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Transcript
NYA Notes Part II Fall 2010 Dr. Charles Berks Periodic Properties Covalent (or bonding) Atomic Radius: The covalent atomic radius is defined differently for nonmetals and metals. Nonmetals: One-half the distance between two atoms bonded together. Metals: One half the distance between 2 atoms of the metal next to each other in a crystal of the metal. The covalent atomic radius is often expressed in picometers (pm). Atomic Radius: An average value for the covalent atomic radius of an element. It is also often expressed in pm. In this course we usually consider the atomic radius rather than the covalent atomic radius. Trends in Atomic Radius: Atomic radii tend to decrease across a row in a period periodic table. Period 2 Element Li Be B C Atomic Radius (pm) 152 112 85 77 29 Across a period in the s and p blocks electrons are added to the same shell while nuclear charge is increasing as protons are added to the nucleus. Electrons in the same shell are not very effective at shielding each other from the nuclear charge. Thus the effective nuclear charge, Zeff, is increasing. As a result the atomic radius tends to decrease. Compare the atomic structure diagrams of lithium and beryllium. The effective nuclear charge of for the valence electron of lithium can be estimated. Zeff of a specific electron = Z – S where Z is the charge of the nucleus and S is the shielding effect of other electons. Zeff must be estimated because while the charge of the nucleus is known the shielding effect of other electrons, S, is in doubt. S depends on whether the other electrons are in the same or different shell, subshell and orbital as the electron being considered and also on the penetration ability of the electron being considered. Penetration: Penetration is the process by which an outer electron moves through the region occupied by inner electrons. In general for electrons of the same shell; the ability of an electron to penetrate depends inversely on its quantum number l. Therefore s electrons penetrate more than p electrons and p electrons more than d electrons. For the outer electron of lithium we can approximate Zeff = Z – S = 3 – 2 = 1 30 For the beryllium the atomic structure diagram would be Thus for beryllium Zeff = Z – S = 4 – 2 = 2. Again this is an estimate of the effective nuclear charge. Therefore it is not surprising that the atomic radius of lithium is greater than that of beryllium since the outer-most electrons of beryllium experience the greater effective nuclear charge. Atomic radius tends to increase down a column in the periodic table. This is because the size of the orbitals increase as one moves down a column. The number of electron shells has also increased and this results in greater repulsions between electrons. Group 1A Element Li Na K Rb Atomic Radius (pm) 152 186 227 248 Group 7A Element F Cl Br I In general, Atomic Radius (pm) 72 99 114 133 Atomic Radius ∞ Period Number Zeff That is; as the number of shells of electrons increases the radius of the atom increases. Whereas as the effective pull (the effective nuclear charge) of the nucleus increases the radius of the atom decreases. As one crosses a row in the periodic table the number of shells of electrons remains constant so it is the effective pull of the nucleus that is the major factor. As one proceeds down a column in the periodic table it is the number of shells of electrons that is the major influence on the radius or size of the atom. 31 Ionic Radius: The ionic radius is the radius of a cation or anion. In order to compare the radius of various ions it is important to be able to predict the electron configuration of the ion. Consider the atoms sodium and fluorine and the monatomic ions they form. Na ([Ne]3s1) Na+ ([Ne]) + e- and F ([He]2s22p5) + e- F- ([Ne]) The sodium cation, Na+, and the fluoride anion are isoelectronic with neon. That is, both ions and the neon atom have the same electron configurations. Each species differs in nuclear charge. In fact, most group A elements produce ions that are isoelectronic with a noble gas. These ions are said to have s2p6 electron configurations. Trends in Ionic Radii: In general the factors affecting the radius of an ion are same factors determining the radius of an atom. Thus ionic radius ∞ n of valence shell of the ion Zeff 1. Consider a series of isoelectronic ions. For example consider the series from O2until Mg2+. Each of these ions is isoelectronic with neon. Thus in the general relationship the value of n for the valence shell is constant at n = 2. However the charge of the nucleus is increasing across the series and therefore Zeff is increasing across the group. As the effective charge of the nucleus increases the radius of the ion must decrease. Ion