Major Concept Area: Electrical Forces and Bonding. Chemists are fascinated by the manner in which atoms aggregate to form molecules and lattices; and molecules aggregate to form condensed phases of matter. Aggregation of atoms, oppositely charged ions, and molecules is a consequence of electrical forces exerted on the electrons of one particle by the nucleus (or nuclei) of the other. In this chapter, we learn that all physical and chemical bonding is the result of molecular forces that are electrical in origin.
Specific Concepts in this Chapter:
The focus of this chapter is the electrical force, which is responsible for all attractive interactions between atoms, molecules, and ions. The electrical force is described quantitatively by Coulomb's Law. A system in which forces act can possess two types of energy. These are kinetic energy, KE, and potential energy, PE. The magnitude of the kinetic energy is a function only of temperature, as we have seen. Potential energy arises from the action of the Coulomb force. We will discuss the various ways in which Coulomb forces operate in atomic/molecular systems, manifesting in both intramolecular and intermolecular contexts. Intramolecular forces include the forces between atoms in covalent molecules; the forces between oppositely charged ions in ionic compounds; and the forces between identical atoms in metals. Intermolecular forces include the forces that operate between one atom or molecule and another, or between the ion of one substance and the molecules of another. Although there are differences among the various types of forces, they are all fundamentally similar: they give rise to potential energy between the interacting particles, described by the potential well. We shall see that an interplay between the depth of this well and the thermal kinetic energies of particles is responsible for many of the physical properties of substances.
6-1 Types of Energy and the Coulomb Force. An atom or a collection of atoms can possess energy of only two types. Kinetic energy (abbreviated KE) is energy of motion. A body of mass, m, moving with velocity, v, has kinetic energy given by equation 6-1-1.
When mass is expressed in kg and velocity in m/s, KE has units of kg m2/s2, or Joules (J). One Joule is defined as 1 kg m2/s2. In Chapter 5, we learned that the kinetic energy of atoms and molecules depends only on temperature, through the simple expression in 6-1-2.
Here Average KE is the average kinetic energy of a single atom or molecule; k is Boltzmann's constant; and T is the Kelvin temperature. Potential Energy (abbreviated PE) is energy of position. A rock poised on top of a hill, even though it is not moving, has the potential to move if given a small push. It therefore has potential energy. This potential energy is converted to kinetic energy as the rock rolls down the hill. All energy is either kinetic or potential. The total energy of any mechanical system is the sum of its kinetic and potential energies:
Potential energy is an abstract concept; it is difficult to understand how something can possess energy simply by virtue of its position. Consider the situation pictured in Figure 6-1. A weight rests on a platform at height h above the ground. The weight is connected by a rope, across a pulley, to another platform that carries a load. If we remove the weight from the platform, it will fall, and in the process will lift the loaded platform. But in lifting the load a distance h, the weight has done work (a form of energy) on the load. To do work, the weight must use energy. It possesses the necessary energy to do the work by virtue of its position at distance h above the earth. Thus the concept of potential energy presupposes a force of attraction or repulsion. The weight falls to earth because there is a gravitational force of attraction between the earth and the weight. To move the weight from the ground to the platform initially, we would have had to use an amount of energy sufficient to overcome this force of attraction. The amount of energy used to lift the weight is stored in the earth-weight system as potential energy. We retrieve this energy by allowing the earth and weight to come together. Potential energy is a consequence of a force.
Our example has been based on the gravitational force because this is a familiar situation. There are 3 forces other than gravitation. One of these is the electromagnetic force. This is the force by which electrons are attracted and held by the nucleus in an atom. The electromagnetic force is responsible for all interactions between atoms and molecules. It is far stronger than the gravitational force, which may be neglected in discussing atomic and molecular interactions (the electromagnetic force of repulsion between two electrons is 1042 times stronger than the gravitational force of attraction between them!). The electromagnetic force operating at the atomic level is responsible for
The magnitude of the electrical force is given by Coulomb's Law, equation 6-1-4.
q1 and q2 are the values of the two interacting electrical charges, in Coulombs, and r is the distance between them, in m. W is a constant that adjusts the units. Its value is 1.113 * 10-10 C2N-1m-2. The force described by Coulomb's Law is responsible for all attractive forces between atoms, molecules, and ions. The potential energy associated with the Coulomb force is called the Coulomb potential energy. Because force is the negative derivative of potential energy with respect to distance (F = -dPE/dr), the Coulomb potential energy can be obtained by integration of Coulomb's Law. It is given in 6-1-5.
When one of the charges is positive and the other negative, the potential energy is negative at all finite values of r.
6-2 Coulomb Forces and the Potential Well. Atoms, molecules, and ions exert attractive forces on one another because the atomic nuclei in one species attract the electrons in the other. These are called molecular forces, and the potential energy that results from the action of these forces is called molecular potential energy. We make several general statements about molecular forces and potential energy, and the relationship between them. The statements are equally valid for a pair of particles or a large collection of particles (here we are using the word, "particle", generically to represent an atom, a molecule, or an ion).
To aid your grasp of Figure 6-2, it has been divided into three regions. We now consider these regions in turn.
Region 1. The particles, shown as two small circles just above the horizontal axis, are separated by large distances. Taking 2 * 10-10 m as a typical particle radius, the distances of separation in region 1 are 10 or more times larger. At large separation distance, molecular forces are nearly zero. For convenience, we take the potential energy to be zero when the forces are zero, so the PE curve coincides with the horizontal axis at large r. This situation of very large separation between atoms, molecules, or ions is a state of deaggregation. The gas phase of a pure substance is an example of a deaggregated state.
Region 2. The particles have approached closely enough to exert noticeable forces on one another. Action of the forces lowers the potential enrgy. The closer the particles approach (i.e., the more aggregated they become), the stronger the forces (equation 6-1-4) and the lower the PE. At the left boundary of region 2, the particles just touch. At this point, attractive forces are maximized and potential energy is at a minimum. This is an aggregated state. The solid phase of a pure substance is an example of an aggregated state.
Region 3. In region 3, PE increases very rapidly as r decreases. In this region, the somewhat deformable electron clouds of the particles are being pushed into one another. Since electrons repel one another, two particles that have interpenetrated exert a repulsive force on one another, and tend to fly apart. Region 3 in Figure 6-2 shows that when particles are forced together so that their centers are separated by less than one molecular diameter, repulsive forces, hence PE, rise very rapidly. This is the force that keeps you from falling through the floor.
The attractive forces are maximized and the PE minimized when the atoms, molecules, or ions just touch, at the bottom of the well. The distance corresponding to the potential energy minimum is the distance of closest approach. We have called this the aggregated state. In an ionic compound, such as KBr, this corresponds to a situation in which positive ions are surrounded by negative ions, and vice versa. In a covalent compound, such as H2O, the bottom of the well represents the situation in which the hydrogen atoms are covalently bonded to the oxygen atom, the bonds are separated by an angle of 104 o, and the distance between the H and O atoms corresponds to r at the bottom of the well. A collection of He atoms or methane molecules can maximize the contact between themselves by aggregating such that a particular molecule touches as many other molecules as possible. This organization is referred to as close packing, and characterizes the solid state of a pure substance. PE is minimized in the aggregated state. Figure 6-2 thus shows how potential energy changes as a substance is converted from the deaggregated to the aggregated state or vice versa. Notice that potential energy increases when an aggregated state is converted to a deaggregated one:
This is an idea that has broad applicability; we will encounter it many times.
The phases (solid, liquid, and gas) of a pure substance fit the aggregation/deaggregation model very well. In the solid phase, the fundamental units of a substance are packed together in an orderly arrangement that minimizes potential energy. This is the aggregated state, in which intermolecular attractive forces are optimized. In the gas phase, the fundamental units are widely separated and experience only very weak forces from one another. This is the deaggregated state. We might ask where the liquid state is on the diagram. From macroscopic observation, we know that the molar volume of a substance, e.g. water, in the liquid phase is very little different from the molar volume of its solid phase. These two aggregated (or condensed) phases have molar volume roughly 1000 times smaller than that of the gas phase. The liquid phase has distance of molecular separation and potential energy only somewhat greater than those of the solid, at the point labelled r2 in Figure 6-2.
The condensed (aggregated) phases of matter and the conversions between the gas, liquid, and solid phases will be discussed in succeeding chapters. Prior to that, it is important that you have an understanding of the types of forces that operate within and between molecules, and their relative strengths. This will enable you to understand, for example, why some substances exist as high-melting solids under normal conditions, while others exist as low melting solids, liquids, or gases.
6-3 Types of Molecular Forces . The nature and structure of a substance determine the type and strength of molecular forces that operate within it. We divide such forces into two broad categories. Intramolecular forces operate within the molecules or fundamental units of a substance. These are the covalent bonds within a molecule of a molecular substance, the forces between ions in an ionic compound, and the forces between atoms in a metal. Intermolecular forces operate between, rather than within, the molecules of a covalent substance; the atoms of a monatomic element; or the ions of one substance and the molecules of another. As a rule, intramolecular forces are much stronger than intermolecular forces. Both types of force, however, give rise to potential energy and a potential well. A listing of the important forces, arranged within these categories, follows:
Intramolecular Forces
In succeeding sections, we will discuss each of these molecular forces, and their relative magnitudes. We will begin with intramolecular forces.
6-4 Intramolecular Forces . The Ionic Bond. When two elements of widely different electronegativities combine, the result is an ionic compound. An example of an ionic compound is potassium bromide, KBr, in which the electronegativity difference between bromine, in group 17, and potassium, in group 1, is 2.14. Ionic bond formation involves transfer of an electron from the potassium atom, with its single valence electron and core charge, Zcore, of 1 unit, to the bromine atom, with its seven valence electrons and large Zcore of 7 units. The electron transfer generates a full valence shell for each species, as indicated here.
The ionic bond can be understood in terms of three energy terms, two of which we have encountered in Chapter 2. To remove an electron from the potassium atom to generate the potassium ion requires input of the first ionization energy of potassium. The process is shown in 6-4-1.
I1 for potassium is relatively small due to the minimum core charge found for group 1 elements. To add an electron to the bromine atom to form the bromide ion releases an amount of energy equal to the first electron affinity of bromine. This process is shown in 6-4-2.
