as a result of the unequal distribution of electrons around a molecule. Although we might imagine them to be equally spread, random movement can create polarization in a molecule.
For a very short period of time, the electrons of a molecule may be gathered toward one side of the molecule. This creates a temporary partial negative charge on that same side of the molecule. The temporary partial negative charge will create a temporary partial positive charge in a neighboring molecule, and so on. These temporary partial charges create an attractive force known as London dispersion forces.
These are the weakest and most temporary forces since they depend on the temporary, unequal distribution of electrons. These forces are also dependent on the distance between molecules—the closer two molecules are to one another, the larger the magnitude of the van der Waals interactions may be. Although these forces are weak, they play an important role in keeping molecules in liquid and solid phases!
Figure 12 An example of van der Waals attractions. Notice how temporary changes in electron distribution causes partial charges, which create intermolecular attractions.
London dispersion forces tend to be stronger between large molecules and molecules of similar shape. Large molecules hold a much larger reservoir of electrons; as a result, any temporary partially negative charges tend to be of larger magnitude. Additionally, molecules of similar shapes are able to maximize the number of bonds that can be formed across a short distance.
London dispersion forces also tend to be the primary intermolecular force at play in hydrocarbon molecules. Hydrocarbons are large molecules composed of carbon and hydrogen (C and H). Due to the relative lack of highly electronegative elements, hydrocarbons do not form strong hydrogen bonds or dipole-dipole interactions. Instead, hydrocarbons are attracted to each other through London dispersion forces. Since London dispersion forces are dependent on size and shape, these attractions tend to be greatest between long, straight hydrocarbons.
Figure 13 Van der Waals interactions are maximized between long molecules of similar shape.
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Part 4: High-yield terms
Ionic bonds: formed between two atoms with vastly different electronegativities
Cation: positively charged ion
Anion: negatively charged ion
Covalent bond: formed between two atoms with similar electronegativities
Ionic character: degree of similarity of a covalent bond to an ionic bond
Dipole moment: quantity that describes the polarity of a bond
Polarity: directionality and magnitude of sharing of electrons between two species
Lewis structure: method of representing covalent bonds; depicts atom’s elemental symbol and its valence electrons
Resonance structure: an alternative model of covalent bonding and electron sharing
Formal charge: describes the amount of “functional” charge an atom would have if all electrons are shared equally
Octet rule: states that an atom is most stable when containing a full valence shell
Bond length: distance between nuclei of two bonded atoms
Bond energy: amount of energy needed to break a bond
Hybridization: combining of two or more atomic orbitals to create a molecular orbital
VSEPR: valence-shell electron pair repulsion theory; used to predict molecular geometries
Hydrogen bond: a result of a hydrogen atom bound to another oxygen, fluorine, or nitrogen atom
Dipole-dipole interactions: intermolecular attractions that form between polar molecules
London dispersion forces/van der Waals interactions: intermolecular attractions that result from the spontaneous and temporary unequal distribution of electrons around a molecule
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Part 5: Passage-based questions and answers
Near-infrared fluorescent proteins (NIR FPs) are large molecules used to image tissues and cell populations in vivo. NIR imaging has been found to be optimal for imaging mammalian tissue, due to the existence of an optical “transparency window” in which light absorption and scattering appear to be minimized. Using near-infrared fluorescence allows optical probing of tissue at multiple depths, enabling whole-body imaging. A subclass of bright, monomeric NIR FPs known as miRFP proteins contains two specially developed domains: a PAS domain and GAF domain, which may form a portion of the protein binding pocket.
Biliverdin is a tetrapyrrole chromophore that fluoresces upon stimulation, and may be covalently linked to NIR FPs to provide fluorescence. The degree, brightness, and quality of the fluorescence are directly affected by the efficiency of biliverdin binding.
Researchers wish to understand the nature of biliverdin binding. To do so, a molecule of unbound biliverdin (BV) is exposed to an miRFP. Several miRFP proteins were selected for binding. In this family, miRFP670 is the most blue-shifted and holds a high significance for multi-color NIR imaging. It is hypothesized that cysteine residues on the molecule contribute to the rigidity of the BV-bound structure, thus increasing the fluorescent yield.