O2FNa+ Mg2+ Radius (pm) 140 136 95 65 2. Cation formation from an atom results in a decrease in radius. When a group A element forms a cation the cation is normally a s2p6 ion. That is, it has an electron configuration that is isoelectronic with a noble gas. The ionization process results in the valence shell electrons in the atom being removed. Thus the cation has a lesser number of shells of electrons than the original atom. This results in a decrease in electron repulsions. At the same time the nuclear charge is constant but the number of shells of screening electrons has been reduced. Thus the effective nuclear charge experienced by the outer electrons of the cation must be less than those experienced by the outer electrons of the original atom. Therefore the radius of the cation is smaller than that of the atom. 3. Anion formation results in an increase in radius. The addition of electrons results in an increase in electron repulsions while the nuclear charge remains constant. Thus the radius must increase. 32 4. As you go down a group ionic radius increases. The increasing number of shells of electrons results in greater repulsions between electrons. Additionally the size of the orbitals is also increasing as the value of n associated with the orbitals increases. 33 Ionization Energy First Ionization Energy (IE1): The first ionization energy is the energy required to remove the most loosely held electron from an isolated gaseous atom in its ground state. The process is endothermic and therefore is associated with a positive enthalpy change, H. The chemical equation related to IE1 can be written as A(g) I+(g) + eIE1 can be understood in terms of the modified Bohr Theory. IE1 is the energy change associated with the electron jump from the ground state to the outermost orbit. Therefore IE1 = E∞ - Eground = Zeff2B n2ground (Since the energy of the n = ∞ orbit is zero.) In the case of the sodium atom the value for IE1 = 496 kJ/mol and the chemical equation is Na(g) Na+(g) + e-. The value of nground for the sodium atom is 3. Thus we can rearrange the above equation to calculate the value of Zeff. Doing so, gives a calculated value of 1.84 for the effective nuclear charge. Trends in First Ionization Energies 1. Across a period IE1’s tend to increase. That is, IE1 ∞ Zeff Consider the period 2 elements. Atom IE1 (MJ/mol) Zeff (calc) Li 0.52 Be 0.90 1.3 1.7 B 0.80a a C 1.09 N 1.40 1.6 1.8 2.1 Exceptions to the general trend O 1.31a F 1.68 Ne 2.08 2.0 2.3 2.5 Exceptions to the general trend can be rationalized. The IE1 for boron is less than the corresponding value for Be. This is because that in the case of B the ionized electron is being removed from a 2p orbital whereas the ionizing electron for beryllium is being removed from a 2s orbital. For boron the 2p electron being ionized is being more efficiently screened from the nucleus by the electrons in the completed 2s subshell and therefore is easier to remove than the 2s electron being ionized in boron. A similar exception is observed in the values for the IE1’s of Al and Ga in periods 3and 4. 34 Another exception occurs between N and O. N has a ½-filled outer 2p subshell from which the ionized electron must be removed whereas for O ionization creates a ½-filled subshell. For O electron-electron repulsion between electrons in the same subshell is being reduced thus ionization of oxygen proceeds with a lower IE1 than the general trend would predict. Similar exceptions are observed in the IE1’s of S and Se in periods 3 and 4. 2. Down a group it is the variation in the value of nground that makes a larger contribution to the value of the IE1. Thus general trend follows the relation IE1 ∞ 1/nground. Successive Ionization Energies Successive ionization can occur for multi-electron systems. The general equation for a second ionization can be represented as the following. I+(g) I2+(g) + eThus the second ionization energy can be defined as the energy required to remove the most loosely held electron from a monatomic cation of 1+ charge. For any one particular system the successive values for the ionization energies become more and more endothermic as the electrons are being removed. Consider the following table illustrating the successive ionization energies for sodium, magnesium and aluminum. Element Na Mg Al Successive Ionization Energies (kJ/mol) for Na, Mg and Al IE1 IE2 IE3 496 4560 6940 738 1450 7730 578 1820 2750 IE4 9540 10500 11600 Note that for each of the three elements the ionization energies increase markedly after the removal of the valence electrons. Thus comparison of successive ionization energies for an element can be used to predict the number of valence electrons an atom of an element possesses. 