The first electron affinity for bromine is substantial, because the large core charge of the bromine atom exerts a reasonable force on the extra electron. Despite this, to accomplish the electron transfer from potassium to bromine requires that 420 - 325 = 95 kJ/mole of energy be supplied. In other words, the electron transfer is substantially unfavorable! Thus we find that, contrary to a commonly held view of ionic bond formation, potassium does NOT "like" to transfer an electron to bromine; the process is unfavorable, and is not responsible for the strength of the ionic bond. What, then, is?
Generation of positive and negative ions results in a strong attraction between them, according to Coulomb's Law. A single positive and negative ion are shown coming together to form an ion pair in 6-4-3.
Formation of one mole of ion pairs releases 589 kJ of energy, much more than was required to generate the ions. We conclude that it is the Coulomb attraction between oppositely charged ions that drives the formation of the ionic bond; this happens despite the need to form ions, which is unfavorable.
In fact, ionic compounds do not consist of ion pairs of the type shown in 6-4-3. Rather, they consist of extended, highly organized aggregates in which positive and negative ions alternate. Figure 6-3 shows both the ion pair and the lattice of potassium bromide. The lattice is more stable than a collection of ion pairs because it surrounds each ion with ions of the opposite charge. This maximizes the attractive forces between positive and negative ions. In the ion pair, the positive charge of the potassium ion is offset on one side by the bromide ion, but is still exposed on all other sides. This exposed positive charge attracts bromide ions of other ion pairs, causing the ion pairs to condense to a lattice. Formation of a lattice containing 1 mole each of potassium and bromide ions as in 6-4-4 releases 672 kJ of energy. This number, about 50% larger than the amount released in formation of the equivalent amount of ion pairs, represents the depth of the potential well for potassium bromide.
The energy change that occurs when 1 mole of ionic substance in the solid lattice is formed from separated positive and negative ions in the gas phase is called the lattice energy, U, of the ionic compound. The lattice energy is given by an expression having the form of 6-4-5:
Here K is a collection of physical constants, Z+ is the cation charge, Z- is the anion charge, and r is the distance of separation of the cation and anion in the lattice. Because the charge of the anion, Z-, is negative, the lattice energy is a negative quantity, as expected. This means simply that the solid lattice is more stable than the separated gas phase ions. Note that this equation is essentially the expression for Coulomb potential energy. Without doing any calculations at all, we can draw some important conclusions based on 6-4-5.
Lattice energies for a number of common ionic compounds are given in Table 6-1.
| Charge Type | Substance | U, kJ/mole |
|---|---|---|
| 1+/1- | LiF | -1009 |
| NaCl | -774 | |
| KBr | -672 | |
| CsI | -584 | |
| 2+/2- | MgO | -2585 |
| BaO | -1978 |
Example 6-2. The lattice energy for NaCl is -774 kJ/mole. The radii of the sodium and chloride ions are 116 pm and 167 pm respectively. Estimate the lattice energy for CsI, which has ion radii of 181 and 206 pm.
Solution. NaCl and CsI are both 1+/1- ionic compounds. They should differ in lattice energy only through the r term of 6-4-5. Because the distance of separation is larger in CsI, the lattice energy should be less negative. We assume that r can be approximated by adding the radii of the ions. Thus for NaCl, r = 283 pm; for CsI, r = 387 pm. The lattice energies should be inversely related to the distance of separation of the ions:
This compares reasonably well with the value in Table 1.
Example 6-3. Arrange the following ionic compounds in order of increasing intramolecular forces: KI, LiCl, MgO, CaO, Al2O3.
Solution. KI and LiCl are 1+/1- ionic compounds. MgO and CaO are 2+/2- ionic compounds. Finally, Al2O3 is a 3+/2- compound. Based on ion charges and equation 6-4-5, we can arrange these in three categories:
The 1+/1- compounds have the least negative lattice energies. That for Al2O3 should be most negative, because the ion charges are largest. Within each category, the compound with the largest ions has the least negative lattice energy. This reasoning leads to the following order:
Here the greater than sign (>) is applied in the algebraic sense, in which numbers are considered to become smaller as they get more negative. In this convention, -50 is greater than -100.
There are very few compounds that are legitimately considered to be ionic; that is, in which electron transfer from metal to non-metal is complete. They are limited to compounds of the group 1 metals (Li, Na, K, Rb, Cs) with fluorine, chlorine, and oxygen. Compounds of the group 2 metals with oxygen, sulfur, and the halogens form typical ionic lattices, and are usually considered to be ionic compounds. However, there is considerable covalent character in the bonding.
Based on our discussion of the ionic bond in this section, it is understandable why ionic compounds are invariably solids at room temperature, with high melting and boiling points. The strong, non-directional Coulomb attractions acting between oppositely charged ions in the lattice can be overcome only at very high temperature.
We will have more to say about the specific arrangements of ions in lattices in Chapter 7.
The Covalent Bond. Two atoms of relatively large and similar electronegativities (i.e., non-metals) tend to form covalent compounds rather than ionic compounds. In covalent compounds, atoms share rather than transfer electrons to achieve full valence shells, because both atoms have similar substantial values of the core charge. Large Zcore tends to make electron transfer unfavorable. The shared electrons are paired, and are usually considered to be localized between the bonded atoms. In this section, we will discuss the chemist's view of the covalent bond. Figure 6-4a shows two separated chlorine atoms, each with 7 valence electrons distributed in the 3s and 3p valence orbitals according to the principles discussed in Chapter 2. Each Cl atom has a single odd electron in a 3p orbital. It is these electrons that will be paired and shared in the covalent bond. For the two atoms to bond by sharing electrons, it is necessary that they establish a mutual region in which the shared pair resides. They accomplish this by merging, or overlapping, the atomic orbitals containing the electrons to be paired, as shown in Figure 6-4b. Because these atomic orbitals have distinct spatial directions, the bonded atoms must maintain a definite orientation relative to one another in order to preserve the overlap of orbitals. Thus covalent bonds are directional.
The operation of Coulomb's Law in the covalent bond is perhaps less obvious than in the ionic bond. Figure 6-4c shows the two chlorine nuclei and the two electrons of the covalent bond, and indicates the various Coulomb forces that operate in this collection of centers. There are several interactions: attraction of nucleus 1 for electron 1; attraction of nucleus 2 for electron 1; attraction of nucleus 1 for electron 2; attraction of nucleus 2 for electron 2; repulsion of nuclei 1 and 2; and repulsion of electrons 1 and 2. Note, however, that because the bonding electrons are required to be between the nuclei so that they can occupy the overlapping atomic orbitals, the nucleus-electron attractive forces are maximized, and the nuclear-nuclear repulsions are minimized, because the electrons shield the two nuclei from each other. Thus the strength of the covalent bond is attributable to the simultaneous attraction of both electrons to both nuclei.
The lengths and strengths of covalent bonds--the position and depth of the covalent bond potential well--vary with the bonded atoms. In general, the strengths are substantial and comparable with those of ionic bonds. Thus to break all of the covalent bonds in a molecule of methane, CH4, requires the input of 1662 kJ of energy. Methane lies in a deep potential well relative to the separated atoms. Generally speaking, the following generalizations hold for covalent bond strengths.
| Bond | Length | Strength |
|---|---|---|
| Cl-Cl | 198 pm | 240 |
| Br-Br | 228 pm | 190 |
| I-I | 266 pm | 149 |
This seems reasonable, because there are more electrons localized between the nuclei and attracted by both.
Strengths and lengths of some typical covalent bonds are presented in Table 6-2. A more complete set of covalent bond strengths is found in Table 9-6.
| Bond | Bond Energy (Strength), kJ/mole | Bond Length, pm |
|---|---|---|
| Single Bonds | ||
| H-H | 432 | 74.2 |
| H-F | 565 | 91.8 |
| H-Cl | 428 | 127.4 |
| H-C | 411 | 109 |
| H-N | 386 | 101 |
| H-O | 459 | 96 |
| C-C | 346 | 154 |
| C-F | 485 | 135 |
| C-Cl | 327 | 177 |
| C-O | 358 | 143 |
| C-N | 305 | 147 |
| N-N | 167 | 145 |
| N-O | 201 | 140 |
| N-F | 283 | 136 |
| O-O | 142 | 148 |
| O-F | 190 | 142 |
| Multiple Bonds | ||
| C=C | 602 | 134 |
| C=O | 799 | 120 |
| C=N | 615 | |
| N=N | 418 | 125 |
| N=O | 607 | 121 |
| O=O | 494 | 120.7 |
| S=S | 425 | 188.7 |
| C(3)C | 835 | 120 |
| C(3)O | 1072 | 113 |
| C(3)N | 887 | 116 |
| N(3)N | 942 | 109.8 |
| P(3)P | 481 | 189 |
| As(3)As | 380 |
Example 6-4. Arrange the following single bonds in order of increasing length. Then arrange them in order of increasing strength.
Solution. Longer bonds are formed by larger atoms; atoms become larger as we descend the periodic table. The sizes of the atoms involved here are ordered as follows:
The bond lengths are then expected to be in this order: H-O < H-Cl < O-Cl < Cl-Cl. The strengths of bonds tend to decrease with increasing length. Thus: Cl-Cl < O-Cl < H-Cl < H-O.
Although simplified, this section gives us an appreciation for the origin and strengths of covalent bonds. We now turn to a problem with our view of the covalent bond having to do with molecular stereochemistry.
Our work in Chapter 3 culminated with a simple, powerful tool for the prediction and rationalization of molecular stereochemistry (shape): the VSEPR theory. Thus VSEPR predicts the tetrahedral geometry for methane, in exact agreement with experiment. The interpretation of the covalent bond in terms of overlap of atomic orbitals of the bonded atoms requires that we understand molecular shape in terms of the distribution in space of the atomic orbitals of the central atom, which are presumably overlapped with suitable terminal atom orbitals in covalent bonds. We pose this question: are the observed shapes of molecules consistent with the distribution in space of the atomic orbitals of the bonded atoms? If we ask this question of methane, the answer must be NO. On one hand, the C-H bonds are known to make 109.5 o angles with one another. On the other hand, the s and p valence orbitals of carbon have spatial orientations that are not consistent with this angle. The s orbital is spherical, hence non-directional; and the p orbitals make mutual angles of 90 o. In fact, the bond angles in most molecules in which 3, 4, or 5 electron groups surround the central atom are inconsistent with this 90 o orbital orientation. This presents a problem for our view of the covalent bond.