Figure 1
Researchers monitored the binding reaction in real-time as the process occurred over a time period of minutes to tens of minutes. The resulting bound structures were analyzed using computational methods. It was determined that BV undergoes isomerization, localization to a binding pocket on the apoprotein, protonation, and a rearrangement of intermolecular bonding.
Question 1: Which of the following bonds are most likely to be formed between two adjacent miRFP molecules?
A) Covalent, hydrogen, ionic, and van der Waals
B) Hydrogen, ionic, van der Waals, and dipole-dipole
C) Covalent, hydrogen, van der Waals, and dipole-dipole
D) Hydrogen, van der Waals, and dipole-dipole
Question 2: According to the data presented in the passage, how many pi bonds exist in the biliverdin molecule?
A) 11
B) 12
C) 14
D) 15
Question 3: Which of the following best describes the bonds within the biliverdin molecule?
A) Covalent; the atoms involved are nonmetals with different electronegativities, resulting in electron exchange
B) Ionic; the atoms involved are metals and nonmetals with different electronegativities, fostering electron exchange
C) Covalent; the atoms involved are nonmetals with similar electronegativities, resulting in electron sharing
D) Ionic; the atoms involved are metals and nonmetals with different electronegativities, resulting in electron sharing
Question 4: In preparation for the experiment described in the passage, scientists have successfully isolated and purified miRFPs. Which of the following should miRFPs be stored in, to minimize molecular damage and denaturing?
A) Hydrochloric acid; HCl will neutralize any bases
B) Sodium hydroxide; NaOH will neutralize any acids that might affect biliverdin
C) Methane; methane won’t interact with biliverdin, preserving its structure
D) Glycerol; glycerol will hydrogen bond with biliverdin
Question 5: What are the possible molecular hybridizations of a carbon atom located within a pyrrole ring labeled as “B”?
A) sp
B) sp2
C) sp2 and sp3
D) sp3
Answers to passage-based questions
1. Answer choice D is correct. Covalent and ionic bonds are interatomic forces, rather than intermolecular forces (choices A, B, and C are incorrect). Hydrogen bonds, van der Waals interactions, and dipole-dipole interactions are all types of intermolecular forces. Van der Waals interactions can be formed between any two molecules; dipole-dipole interactions and hydrogen bonds are formed when two atoms of differing electronegativity are bonded together on the same molecule. According to the structure of the biliverdin molecule, there is potential for all of these intermolecular forces to exist (choice D is correct).
2. Answer choice D is correct. Pi bonds exist as the second and third bonds in double and triple bonds. The first bond that is formed in double and triple bonds must be a sigma bond. Thus, to answer this question, it is sufficient to count the number of pi bonds in the double bonds of the molecule. Don’t forget the carboxyl groups that are shown as COOH! Careful counting gets us to 15 pi bonds (choice D is correct).
3. Answer choice C is correct. Recall that covalent bonds exist between nonmetals while metals and nonmetals are needed for ionic bonds. According to the provided structure of biliverdin, all of the atoms within the structure are nonmetals. Thus, biliverdin must be linked together by covalent bonds (choices B and D are incorrect). Further, recall that nonmetals have similar electronegativities, which results in electron sharing instead of electron transfer (choice C is correct).
4. Answer choice D is correct. miRFPs are proteins that must be carefully stored. Proteins tend to denature in extreme conditions, including conditions with excessively low or high pH (choices A and B are incorrect). To preserve stability, the protein must also be in a confirmation that helps reduce its spontaneous activity. Glycerol is able to stabilize the protein by encouraging hydrogen bonding between the protein and solvent (choice D is correct).
5. Answer choice B is correct. Any carbon atom located within the indicated pyrrole ring is bonded to three other atoms. Thus, three atomic orbitals must be involved with bonding: one s orbital, and two p orbitals. These atomic orbitals hybridize to form a single sp2 molecular orbital (choice B is correct).
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Part 6: Standalone questions and answers
Question 1: Which of the following statements best describes the primary differences between ionic and covalent bonds?
A) Ionic bonds involve atoms sharing electrons, while covalent bonds transfer electrons
B) Covalent bonds share electrons, while ionic bonds transfer them
C) Ionic bonds only occur between nonmetals, while covalent bonds occur between metals
D) Covalent bonds solely occur between nonmetals, while ionic bonds occur between metals
Question 2: Water has a relatively high boiling point of 100 degrees Celsius. Which of the following statements is the best explanation for this phenomenon?
A) Covalent bonds between hydrogen and oxygen provide high stability in water atoms
B) Ionic bonds between water and dissolved salts increase the number of intermolecular attractions
C) Hydrogen bonds between adjacent water molecules provides strong intermolecular bonding
D) Molecular hybridization of atomic orbitals requires high amounts of energy to disrupt
Question 3: From left to right, what are the formal charges of each atom in the following molecule?
A) 0, 0, 0
B) -2, +1, -2
C) +1, -2, +1
D) -1, +2, +1
Question 4: A triglyceride molecule is formed by a single glycerol head group and three hydrocarbon tails. Which of the following best describes the interactions between two individual triglyceride molecules?
A) Hydrocarbon tails on a molecule are attracted to each other by dipole-dipole interactions
B) Glycerol heads on two adjacent molecules are primarily attracted to each other by van der Waals interactions
C) Interactions between fully saturated triglycerides are stronger than interactions between unsaturated triglycerides
D) Interactions between shorter triglycerides are stronger than interactions between longer triglycerides
Answers to standalone questions
1. Answer choice B is correct. Ionic bonds form through the transfer of electrons between two elements of differing electronegativity (choice A is incorrect). Covalent bonds form through the sharing of electrons between two elements of similar electronegativity (choice B is correct).
2. Answer choice C is correct. Recall that water is composed of both oxygen and hydrogen. As hydrogen and oxygen have vastly different electronegativities, the bond between them is polar and forms very strong hydrogen bonds with other water molecules. These hydrogen bonds form a superstrong lattice, which requires high amounts of thermal energy to break and disrupt (choice C is correct). The dissolution of salts may result in the formation of ionic “salt bridges” between molecules; however, this would serve to depress the boiling point instead of increasing it (choice B is incorrect). While covalent bonds between atoms lead to high intramolecular stability, it is not the primary reason for intermolecular stability (choice A is incorrect).
3. Answer choice A is correct. To determine the formal charges, apply the following formula:
formal charge = # valence electrons - nonbonding electrons - bonding electrons ÷ 2
First, determine the number of valence electrons each element has. Oxygen has six valence electrons, while carbon has four. Then, subtract the number of nonbonding electrons (4 nonbonding electrons in the case of oxygen) and half of the bonding electrons for each element (2 bonding electrons for each oxygen and 4 bonding electrons for each carbon atom). Thus, the formal charge of each carbon atom is: 4-0-(8÷2)=0, and the formal charge for each oxygen atom is 6-4-(4÷2)=0.
4. Answer choice C is correct. Triglycerides contain hydrocarbon tails, which are unable to form dipole-dipole interactions or hydrogen bonds (choice A is incorrect). A molecule that contains a hydrogen atom bonded to nitrogen, oxygen, or fluorine may form relatively strong hydrogen bonds (choice B is incorrect). The attraction between hydrocarbon tails is formed through van der Waals interactions. Recall that van der Waals interactions are highly dependent on the constituent molecules’ shape and size; molecules of longer length tend to form stronger van der Waals interactions (choice D is incorrect). Saturation refers to the degree of bonding between carbon and other atoms; a fully saturated triglyceride has only single bonds between carbons. An unsaturated triglyceride may have double or triple bonds, which form a “kink” in the hydrocarbon chain. This disrupts the regular linear structure of the hydrocarbon and reduces the ability of the molecule to form van der Waals interactions with other molecules (choice C is correct).