35 Electron Affinity First Electron Affinity (EA1) The first electron affinity is defined as the energy change associated with the addition of an electron to an isolated gaseous atom. A(g) + e- I-(g) If the anion formed by the reaction is more stable than the original atom then the process is exothermic and the EA1 is a negative quantity. If the atom is more stable than the anion then the process is endothermic and the EA1 is a positive quantity. General Trends in First Electron Affinities 1. Across a period first electron affinities become more negative in value. This is understandable since effective nuclear charge for the outer-most electrons tends to increase as one crosses a period. However exceptions to the general trend can be observed. Consider the following table for the period 2 elements. Element EA1 Li -60 First Electron Affinities (kJ/mol) for Period 2 Elements Be B C N O F >0 -27 -122 >0 -141 -328 Ne >0 Lithium as a typical alkali (group IA) element has a rather exothermic EA1. This is because the added electron completes the s subshell of the valence shell. Elements such as Be, N and Ne which already have rather stable electron configuration form anions of lesser stability and therefore have endothermic EA1’s. 2. First electron affinities tend to become less exothermic down a group. Down a group the added electron is being added to shells which are further and further away from the positively charged nucleus. Period 2 elements often have less exothermic first electron affinities than the general trend would predict. This is because the period 2 elements are of relatively small radius and electron-electron repulsions tend to resist the addition of an extra electron. Li -60 Na -53 K -48 Rb -47 First Electron Affinites (kJ/mol) O -141 S -200 Se -195 Te -190 F -328 Cl -349 Br -325 I -295 Similar to ionization energies successive electron affinities are possible. Second and higher electron affinities are always endothermic processes. For example, the second electron affinity of oxygen is 844 kJ/mol. 36 Ionic Bonding and the Born-Haber Cycle Ionic Bonding Binary ionic bonding forms between a metal and a nonmetal whose electronegativities widely differ. This type of bonding produces clusters or aggregates of ions that are more stable than the isolated atoms from which they are formed. This stability is due to the electrostatic attractions that exist between the cations and the anions of opposite charge. Ionic bonding is associated with crystalline solids of high melting and boiling points. When dissolved in a solvent the ions can conduct electricity. Crystal Lattice and Lattice Energy. Sodium metal reacts with chlorine gas to form the most famous example of an ionic compound, sodium chloride. Whereas metallic sodium and chlorine gas are very reactive chemicals sodium chloride is very stable. Let us see if we can discover the source of this stability of sodium chloride. Consider the reaction between an unstable gaseous sodium atom and an unstable chlorine atom to produce an ion-pair of gaseous sodium chloride. Is this an exothermic process? The overall reaction is Na(g) + Cl(g) Na+(g) + Cl-(g) Can we predict the enthalpy change for this reaction? Yes we can if we use the Law of Hess and realize that the above process can be thought of happening in two stages. What we are doing is creating a thermochemical cycle leading to the formation of a sodium chloride ion pair. Stage 1: IE1 of sodium Na(g) + Na+(g) + e-; ΔH = IE1 = 496 kJ Stage 2: EA1 of chlorine Cl(g) + e- Cl-(g); ΔH = EA1 = -349 kJ Combining Stage 1 with Stage 2 Na(g) + Cl(g) Na+(g) + Cl-(g); ΔH = IE1 + EA1 = 496 kJ + (-349) kJ = 147 kJ Thus we can see that the process of forming an ion-pair from the gaseous atoms is endothermic! Where then does the great stability of the ionic compound come from? To truly appreciate the stability of sodium chloride we must construct a realistic cycle of its formation from sodium and chlorine. 37 The true thermochemical cycle we require must feature 3 stages each containing 1 or more parts. Stage 1: Preparation of the gaseous metal cation: Metallic sodium is a metallic solid! The solid must first be sublimed to form the sodium atoms in the gaseous state. These gaseous atoms must then be ionized to form sodium cations. Stage 2: Preparation of the gaseous nonmetal anion: Chlorine as an element is of course bimolecular chlorine gas. The chlorine molecule must be first converted into individual gaseous chlorine atoms. Then the chlorine atoms can be converted to gaseous chloride ions. Stage 3: Formation of the Crystal Lattice: Lastly the sodium cations and the chlorine anions combine together not to form gaseous ion-pairs of sodium chloride but combine to form the crystalline solid sodium chloride. It is the crystalline sodium chloride that is the chemical we call sodium chloride. The preparation of any binary ionic compound can be evaluated in terms of the three above stages. The stages when combined result in a thermochemical cycle leading from the individual elements in their standard states to the binary ionic compound in its standard state, the crystal lattice. This thermochemical chemical cycle is often called a Born-Haber cycle. Using the above information we can now calculate the enthalpy of formation of solid sodium chloride to be -411 kJ per mol of sodium chloride formed. 