This problem was recognized in the 1920s, when the modern quantum theory was applied to the hydrogen atom and the spatial shapes of the atomic orbitals were discovered. Tetrahedral stereochemistry at carbon in methane had been proposed in 1874 by Vant Hoff on the basis of the number of isomers produced when hydrogen atoms were replaced by different atoms. At that time, though, people knew nothing of atomic orbitals. Thus the new quantum theory was faced almost immediately with a challenge: explain the shape of methane. This problem was tackled and resolved by Linus Pauling, who made during his life almost innumerable substantial contributions to science. Pauling proposed that the carbon atom uses hybrid orbitals to form four identical bonds to hydrogen atoms, directed to the vertices of a tetrahedron. His theory of hybrid atomic orbitals is still widely used today to rationalize (not predict) molecular stereochemistry. Applied to methane, the theory goes as follows.
In its most stable electron configuration, the carbon atom is shown in Figure 6-5a. The s orbital is fully occupied with an electron pair, and two of the three p orbitals are singly occupied. Based on this arrangement, we might expect carbon to form two bonds using the two odd electrons. However, this would fail to give it the octet. Thus we imagine that carbon moves, or "promotes", one of the 2s electrons to the empty p orbital, as in Figure 6-5b. This prepares it to form 4 bonds. However, bond formation obviously does not occur using the pure atomic orbitals, because these can not produce a tetrahedral stereochemistry when overlapped with 1s orbitals of hydrogen atoms. Pauling proposed that the carbon atom creates from its 4 valence orbitals four new, equivalent orbitals called "hybrids." The word hybrid means the same thing here as in agriculture: a blend or combination of two strains of a species. He called these sp3 hybrids to indicate that they are composed of a mixture of a single s and three p orbitals. Hybrid formation is shown schematically in Figure 6-5c. Indeed, he showed mathematically that it is possible to combine the four s and p atomic orbitals to produce four new orbitals, and that these new orbitals are directed to the corners of a tetrahedron. The four hybrids are shown in Figure 6-5d. Pauling thus "squared" fact with quantum theory by proposing hybrid orbitals. Let's apply these ideas to a few more situations.
The phosgene (COCl2) molecule is pictured in Figure 6-6a. It has been experimentally determined that the molecule is trigonal planar (as expected from VSEPR), with a double bond between carbon and oxygen. How is the trigonal planar stereochemistry, involving 120 o bond angles, to be rationalized with the 90 o spatial distribution of carbon 2p orbitals? Pauling showed that the 120 o bond angle results if one assumes that C hybridizes the s and two of the three p orbitals to produce sp2 hybrids. Mathematically, the mixing of an s and two p's does indeed produce three equivalent new orbitals, directed to the corners of an equilateral triangle. These are shown in Figure 6-6b. The plane in which the 3 hybrids lie depends upon which two p orbitals are used in the process. By convention, chemists normally choose to use px and py in the hybrids, which requires that the hybrids lie in the x-y plane of a Cartesian coordinate system. The pz orbital, not used in the hybrids, retains its usual shape and orientation along the z axis. Note, however, that it still contains an odd electron. Once the hybrids are formed, each containing a single electron, carbon forms covalent bonds to the two hydrogen atoms and the oxygen atom. Each bond is formed by overlap of one carbon hybrid with one atomic orbital of the bonded atom (H or O), as shown in Figure 6-6c. In forming these three bonds, orbitals are overlapped along the line joining the bonded nuclei. Bonds involving overlap of this type are called sigma bonds. At this point, bonding is not complete because carbon shares in only 7 electrons, as does oxygen. Each can complete the octet by forming one more bond, using the p orbitals oriented along the z axis. Because these do not point at one another, but instead are parallel, overlap must occur in sideways fashion as shown in Figure 6-6d. Bonds formed by sideways overlap similar to this are called pi bonds. At this point, bonding is complete. We say that in phosgene, carbon uses sp2 hybrids to sigma bond to H and O, and in addition forms a pi bond to O. The sp2 hybrids are consistent with the trigonal planar sterochemistry.
We are ready to make a few useful generalizations about hybrid orbital theory. Having done that, we will look at an example.
The flow of logic in determining hybrids used by the central atom is as follows. First, count the number of electron groups around the central atom; then apply VSEPR to determine the stereochemistry at that atom. The number of hybrids is the same as the number of electron groups counted for VSEPR. A number of atomic orbitals equal to the required number of hybrids is used in forming the hybrids. The sum of the exponents in the hybrid designation indicates the number of hybrids of that type, and the number of atomic orbitals of each type used to make them.
Example 6-5. For each molecule, determine the Lewis structure; use VSEPR to determine molecular shape; and state the hybrid orbitals used by the central atom in bonding to the terminal atoms: SiCl4; NH3; BF3; CO2.
Solution. Lewis structures, determined systematically by the methods of Chapter 3, are shown in Figure 6-7. Then
| Molecule | #groups (hybrids) | Stereochemistry | Hybrids |
|---|---|---|---|
| SiCl4 | 4 | tetrahedral | sp3 |
| NH3 | 4 | trigonal pyramidal | sp3 |
| BF3 | 3 | trigonal planar | sp2 |
| CO2 | 2 | linear | sp |
Note that the number of pi bonds in BF3 and CO2 is the same as the number of unhybridized p orbitals on the central atom.
It is relatively simple to extend the hybrid orbital concept to rationalize the stereochemistries of molecules in which the central atom has an expanded valence shell (more than 8 electrons). Essentially, this is done by incorporating in the hybrids as many valence d orbitals as required to give the necessary number of hybrid orbitals.
Example 6-6. Determine the hybrid orbitals used by the central atom in each molecule: PF5, ClF3, XeOF4.
Solution. Using the methods of Chapter 3,
| Molecule | # Electron Groups | Shape |
|---|---|---|
| PF5 | 5 | trigonal bipyramidal |
| ClF3 | 5 | T |
| XeOF4 | 6 | square pyramidal |
Based on the numbers of groups, we deduce the hybrids. Five groups require 5 hybrids, which in turn require 5 atomic orbitals: one s, three p, and one d. Thus sp3d. Six groups require 6 hybrids, which require the input of 6 atomic orbitals: one s, three p, and two d. Thus sp3d2. It can be mathematically demonstrated that 5 sp3d hybrids point to the vertices of a trigonal bipyramid; and that 6 sp3d2 hybrids point to the vertices of an octahedron.
Before we leave the covalent bond and the hybrid orbital idea, something must be made clear. What is experimentally known is that carbon and hydrogen react to form methane, and that the methane molecule involves four carbon-hydrogen bonds that are perfectly tetrahedrally distributed in space. We introduced the ideas of electron promotion and orbital hybridization above as a way of rationalizing for ourselves, in terms of the output of quantum theory, what is experimentally observed. Thus promotion and hybridization are mental constructs. It is important that you realize that the carbon atom does not actually do these things. It is simply convenient for us to think of the carbon atom as doing these things.
There are numerous elements and compounds that consist of huge numbers of atoms covalently bonded together to give a large, 2- or 3-dimensional network. These substances are therefore similar to ionic compounds in that they do not consist of small, discrete molecules; however, they do not consist of discrete ions either. Examples of such substances are silicon dioxide (SiO2), arsenic trioxide (As2O3), carbon in the diamond form, and phosphorus in the red form. The large numbers of covalent bonds in these structures leads to potential well depths comparable to those found in ionic compounds. Most such materials exist as solids at ordinary temperature, and melt and boil only at very elevated temperature.
Ionic and Covalent Bonding--A Continuum. The bond between two identical non-metal atoms, as in Cl2, is purely covalent; that is, each atom has an exactly equal share in the bonding electron pair. Pure covalent bonds exist in the elemental forms of many of the non-metals (for example, all of the diatomic halogens of group 7; O2 and S8 of group 6; N2 and P4 of group 5; C and Si of group 4; and others). Bonds between different non-metal atoms are covalent, in that they involve electron sharing, but polar, meaning that they have some ionic character. The bonds between chlorine and the other elements of period 3 are shown below. The transition from pure covalent to polar covalent to ionic as we move from right to left in the period is clear.
Although the Na-Cl and Mg-Cl bonds are usually considered to be ionic, they are not purely ionic; electron transfer from the sodium or magnesium atom to the chlorine atom is not complete. There is no such thing as a pure ionic bond. The process of electron transfer from a metal atom to a non-metal atom generates a positive ion and a negative ion. The positive ion is electron deficient, and attracts an electron pair of the negative ion strongly. Thus the very act of electron transfer creates a disparity that must then be offset by partial electron sharing. If the anion is one of the very electronegative non-metals (fluorine or oxygen), it can resist the pull of the cation effectively, and hold on to the electron pair. Anions other than F- and O2-, however, cannot completely resist the attraction of the cation for their lone pairs. These atoms share a lone pair with the cation to form a bond that, although it is very polar, is nonetheless partially covalent. As a general rule, the higher the charges on the cation and anion, the more covalency in the bond. A highly charged cation (3+ or 4+) is small and compact, and exerts a powerful attractive force on electrons. A highly charged anion (2-, 3-) holds its extra electrons weakly (recall the electron affinity from Chapter 2) and readily shares them to some extent with the cation. The following series of compounds illustrates nicely the evolution from largely ionic to covalent bonding as cation charge increases:
| Compound | Physical State, 298 K | Melting Point oC | |
|---|---|---|---|
| NaCl | solid | 801 | |
| MgCl2 | solid | 714 | |
| AlCl3 | solid | 190 | |
| TiCl4 | liquid | -25 |
The decrease in melting point indicates increasing covalency in the bonding. Titanium tetrachloride, TiCl4, is a liquid at room temperature, indicating that it is molecular, not ionic.
