MCAT General Chemistry Review
Part I Review
Chapter 1: Atomic Structure
“Salting the water makes the pasta cook faster because the salt raises the boiling point. It’s chemistry—it’s a colligative property.” My colleague stared me down, as if the intensity of his gaze gave further credence to his pronouncement. It was late on a Friday afternoon, about 5:00 P.M., and the science of cooking had become the topic of discussion that would carry us through the final hour of the workweek. However, here we were, arguing over the relative merits of culinary and scientific justifications for adding salt to pasta water. We were two opposing camps: one side arguing that adding salt to the water served no real purpose other than to flavor the pasta as it absorbed the water; the other side—the one of my colleague and his compatriots—fighting in the name of science. Finally, we agreed on a plan toward resolution and a lasting peace. We would calculate the actual chemical impact of the salt in the water and answer the question Why add salt to pasta water? once and for all.
Well, you can figure out the details of the calculation if you are so inclined, but suffice it to say that I wouldn’t be telling you this story if I hadn’t been proven correct. Our calculation showed that the amount of salt that most of us would probably consider a reasonably sufficient quantity for a large pot of boiling water had minimal impact on the boiling point temperature of the resulting solution. To be clear, the addition of the salt did raise the boiling point temperature—and even those of us arguing against the “scientific justification” never denied that it would. Nevertheless, the calculated rise in boiling point was exceedingly small, because the solution we imagined had such a low molality. The truth of the matter is this: Adding salt to cooking water in reasonable amounts does not measurably increase the boiling point temperature or decrease the cooking time. The salt merely flavors the food as it cooks in the dilute solution.
Why introduce a book of general chemistry for the MCAT review by telling you a story about food preparation? Am I telling you this simply to showcase an instance in which I won an argument? No, it’s because chemistry is the study of the stuff of life or, to put it more properly, the nature and behavior of matter. Chemistry is the investigation of the atoms and molecules that make up our bodies, our possessions, the world around us, and of course, the food that we eat. There are different branches of chemistry, two of which are tested directly on the MCAT—general inorganic chemistry and organic chemistry—but ultimately all investigations in the realm of chemistry are seeking to answer the questions that confront us in the form (literally, “form”: the shape, structure, mode, and essence) of the physical world that surrounds us.
At this point, you’re probably saying the same thing you say when talking about physics: But I’m premed. Why do I need to know any of this? What good will this do me as a doctor? Do I only need to know this for the MCAT? Let me make it clearer for you: How can you expect to be an effective doctor for your patients—who are made of the organic and inorganic stuff—unless you understand how this stuff makes up and affects the human body?
So, let’s get down to the business of learning and remembering the principles of the physical world that help us understand what all this stuff is, and how it works, and why it behaves the way it does, at both the molecular and macroscopic levels. In the process of reading through these chapters and applying your knowledge to practice questions, you’ll prepare yourself for success not only on the Physical Science section of the MCAT but also in the medical care of your patients and the larger communities you will serve as a trained physician.
The building blocks of the atom are also the building blocks of knowledge for the General Chemistry concepts tested on the MCAT. By understanding these interactions, we will be able to use that knowledge as the “nucleus” of understanding to all of General Chemistry.
This first chapter starts our review of general chemistry with a consideration of the fundamental unit of matter, the atom, and the even smaller particles that constitute the atom: protons, neutrons, and electrons. We will also review the two models of the atom with a particular focus on how the two models are similar and different.
Although you may have encountered in your university-level chemistry classes such subatomic particles as quarks, leptons, gluons, and other particles whose names sound as if they were picked up from a Star Trek episode, the MCAT’s approach to atomic structure is much simpler. There are three subatomic particles that you must understand.
Protons are found, along with neutrons, in the nucleus of an atom. Each proton has an amount of charge equal to the fundamental unit of charge (1.6 × 10-19 C), and we denote this fundamental unit of charge as “+1” for the proton. Protons have a mass of approximately one atomic mass unit, oramu. The atomic number (Z) of an element is equal to the number of protons found in an atom of that element. The atomic number is like your Social Security number; it acts as a unique identifier for each element because no two elements have the same one. All atoms of a given element have the same atomic number, although, as we will see, they do not necessarily have the same atomic mass.
Neutrons are the Switzerland of an atom; they are neutral, which means that they have no charge. A neutron’s mass is only slightly larger than that of the proton, and together, the protons and the neutrons of the nucleus make up almost the total mass of an atom. Every atom has a characteristicmass number, which is the sum of the protons and neutrons in the atom’s nucleus. The number of neutrons in the nuclei of atoms of a given element may vary; thus, atoms of the same element will always have the same atomic number but will not necessarily have the same mass number. Atoms that share an atomic number but have different mass numbers are known as isotopes of the element. The convention is used to show both the atomic number (Z) and the mass number (A) of atom X.
If you think of the nucleus as a game of checkers, the electrons would be children who express varying degrees of interest in playing or watching the game. Electrons move around in pathways in the space surrounding the nucleus and are associated with varying levels of energy. Each electron has a charge equal to that of a proton but with the opposite (negative) charge, denoted by “-1.” The mass of an electron is approximately that of a proton. Because subatomic particle masses are so small, the electrostatic force of attraction between the unlike charges of the proton and electron is far greater than the gravitational force of attraction based on their respective masses.
The valence, or outer, electrons will be very important to us in both General and Organic Chemistry. Knowing how tightly held those electrons are will allow us to understand many of an atom’s properties and how it interacts with other atoms.
Going back to our checkers analogy, consider how children form rough circles surrounding a game of checkers; the children sitting closer to the game are more interested in it than the children who are sitting on the periphery. Similarly, electrons are placed in pathways of movement that are progressively farther and farther from the nucleus. The electrons closer to the nucleus are at lower (electric potential) energy levels, while those that are in the outer regions (or shells) have higher energy. Furthermore, if you’ve ever seen children sitting around a game, you know that the “troublemakers” are more likely to sit on the periphery, which allows them to take advantage of an opportunity for mischief when it arises—and so it is also for electrons. Those in the outermost energy level, or shell, called the valence electrons, experience the least electrostatic draw to their nucleus and so are much more likely to become involved in bonds with other atoms (filling empty spaces in other atoms’ valence shells). Generally speaking, the valence electrons determine the reactivity of an atom. In the neutral state, there are an equal number of protons and electrons; a gain of electron(s) results in the atom gaining a negative charge, while a loss of electron(s) results in the atom gaining a positive charge. A positively charged atom is a cation, and a negatively charged atom is an anion.