38 In order to appreciate the Born-Haber cycle we must review the concept of enthalpy of formation. By definition, the enthalpy of formation is the enthalpy change associated with the formation of 1 mole of a chemical in a particular state from its elements in their standard states. The enthalpy of formation thus is a comparison of the relative stability of the desired chemical in a given state to its elements in their standard states. As we can see by the large negative value for the enthalpy of formation of solid sodium chloride it is much more stable than its elements. The cycle allows us to see where this stability actually comes from. Most steps in the cycle are actually endothermic! Only 2 steps are in fact exothermic; the first electron affinity of chlorine and the final step leading to the formation of the crystal lattice. It is this final step that is by far the more exothermic step and is the real source of the stability of the solid sodium chloride! Lattice Energy (U) Energy released when isolated gaseous ions combines to form an ionic crystal lattice. M+(g) + X-(g) MX(s) The lattice energy is by far the most exothermic stage of the Born-Haber cycle. The stability of ionic compounds is directly attributable to the highly exothermic nature of the lattice energy. There exists some confusion concerning the term lattice energy. Some chemists define the lattice energy as the amount of energy required to separate the ions in one mole of crystal from their positions in the lattice to an infinite separation in the gaseous state. That is MX(s) M+(g) + X-(g). Both definitions result in a lattice energy of the same magnitude but of opposite sign. The definition used by our textbook is that given above and always results in a negative value for the lattice energy. The value for the lattice energy is hard to measure experimentally. Very often the value is obtained with the aid of the Born-Haber cycle. However, in order to do so the other required thermochemical values such as electron affinities must be available. In 1956 Kapustinskii published an equation which allows for the calculation of the lattice energy for certain ionic compounds. The Kapustinskii equation has the form where K = 1.2025x10-4 J m mol-1; d = 3.45x10-11 m; ν is the # of ions in the empirical formula; z+ is the charge of the cation; z- is the charge of the anion; r+ is the radius of the cation in m; r- is the radius of the anion. Applying the Kapustinskii equation to sodium chloride gives a calculated value of -762 kJ/mol for the lattice energy. The Born-Haber cycle value is sometimes reported as -771 kJ/mol. 39 Standard States of the Halogens Many important ionic compounds are halides. Therefore knowledge of the different standard states for fluorine, chlorine, bromine and iodine is crucial in order to correctly prepare a Born-Haber cycle for these compounds. Both fluorine and chlorine and gases and therefore stage 1 for sodium fluoride would be analogous to stage 1 in the cycle of sodium chloride provided above. Bromine is a liquid in its standard state. Therefore the first stage must include a thermochemical step for the vapourization of the liquid bromine. This enthalpy change can be referred to as the enthalpy of vapourization of liquid bromine or as the enthalpy of formation of gaseous bromine. Br2(l) Br2(g); ΔH = ΔHvap Iodine is solid. The Born0-Haber cycle of an iodide must therefore feature a step where the solid is sublimed. This thermochemical step can also be thought of as the enthalpy of formation of gaseous iodine. I2(s) I2(g); ΔH = ΔHsub Properties Associated with Lattice Energy As the magnitude of the lattice energy increases the following properties become more and more apparent in ionic compounds. 1. Melting points become higher and higher 2. Solubility in water often decreases ( Review solubility rules) The relative magnitude of the lattice energy, [U], for two ionic compounds can be predicted using the following relationship. This equation is based on the Kapustinskii equation. [U] ∞ z+z-/r where z+ and z- represent the charge magnitudes of the ions and r is the sum of the ionic radii. That is r = r+ + rLet’s compare lithium chloride (LiCl) and calcium oxide (CaO). Chemical LiCl CaO z+ 1 2 z1 2 r+ (pm) 60 99 r- (pm) 181 140 r (pm) 241 239 For both chemicals the value of r is roughly the same. CaO however can a much larger charge factor. Thus we can predict that calcium oxide has the lattice energy of greater magnitude. The following properties have been reported for these two chemicals. 