The Metallic Bond. Elements of similar but relatively low electronegativities (i.e., metals) form metallic bonds. It is these bonds that are operative in a typical metal and that are responsible for metallic properties: reflectivity, conductivity, malleability, and strength. Metallic bonds form between atoms that have fewer valence electrons than they have valence orbitals. We shall try to develop a picture of the metallic bond by considering sodium metal, which exists as a very low melting, highly reactive solid at room temperature. Figure 6-8a shows the Lewis dot symbols for two sodium atoms. The single odd electron is in the 3s valence atomic orbital of the sodium atom. At first glance, it might be expected that two sodium atoms, each with a single unpaired valence electron, would form a single covalent bond by overlapping the 3s valence orbitals and pairing the odd electrons in the resulting common region of space. This hypothetical bond is shown in Figure 6-8b. There are problems with this bonding situation however. First, bonding of this type would give each sodium atom only a duet of electrons, nowhere near the required octet. Each atom would have three empty 3p orbitals protruding from it into space. Second, pairing the 3s valence electrons localizes them in the region between the two nuclei. They no longer have probability of being found in the peripheral regions opposite the bond (these regions are indicated in Figure 6-8b). Because these are the only valence electrons, the regions marked with + symbols in the figure are exposed to the full core charge (+1) of the sodium atom. (The + signs in the figure do not represent actual positive charges; instead, they are intended to indicate that another atom located in this region would experience the core charge of the sodium atom near it.) Here is the key point: another sodium atom in the region of the bonded pair would be drawn, via its valence electron, to the exposed core charge of the 2-atom cluster. It would contribute its valence electron to a bond with one or the other of the pair. The sodium atoms are therefore driven to aggregate, using their relatively few valence electrons to bond together as many nuclei as possible via the Coulomb attraction. The aggregation is very regular in nature, resulting in an ordered arrangement of metal atoms called a lattice. As a result of aggregation, each sodium atom uses all of its valence orbitals in forming partial bonds with a number of neighbors.
Contrast this with the covalent bonding between two chlorine atoms, shown in Figure 6-4c. The covalent bond is again localized, but the remaining valence electrons are distributed evenly around each chlorine atom, effectively shielding the core charge from chlorine atoms in other molecules. Further, all of the s and p valence orbitals are full. In order to aggregate as sodium does, chlorine would have to use either the 3d orbitals or the orbitals from the n = 4 main shell. But the core charge experienced by an electron in either of these orbital sets is essentially zero (17-17 = 0). Consequently, chlorine is limited in its aggregation to the formation of covalently bonded diatomic molecules. As we will discuss shortly, there is only a very weak tendency for the diatomic molecules to aggregate via intermolecular forces. Interestingly, although hydrogen, like sodium, has only a single valence electron, it is not metallic. Instead, hydrogen atoms form covalently bonded diatomic molecules. Why does hydrogen behave so differently from sodium and the other metals? Formation of a localized covalent bond between two hydrogen atoms results in exposure of the peripheral regions of the molecule to the core charge of the hydrogen atoms. In this case, because there are no core electrons, the H atom core charge is the nuclear charge of 1+. In the case of sodium and other metals, this situation leads to aggregation using the empty valence p orbitals. However, hydrogen has no valence p orbitals. The single orbital in the n=1 shell is full; no further aggregation is possible. The formation of metallic bonds can occur only if empty valence orbitals are available. Experiments have shown that if hydrogen is subjected to extremely high pressure, which forces the diatomic molecules to aggregate, it exhibits metallic characteristics.
In sodium, as in most metals, there are too few electrons to form normal two-electron bonds between all pairs of bonded atoms. To achieve stability, the valence electrons are spread over many nuclei, simultaneously bonding them all together. This spreading of electrons is called delocalization, and in some ways is similar to the delocalization that occurs in resonance (Chapter 3). In metals, however, delocalization is much more extensive, and it is best to think of the electrons as waves that extend over the entire dimensions of the metallic crystal. Figure 6-8c shows a schematic structure of metallic bonding. The nuclei are represented as points, and the electrons as waves. This picture enables us to understand the properties of metals. That the electrons are delocalized waves is consistent with the conductivity of metals, and with their flexibility and strength. Putting stress on metals moves the nuclei, but the waves adjust, maintaining bonding. Because the electrons are waves, they are everywhere in the crystal at once. They are mobile, rationalizing the electrical and thermal conductivity of metals. Finally, the pooling of valence orbitals by aggregated sodium atoms leads to a large number of closely spaced orbitals in the metallic crystal. Thus the electrons absorb and reemit photons of all energies in the visible region of the spectrum, giving metals their highly reflective appearance.
There are so few bonding electrons in metallic sodium that only about 1/8 of an electron pair bond holds each pair of adjacent sodium atoms together. It is little surprise, then, that sodium is very soft (weak bonding), very reactive (deficiency of bonding electrons), with low density, and low melting and boiling points.
Example 6-7. In magnesium, how many metallic bonds form per Mg atom?
Solution. The electron configuration of the magnesium atom is 1s2 2s2 2p6 3s2. We assume that metallic bonds are formed using the valence electrons only. Because each Mg atom contributes 2 valence electrons to the metallic bonding, and because 2 electrons constitute one bond in Lewis terms, there is one bond per Mg atom in the metallic lattice.
6-5 Intermolecular Forces . Having described the forces that operate within molecules (including ionic crystals and metals), we are ready to discuss the interactions that occur between species (atoms, molecules, and ions) that are not covalently bonded to each other. These are called intermolecular forces; they can originate in a number of ways.
Dipole-Dipole Forces. Intermolecular forces that operate between neutral molecules having molecular dipole moments are called dipole-dipole forces. We saw in Chapter 3 how the structure of a molecule may cause its centers of positive and negative charge to be displaced from one another, giving the molecule a dipole moment. The dipole moments of two neighboring molecules tend to align with the + end of one dipole near the - end of the other, so that forces of attraction between them are maximized. The alignment of two dipoles is shown in Figure 6-9.
The maximum force of attraction between two dipoles, m1 and m2, separated by a distance r is given by equation 6-5-1.
Equation 6-5-1 is Coulomb's Law, in disguised form. It is necessary to raise the distance of separation to the 4th power in the denominator to cancel the distance units that appear in both dipole moments in the numerator. Physically, the 4th power dependence means that the attraction between dipoles falls off much more rapidly with distance than would the attraction between isolated charges. This is because the + and - ends of dipole 2 are almost equidistant from dipole 1 when the dipoles are separated by a distance that is much greater than the dipole length. The force of attraction between the + end of dipole 1 and the - end of dipole 2 is almost cancelled by the force of repulsion between the two + ends. Dipole-dipole forces operate between molecules of water, and between molecules of the substances pictured in Figure 6-10.
Such forces are obviously much weaker than those operating in ionic or covalent network solids, and give rise to potential wells having depths in the approximate range 5-20 kJ/mole. Many molecular substances with dipolar molecules exist as liquids at ambient temperature, and have relatively low boiling points. In particular, many organic (carbon-containing) compounds are of this type.
Example 6-8. Which substances experience dipole-dipole intermolecular forces?
Solution. Using the methods of Chapter 3, we can draw the following conclusions about the molecules of these substances.
| SiF4 | tetrahedral | Si-F bonds are polar, but no molecular dipole; bond dipoles cancel |
| CO2 | linear | C-O bonds are polar, but no molecular dipole; bond dipoles cancel |
| SO2 | bent | S-O bonds are polar and do not cancel. Sulfur lone pair dipole only partially offsets net bond dipole. |
| CHCl3 | tetrahedral | C-H and C-Cl bonds are polar and do not cancel |
SO2 and CHCl3 experience dipole-dipole intermolecular forces.
Hydrogen Bonding. Hydrogen bonding (H-bonding) is a special kind of dipole-dipole force that occurs when a hydrogen atom is bonded to one of the very electronegative atoms, F, O, or N. The H-F, H-O, and H-N bonds are very polar, because the electronegative atom draws the bonding electron pair strongly to itself. This leaves the hydrogen nucleus exposed, as indicated in Figure 6-11a. The resulting bond dipole is substantial. The positive end of this dipole, consisting essentially of an exposed hydrogen nucleus, has a high charge density because the proton is so small. Consequently, it strongly attracts the negative end of a similar bond dipole in another molecule to give a structure like that pictured in Figure 6-11b. The hydrogen atom is sandwiched between the two electronegative atoms in a linear arrangement. Hydrogen bond forces cause potential wells of depth in the range 5-50 kJ/mole. Note that these are on average deeper than those resulting from ordinary dipole-dipole forces. The intermolecular forces between water molecules are of the hydrogen bonding type. They are responsible for the abnormally high boiling point of water, which, with its small molecular weight, would otherwise be a gas at room temperature. Compare the boiling point of water (100 oC) with that of methane (-162 oC), which has a similar molecular weight, but lacks hydrogen bonding and ordinary dipole-dipole forces. Hydrogen bonding is responsible for the formation of genes in the DNA molecule; for the helical structure of proteins; and for the incredible strength of Kevlar, a DuPont polymer used for canoe hulls and bullet-proof vests.
Example 6-9. In which substances do hydrogen bonding forces operate between molecules?
Solution. Hydrogen is bonded to one of the very electronegative atoms in CH3CH2OH and HNO3. Hydrogen bonding should occur in both of these substances.
Instantaneous Dipole-Induced Dipole (Dispersion) Forces. It may seem that there would be no forces of attraction between atoms of helium, because they are uncharged and have no permanent dipole moment. Yet He can be liquified at the very low temperature of 4 K. That this is possible indicates that intermolecular forces do exist, though they are very weak. What is their origin? We portray the He atom as if the electron distribution about the nucleus is perfectly spherical. However, the electrons are constantly in motion and it is possible at any given instant for the electron cloud to be skewed slightly to one side of the nucleus. This generates an instantaneous dipole moment in the He atom. During its transitory existence, this dipole induces a dipole in a neighboring atom and the two instantaneous dipoles attract. This is pictured in Figure 6-12. Forces between transitory dipoles are called instantaneous dipole-induced dipole forces, or alternately, London dispersion forces after the scientist who first proposed them. We will refer to them as dispersion forces. It is these forces that are responsible for the liquefaction and/or solidification of a number of substances whose molecules do not possess permanent dipole moments. For example, dry ice is solid CO2. The CO2 molecule is linear and, though it has non-zero bond dipoles, has no overall dipole moment. Similarly, CCl4, with perfectly tetrahedral molecules and no net dipole moment, is a liquid at room temperature and up to 78 oC. And iodine, which consist of I2 molecules, is a solid at room temperature.