Some basic features of the three subparticles are shown in Table 1.1.
Example: Determine the number of protons, neutrons, and electrons in a nickel-58 atom and in a nickel-60 2+ cation.
Solution: 58Ni has an atomic number of 28 and a mass number of 58. Therefore, 58Ni will have 28 protons, 28 electrons, and 58 - 28, or 30, neutrons.
In the 60Ni2+ species, the number of protons is thet same as in the neutral 58Ni atom. However, 60Ni2+ has a positive charge because it has lost two electrons; thus, Ni2+ will have 26 electrons. Also the mass number is two units higher than for the 58Ni atom, and this difference in mass must be due to two extra neutrons; thus, it has a total of 32 neutrons.
Atomic Weights and Isotopes
• Atomic number (Z) = number of protons.
• Mass number (A) = number of protons + number of neutrons.
• Number of protons = number of electrons (in a neutral atom).
As we’ve seen, the mass of one proton is defined as approximately one amu. The size of the atomic mass unit is defined as exactly the mass of the carbon-12 atom, approximately 1.66 × 10-24 grams (g). Because the carbon-12 nucleus has six protons and six neutrons, an amu is really the average of the mass of a proton and a neutron. Because the difference in mass between the proton and the neutron is so small, the mass of the proton and the neutron are each about equal to 1 amu. Thus, the atomic mass of any atom is simply equal to the mass number (sum of protons and neutrons) of the atom. A more common convention used to define the mass of an atom is the atomic weight. The atomic weight is the mass in grams of one mole of atoms of a given element and is expressed as a ratio of grams per mole (g/mol). A mole is the number of “things” equal toAvogadro’s number: 6.022 × 1023. For example, the atomic weight of carbon is 12 g/mol, which means that 6.022 × 1023 carbon atoms (1 mole of carbon atoms) have a combined mass of 12 grams (see Chapter 4, Compounds and Stoichiometry). One gram is then equal to one mole of amu.
Mole Day is celebrated at 6:02 on October 23 (6:02 on 10/23) because of Avogadro’s number (6.02 × 1023). We will revisit this number in Chapter 4 when we discuss moles in more detail.
The term isotope comes from the Greek, meaning “the same place.” Isotopes are atoms of the same element (hence, occupying the same place on the periodic table of the elements) that have different numbers of neutrons (which means that these atoms of the same element have different mass numbers). Isotopes are referred to by the name of the element followed by the mass number (e.g., carbon-12 has six neutrons, carbon-13 has seven neutrons, etc.). Only the three isotopes of hydrogen are given unique names: protium (Greek protos; first) has one proton and an atomic mass of 1 amu; deuterium (Greek deuteros; second) has one proton and one neutron and an atomic mass of 2 amu; tritium (Greek tritos; third) has one proton and two neutrons and an atomic mass of 3 amu. Because isotopes have the same number of protons and electrons, they generally exhibit the samechemical properties.
In nature, almost all elements exist as two or more isotopes, and these isotopes are usually present in the same proportions in any sample of a naturally occurring element. The presence of these isotopes accounts for the fact that the accepted atomic weight for most elements is not a whole number. The masses listed in the periodic table are weighted averages that account for the relative abundance of various isotopes. See Figure 1.2 for the relative abundances in nature of the first several elements. Hydrogen, which is very abundant, has three isotopes.
Bromine is listed in the periodic table as having a mass of 79.9 amu. This is an average of the two naturally occurring isotopes, bromine-79 and bromine-81, which occur in almost equal proportions. There are no bromine atoms with an actual mass of 79.9 amu.
Example: Element Q consists of three different isotopes, A, B, and C. Isotope A has an atomic mass of 40 amu and accounts for 60 percent of naturally occurring Q. The atomic mass of isotope B is 44 amu and accounts for 25 percent of Q. Finally, isotope C has an atomic mass of 41 amu and a natural abundance of 15 percent. What is the atomic weight of element Q?
Solution: 0.60(40 amu) + 0.25(44 amu) + 0.15(41 amu) = 24.00 amu + 11.00 amu + 6.15 amu = 41.15 amu
The atomic weight of element Q is 41.15 g/mol.
Bohr’s Model of the Hydrogen Atom
We’ve come a long way from J. J. Thomson’s 1904 “plum pudding” model of the atom as negatively charged “corpuscles” (what we call electrons) surrounded by a density of positive charge that others (but not Thomson himself) likened to free-moving negatively charged “plums” suspended in a positively charged “pudding.” We kid you not; we couldn’t make this stuff up if we tried. In 1910, Ernest Rutherford provided experimental evidence that an atom has a dense, positively charged nucleus that accounts for only a small portion of the atom’s volume. Eleven years earlier, Max Planck developed the first quantum theory, proposing that energy emitted as electromagnetic radiation from matter comes in discrete bundles called quanta. The energy value of a quantum, he determined, is given by this equation:
E = hf
where h is a proportionality constant known as Planck’s constant, equal to 6.626 × 10-34 J • s, and f (sometimes designated as the Greek letter , which looks a lot like the English letter v; sometimes you’ll hear students say, “Eee equals eich-vee”) is the frequency of the radiation.
BOHR MODEL OF THE HYDROGEN ATOM
Danish physicist Niels Bohr, in 1913, used the work of Rutherford and Planck to develop his model of the electronic structure of the hydrogen atom. Starting from Rutherford’s findings, Bohr assumed that the hydrogen atom consisted of a central proton around which an electron traveled in a circular orbit and that the centripetal force acting on the electron as it revolved around the nucleus was the electrical force between the positively charged proton and the negatively charged electron.
When we see a formula in our review or on Test Day, we need to focus on ratios and relationships rather than the equation as a whole. This simplifies our “calculations” to a conceptual understanding, which is usually enough to lead us to the right answer.
Bohr used Planck’s quantum theory to correct certain assumptions that classical physics made about the pathways of electrons. Classical mechanics postulates that an object revolving in a circle, such as an electron, may assume an infinite number of values for its radius and velocity. The angular momentum (L = mvr) and kinetic energy (KE = ½mv2) of the object, therefore, can take on any value. However, by incorporating Planck’s quantum theory into his model, Bohr placed restrictions on the value of the angular momentum of the electron revolving around the hydrogen nucleus. Analogous to quantized energy, the angular momentum of an electron, Bohr predicted, is quantized according to the following equation:
where h is Planck’s constant and n is the quantum number, which can be any positive integer. Because the only variable is the quantum number, n, the angular momentum of an electron changes only in discrete amounts with respect to the quantum number.
Bohr then related the permitted angular momentum values to the energy of the electron to obtain the following equation:
where RH is the experimentally determined Rydberg constant, equal to 2.18 × 10-18 J/electron. Therefore, like angular momentum, the energy of the electron changes in discrete amounts with respect to the quantum number. A value of zero energy was assigned to the state in which the proton and electron are separated completely, meaning that there is no attractive force between them. Therefore, the electron in any of its quantized states in the atom will have a negative energy as a result of the attractive forces between the electron and proton; hence the negative sign in the previous energy equation. Now, don’t let this confuse you, because ultimately, the only thing the energy equation is saying is that the energy of an electron increases the further out from the nucleus that it is located. Remember that as the denominator (n2, in this case) increases, the fraction gets smaller. However, here we are working with negative fractions that get smaller as n2 increases. As negative numbers get smaller, they move to the right on the number line, toward zero. So, even though the absolute value is getting smaller (e.g., -8, -7, -6, etc.), its true value is increasing. Think of the concept of quantized energy as the change in gravitational potential energy that you experience when you ascend or descend a flight of stairs. Unlike a ramp, on which you could take an infinite number of steps associated with a continuum of potential energy changes, a staircase only allows you certain changes in height and, as a result, allows only certain discrete (quantized) changes of potential energy.
At first glance, it may not be clear that the energy (E ) is directly proportional to the principle quantum number (n) in this equation. Take notice of the negative charge, which causes the values to approach zero from a greater negative value as n increases (thereby increasing the energy). The negative sign is as important as n’s place in the fraction when it comes to determining proportionality.
Bohr came to describe the structure of the hydrogen atom as a nucleus with one proton forming a dense core around which a single electron revolved in a defined pathway of a discrete energy value. Transferring an amount of energy exactly equal to the difference in energy between one pathway, or orbit, and another, resulted in the electron “jumping” from one pathway to a higher energy one. These pathways or orbits had increasing radii, and the orbit with the smallest radius in which hydrogen’s electron could be found was called the ground state and corresponded to n = 1. When the electron was promoted to a higher energy orbit (one with a larger radius), the atom was said to be in the excited state. Bohr likened his model of the hydrogen atom to the planets orbiting the sun, in which each planet traveled along a (roughly) circular pathway at set distances (and energy values) with respect to the sun. In spite of the fact that Bohr’s model of the atom was overturned within the span of two decades, he was awarded the Nobel Prize in Physics in 1922 for his work on the structure of the atom and to this day is considered one of the greatest scientists of the 20th century.
Note that all systems tend toward minimal energy; thus on the MCAT, atoms of any element will generally exist in the ground state unless subjected to extremely high temperatures or irradiation.
APPLICATIONS OF THE BOHR MODEL
The Bohr model of the hydrogen atom (and other one-electron systems, such as He+ and Li2+) is useful for explaining the atomic emission spectrum and atomic absorption spectrum of hydrogen, and it is helpful in the interpretation of the spectra of other atoms.
Atomic Emission Spectra
At room temperature, the majority of atoms in a sample are in the ground state. However, electrons can be excited to higher energy levels by heat or other energy forms to yield the excited state of the atom. Because the lifetime of the excited state is brief, the electrons will return rapidly to the ground state, resulting in the emission of discrete amounts of energy in the form of photons. The electromagnetic energy of these photons can be determined using the following equation:
where h is Planck’s constant, c is the speed of light in a vacuum (3.00 × 108 m/s), and is the wavelength of the radiation.
E = hf for photons in physics. This also holds true here because we know that c = f. This is based on the formula v = f for photons.
The different electrons in an atom can be excited to different energy levels. When these electrons return to their ground states, each will emit a photon with a wavelength characteristic of the specific energy transition it undergoes. The quantized energies of light emitted under these conditions do not produce a continuous spectrum (as expected from classical physics). Rather, the spectrum is composed of light at specified frequencies and is thus known as a line spectrum, where each line on the emission spectrum corresponds to a specific electronic transition. Because each element can have its electrons excited to different distinct energy levels, each one possesses a unique atomic emission spectrum, which can be used as a fingerprint for the element. One particular application of atomic emission spectroscopy is in the analysis of stars and planets: While a physical sample may be impossible to procure, the light from a star can be resolved into its component wavelengths, which are then matched to the known line spectra of the elements.
Emissions from electrons in molecules, or atoms dropping from an excited state to a ground state, give rise to fluorescence. We see the color of the emitted light.
The Bohr model of the hydrogen atom explained the atomic emission spectrum of hydrogen, which is the simplest emission spectrum among all the elements. The group of hydrogen emission lines corresponding to transitions from the upper energy levels n > 2 to n = 2 (that is to say, the pattern of photon emissions from the electron falling from the n > 2 energy level to the n = 2 energy level) is known as the Balmer series and includes four wavelengths in the visible region. The group corresponding to transitions from the upper levels n > 1 to n = 1 (that is to say, the emissions of photons from the electron falling from the higher energy levels to the ground state) is called the Lyman series, which includes larger energy transitions and therefore shorter photon wavelengths in the UV region of the electromagnetic spectrum.
When the energy of each frequency of light observed in the emission spectrum of hydrogen was calculated according to Planck’s quantum theory, the values obtained closely matched those expected from energy level transitions in the Bohr model. That is, the energy associated with a change in the quantum number from an initial higher value ni to a final lower value nf is equal to the energy of the photon predicted by Planck’s quantum theory. Combining Bohr’s and Planck’s calculations, we arrive at
The energy of the emitted photon corresponds to the precise difference in energy between the higher-energy initial state and the lower-energy final state.
Atomic Absorption Spectra
When an electron is excited to a higher energy level, it must absorb energy. The energy absorbed that enables an electron to jump from a lower-energy level to a higher one is characteristic of that transition. This means that the excitation of electrons in the atoms of a particular element results in energy absorption at specific wavelengths. Thus, in addition to a unique emission spectrum, every element possesses a characteristic absorption spectrum. Not surprisingly, the wavelengths of absorption correspond directly to the wavelengths of emission because the difference in energy between levels remains unchanged. Identification of elements present in a gas phase sample requires absorption spectra.
Absorption is the basis for color of compounds. We see the color of the light that is NOT absorbed by the compound.
You’ve just been put through a series of paragraphs crammed with technical language (mumbo jumbo is too strong a term; after all, you are sufficiently intelligent to grasp these concepts). That said, at least a few pairs of eyes reading this book will have gone glassy by this point. Therefore, let’s bring this back to the realm of experience by way of analogy. We’ve already discussed equating the energy levels available to electrons to stairs on a staircase. Taking this analogy one step further, so to speak: Let’s imagine that you and your friend are walking side-by-side up a set of stairs. You have very long legs, so it is your habit to take two, sometimes even three, steps at a time; your friend has short legs and so takes one, or at most two, steps at a time. The pattern by which you jump from a lower step to a higher one will be characteristic to you and you alone and will be quite different from the pattern by which your friend jumps from a lower step to a higher one, which will be unique to her. Furthermore, you have to invest energy into the process of ascending the staircase. This in a nutshell is the significance of the atomic absorption spectrum. The atomic emission spectrum is simply a record of the process in reverse.
ΔE is the same for absorption or emission between any two energy levels. The sign (positive or negative) of ΔE indicates whether the energy goes in or out and, therefore, whether the electron is going to an excited state (absorption) or to the ground state (emission), respectively.
Quantum Mechanical Model of Atoms
While Bohr’s model marked a significant advancement in the understanding of the structure of atoms (at least we were no longer talking about plum pudding), his model ultimately proved inadequate to explain the structure and behavior of atoms containing more than one electron. The model’s failure was a result of Bohr’s not taking into account the repulsion between multiple electrons surrounding one nucleus. Modern quantum mechanics has led to a more rigorous and generalized study of the electronic structure of atoms. The most important difference between Bohr’s model and the modern quantum mechanical model is that Bohr postulated that electrons follow a clearly defined circular pathway or orbit at a fixed distance from the nucleus, whereas modern quantum mechanics has shown that this is not the case. Rather, we now understand that electrons move rapidly in extraordinarily complex patterns within regions of space around the nucleus called orbitals. The confidence by which those in Bohr’s time believed they could identify the location (or pathway) of the electron is now replaced by a more modest suggestion that the best we can do is describe the probability of finding an electron within a given region of space surrounding the nucleus. In the current quantum mechanical model, it is impossible to pinpoint exactly where an electron is at any given moment in time, and this is expressed best by the Heisenberg uncertainty principle: It is impossible to simultaneously determine, with perfect accuracy, the momentum and the position of an electron. If we want to assess the position of an electron, the electron has to stop (thereby changing its momentum); if we want to assess its momentum, the electron has to be moving (thereby changing its position).
Modern atomic theory postulates that any electron in an atom can be completely described by four quantum numbers: n, l, ml, ms. Furthermore, according to the Pauli exclusion principle, no two electrons in a given atom can possess the same set of four quantum numbers. The position and energy of an electron described by its quantum numbers is known as its energy state. The value of n limits the value of l, which in turn limits the values of ml. Think of this like a country: A country has a defined number of states, and each state has a defined number of cities or towns. The values of the quantum numbers qualitatively give information about the orientation of the orbital. As we examine the four quantum numbers more closely, pay attention especially to l and ml, as these two tend to give students the greatest difficulty.
Principal Quantum Number
The first quantum number is commonly known as the principal quantum number and is denoted by the letter n. This is the quantum number used in Bohr’s model that can theoretically take on any positive integer value. The larger the integer value of n, the higher the energy level and radius of the electron’s orbit(al). Within each shell of some n value, there is a capacity to hold a certain number of electrons equal to 2n2, and the capacity to hold electrons increases as the n value increases. The difference in energy between two shells decreases as the distance from the nucleus increases because the energy difference is a function of [1/ni2 - 1/nf2]. For example, the energy difference between the n = 3 and the n = 4 shells is less than the energy difference between the n = 1 and the n = 2 shells. The term shell brings to mind the notion of eggshells, and you’ve probably heard the analogy between n values and eggshells of increasing size. This is fine as long as you don’t extend the analogy to the point that you are thinking about electron pathways as precisely defined orbits. Nevertheless, if thinking about eggshells helps you to remember that the principal quantum number says something about the overall energy of the electron orbitals as a function of distance from the nucleus, then go with it.
A larger integer value for the principal quantum number indicates a larger radius and higher energy. This is similar to gravitational potential energy, where the higher the object is above the earth, the higher its potential energy will be.
Azimuthal Quantum Number
The second quantum number is called the azimuthal (angular momentum) quantum number and is designated by the letter l. The second quantum number refers to the shape and number of subshells within a given principal energy level (shell). The azimuthal quantum number is very important because it has important implications for chemical bonding and bond angles. The value of n limits the value of l in the following way: For any given value of n, the range of possible values for l is 0 to (n–1). For example, within the first principal energy level, n = 1, the only possible value for l is 0; within the second principal energy level, n = 2, the possible values for l are 0 and 1. A simpler way to remember this relationship is that the n-value also tells you the number of possible subshells.
Therefore, there’s only one subshell in the first principal energy level; there are two subshells within the second principal energy level; there are three subshells within the third principal energy level, and so on. The subshells also go by names other than the integer value of l: The l = 0 subshell is also known as the s subshell; the l = 1 subshell is also known as the p subshell; the l = 2 subshell is known as the d subshell; and finally, the l = 3 subshell is the f subshell. You’re probably more used to working with these letter names than with the integer values.
For any principal quantum number n, there will be n possible values for l.
The maximum number of electrons that can exist within a given subshell is equal to 4l + 2. The energies of the subshells increase with increasing l value; however, the energies of subshells from different principal energy levels may overlap. For example, the 4s subshell will have a lower energy than the 3d subshell. This is why, ultimately, the image of increasingly larger eggshells falls short of adequately serving as an analogy.
Magnetic Quantum Number
The third quantum number is the magnetic quantum number and is designated ml. The magnetic quantum number specifies the particular orbital within a subshell where an electron is highly likely to be found at a given moment in time. Each orbital can hold a maximum of two electrons. The possible values of ml are the integers between -l and +l, including 0. For example, the s subshell, with its l value = 0, limits the possible ml value to 0, and since there is a single value of ml for the s subshell, there is only one orbital in the s subshell. The p subshell, with its l value = 1, limits the possible ml values to -1, 0, +1, and since there are three values for ml for the p subshell, there are three orbitals in the p subshell. The d subshell has five orbitals, and the f subshell has seven orbitals. The shape of the orbitals, as the number of orbitals, is dependent upon the subshell in which they are found. The s subshell orbital is spherical, while the three p subshell orbitals are each dumbbell shaped along the x-, y-, and z-axes. In fact, the p orbitals are often referred to as px, py, and pz. The shapes of the orbitals in the d and f subshells are much more complex, and the MCAT will not expect you to answer questions about their appearance. Of course, any discussion of orbital shape must not allow for a literal interpretation of the term, since we are using the term to describe “densities of probabilities” for finding electrons in regions of space surrounding the nucleus.
For any value of l, there will be 2l + 1 possible values for ml. For any n, this produces n2 possible values of ml (i.e., n2 orbitals).
Spin Quantum Number
The fourth quantum number is called the spin quantum number and is denoted by ms. In classical mechanics, an object spinning about its axis has an infinite number of possible values for its angular momentum. However, this does not apply to the electron, which has two spin orientations designated +½ and -½. Whenever two electrons are in the same orbital, they must have opposite spins.
For any value of n, there will be a maximum of 2n2 electrons (i.e., two per orbital).
In this case, they are often referred to as paired. Electrons in different orbitals with the same ms values are said to have parallel spins.
The quantum numbers for the orbitals in the second principal energy level, with their maximum number of electrons noted in parentheses, are shown in Table 1.2.
ELECTRON CONFIGURATION AND ORBITAL FILLING
For a given atom or ion, the pattern by which subshells are filled and the number of electrons within each principal energy level and subshell are designated by its electron configuration. In this notation, the first number denotes the principal energy level, the letter designates the subshell, and the superscript gives the number of electrons in that subshell. For example, 2p4 indicates that there are four electrons in the second ( p) subshell of the second principal energy level. By definition, this also implies that the energy levels below 2p (that is, 1s and 2s) have already been filled (see Figure 1.3).
Remember that the shorthand used to describe the electron configuration is derived directly from the quantum numbers.
To write out an atom’s electron configuration, you need to know the order in which subshells are filled. They are filled from lower to higher energy, and each subshell will fill completely before electrons begin to enter the next one. You don’t really need to memorize this ordering because there are two very helpful ways of recalling this. The (n + l ) rule can be used to rank subshells by increasing energy. This rule states that the lower the sum of the values of the first and second quantum numbers (n + l ), the lower the energy of the subshell. This is a very helpful rule to remember for Test Day. If two subshells possess the same (n + l ) value, the subshell with the lower n value has a lower energy and will fill with electrons first. The other helpful way to recall this ordering is the flow diagram in Figure 1.3, which also gives the order in which electrons will fill the shells and subshells. We recommend that you quickly write out this flow diagram on the scratch material provided at the testing center at the start of your Physical Science section for quick and easy reference throughout the section.
Remember this chart for Test Day—being able to re-create it quickly on your scratch paper may save you some time and get you that higher score!
Example: Which will fill first, the 3d subshell or the 4s subshell?
Solution: For 3d, n = 3 and = 2, so (n + ) = 5. For 4s, n = 4 and = 0, so (n + ) = 4. Therefore, the 4s subshell has lower energy and will fill first. This can also be determined from the chart by examination.
If you are asked to determine which subshells are filled, you must know the number of electrons in the atom. In the case of a neutral (uncharged) atom, the number of electrons equals the number of protons, which can be found by the atomic number of the element. If the atom is charged, the number of electrons is
• equal to the atomic number plus the extra electrons if the ion is negatively charged; or
• equal to the atomic number minus the missing electrons if the ion is positively charged.
In subshells that contain more than one orbital, such as the 2p subshell with its three orbitals, the orbitals will fill according to Hund’s rule, which states that within a given subshell, orbitals are filled such that there are a maximum number of half-filled orbitals with parallel spins. Electrons are somewhat curmudgeonly and misanthropic in that they don’t like the company of other electrons. They prefer to spend as much time alone as possible. Therefore, an electron will be happier (at a lower energy level) if it is placed into an empty orbital rather than being forced to share its living quarters with another electron. (If you ever had to share a bedroom with a sibling, you know exactly what we’re talking about.) Of course, the basis for this preference is the fact that all electrons have a negative charge and, as like charges, they exert repelling forces against each other; these forces must be overcome for two electrons to be in the same orbital.
Hund’s rule is like sitting down on a bus. Everyone wants an empty seat, but we have to start pairing up after all the seats are half filled.
Example: What are the written electron configurations for nitrogen (N) and iron (Fe) according to Hund’s rule?
Solution: Nitrogen has an atomic number of 7. Thus, its electron configuration is 1s2 2s2 2p3. According to Hund’s rule, the two s-orbitals will fill completely, while the three p-orbitals will each contain one electron, all with parallel spins.
Iron has an atomic number of 26, and its 4s subshell fills before the 3d. Using Hund’s rule, the electron configuration will be as follows:
Iron’s electron configuration is written as 1s2 2s2 2p6 3s2 3p6 3d6 4s2. Subshells may be listed either in the order in which they fill (e.g., 4s before 3d) or with subshells of the same principal quantum number grouped together, as shown here. Both methods are correct.
Half-filled and fully filled orbitals have lower energies than intermediate states. We will see this again with transition metals.
The presence of paired or unpaired electrons affects the chemical and magnetic properties of an atom or molecule. The magnetic field of materials made of atoms with unpaired electrons will cause the unpaired electrons to orient their spins in alignment with the magnetic field, and the material will be weakly attracted to the magnetic field. These materials are considered paramagnetic. Materials consisting of atoms that have all paired electrons will be slightly repelled by a magnetic field and are said to be diamagnetic.
The valence electrons of an atom are those electrons that are in its outermost energy shell, most easily removed, and available for bonding. As with the unruly children sitting farthest from the checkers game who are most likely to get themselves into trouble, the valence electrons are the “active” electrons of an atom and to a large extent dominate the chemical behavior of the atom. For elements in Groups IA and IIA (Groups 1 and 2), only the outermost s subshell electrons are valence electrons. For elements in Groups IIIA through VIIIA (Groups 13 through 18), the outermost s and psubshell electrons in the highest principal energy level are valence electrons. For transition elements, the valence electrons are those in the outermost s subshell and in the d subshell of the next-to-outermost energy shell. For the inner transition elements (that is, those in the lanthanide and actinide series), the valence electrons include those in the s subshell of the outermost energy level, the d subshell of the next-to-outermost energy shell, and the f subshell of the energy shell two levels below the outermost shell. All elements in period 3 and below may accept electrons into their d subshell, which allows them to hold more than eight electrons in their valence shell, in apparent “violation” of the octet rule (see Chapters 2 and 3). We’ll learn that this is perfectly acceptable for these elements, however, and is occasionally preferred.
Remember that paramagnetic means that a magnetic field will cause parallel spins in unpaired electrons and therefore cause an attraction.
The valence electron configuration of an atom helps us understand its properties and is ascertainable from the periodic table (the only “cheat sheet” available on the MCAT!). The “EXHIBIT” button on the bottom of the screen on Test Day will bring up a window with the periodic table. Use it often!
Example: Which are the valence electrons of elemental iron, elemental selenium, and the sulfur atom in a sulfate ion?
Solution: Iron has 8 valence electrons: 2 in its 4s subshell and 6 in its 3d subshell.
Selenium has 6 valence electrons: 2 in its 4s subshell and 4 in its 4p subshell. Selenium’s 3d electrons are not part of its valence shell.
Sulfur in a sulfate ion has 12 valence electrons: its original 6 plus 6 more from the oxygens to which it is bonded. Sulfur’s 3s and 3p subshells can contain only 8 of these 12 electrons; the other 4 electrons have entered the sulfur atom’s 3d subshell, which in elemental sulfur is empty.
Congratulations! You’ve made it through the first chapter, and as far as we can tell, you’re still alive and there seems to have been minimal shedding of tears and blood. Good! Now that we have covered topics related to the most fundamental unit of matter, the atom, you’re set to advance your understanding of the physical world in more complex ways. This chapter described the characteristics and behavior of the three subatomic particles—the proton, the neutron, and the electron. In addition, it compared and contrasted the two most recent models of the atom. The Bohr model is adequate for describing the structure of one-electron systems, such as the hydrogen atom or the helium ion, but fails to describe adequately the structure of more complex atoms. The quantum mechanical model theorizes that electrons are found, not in discrete-pathway orbits, but in “clouds of probability,” or orbitals, by which we can predict the likelihood of finding electrons within given regions of space surrounding the nucleus. Both theories tell us that the energy levels available to electrons are not infinite but discrete and that the energy between levels is a precise amount called aquantum. The four quantum numbers completely describe the position and energy of any electron within a given atom. Finally, we learned two simple recall methods for the order in which electrons fill the shells and subshells of an atom and that the valence electrons are the troublemakers in an atom—and the ones we need to keep our eyes on.
CONCEPTS TO REMEMBER
The subatomic particles include the proton, which has a positive charge; the neutron, which has no charge; and the electron, which has a negative charge. The nucleus contains the protons and neutrons, while the electrons reside in regions of space. The element’s atomic number is its number of protons, while the sum of an electron’s protons and neutrons is its mass number.
Isotopes are atoms of a given element that have different mass numbers because they have different numbers of neutrons in their nuclei. Because they have the same atomic number, they are all of the same elemental type. Most isotopes of elements are identified by the element followed by the mass number (e.g., carbon-12, carbon-13, carbon-14). The three isotopes of hydrogen go by different names: protium, deuterium, and tritium.
Bohr proposed a model of the atom with a dense, positively charged nucleus surrounded by electrons revolving around the nucleus in defined pathways of distinct energy levels called orbits.
The energy of an electron is quantized, which is to say that there is not an infinite range of energy levels available to an electron. Electrons can exist only at certain energy levels, and the energy of an electron increases the farther it is from the nucleus. The energy difference between energy levels is called a quantum.
For an electron to jump from a lower energy level to a higher one, it must absorb an amount of energy precisely equal to the energy difference between the two levels. Every element has a characteristic atomic absorption spectrum. When electrons return from the excited state to the ground state, they emit an amount of energy that is exactly equal to the energy difference between the two levels. Every element has a characteristic atomic emission spectrum. Sometimes the electromagnetic energy emitted corresponds to a frequency in the visible light range.
The quantum mechanical model posits that electrons do not travel in defined orbits but rather in complex patterns called orbitals. An orbital is a region of space around the nucleus defined by the probabilities of finding an electron in that region of space. The Heisenberg uncertainty principle states that it is impossible to know at the same time both an electron’s position and its momentum.
There are four quantum numbers. These numbers completely describe any electron in an atom. The principal quantum number, n, describes the average energy of an orbital. The azimuthal quantum number, l, describes the subshells within a given principal energy level. The magnetic quantum number, ml , specifies the particular orbital within a subshell where an electron is likely to be found at a given moment in time. The spin quantum number, ms, indicates the spin orientation of an electron in an orbital.
The system of designating the placement of electrons into the principal energy levels, subshells, and orbitals is electron configuration. For example, 1s22s22p63s2 is the electron configuration for magnesium. A neutral magnesium atom has 12 electrons: two in the s-orbital of the first energy level, two in the s-orbital of the second energy level, 6 in the p-orbitals of the second energy level, and 2 in the s-orbital of the third energy level. The two electrons in the s-orbital of the third energy level are the valence electrons for the magnesium atom.
Electrons fill the principle energy levels and subshells according to increasing energy, which can be determined by the (n + l) rule. Electrons fill orbitals according to Hund’s rule, which states that electrons prefer to be unpaired with parallel spins.
Valence electrons are those electrons in the outermost shell and/or those available for interaction (bonding) with other atoms. For the representative elements, the valence electrons are found in s- and/or p-orbitals. For the transition elements, the valence electrons are found in s-, d-, and f-orbitals. Many atoms interact with other atoms to form bonds so as to complete the octet in the valence shell.
EQUATIONS TO REMEMBER
1. Which of the following is the correct electron configuration for Zn2+?
2. Which of the following quantum number sets describes a possible element?
A. n = 2; l = 2; ml = 1; ms = +½
B. n = 2; l = 1; ml = -1; ms = +½
C. n = 2; l = 0; ml = -1; ms = -½
D. n = 2 ; l = 0; ml = 1; ms = -½
3. What is the maximum number of electrons allowed in a single atomic energy level in terms of the principal quantum number n ?
B. 2n + 2
D. 2n2 + 2
4. Which of the following equations describes the maximum number of electrons that can fill a subshell?
A. 2l + 2
B. 4l + 2
D. 2l2 + 2
5. Which of the following substances is most likely to be diamagnetic?
6. An electron returns from an excited state to its ground state, emitting a photon at = 500 nm. What would be the magnitude of the energy change if this process were repeated such that a mole of these photons were emitted?
A. 3.98 × 10-19 J
B. 3.98 × 10-21 J
C. 2.39 × 105 J
D. 2.39 × 103 J
7. Suppose an electron falls from n = 4 to its ground state, n = 1. Which of the following effects is most likely?
A. A photon is absorbed.
B. A photon is emitted.
C. The electron gains velocity.
D. The electron loses velocity.
8. Which of the following compounds is not a possible isotope of carbon?
9. According to the Heisenberg uncertainty principle, which of the following properties of a particle can an observer measure simultaneously?
A. I and II
B. I and III
C. II and III
D. I, II, and III
10. Which of the following electronic transitions would result in the greatest gain in energy for a single hydrogen electron, assuming that its ground state is n = 1?
A. An electron moves from n = 6 to n = 2.
B. An electron moves from n = 2 to n = 6.
C. An electron moves from n = 3 to n = 4.
D. An electron moves from n = 4 to n = 3.
11. Suppose that a chemical species fills its orbitals as shown.
This compound could be said to obey which of the following laws of atomic physics?
A. Hund’s rule
B. Heisenberg uncertainty principle
C. Bohr’s model
D. Pauli exclusion principle
12. How many total electrons are in a 133Cs cation?
13. The atomic mass of hydrogen is 1.008 amu. What is the percent composition of hydrogen by isotope, assuming that hydrogen’s only isotopes are 1H and 2D?
A. 92% H, 8% D
B. 99.2% H, 0.8% D
C. 99.92% H, 0.08% D
D. 99.992% H, 0.008% D
14. Consider the two sets of quantum numbers shown in the table, which describe two different electrons in the same atom.
Which of the following terms best describes these two electrons?
15. Which of the following species is represented by the electron configuration 1s22s22p63s23p64s13d5 ?
D. both (A) and (B)
16. Which of the following statements is NOT true of an electron’s ground state?
A. The electron is at its lowest possible energy level.
B. The electron is in a quantized energy level.
C. The electron is traveling along its smallest possible orbital radius.
D. The electron is static.
17. Which of the following experimental conditions would NOT excite an electron out of the ground state?
B. High temperature
C. High pressure
D. None of the above
Small Group Questions
1. Which orbital fills first: 4s or 3d ? Which is first to give up electrons?
2. Why do atoms have neutrons? What purpose do they serve?
Explanations to Practice Questions
Because an electron was pulled off the neutral parent atom, consider how that parent atom was formed and which electron it would be willing to give up. Zn° has 30 electrons, so it would have an electron configuration of 1s22s22p63s23p64s23d10. Subshells strongly prefer to be completely empty, exactly half full, or completely full. Therefore, the best way to lose two electrons and form Zn2+ is to pull them both out of the 4s-orbital. (B) implies that electrons are pulled out of the d-orbital, (C) presents the configuration of the uncharged zinc atom, and (D) shows the configuration that would exist if four electrons were removed.
The azimuthal quantum number l cannot be higher than n–1, ruling out (A). The ml number, which describes the chemical’s magnetic properties, can only be an integer value between –l and l, and it cannot be equal to 1 if l = 0, ruling out (C) and (D).
The correct answer is (C). For any value of n there will be a maximum of 2n2 electrons, i.e., two per orbital.
This formula describes the number of electrons in terms of the azimuthal quantum number l, which ranges from 0 to n–1, n being the principal quantum number.
Azimuthal Quantum Number (l)
Number of Electrons
Using the data in the table, the quickest way to solve this problem is to choose a subshell and plug in its l value to solve for the number of electrons. If more than one option works, discard the others and test the remaining possibilities using another subshell.
Sulfur is diamagnetic as opposed to ferromagnetic (iron, cobalt) or paramagnetic (hydrogen). Ferromagnetism refers, loosely, to the ability of a surface to attract an external magnetic field. It is characteristic of iron (Fe), from which it derives its name. More specifically, paramagnetism describes the tendency of valence electrons to align with the same spin in the presence of a strong magnetic field. Strongly paramagnetic materials, including transition metals, are usually called ferromagnetic. Transition metals like iron are characterized by a “sea” of electrons moving freely about the surface, which makes it easier for all these electrons to align in one direction. (This electron “sea” is an imprecise model, but it’s good enough for the MCAT.) It is harder for more stable elements (e.g., oxygen, halogens, noble gases) to align their electrons in one orientation because their orbitals are nearly filled; these substances are known as diamagnetic. Sulfur has a similar atomic structure to oxygen, so it is also diamagnetic.