40 Chemical LiCl CaO U (kJ/mol) Melting Point (oC) Solubility -845 614 vs -3460 2850 ss* *Slightly soluble with the formation of calcium hydroxide Drawing Lewis Structures Example 1: Carbon Dioxide, CO2 1. Calculate the number of valence electrons. Carbon = 4 2 Oxygen = 2x6=12 Total = 16 Note: If the example is an anion then increase the number by the charge. If the example is a cation then decrease the number by the charge. 2. Octet Rule: The number of valence electrons needed for octets = 2(# H atoms in the formula) + 8(# of other atoms in the formula) 8(3) = 24 3. Number of valence electrons shared = (result of calculation 2) – (result of calculation 1) = 24 – 16 = 8 4. Number of Covalent Bonds = (Result of calculation 3)/2 = 8/2 = 4 5. Construct a Skeleton Structure a) Central atom tendency increases towards the center and towards the bottom of the periodic table. b) Oxyacids such as nitric acid or sulfuric acid have their acidic protons bonded to oxygen. c) Avoid O-O single bonds since peroxides tend to be unstable. O C O 6. Complete the octets about each atom by adding non-bonding electrons (lone pairs). O C O 7. Calculate the formal charge for each atom. If the formal charge for an atom is not zero then the value (including the sign) must be added to the atom in the Lewis structure. 41 Formal Charge Formal charge is used as a guide in drawing Lewis structures. Let x = # of valence electrons associated with the isolated atom of an element And let y = # of valence electrons associated with an atom of the element in a bonding situation Then the formal charge of the atom in the bond = x – y When the formal charge of an atom is not zero than the magnitude and the sign of the formal charge are added to the atom in the Lewis structure. Consider the ammonium cation. H H N H H The formal charge of the N = 5 – 4 = 1+. This is because each N atom has 5 valence electrons and the N in the cation has 4 valence electrons associated with it. The formal charge of each H = 1 – 1= 0. This because a hydrogen atom has 1 valence electron and each H is associated with only 1 valence electron. Thus the Lewis structure of the ammonium cation is H H N H H Formal charge calculation assumes that bonding electrons are equally shared. For a neutral molecule the sum of all the formal charges = 0. For an ion the sum of all the formal charges = the charge of the ion. 42 Principle of Electroneutrality Formal charges ought to be avoided, if possible, especially those of magnitude greater than one. For example, hydrogen cyanide, could have two possible skeleton structures but only one is correct. H C N H N C more likely Resonance Lewis structures were developed following very simplistic rules. It is a wonder that such simple rules allow for the writing of structures that could give us any kind of valuable insight into the structure of real molecules or polyatomic ions. Yet Lewis structures do have real value! It must not be surprising that there are molecules and polyatomic ions for which Lewis structures do not give a very complete understanding of the chemistry of the chemical. A Lewis structure is like a photograph. For a very simple molecule a single photograph might be sufficient to give us some idea into its character. For many molecules, much like humans, a single photo is not adequate. Such a molecule or polyatomic ion is called a resonance hybrid and the resonance hybrid must be represented by several photos to get a better understanding of its nature just as an individual may place several photos into a photo album to fully represent the person’s interests and character. 43 Failure of the Octet Rule - 3 cases 1. Species having incomplete octets. Central atom has less than 4 valence electrons, such as, Be, B or Al. Consider boron trifluoride. Boron coming from Group III A has 3 valence electrons. First try to arrive at a Lewis structure following the rules assuming the octet rule is followed. 1. 24 valence electrons; 2. 4(8) = 32 valence electrons; 3. 32-24 = 8 shared 4. 4 bonds However, 4 bonds are not reasonable since fluorine normally only forms 1 bond. Thus the boron must have an incomplete octet! In the Lewis structure it is important to remember that non-central atoms must follow the octet rule. Thus a reasonable structure would be F F B F Note that the correct structure still contains 24 valence electrons! 44 2 Octet Expansion Only may occur for central atoms having atomic number equal to or greater than 14 (Si). That is, the central atom must have a d subshell in its valence shell. The number of valence electrons in an expanded octet is usually limited to a maximum of 12 (sometimes 14). Thus the maximum number of covalent bonds is 6 (or 7). Consider each of the following examples. a) Bromine trifluoride 1. 28 2. 32 3. 4 4. 2 bonds But at least 3 bonds are required! Therefore expand the octet of the central atom. F Br F F Note that the accepted Lewis structure still contains 28 valence electrons. b) Sulfate anion 1. 32 2. 40 3. 8 4. 4 bonds Based on the octet rule the Lewis structure would be the following. O 2 O S O O The Principle of Electroneutrality must now be applied. S has an atomic number greater than 14 and therefore the octet of the sulfur can be expanded. Expansion of the sulfur can take place by converting S-O single bonds into double bonds. This is accomplished by converting a lone pair on one of the oxygen into the electrons of a second bond between that oxygen and the sulfur. Note this lowers the formal charge of the sulfur by one unit while increasing the formal charge of the oxygen by one unit. O 2 O S O O O S O O 45 O O O S O O 3. Odd-Electron Species (Free Radicals) Free radicals are normally unstable but a few examples may have a limited lifetime in nature. Common free radicals include nitrogen monoxide and nitrogen dioxide. The Lewis structure for nitrogen monoxide is the following. N O Electronegativity Electronegativity is the ability of a bonded atom to attract the electrons of the bond. Thus electronegativity is different from the electron affinity of an atom, which is a characteristic of an isolated gaseous atom. Since electronegativity is a property of a bonded atom its value depends on the identity of the atom and the identity of the atom to which the bond is formed. Linus Pauling: Established the Pauling Scale of electronegativity values. Arbitrarily set the electronegativity value of fluorine at 4.0. Periodic trends in electronegativity values: Generally increase across a period. Tend to decrease down a column. Fluorine is the most electronegative element and cesium (or francium) is the least. Most metals have electronegativity values less than 2. Most nonmetals have electronegativity values greater than 2. Metalloids tend to have electronegativity values between 1.8 and 2. Electronegativity Differences and Bond Type. a) Covalent bond. Electronegativity difference less than 0.5 Cl2 EN(Cl) = 3.0. Obviously the electronegativity difference between any two identical atoms must be zero. Thus for any homonuclear diatomic molecule the bond type must be covalent. b) Polar Covalent Bond. Electronengativity difference between 0.4 and 2.0 HCl EN(Cl) = 3.0; EN(H) = 2.1. Thus ΔEN = 0.9 c) Ionic Bond. Electronegativity difference is greater than 2.0. NaCl; EN(Cl) = 3.0; EN(Na) = 0.9. Thus ΔEN = 2.1 46 Percent Ionic Character A pure covalent bond only exists between two identical atoms such as in a homonuclear diatomic molecule. Most covalent bonds are at least somewhat polar covalent. Thus most covalent bonds have some degree of ionic character. The extent of the ionic character of a bond can be expressed in terms of the bond’s percent ionic character. Consider the H-Cl bond in hydrogen chloride. As shown on the previous page this bond can be classified as a polar covalent bond based on the electronegativity difference between the two atoms. It is possible to measure the Dipole Moment for the H-Cl bond in hydrogen chloride. The experimentally determined value for the dipole moment is 1.03 D. The debye unit, D, is 3.34x10-34 C.m. If hydrogen chloride were ionic it would contain the H+ cation and the Cl- anion. The dipole moment of this imaginary ionic compound can be calculated since the bond length is known to be 1.27x10-10 m. Thus the dipole moment of imaginary ionic HCl = (charge) (distance) = (1.60x10-19 C) (1.27x10-10 m) = (2.03x10-29 C.m) (1 D/3.34x10-34 C.m) = 6.08 D The ionic character of the bond = (1.03 D / 6.08 D) = 0.169 and The percent ionic character is 16.9%. In general the percent ionic character is the ratio of the measure dipole moment of a bond to the dipole moment of the imaginary ionic bond expressed as a percentage. Percent ionic character of a bond increases as the dipole moment difference between the bonded atoms increases. No bond is completely ionic. Sodium chloride, for example, has a percent ionic character of approximately 75%. In general, bonds with greater than 50% ionic character are classified as ionic. It is conceivable that bond type as predicted by electronegativity difference be in disagreement with the type of bond predicted by percent ionic character. For example the bond in LiI is classified as polar covalent based upon the electronegativity difference between Li and I ΔEN = 1.5) whereas the percent ionic character is above 50%. In such cases more credence is usually given the percent ionic character since it is calculated using data specific to the particular bond. 47 Modern Theories of Covalent Bonding 1. Valence Shell Electron Pair Repulsion Theory (VSEPR) 2. Valence Bond Theory (Hybridization) 1. Valence Shell Electron Pair Repulsion Theory (VSEPR) Basic assumption: Valence shell electron pairs are arranged about central atoms so that repulsions between the electron pairs are minimized. 3 types of electron pair repulsions: a) Lone pair-lone pair repulsions. The strongest type. b) Lone pair-bonded pair repulsions. c) Bonded pair-bonded pair repulsions. The weakest type. Molecular geometry: The arrangement of atoms about a central atom. The molecular geometry is also known as the shape. Central atom: Any atom bonded to 2 or more other atoms. Bond angle. Angle between 2 atoms bonded to the same central atom. Procedure for applying VSEPR: a) Draw or have access to a proper Lewis structure. For a resonance hybrid use an important contributing structure. b) Identify central atom(s). c) About each central atom sum the number of bonds and the number of lone pairs. Count multiple bonds as single bonds for this calculation. This sum of the number of bonds and lone pairs is sometimes called the steric number. d) Consult the following table to assign the orientation of electrons (orbital orientation), molecular geometry and the bond angles about each central atom. 48 Predicting Molecular Shape from VSEPR Theory Steric Basic Geometry 1 lone pair 2 lone pairs No. 0 lone pair 2 3 lone pairs Linear 3 Trigonal planar Bent 4 Bent Tetrahedral Trigonal pyramid Trigonal bipyramid Seesaw T-shaped Octahedral Square pyramid Square planar 5 6 49 Linear Applying VSEPR Steric number # of bonded pairs 2 2 Orbital Orientation Ideal Bond Angle Molecular geometry Example 180o Linear BeH2 Trigonal Planar BH3 SO2 Linear o 120 3 3 Trigonal Planar 3 2 Trigonal Planar Bent 3 1 Trigonal Planar Linear 4 4 Tetrahedral 4 3 4 4 109.5o Tetrahedral CH4 Tetrahedral Trigonal Pyramidal NH3 2 Tetrahedral Bent H2O 1 Tetrahedral Linear 5 5 Trigonal Bipyramidal 5 * 4 5 120o 900 Trigonal Bipyramidal PCl5 Trigonal Bipyramidal See-Saw SF4 3 Trigonal Bipyramidal T-shape ClF3 5 2 Trigonal Bipyramidal Linear I3 - 5 1 Trigonal Bipyramidal Linear 6 6 Octahedral 6 5 6 4 900 Octahedral SF6 Octahedral Square Pyramidal BrF5 Octahedral Square Planar XeF4 *When adding lone pairs to a trigonal bipyramidal orientation add lone pairs initially to the equatorial positions in order to minimize 90o repulsions. 50 Molecular Geometry and Dipole Moment A dipole exists if there is a separation of charge. The magnitude of the dipole depends on the magnitude of the charge being separated and the distance of the separation. Consider the simple binary molecule, hydrogen chloride. The electronegativity of chlorine (3.0) is greater enough than that of hydrogen (2.1) for this bond to be classified as polar covalent (ΔEN = 0.9). The bond dipole moment may now be calculated as bond dipole moment = δ x (bond distance) The molecular dipole moment is the vector sum of the bond dipole moment. For a binary compound such as hydrogen chloride where there is only a single bond dipole moment the molecular dipole moment is the same as the bond dipole moment. For structures containing more than one bond the individual dipole moments must be estimated and then the vector sum taken. This requires knowledge of molecular shape as well as knowledge of electronegativity. In our course we usually do not calculate the bond dipole moments. We simply use electronegativity differences between bonded atoms to estimate whether covalent bonds can be classified as polar or nonpolar. Obviously nonpolar bonds do not have a bond dipole moment. For example let us consider water. Water is known to be a polar molecule. This is obvious if one considers that each bond dipole moment would add together to produce a vector sum in the direction of the oxygen. 51 Sometimes bond dipole moments oppose each other. When this occurs the effect of the individual bond dipole moments cancel each other out. Thus carbon dioxide is a nonpolar molecule even though each bond is polar covalent. We all know the old adage like dissolves like is often to predict solubility of one chemical in another. Thus it is not surprising to learn that the nonpolar gas carbon dioxide has very limited so lubility in water. 3. Valence Bond Theory (Hybridization) VSEPR provides a simple and fast method for predicting molecular shape and bond angles. However it does not explain anything concerning the nature of covalent bonding. There is no consideration of the types of atomic orbitals that overlap to form covalent bonds. A more satisfactory theory is valence bond theory (VB theory). VB considers that a covalent bond is formed by the overlap of atomic orbitals containing unpaired electrons. The covalent bond is formed due to orbital overlap. Consider the hydrogen molecule. VB Theory considers the covalent bond in hydrogen to be formed by the orbital overlap of the two 1s orbitals of the hydrogen atoms. Remember that the ground state electron configuration of the hydrogen atom is 1s1. Thus each atom is donating one electron to the bond. To picture how the bond would form let us start by considering the two atoms as being independent of one another. In such a situation the two atoms would exhibit no orbital overlap. As the atoms approach each other orbital overlap would commence. This results in a net stabilization of the system as the positively charged nuclei become attracted to 52 the negatively charged electron density between them. Note however, that stabilization can only come about if the atomic orbitals overlap in such as way so that spins of the electrons are spin-paired. That is, the paired electrons have opposite signs for ms. As the degree of orbital overlap increases the net stabilization increases until the nuclei begin to come too close to each other. Then net stabilization begins to decline. The above graph shows that for hydrogen maximum stabilization occurs when the internuclear distance is 74 pm. Thus the reported bond length for the H-H bond is 74 pm. VB theory is simple to apply and easy to understand. Unfortunately the basic model works with only a few examples. Consider the molecule BeF2. The central atom in this molecule is beryllium and the ground state of beryllium contains no unpaired electrons (a requirement for covalent bond formation according to VB). Be: [He]2s2 – no unpaired electron In terms of an orbital box diagram we can represent beryllium in its ground state as Yet we know that the molecule BeF2 does exist and moreover it is found that both Be-F bonds are identical (that is they have the same bond energy. For both of the bonds to be identical implies that the bonds are formed by the orbital overlap of the same type of atomic orbitals. A solution to this dilemma comes from the notion of hybridization (an add-on to VB theory). Hybridization is based on the idea that covalent bonds are sometimes formed by the orbital overlap of non-ground state atomic orbitals. The excited state atomic orbitals are formed on central atoms as they enter into bonding and are not observed in isolated gaseous atoms. Hybridization involves the mixing of atomic orbitals of the central atom to produce new hybrid atomic orbitals. These new orbitals are part of a single hybrid atomic orbital subshell and are therefore degenerate. # of hybrid atomic orbitals formed = # of atomic orbitals combined = steric number (SN) of the central atom 53 It is important to remember that hybridization is part of the VB theory and that VB theory requires that every atom forming a bond with another atom must provide one electron. Therefore we sometimes most promote one or more ground state electrons to ensure the correct number of unpaired electrons. Remember that promotion takes place prior to hybridization. In the case of BeF2 promotion is required. The beryllium must form 2 bonds yet the ground state of beryllium has no unpaired valence electrons. Thus we must promote one of the 2s electrons into one of the vacant 2p orbitals. This would create an excited state beryllium atom having two unpaired valence electrons. Now that we have the correct number of unpaired electrons you might think that this is the excited state that the beryllium uses to form its bond. However, we know that the beryllium must form two equivalent bonds resulting in a bond angle of 180 degrees (as predicted by VSEPR). Thus hybridization is still necessary. The steric number of the beryllium is 2. Hybridization requires us to combine 2 atomic orbitals. We thus combine one 2s orbital and one 2p orbital to form a hybrid sp subshell containing 2 hybrid atomic orbitals. We then use the electron that were in these orbitals in the excited state (formed due to electron promotion) to populate the sp subshell. The orbitals of the hybrid atomic orbital subshell are filled following Hund’s rule. 54 The two sp hybrid orbitals centered about the central beryllium result in the desired bond angle of 180 degrees. Properties of Hybrid Atomic Orbitals Hybrid orbitals are made up of 2 lobes of which one lobe is much larger than the other. It is the larger lobe that is used to form a bond. (The smaller lobe is so small that it is often ignored in the drawing of the orbital.) The hybrid orbital only overlaps well in one direction and therefore is the ideal covalent bond forming orbital. The two sp orbitals about the beryllium in BeF2 occupy a linear geometry about the central. Each overlaps with a 2p orbital of fluorine to create a 180 degree bond angle. 55