Dispersion forces depend on two features of molecular structure. First, they increase in magnitude with the size and distortability (usually called the polarizability) of the electron clouds of the interacting particles. Size and polarizability increase as molecular weight increases. It follows that dispersion forces increase as molar mass increases. For substances of large atomic or molecular mass, dispersion forces are strong enough that the substances are solid or liquid at room temperature. We saw this above for CCl4 and I2. A striking example is provided by the diatomic halogens of group 17, which progress from F2, a gas at room temperature and 1 atm pressure, to Cl2, also a gas, to Br2, a liquid, and finally to solid I2. Only dispersion forces are operative in this series of substances. Second, dispersion forces depend upon molecular shape via the surface area over which two molecules can be in contact. The larger the surface area of contact, the stronger the dispersion forces. Molecules that are roughly spherical in shape are able to contact each other only minimally, as shown in Figure 6-13a. In contrast, molecules that are planar or linear in shape can maintain a large surface area of contact, with correspondingly larger dispersion forces. The classic example of the effect of molecular symmetry (shape) on the magnitude of dispersion forces is provided by the series of isomeric pentanes (C5H12) shown in Figure 6-13b. Neopentane is highly symmetrical and nearly spherical in shape. Intermolecular forces are relatively small and the boiling point is correspondingly low. Pentane is linear, so two pentane molecules are able to contact each other along the entirety of their length. Intermolecular forces are relatively large, and the boiling point is correspondingly high. Isopentane is intermediate structurally, and the least symmetrical of the three isomers. Intermolecular forces are less than in pentane, but greater than in neopentane. The boiling point is intermediate. Graphite, the most stable form of elemental carbon at ordinary temperature and pressure, has the structure shown in Figure 6-13c. Each carbon atom is bonded to three others in trigonal planar fashion, using sp2 hybrid orbitals. The unhybridized p orbitals are used to form a network of pi bonds that extends over the entire array of bonded carbons, producing a large planar sheet of carbon atoms. These sheets are then stacked one atop the other to produce the three dimensional structure of graphite. Interestingly, there are no covalent bonds holding one sheet to the next. The sheets attract one another via dispersion forces, which are very strong because of the large surface area of contact.
It is important to realize that dispersion forces operate between all molecules, whether or not other forces also operate. Molecules of chloroform, CHCl3, are atttracted by a combination of dipole-dipole and dispersion forces. The magnitude of dispersion forces varies with molar mass; however, they are generally weaker than dipole-dipole forces, giving potential wells in the range 0.1 to 5 kJ/mole.
Example 6-10. Arrange the following non-polar molecules in order of increasing melting point.
Solution. None of these molecules is expected to have a dipole moment. Only dispersion forces will operate, and these increase with increasing molecular mass. The molar masses of these substances follow:
| Substance | MM |
|---|---|
| SiF4 | 104.077 |
| CS2 | 76.131 |
| CI4 | 519.631 |
| GeCl4 | 214.402 |
The intermolecular forces, and the melting points, should increase in the following order:
The experimentally determined melting points are -110.8, -90, -49.5, and 171 oC, respectively.
Example 6-11. Molar masses, dipole moments, and boiling points for several covalently-bonded substances are given below. Rationalize the observed boiling points in terms of intermolecular forces.
| Substance | MM | m | Boiling Point, K |
|---|---|---|---|
| He | 4.003 | 0 | 4 |
| H2 | 2.016 | 0 | 20 |
| N2 | 28.014 | 0 | 77 |
| CO | 28.010 | 0.33 | 83 |
| HCl | 36.461 | 3.60 | 188 |
| HBr | 80.912 | 2.67 | 206 |
| SO2 | 64.058 | 5.42 | 263 |
| HF | 20.006 | 6.37 | 292 |
| CCl4 | 153.823 | 0 | 350 |
| H2O | 18.015 | 6.17 | 373 |
Solution. The substances fall into two categories, those with dipole moments and those without. Only dispersion forces operate in substances without dipole moments. Selecting from the table those substances with no dipole moment, shown below, we see that, with the exception of hydrogen, boiling point increases with increasing MM, as expected from our discussion of dispersion forces above. The dipole moment of CO is so small that it seems reasonable to include it in this list. Its boiling point is only slightly larger than that of N2, due to weak dipole-dipole forces in addition to the dispersion forces.
| Substance | MM | m | Boiling Point, K |
|---|---|---|---|
| He | 4.003 | 0 | 4 |
| H2 | 2.016 | 0 | 20 |
| N2 | 28.014 | 0 | 77 |
| CCl4 | 153.823 | 0 | 350 |
Substances having substantial dipole moments are listed below, now in order of increasing dipole moment.
| Substance | MM | m | Boiling Point, K |
|---|---|---|---|
| HBr | 80.912 | 2.67 | 206 |
| HCl | 36.461 | 3.60 | 188 |
| SO2 | 64.058 | 5.42 | 263 |
| H2O | 18.015 | 6.17 | 373 |
| HF | 20.006 | 6.37 | 292 |
We see only a rough correlation of boiling point with dipole moment: HBr and HCl have boiling points on the low end of the scale; SO2 is intermediate; and H2O and HF have high-end boiling points. However, the boiling points of HBr and HCl are inverted based on the size of the dipole moment. HCl, with the larger dipole moment, has the lowest boiling point in the group of five. Here is clear evidence that dispersion forces operate in addition to dipole-dipole forces. Dispersion forces are substantially larger for HBr (MM 80.9) than for HCl (MM 36.5). This is sufficient to raise the boiling point for HBr above that for HCl.
The remaining 3 substances do not yield to the same argument, however. The MM of SO2 is more than three times larger than those of HF and H2O, and its dipole moment is only somewhat less than that for water. Yet the boiling points of HF and particularly of H2O are strikingly higher than that for SO2. Here we have evidence of the potency of hydrogen bonding intermolecular forces, which operate in both HF and H2O. Despite its small MM, HF boils 30 o higher than SO2 because hydrogen bonding causes the potential well of the liquid to be relatively deep. The boiling point for water is about 80 o higher than that for HF because each water molecule can engage in up to four hydrogen bonds (one for each H atom, and one for each lone pair on the O atom).
Mixed Forces in Solutions. All of the forces that can operate at the molecular level within a single pure substance are discussed above. It is possible to have what we might call hydrid forces that operate when pure substances are mixed to form solutions. When an ionic solid such as sodium chloride dissolves in water, ion-dipole forces are responsible. The maximum magnitude of ion-dipole forces is given by a form of Coulomb's Law that is intermediate between equations 6-1-4 and 6-5-1.
Here q is the charge on the ion, m is the dipole moment of the solvent, and r is the distance between the ion and the center of the solvent dipole.
Example 6-12. Consider the following three situations:
Compare the magnitudes of the attractive forces operating in these situations. This will provide an idea of the relative magnitudes of ion-ion, ion-dipole, and dipole-dipole forces. Then, compare the forces when the distance of separation of the interacting entities is increased by a factor of 4 to 8.0 x 10-10 m.
Solution. We apply equations 6-1-4, 6-5-1, and 6-5-2:
The ion-ion force is stronger by a factor of 4 than the dipole-dipole force. The ion-dipole force is intermediate.
When the separation distance is increased by a factor of 4, the magnitudes of all forces decrease because they depend inversely on a power of the distance. The ion-ion force is reduced by a factor of 42; the ion-dipole force by a factor of 43; and the dipole-dipole force by a factor of 44:
The forces involving dipoles clearly decrease much more rapidly with increasing distance of separation than do ion-ion forces.
Whether or not two liquids are mutually soluble (miscible) depends on whether the mixed intermolecular forces can compete with the forces operating within each pure liquid alone. The intermolecular forces in pure water are primarily hydrogen bonding (dipole-dipole) forces, superimposed on weak dispersion forces. Those in pure acetone are dipole-dipole forces superimposed on dispersion forces that are stronger than the dispersion forces in water. When acetone and water are mixed, they readily dissolve in one another in all proportions, apparently because the dipole-dipole forces between water and acetone molecules are of strengths comparable to those operating in the two separate liquids. On the other hand, only dispersion forces operate within pure carbon tetrachloride, because the CCl4 molecule has a molecular dipole moment of zero. Because the molecules are spherical, giving small surface area of contact, these forces are fairly weak. Consequently, carbon tetrachloride and water are not mutually soluble. They are said to be immiscible. When CCl4 and water are mixed, the two liquids segregate into layers, with the denser liquid, chloroform, on the bottom. The liquid-liquid interface (boundary) is clearly visible. Finally, a non-polar solid like I2 dissolves readily in a non-polar liquid like CCl4, because the mixed dispersion forces are comparable in magnitude to the dispersion forces in pure CCl4 and pure I2. However, the solubility of I2 in water is essentially zero, because mixed forces cannot compete with the strong dipole-dipole forces in water. Water molecules are unwilling to separate in order to make room for the non-polar I2 molecules. In this way, we can extend our intermolecular force ideas to provide an understanding of the formation of mixtures.
Summary of Mixed-Force Concepts:
Example 6-13. For each pair, predict whether substance A will dissolve in liquid B, and explain your reasoning.
| Substance A | Liquid B |
|---|---|
| C2H5OH | CH3OH |
| Br2(l) | C5H12 |
| Br2(l) | H2O |
| C2H5OH | H2O |
| C2H5OH | CHCl3 |
| NaCl | CHCl3 |
| NaCl | CH3CH2OH |
| H2O | CH2Cl2 |
Solution.
C2H5OH (ethanol) and CH3OH (methanol) are both liquids due to fairly strong H-bonding intermolecular forces. Ethanol is expected to dissolve in methanol.
Br2(l) and C5H12 (pentane) are both non-polar liquids with dispersion forces operating between molecules. Thus they are "like" and we expect bromine to dissolve in pentane.
Br2 and H2O are "unlike." Bromine will not dissolve readily in water.
Both liquids exert H-bonding forces at the molecular level. Ethanol is expected to dissolve in water.
This is not obvious. At first glance, we would call ethanol and chloroform "unlike" because H-bonding forces operate in ethanol, whereas dispersion forces operate in chloroform. However, in addition to its polar -OH group, ethanol has a non-polar hydrocarbon portion that is capable of exerting dispersion forces on molecules of chloroform. These two liquids are therefore miscible.