The problem requires the MCAT favorite equation E = hf, where h = 6.626 × 10-34 (Planck’s constant) and f is the frequency of the photon. (Memorize Planck’s constant!) One can calculate the frequency of the photon using the provided wavelength, 500 nm, with the equation f = c/ , where c = 3 × 108 m/s, the speed of light. Here, f = (3 × 108 m/s)/500 × 10-9 m, or 6 × 1014 s-1 (1 Hz = 1 s-1). That leads to E = hf, or E = (6.626 × 10-34) × (6 × 1014 Hz) = 3.98 × 10-19 J. (Don’t worry about memorizing the units of Planck’s constant—energy is always in joules!) However, the problem includes an additional trick, in that the answer must account for a mole of photons. The E = hf equation works for a single photon only. Thus, the answer must account for this using Avogadro’s number, 6.022 × 1023 photons. Multiply: (3.98 × 10-19 J/photon) × (6.022 × 1023 photons) = 2.39 × 105J.
There is not enough information in the problem to determine how the velocity of the electron will change. There will be some energy change, however, as the electron must lose energy to return to the minimum energy ground state. That will require emitting radiation in the form of a photon, (B).
Recall that the superscript (i.e., the A in AC) refers to the mass number of an atom, which is equal to the number of protons plus the number of neutrons present in an element. (Sometimes a text will list the atomic number, Z, or total number of protons, under the mass number A.) According to the periodic table, carbon contains 6 protons; therefore, its atomic number (Z) = 6. An isotope contains the same number of protons and a different number of neutrons as the element. Carbon is most likely to have an atomic number of 12, for 6 protons and 6 neutrons. (C) and (D) are possible isotopes that would have more neutrons than does 12C. The 6C isotope is unlikely. It would mean that there were 6 protons and 0 neutrons, and it would probably collapse under the stress of the positive charge.
The Heisenberg uncertainty principle states that you cannot know the position and momentum of a particle simultaneously, which eliminates (A) and (D). Momentum depends on velocity (recall from Newtonian mechanics that p = mv), so in order to calculate momentum, velocity must be known. (C) explains this.
For the electron to gain energy, it must absorb photons to jump up to a higher energy level. This eliminates (A) and (D). Between (B) and (C), there is a bigger jump between n = 2 and n = 6 than there is between n = 3 and n = 4. Therefore, (B) represents the greatest energy gain.
The MCAT covers qualitative topics more often than quantitative topics in this unit. It is critical to be able to distinguish the fundamental principles that determine electron organization, which are usually known by the names of the scientists who discovered them. The Heisenberg uncertainty principle refers to the momentum and position of a single electron, and the Bohr model was an early attempt to describe the behavior of the single electron in a hydrogen atom. (D) is tempting, but (A) is more complete and therefore the correct answer. The element shown here, nitrogen, is often used to demonstrate Hund’s rule because it is the smallest element with a half-filled p subshell. Hund’s rule explains that electrons fill empty orbitals first, and in fact, the three p-electrons in this image each occupy a separate orbital. Hund’s rule is really a corollary of the Pauli exclusion principle, in that the Pauli exclusion principle suggests that each orbital contains two electrons of opposite spin. Additional electrons must fill new orbitals so the compound remains stable in its ground state.
The quickest way to solve this problem is to use the periodic table and find out how many protons are in Cs atoms; there are 55. Neutral Cs atoms would also have 55 electrons. A Cs cation is most likely to have a single positive charge because it has one unpaired s-electron. This translates to one fewer electron than the number or protons, or 54 electrons.
The easiest way to approach this problem is to set up a system of two algebraic equations, where x and y are the percentages of H (mass = 1 amu) and D (mass = 2 amu), respectively. Your setup should look like the following system:
x + y = 1 (proportion H (x) + proportion D ( y) in whole, x% + y% = 100%).
1x + 2y = 1.008 (the total atomic mass).
Substitute one variable for the other so the atomic mass is in terms of one variable (1–y = x), then solve for the other percentage ([1–y] + 2y = 1.008 simplifies to 0.008 = y, or 0.8% D). That plus 99.2 percent H makes 100 percent. Another way to examine this problem is if we had 50 percent of H and 50 percent of D, we could probably imagine an atomic mass of 1.5 amu. Therefore, if 50 percent D gave a mass of 1.5 amu, 80 percent D would yield an atomic mass of 1.8 amu. Eight percent would result in 1.08 amu, and 0.8 percent in 1.008 amu, which is the desired mass.
The terms in the answer choices refer to the magnetic spin of the two electrons. The quantum number ms represents this property as a measure of the electrons’ relative intrinsic angular momentum. These electrons’ spins are parallel, in that their spins are aligned in the same direction (ms = +½ for both species). (B), (C), and (D) suppose that ms = +½ for one electron and -½ for the other. (D) implies that the two electrons have opposite spins but lie within the same orbital.
When dealing with ions, you cannot directly approach electronic configurations based on the number of electrons they currently hold. First examine the neutral atom’s configuration, and then determine which electrons would have been removed.
Neutral Atom’s Configuration
Cr0: [Ar] 4s13d5
Mn0: [Ar] 4s23d5
Mn+: [Ar] 4s13d5
Fe0: [Ar] 4s23d6
Fe2+: [Ar] 4s03d6
Due to the stability of half-filled d-orbitals, neutral chromium assumes the electron configuration of [Ar] 4s13d5. Mn must lose one electron from its initial configuration to become the Mn+ cation. That electron would come from the 4s-orbital, because the 3d-orbital would strongly prefer to remain half-full. This d-orbital’s desire to be half-full trumps the s-orbital’s desire to be completely full, because it is at a higher energy. Fe must lose two electrons to become Fe2+. They’ll both be lost from the same orbital, because the scientist can’t pick and choose which electrons to take! Moreover, the only way Fe2+ could hold the configuration in the question stem would be if one d-electron and one s-electron were lost together. Therefore, both answer choices (A) and (B) match the electron configuration in the question stem.
(D) is the only incorrect statement in the set. Electrons are assumed to be in motion in any energy level, even in the ground state. All of the other statements are true. An electron’s ground state describes its position at the lowest and most stable energy level, i.e., lowest n value. It must be an integer value in the ground state and in the excited state. This number gives a relative indication of the electron’s distance from the nucleus, so it must have the smallest radius of all the energy levels. The farther the electron moves from the nucleus, the greater energy it needs to overcome its attractive forces.
High pressure is unlikely to excite an electron out of the ground state unless it causes an extreme change in temperature, which would add enough energy to the system to promote the electron. There is not enough information in the problem to determine whether the stated high pressure would be high enough, as the answer is too generic. Irradiation has the same effect as temperature.