Ionic NaCl will certainly not dissolve in non-polar chloroform.
Because ethanol is polar at the -OH end, we might expect NaCl to dissolve to some extent in ethanol, but not to the same extent as in water. This is indeed found.
Polar water will not mix with non-polar dichloromethane.
Example 6-14. Describe what the system will look like during and following each process:
Solution.
Distribution of a solute Between Two Immiscible Liquids in Contact. When two immiscible liquids, such as water and chloroform (CHCl3), are placed in contact, they will mutually exclude each other, forming two liquid layers. The less dense liquid will float on the denser one, and a boundary surface will be clearly visible. It is interesting to think about what will happen if a third substance, solid or liquid, is added to the system consisting of two immiscible liquids, A and B. If the third substance, C, is soluble in one of the liquids (say A) and insoluble in the other (B), it will dissolve exclusively in the A layer, as in example 6-14. More interesting is the situation in which the C is soluble in both liquids to at least some extent. In this case, we find that the substance C partitions, or distributes, itself between the two liquid layers. We further find that the concentration of the third substance is higher in the liquid with which its intermolecular forces are most compatible. The ratio of the concentration of C in liquid A to that in liquid B is found to attain a constant value Kd, defined as in (6-5-3):
This constant ratio is established no matter how the 3-component mixture is prepared. Three such ways follow:
In addition, the ratio is reestablished if the mixture is altered in any of the following ways:
We can understand this behavior in terms of a simple diagram showing layers of A and B in contact at a boundary, and molecules of C initially dissolved in liquid A. The boundary between the two liquids provides a "doorway" from one to the other through which C molecules, which have reasonable interactions with both A and B molecules, can pass. Thus molecules of C will tend to move across the boundary from liquid A, where they are initially dissolved, into liquid B, in which there are initially no molecules of C. In other words, they begin to distribute themselves between the two liquids. After some time, enough C molecules will have passed into liquid B that the rates of passage of C from A to B and back will be the same. Once this situation is attained, there will be no further change in the concentration of C in the two layers. In other words, the constant ratio in (6-5-3) will be attained. From the magnitude of Kd, we may easily deduce which liquid C prefers. If Kd < 1, C prefers B; if > 1, C prefers A.
Distribution of a substance between two liquid phases is an example of dynamic equilibrium, a situation in which two opposing processes (here, movement of C both ways across the boundary) occur at equal rates. We will have much more to say about dynamic equilibrium later.
A Nod to Other Classification Systems. Classification of the various types of forces as "inter-" or "intramolecular" is not absolute. Thus you may see the forces classified as follows:
In this classification scheme, ionic and metallic bonds are categorized separately rather than as intermolecular forces, because there are no discreet molecules in ionic compounds or metals.
Another classification system has recently become popular and is experiencing increasing use. This goes as follows.
Here there are only 2 major categories: covalent, and not covalent! It is important to realize that the distinction between covalent bonding and some of the non-covalent interactions is blurred. For example, many people believe that metal-ligand interactions and even hydrogen bonding are weak covalent interactions.
The bottom line is that it is not particularly important how the classification is done, as long as the various types of interactions are understood. Pick your own favorite classification system, and use it!
6-6 The Interplay Between Molecular Kinetic and Potential Energies. The concept of temperature is of profound significance in science. We take it for granted in our daily lives, without thinking much about what it is. We discovered in Chapter 5 that temperature is the manifestation of molecular kinetic energy. The steam rising from a pan of boiling water feels hot because the molecules of water vapor are moving rapidly. Their collisions with the skin transfer kinetic energy, which we interpret as heat. The relationship between temperature and the average translational kinetic energy of an atomic, ionic, or molecular unit is very simple. It is given in equation 6-1-2.
T is the absolute (Kelvin) temperature, and k is a fundamental constant of nature called Boltzmann's constant. Its value is 1.381 * 10-23 J/K. By KEmolecule, we signify the kinetic energy of any microscopic unit of matter. This can be anything from an electron to a DNA molecule. Keep in mind in using equation 6-1-2 that the kinetic energy referred to is the translational kinetic energy--the energy of motion of the particle as a whole through space. Translational kinetic energy is often called thermal energy.
Example 6-15. Calculate the thermal energy of one mole of nitrogen gas at room temperature, 293 K.
Solution. Calculate the kinetic energy of an average molecule; then multiply by Avogadro's Number.
As we saw in Chapter 5, the product of Boltzmann's constant and No is a constant, R, which occurs in the ideal gas law. Its value is 8.314 J/K-mole. Expressed in terms of R, the kinetic energy per mole of gas is 3RT/2.
We have seen that molecular forces promote aggregation by lowering molecular potential energy. Kinetic energy counteracts the effects of intermolecular forces. These ideas lead to a very useful generalization:
The stronger the molecular forces, the more rapidly atoms, molecules, or ions must move in order to overcome them. Even the strongest forces, however, may be overcome by the vigor of molecular motion if the temperature is high enough. At a temperature that gives a kinetic energy comparable in magnitude to the depth of the potential well, molecules begin to escape the potential well, and the substance undergoes a phase change (that is, sublimes, melts, or boils). A phase change may therefore by viewed as an exchange of KE for PE. If the potential well is shallow, as in liquid helium (where the only forces are the very weak dispersion type) the required kinetic energy is achieved at very low temperature (4 K). If the potential well is deep, as in the ionic compound MgO (where the forces are the strong ion-ion type), very high temperature is required for particles to escape (3125 K). Table 6-4 shows the parallel between the potential well depth and the temperature of melting for a number of substances.
| Substance | Potential Well Depth, J/mole | Melting T, K |
|---|---|---|
| He | 0.105 | 3.5 |
| H2 | 1.04 | 14.0 |
| Xe | 14.7 | 161 |
| N2 | 6.30 | 63.2 |
| Cl2 | 26.8 | 172.1 |
| I2 | 56.3 | 387 |
| H2O | 46.7 | 273 |
| CO2 | 33.6 | 217 |
| NaCl | 771 | 1074 |
Try plotting melting point against potential well depth to see how good the correlation is.
6-7 Non-Ideality of Gases . The postulates of KMT define an ideal gas. An ideal gas obeys the ideal gas law under all conditions of T and P. There is in fact no real gas that satisfies this definition. The ideal gas is a concept that we can think about but which does not actually exist.
Consider a sample of N2 gas at 25 oC and 1 atm pressure. Imagine measuring the volume of the gas periodically as we cool it at constant pressure. If N2 were ideal, it would follow Charles' Law exactly even to absolute zero. A plot of V versus T for ideal N2 would follow curve A in Figure 6-14. The actual V-T plot for N2 follows curve A to a temperature of about 100 K. At this point the plot starts to curve down (curve B). At 77 K the volume sharply decreases while the temperature remains constant, and liquid appears in the container. When all gas has converted to liquid, further decrease in T has little effect on the volume occupied by the liquid until T = 63 K. At this point, N2 solidifies with a slight decrease in volume. Further cooling causes negligible change in V.
Because all real substances liquify at some T, no real substance can obey the ideal gas law under all P,T conditions. Deviation of real gases from ideal behavior is shown conveniently in terms of the compressibility factor, Z, in equation 6-7-1.
Z is plotted against P in Figure 6-15. For an ideal gas, Z = 1 at all pressures (the horizontal dashed line). Two deviations from ideality are commonly observed. In one type, Z always exceeds 1. H2 is an example. In the second type, Z < 1 at moderate pressure but becomes > 1 at high pressure. N2 is an example. Most gases show behavior similar to that of N2. Real gases deviate from ideality because they violate two postulates of the KMT.
First, real molecules exert attractive forces on one another, as we have seen earlier in this chapter. The forces cause the molecules to temporarily stick together in pairs when they collide, reducing the effective number of free particles in the gas. This is shown in Figure 6-16. Although there are 7 gas molecules in the container, 4 of them are temporarily stuck together in pairs, reducing the effective number of independent gas particles to 5. Sticking together reduces the pressure below the ideal value because there are fewer collisions with the container walls per unit time. Thus
Here Pact is the actual pressure of the gas, Pideal is the pressure it would exert in the absence of intermolecular forces, and D P is the pressure reduction. If there are n moles of gas in a container of volume V, the pressure reduction, D P, is proportional to (n/V)2. The following argument demonstrates this. First, we assume that D P is proportional to the number of pairs of stuck-together molecules:
How many pairs are possible for N molecules? If N = 2, there is only 1 pair. If N = 3, 3 pairs are possible. For N molecules, there are N(N-1)/2 possible pairs, where division by 2 avoids double counting. If N is very large, as in a macroscopic gas sample, N-1 = N and the number of pairs is N2/2. Dividing by volume to convert to concentration and by No to convert molecules to moles gives
where "a" is a proportionality constant. Substitution in equation 6-7-2 gives
Second, real molecules are not points, but instead occupy finite volume. Thus a small fraction of available container space is occupied by the molecules themselves, and is not available to accommodate motion of the molecules. The total container volume is the sum of the volume of the molecules and the free volume available for motion:
The volume of the molecules is constant for a given gas, and depends on molecular size. As Vcontainer is made smaller, Vavailable becomes a smaller fraction of Vcontainer. Since the molecules have smaller free volume than they would if ideal (when Vmolecules = 0), they exert a larger than ideal pressure. Equating Vcontainer with Vactual and Vavailable with Videal, we obtain
The volume of the molecules is proportional to the number of molecules, which is in turn proportional to the moles of gas:
where b is a proportionality constant roughly equal to the volume per mole of substance. (The volume of the molecules can be determined by crowding the molecules together--i.e., by liquifying the substance. The molar volume of the liquid phase gives an estimate of "b". The molar volume of liquid is the product of the MW and the reciprocal of density:
Using this approach, b should be 18/1.0 = 18 mL/mole for water.) Substituting equation 6-7-6 into 6-7-5 and rearranging for Videal gives
Finally we substitute equations 6-7-7 and 6-7-3 into the ideal gas equation, PidealVideal = nRT to obtain equation 6-7-8:
This equation was first proposed by van der Waals in 1873 to explain deviations of gases from ideal behavior. Its success earned him the Nobel Prize for Physics in 1910.
In real gases the effects of non-zero molecular volume and intermolecular forces oppose. The former causes pressure to be higher than ideal, and the latter causes it to be lower than ideal. For N2 and many other gases having Z < 1 at moderate pressure and Z > 1 at high pressure, first one effect, then the other, takes precedence. At low pressure (10-100 atm), the intermolecular forces dominate because the molecules are not yet compressed enough for the volume effect to be felt. This makes Pact < Pideal, and Z < 1. As P becomes very large (> 100 atm), non-zero molecular volume becomes important, making Pact > Pideal and Z > 1. For hydrogen, intermolecular forces are small and the volume effect dominates at all pressures. At low pressure (approximately 1 atm) and temperature much higher than the boiling point of the substance, both non-ideal effects are small and real gases approach ideal behavior.
Values of the van der Waals constants a and b for several gases are in Table 6-5.
| Gas | a, L2-atm/mole | b, L/mole | |
|---|---|---|---|
| Ar | 1.35 | 0.0322 | |
| Cl2 | 6.49 | 0.0562 | |
| CO2 | 3.59 | 0.0427 | |
| H2 | 0.244 | 0.0266 | |
| He | 0.034 | 0.0237 | |
| N2 | 1.39 | 0.0391 | |
| O2 | 1.36 | 0.0318 |
Example 6-16. Discuss the Van der Waals constants in Table 6-5 in terms of molecular structure.
Solution. The constant, a, is related to the magnitude of intermolecular forces; b is related to molecular volume. As a first guess, we might expect both of these to get larger with increasing molar mass of the gaseous substance. Let us predict that the constants will parallel molar mass. The substances and their constants are arranged in order of increasing MM below:
| Gas | MM | a, L2-atm/mole | b, L/mole |
|---|---|---|---|
| H2 | 2.016 | 0.244 | 0.0266 |
| He | 4.003 | 0.034 | 0.0237 |
| N2 | 28.014 | 1.39 | 0.0391 |
| O2 | 31.998 | 1.36 | 0.0318 |
| Ar | 39.948 | 1.35 | 0.0322 |
| CO2 | 44.009 | 3.59 | 0.0427 |
| Cl2 | 70.906 | 6.49 | 0.0562 |
In a rough sense, our prediction is borne out. Both a and b trend upward as molar mass increases. However, we are wrong in the details. The value of a for He is a clear exception; and the constants for nitrogen, oxygen, and argon do not parallel the molar masses. Having predicted, somewhat but not entirely successfully, we now must rationalize (i.e., try to think of explanations for deviations from the trend).
First, we examine the hydrogen, helium pair. The difference between these is that hydrogen is molecular, and helium is atomic. Hydrogen molecules are highly symmetrical, but elongated from a perfect sphere. The hydrogen molecules are more susceptible to dispersion forces than are the spherical, compact (high core charge) helium atoms. Consequently, forces (and the constant a) are larger for H2, despite its lower molar mass. Incidentally, because there are 2 atoms per particle in hydrogen, the volume constant, b, is slightly larger for hydrogen.
Next, we consider nitrogen, oxygen, and argon, which have roughly similar molar masses. Again, Ar is atomic and spherical, whereas nitrogen and oxygen are molecular, with elongated electron distributions. Despite their greater mass, the argon atoms are less susceptible to dispersion forces because they are spherically shaped. For the same reason, the volume constant for Ar is comparable to those for nitrogen and oxygen, despite the greater number of electrons. Recall from our discussion in Chapter 2 that atomic size decreases left to right in a period; oxygen atoms are actually smaller than nitrogen atoms, despite their larger atomic weight. Thus the volume constant, b, is smaller for oxygen than for nitrogen. Oxygen is more electronegative than nitrogen, making it somewhat less distortable. Intermolecular forces, measured by a, are somewhat less in O2.
This is rationalization. When a chemist rationalizes, it is somewhat similar to a quarterback scrambling.
6-8 Some Consequences of Intermolecular Forces . The operation of the types of forces discussed in this chapter between molecules, atoms, and ions has profound implications for the properties, function, and reactivity of chemical substances. By way of illustration, we discuss intermolecular forces in that most fundamental and essential of all molecules, water. The water molecule is bent, with a bond angle of about 104 o. This shape is understandable in terms of the VSEPR theory developed in Chapter 3. A consequence of this shape is that the molecule is polar (i.e., has a molecular dipole, Chapter 3). The molecular dipole moment and the fact that water contains hydrogen bonded to the very electronegative oxygen atom leads to strong intermolecular forces of the hydrogen bonding (dipole-dipole) type. A consequence of the existence of these forces is that the melting and boiling points of water are much higher than would be expected on the basis of molar mass (e.g., the boiling point of methane, with molar mass 16, is -162 oC). Thus water is a liquid or solid under temperature conditions found on the earth's surface. As a liquid, it provides a medium for the genesis and sustenance of life. Another consequence of polarity is that water has excellent solvating properties: it dissolves substances as diverse in structure as salts, sugars, and huge protein molecules that function as enzymes. Due to the shape of the water molecule and the intermolecular hydrogen bonding, water adopts an open lattice structure when it freezes, in which each oxygen atom is covalently bonded to two hydrogen atoms and hydrogen bonded to two more via its two lone pairs of electrons. A consequence of this open lattice is that ice is less dense than liquid water, and floats. Lakes therefore freeze top down, the layer of ice on top serving to insulate the unfrozen liquid below and thereby to protect aquatic life. If, like most substances, water were more dense as a solid, ice would sink, entire lakes would freeze, and life as we know it would not have arisen. (Of course, the density inversion of water has negative consequences, too, with which we are all familiar: potholes in roads, erosion of rock formations, and the like.) Finally, the structure of the water molecule and the resulting intermolecular interactions are at the root of the ability of water to function as both an acid and a base. This aspect of the reactivity of water is crucial in many biochemical and geological processes. We will have much more to say about the acid-base aspects of water in Chapter 13.
Supplement: Chromatography Separation and analysis of mixtures into pure components is an important problem in chemistry, with practical application in the forensics, cosmetics, food, liquor, petroleum, and paint industries. Modern chemists often use chromatography for such separations. Chromatography is a word used to encompass a range of techniques in which mixtures of pure substances are separated into the individual substances by using a mobile phase (usually a liquid or gas) to push the mixture along a stationary phase (usually a solid or liquid coated on a solid). Because the individual substances have different molecular structures, they interact differently with both the stationary and mobile phases, and consequently are "pushed" at different rates by the mobile phase. A number of chromatographic techniques are summarized in Table 6S-1.
Table 6S-1: Types of Chromatography
|
Type |
Stationary phase |
Mobile phase |
|
Paper chromatography (GC) |
Filter paper |
Liquid, often water |
|
Gas chromatography (GC) |
Polar or non-polar liquid |
Helium gas |
|
High Performance liquid Chromatography (HPLC) |
Solid |
Liquid |
|
Gel Permeation Chromatography (GPC) |
|
|
|
Thin Layer Chromatography (TLC) |
Solid on glass or plastic plate |
Liquid |
Thin Layer Chromatography Thin-Layer Chromatography (TLC) is a simple and inexpensive technique that is often used to judge the purity of a synthesized compound or to indicate the extent of progress of a chemical reaction. In this technique, a small quantity of a solution of the mixture to be analyzed is deposited as a small spot on a TLC plate, which consists of a thin layer of silica gel (SiO2) or alumina (Al2O3) coated on a glass or plastic sheet. The plate constitutes the stationary phase. The sheet is then placed in a chamber containing a small amount of solvent, which is the mobile phase. The solvent gradually moves up the plate via capillary action, and it carries the deposited substances along with it at different rates. The desired result is that each component of the deposited mixture is moved a different distance up the plate by the solvent. The components then appear as a series of spots at different locations up the plate. Substances can be identified from their so-called Rf values. The Rf value for a substance is the ratio of the distance that the substance travels to the distance that the solvent travels up the plate. For example, an Rf value of 0.5 means that the spot corresponding to the substance travels exactly half as far as the solvent travels along the plate.
The Process of TLC. Performing a TLC analysis consists of a number of steps: preparing a spotting capillary; spotting the TLC plate; developing the TLC plate; drying the plate; visualizing the substance spots, and measuring the Rf values. We consider these steps in turn.
Preparing a spotting capillary. Glass capillaries used for spotting TLC plates are commercially available. However, it is occasionally necessary to make your own capillaries. To accomplish this, light a Bunsen burner and adjust for a medium flame. Hold a melting point capillary in the flame until it just begins to soften, then quickly pull the two ends of the capillary in opposite directions. The central, soft part of the glass will elongate and thin down to a capillary with very small diameter. Break the two pieces apart at the center of the thin portion to obtain two TLC spotting capillaries.
Spotting the TLC plate. Obtain a silica gel TLC plate that is approximately 2 cm wide and 5 cm long. Using a pencil, draw a straight line parallel to the short dimension of the plate, about 1 cm from one end of the plate. This line will serve as a guide for placing the substance spots, and as a point from which to measure Rf values. Be sure to use pencil to draw this line, rather than pen, because inks will be moved by many developing solvents. Place the narrow end of one of your capillaries into a vial containing a solution of the substance to be analyzed, and allow the solution to rise in the capillary; this will happen spontaneously. Once the capillary is loaded, hold it vertically just above the pencil line on the plate. Lower it until the narrow end of the capillary just touches the plate. You will observe that some of the solution leaves the capillary and deposits on the plate. Leave the capillary in contact with the plate only briefly so that the spot is no larger than 1 mm in diameter, then raise the capillary. Allow the solvent to completely evaporate from the spot. Then, if desired, make a second deposit on the same spot with the capillary. Again allow the solvent to completely evaporate.
Developing the TLC plate. Pour some of the desired developing solvent into a small wide-mouth glass or plastic bottle to a depth of about 4-5 mm. Using tweezers, pick up the TLC plate at the top, which is the end opposite where the pencil line is drawn. Place it carefully in the developing bottle so that it stands as nearly vertical as possible. It is important that you not allow the TLC slide to tilt too much when in the developing bottle. If the slide is tilted, solvent will not advance uniformly along the plate and development will not take place properly. Leave the slide in the chamber until solvent has advanced to within 8-10 mm of the top of the slide. Then use the tweezers to withdraw the slide from the chamber and quickly draw a pencil line marking the solvent front.
Drying the Plate. Place the plate flat on a clean dry surface and allow the solvent to completely evaporate. If the solvent is not highly volatile, this can be facilitated by placing the slide on a flat surface in an oven at a temperature of 50-60oC (higher temperatures will melt the plastic substrate material). When the plate is completely dry, it is ready for visualization.
Visualization of the TLC Plate. If the substances being separated are colored, the spots can be seen without any further effort. Using a pencil, draw a boundary around each spot that matches the shape of the spot. Many substances are colorless (white) and do not show up on the white silica gel unless steps are taken to make them visible. There are a number of techniques for doing this. First is the technique of iodination. The dry plate is placed in a chamber containing a few crystals of iodine. The iodine vapor in the chamber oxidizes the substances in the various spots, making them visible to the eye. Once the spots are visible, they may be outlined with a pencil before the iodine coloration fades. Second is the ninhydrin technique, which is particularly effective for visualizing amino acid spots. In this method, a solution containing 0.2% ninhydrin in ethanol is sprayed on the dry plate. Alternately, the plate can be dipped in ninhydrin solution. In contact with an amino acid, ninhydrin displays a purple coloration that is easily seen. It usually takes a few minutes for this color to develop, so after the plate is sprayed, it is allowed to sit for several minutes. Placing it in a 50oC oven will hasten the appearance of the purple color. Once this appears, the spots may be circled in pencil to permanently mark their positions. Third, one may use TLC plates that have been loaded with a fluorescent substance that is uniformly distributed in the silica gel. Substances moving up the plate block this fluorescence at their locations. When the dried plate is viewed using a special UV light, the substance spots are visible for their lack of fluorescence on an otherwise uniformly fluorescing field. The spots may be outlined with pencil while being viewed in the light.
Measurement of Rf. The distance between the 2 horizontal pencil lines is the distance of solvent advance. The distance from the bottom pencil line to the center of a substance spot is the distance of advance of the substance. The ratio of substance advance distance to solvent advance distance is the Rf value for the substance. The figure shows a developed plate for a mixture consisting of 4 components with Rf values of about 0.05, 0.2, 0.5, and 0.9
Finally, it is very important to be observant of detail in doing TLC. In addition to the Rf value for a substance, the shape of the spot produced by a particular developing solvent and the shade of color produced by iodine or ninhydrin can be characteristic of the substance. Please note all of these things.
TLC of Inks. Before attempting to apply TLC to a real problem, it is advisable to learn and practice the technique by applying it to mixtures that are easily visualized and separated. Inks provide an ideal practice vehicle for TLC because they normally contain several colored components that separate nicely in common solvents such as ethanol, acetone, or chloroform. Spotting the plate is also easy: it may be done simply by making a VERY small mark on the plate with the tip of a pen, just above the pencil line drawn across the bottom edge of the plate. Inks are of different types and colors, of course. Some are washable (water-soluble), others are permanent. Different types require different solvents for development. Common sources of ink are ball point pens, felt-tip markers, and roller ball pens. Bottled ink is still available for people using fountain pens. Particularly good ink sources are bottled inks by Parker, Sheaffer, and Mont Blanc; Sharpie marker inks (black, red, blue, orange, brown, yellow, green); Marks-a-Lot Stay sharp markers; and Sanford calligraphy pens.
TLC of Amino Acids. TLC of amino acids is more difficult than TLC of inks, because amino acids are colorless. Therefore, not only can you not monitor their progress up the plate, but you cannot see the spots with the naked eye once the plate is fully developed and dried. To see the spots, it is necessary to use either the ninhydrin or the black-light visualization techniques. Of course, the latter works only if you use fluorescent TLC plates. Until you see the spots, you will not know whether or not a chosen solvent system has been effective in moving an amino acid or in separating a mixture. Therefore the process of finding an effective solvent system can be long and painstaking. As points of general information, amino acids are quite polar and tend to move on silica gel plates with polar solvents. They have Rf values close to 1 when water or concentrated ammonia is used as the developing solvent, probably because of their high solubility in water. Diluting a polar solvent with a less polar one results in smaller Rf values, roughly in proportion to the amount of less polar solvent used, Thus, alanine, glycine, threonine, and proline all have Rf values of around 0.60 when developed with a 50/50 mixture of water and n-propanol, and around 0.40 when developed with a 30/70 mixture of concentrated NH3 and n-propanol.
When alanine, glycine, threonine, and proline are spotted side-by-side on a plate and developed with 70% n-propanol/30% conc NH3, the following observations can be made:
|
Amino Acid |
Solvent |
Spot Color after Iodination |
Spot Color with Ninhydrin |
Rf Value |
Spot Shape |
|
alanine |
30/70 conc NH3/n-propanol |
white on brown bkgrnd |
purple |
elongated oval |
|
|
alanine |
50/50 water/n-propanol |
white on brown bkgrnd |
purple |
circle |
|
|
glycine |
30/70 conc NH3/n-propanol |
white on brown bkgrnd |
pink |
elongated oval |
|
|
glycine |
50/50 water/n-propanol |
white on brown bkgrnd |
pink |
circle |
|
|
threonine |
30/70 conc NH3/n-propanol |
white on brown bkgrnd |
purple |
elongated oval |
|
|
threonine |
50/50 water/n-propanol |
white on brown bkgrnd |
purple |
circle |
|
|
proline |
30/70 conc NH3/n-propanol |
dark brown on brown bkgrnd |
yellow with pink border |
elongated oval |
|
|
proline |
50/50 water/n-propanol |
white on brown bkgrnd |
yellow with pink border |
circle |
Problems in TLC.
Over-large Spots. Sample spots made using TLC capillaries should be no larger than 1-2 mm in diameter, because component spots in the developed plate will be no smaller than, and will usually be larger than, the size of the initial spot. If the initial spot is larger than 2 mm in diameter, then components with similar Rf values may not be resolved because their spots will be so large that they will overlap considerably and may appear to be one large spot. Small initial spots, on the other hand, maximize the potential of complete separation of components.
Uneven Advance of Solvent Front. A common problem in TLC is uneven advance of solvent along the plate. Instead of a straight line, the solvent front may appear to bow either up or down in the center. Uneven advance of solvent leads to uneven advance of substance spots, and inaccurate Rf values result. A frequent cause of uneven solvent advance is the use of a developing chamber that does not have a flat bottom. Glass bottles usually have bottoms that curve upward from the edges to the center. If the bottom of the TLC plate is placed on this curved surface, the shape of the solvent advance line may mirror the shape of the container bottom. It is therefore important to use flat-bottomed developing tanks in TLC. A bowed solvent front may also result if too little developing solvent is placed in the chamber; if the plate is cut improperly, so that the sides are not exactly perpendicular to the bottom edge; and if the slide is allowed to deviate excessively from a vertical position in the chamber. Care in choosing and using a developing chamber is the best defense against curved solvent fronts.
Water is seldom used as a developing solvent because it has a tendency to produce a dramatically curved front. This may be due to its unusually high surface tension.
Streaking. Sometimes a substance will move along a TLC plate as a long streak, rather than as a single discrete spot. This is the result of spotting the plate with too much substance, more than the moving solvent can handle. The solvent moves as much as it can, but a substantial amount of substance is left behind. The substance is dragged along by the solvent leaving a trail of substance that may sometimes span the entire distance between the starting line and the solvent front. Streaking can be eliminated by systematically diluting the spotting solution until development and visualization show the substances moving as single spots, rather than elongated streaks.
Gas Chromatography. Gas chromatography is a bit more sophisticated than TLC. It is performed using a gas chromatograph, an instrument in which the mobile phase, usually helium gas, is passed through a column (a long narrow tube) containing the stationary phase. The essential design of a gas chromatograph is shown in the figure.
The three main sections of the gas chromatograph are the injector, the column, and the detector. Each of these may be separately thermostatted to any desired T between room T and about 250 oC. At the injector, a precise volume of the sample to be analyzed is added to the gas chromatograph using a syringe. A typical sample volume is 1-10 mL, where 1 mL = 1 mm3). The injector must be hot enough to vaporize the sample. The sample vapor is picked up by the flow of helium carrier gas and swept onto the column. The column is a long tube coated with a material (the stationary phase) which attracts molecules of the sample according to their structures. Molecules that are strongly attracted "stick" to (or are retained by) the column more tightly and take more time to pass through. Molecules that are weakly attracted take less time to pass thru. The total amount of time it takes after injection for a substance to pass through the column is called the retention time for the substance. The detector is a device that detects molecules of the sample and produces an electrical signal when they pass through. The electrical signal is sent to a recording device (often a computer) that shows a trace corresponding to the electrical signal produced by the detector.
The result of doing gas chromatograph of a sample is a chromatogram, a plot of detector signal output versus time elapsed since sample injection. It typically appears as shown in figure.
Substances may be identified by their retention times, which are characteristic and reproducible for a given set of injector, column, and detector temperatures, column stationary phase material, carrier gas flow rate, and so on. The signal intensities (measured as the area under a signal) are proportional to the molar amounts of the substances in the sample injected.
A number of simple and interesting problems can be solved using gas chromatography. An example is the determination of the active ingredient in fingernail polish remover. One might approach this problem by first listing some likely candidates, simple inexpensive non-toxic chemical substances that might do the job. Clearly water is not a good choice, because fingernail polish is untouched by it. However, some reasonable candidates are listed below:
Having identified some likely candidates, one would proceed to use a gas chromatograph to determine their retention times on a particular column and with a particular set of injector, column, and detector temperatures. Some trial and error might be required to establish effective values for these temperatures. Once the retention times for the candidates were known, it would be a simple matter to inject a sample of each of several commercial fingernail polish removers to the chromatograph, using the same set of temperatures and carrier gas flow rate. If observed retention times for the removers matched any of the tested candidates, the problem would be solved. If not, other potential candidates would have to be identified and tested.
(P.S. The gas chromatographic experiment would show that commercial fingernail polish removers contain either acetone or ethyl acetate as the active ingredient. Ethyl acetate was chosen as a less toxic alternative for acetone when environmental concerns about the use of acetone were raised.)
6-1. Arrange the following ionic substances in order of increasing magnitude of the lattice energy, U, and explain your ordering: