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Homework Problems Chapter 8 Homework Problems: 4, 6, 18, 20, 21, 22, 28, 30, 38, 42, 48, 54, 58, 68, 71, 76, 88, 90, 106 a-c, 115.

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Presentation on theme: "Homework Problems Chapter 8 Homework Problems: 4, 6, 18, 20, 21, 22, 28, 30, 38, 42, 48, 54, 58, 68, 71, 76, 88, 90, 106 a-c, 115."— Presentation transcript:

1 Homework Problems Chapter 8 Homework Problems: 4, 6, 18, 20, 21, 22, 28, 30, 38, 42, 48, 54, 58, 68, 71, 76, 88, 90, 106 a-c, 115

2 CHAPTER 8 Chemical Bonding I: Basic Concepts

3 Types of Chemical Bonds There are three general types of chemical bonds that substances form, two of which we discuss in this chapter. Ionic bonding - Electrons are transferred; bonding is due to electrostatic attraction between cations (positive ions) and anions (negative ions). Covalent bonding - Pairs of electrons are shared between atoms. Metallic bonding - Electrons are pooled (spread out) among positively charged metal cations.

4 Lewis Dot Structure In ionic and covalent bonding it is the valence electrons that are usually involved in bond formation. One method used to focus on changes that occur when bonds formed is to represent atoms and ions in terms of dot structures. The method was developed by the American chemist G. N. Lewis, and so is often called the Lewis dot structure method. In the dot structure method valence electrons are represented as dots around the symbol for the element. For simple cases a maximum of 8 valence electrons can occur (2 for hydrogen and helium). The electrons are placed around the atom or ion, in pairs when necessary. Example: What are the Lewis structures for H, N, and F? H 1s 1 N 1s 2 2s 2 2p 3 F 1s 2 2s 2 2p 5

5 Dot Structure For Main Group Elements The dot structures for atoms of main group elements are given below.

6 Dot Structure For Atomic Ions In addition to dot structures for atoms, we can also write dot structures for ions formed from atoms by adding or removing the appropriate number of electrons. Example: Give the dot structures for O, O 2-, and O 2+. O1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 O 2+ 1s 2 2s 2 2p 2 O O 2- O 2+

7 Octet Rule The general behavior observed by main group elements in ionic compounds may be summarized in terms of a general principle called the octet rule. Octet rule - Main group elements tend to gain or lose electrons so that they end up with a noble gas electron configuration. This will either completely fill or completely empty the valence shell of electrons. For most atoms, filling the valence shell means having a total of eight electrons - hence, the name octet rule. Note that for two elements (Li, Be) the octet rule predicts that atoms of these elements will lose one (Li) or two (Be) electrons to obtain the same electron configuration as He. Hydrogen (H) will tend to add one electron to obtain the same electron configuration as He.

8 Ionic Bonding Ionic bonding refers to the bonding that occurs between cations and anions due to the attractive force that acts between particles of opposite charge. We may say the following: 1) Ionic bonding usually occurs between a metal cation and a nonmetal anion (or anion group). 2) The attractive forces in ionic bonding are isotropic (the same in all directions). 3) Ionic compounds are usually solids at room temperature. 4) Ionic compounds usually have high melting points and high boiling points relative to those observed for other substances. 5) Ionic solids are usually hard and brittle, and easily cleaved.

9 Formation of Ionic Compounds To form a binary ionic compound one or more electrons are transferred between a metal atom and a nonmetal atom to form ions. NaClNa1s 2 2s 2 2p 6 3s 1 Na + 1s 2 2s 2 2p 6 Cl1s 2 2s 2 2p 6 3s 2 3p 5 Cl - 1s 2 2s 2 2p 6 3s 2 3p 6 MgCl 2 Mg1s 2 2s 2 2p 6 3s 2 Mg 2+ 1s 2 2s 2 2p 6 Cl1s 2 2s 2 2p 6 3s 2 3p 5 Cl - 1s 2 2s 2 2p 6 3s 2 3p 6 The general tendency is for atoms to gain or lose electrons to either completely fill or completely empty the ns np valence orbitals. This accounts for the characteristic charges observed for main group ions. Note that the formation of ions is similar to the processes of ionization (for formation of cations) and electron affinity (for the formation of anions).

10 Ionic Bonding With Dot Structures Formation of ionic bonds can be pictured in terms of dot structure as the transfer of electrons from metal atoms to nonmetal atoms. Example: Na + Cl  NaCl Note that we place ions within brackets, with the charge of the ion indicated outside of the bracket. This dot structure notation is not particularly useful for discussing ionic bonding, but is extremely useful in discussing covalent bonding.

11 Lattice Energy Lattice energy (  H  lattice ) is defined as the energy required to convert exactly one mole of an ionic compound into gas phase ions. NaCl(s)  Na + (g) + Cl - (g)  H  lattice (NaCl) = 787. kJ/mol Lattice energy is expected to be positive when defined in this way, as we must add energy to break apart an ionic solid

12 Lattice Energy and Coulomb’s Law We may use Coulomb’s law to predict general trends in lattice energy. F ~ Q + Q - Q + = charge of cation Q - = charge of anion d 2 d = distance between ion centers d = r + + r - r + = radius of cation r - = radius of anion The larger in magnitude the attractive force between ions the larger the value for lattice energy.

13 Trends in Lattice Energy We may use Coulomb’s law to predict general trends in lattice energy. F ~ Q + Q - d 2 The following follows from the above relationship. 1) As Q + increases in magnitude the size of the lattice energy increases. 2) As Q - increases in magnitude the size of the lattice energy increases. 3) As the sizes of the ions decreases the size of the lattice energy increases (since smaller ion size means a smaller value for d). So the lattice energy increases as the magnitude of the charges of the ions increases and as the size of the ions decreases.

14 Examples of Trends (1) 1) Cations in same group forming ionic compounds with the same anion. Lattice energy decreases in size in moving from top to bottom within a group. Ionic compound  H  lattice (kJ/mol) LiCl 834. NaCl 787. KCl 701. CsCl 657.

15 Examples of Trends (2) 2) Anions in the same group forming ionic compounds with the same cation. Lattice energy decreases in size in moving from top to bottom within a group. Ionic compound  H  lattice (kJ/mol) LiF 1017. LiCl 860. LiBr 787. LiI 632.

16 Examples of Trends (3) 3) As |Q + Q - | increases, the size of the lattice energy increases.  H  lattice = 910. kJ/mol  H  lattice = 3414. kJ/mol

17 Example Arrange the following ionic compounds in order from largest to smallest size of lattice energy. 1) KF, KCl, KBr, KI 2) MgO, CaO, SrO, BaO 3) KF, K 2 O, K 3 N

18 Example: Arrange the following ionic compounds in order from largest to smallest size for lattice energy. 1) KF, KCl, KBr, KI KF > KCl > KBr > KI Cation is the same, anion is larger as we go from F - to I -. 2) MgO, CaO, SrO, BaO MgO > CaO > SrO > BaO Anion is the same, cation is larger as we go from Mg 2+ to Ba 2+. 3) KF, K 2 O, K 3 N K 3 N > K 2 O > KF Cation is the same, anion has a larger charge as we go from F - to N 3-.

19 Determination of the Value for Lattice Energy We may use our previous discussion of thermodynamics to find an experimental method for determining a value for lattice energy. We illustrate the method for the ionic compound sodium chloride Na(s)  Na(g)  H  f (Na(g)) 1 / 2 Cl 2 (g)  Cl(g)  H  f (Cl(g)) Na(g)  Na + (g) + e - IE 1 (Na) Cl(g) + e -  Cl - (g)EA(Cl) Na + (g) + Cl - (g)  NaCl(s) -  H  lattice (NaCl) Na(s) + 1 / 2 Cl 2 (g)  NaCl(s)  H  f (NaCl(s)) The above collection of processes is called a Born-Haber cycle.

20 Finding the Lattice Energy Na(s)  Na(g)  H  f (Na(g)) 1 / 2 Cl 2 (g)  Cl(g)  H  f (Cl(g)) Na(g)  Na + (g) + e - IE 1 (Na) Cl(g) + e -  Cl - (g)EA(Cl) Na + (g) + Cl - (g)  NaCl(s) -  H  lattice (NaCl) Na(s) + 1 / 2 Cl 2 (g)  NaCl(s)  H  f (NaCl(s))  H  f (Na(g)) +  H  f (Cl(g)) + IE 1 (Na) + EA(Cl) + (-  H  lattice (NaCl)) =  H  f (NaCl(s))  H  lattice (NaCl) = [  H  f (Na(g)) +  H  f (Cl(g)) + IE 1 (Na) + EA(Cl)] -  H  f (NaCl(s))

21 “Ionic Bonding” in Nonmetals Given the success of ionic bonding in explaining substances composed of metals + nonmetals, it is reasonable to attempt to apply the model for bonding between nonmetal atoms. Cl 2 Cl [Ne] 3s 2 3p 5 Cl - [Ne] 3s 2 3p 6 Cl [Ne] 3s 2 3p 5 Cl + [Ne] 3s 2 3p 4 Problems: 1) While the Cl - anion now satisfies the octet rule, the Cl + cation is worse off than before. 2) Experimentally, both Cl atoms in Cl 2 are the same. 3) Cl 2 exists as molecules and is a gas at room temperature. We would expect ionic substances to exist as crystalline solids.

22 Covalent Bonding We can get around the problems associated with transferring electrons by sharing one or more pairs of electrons. The shared electrons can be counted by both atoms in satisfying the octet rule. Bonding by sharing of one or more electron pairs is called covalent bonding. Notation 1) Bonding electron pairs are indicated by lines. 2) Nonbonding electrons, called lone pair electrons, are indicated by dots.

23 Multiple Bonds It is possible for atoms to share more than one pair of electrons. Consider the diatomic molecules F 2, O 2, and N 2. F 1s 2 2s 2 2p 5 O 1s 2 2s 2 2p 4 N 1s 2 2s 2 2p 3 Bond order = number of pairs of shared electrons.

24 Polyatomic Molecules and Ions Covalent bonding can take place in polyatomic molecules or ions. Lewis structure - Indicates which atoms are bonded together, the bond orders for these bonds, and the number of lone pairs electrons. Lewis structures do not directly indicate molecular geometry (the arrangement of atoms in three dimensions). As previously noted, we often write molecular formulas in such a way that they provide information about the arrangement of atoms in a polyatomic molecule.

25 There are often several ways in which a Lewis structure may be drawn for a particular molecule. Example: CH 3 COOH (acetic acid) Although these Lewis structures look different, they are all the same, and communicate the same information about the bonding in acetic acid.

26 Average Bond Length The average bond length is the average length of a particular type of bond (single, double, triple) between two atoms. In general, the average bond length decreases as we go from a single to a double to a triple bond. bond N-N N=N N  N length (nm)0.1470.1240.110

27 Average Bond Lengths (Table)

28 Lewis Structures For Ions Lewis structures for ions are drawn the same way as for molecules, except that the ion is placed within brackets, with the charge of the ion shown outside the brackets. NO 2 + NH 4 +

29 Comparison of Ionic and Covalent Compounds IonicCovalent Bonding by transfer of electrons Bonding by sharing of electrons Bonding is isotropic Bonding is directional Do not exist as molecules Exists as molecules Solids at room temperature May be solid, liquid, or gas at room temperature High melting point Low melting point High boiling point Low boiling point Strong electrolytes Usually nonelectrolytes

30 Bond Polarity A covalent bond represents the sharing of one or more pairs of electrons. However, the electron pairs are not necessarily equally shared between the two bonded atoms. We call a bond where there is an unequal sharing of electrons a polar covalent bond. There are three general cases: 1) Equal sharing - Atoms bonded by electrons that are equally shared 2) Unequal sharing - Atoms bonded by electrons that are unequally shared. 3) Ionic bonding - Ions are formed from the transfer of one or more electrons to form cations and anions

31 Bond Polarity - Examples H - H H  + - F  - [ Na + ] [F - ]

32 Electronegativity Electronegativity (EN) is a number (between 0 - 4) that is assigned to an element that represents the tendency of atoms of that element to attract electrons. The larger the value for electronegativity the greater the tendency for atoms of that element to attract electrons. Note that a value for electronegativity is not assigned to most noble gases. The general trends for electronegativity are as follows: 1) Within a group of elements electronegativity increases from bottom to top. 2) Within a period electronegativity increases from left to right. For a polar covalent bond between two atoms, the atom with the larger value for electronegativity will have a partial negative charge (  -) and the atom with the smaller value for electronegativity will have a partial positive charge (  +).

33 Electronegativity Chart

34 Use of Electronegativity The difference in electronegativity between two atoms can be used to predict the type of bonding that exists between the atoms.  EN < 0.5Bond is nonpolar or only slightly polar.  EN 0.5 - 2.0Bond is polar covalent, more electronegative atom has a partial negative charge  EN > 2.0Bond is ionic; more electronegative atom forms an anion Above guidelines are approximate.

35  EN = 2.1  EN = 0.9  EN = 0.0 ionic polar covalent nonpolar covalent

36 Dipole Moment (  ) Dipole moment (  ) is a measure of how polar a bond or a molecule is. By definition, dipole moment is given by the expression  = QrQ = size of separated +/- charge r = distance of separation of charge As q increases and r increases  also increases. Dipole moment is measured in units of Debye 1 Debye = 3.336 x 10 -30 C  m|e - | = 1.602 x 10 -19 C Molecule  EN  (Debye) F 2 0.0 0.00 HF 1.9 1.82 LiF 3.0 6.33

37 Percent Ionic Character The percent ionic character is a measure of the approximate amount of ionic character in a bond between two atoms. It is defined as % ionic character =  (observed) 100 %  (assuming discrete charges) where  (observed) = experimentally observed dipole moment  (assuming discrete charges) = dipole moment calculated using the bond distance and assuming complete electron transfer

38 Percent Ionic Character - Example The experimental values for dipole moment and bond distance for the molecule HCl are  = 1.08 D r = 0.127 nm What is the percent ionic character of the bond?

39 The experimental values for dipole moment and bond distance for the molecule HCl are  = 1.08 D r = 0.127 nm What is the percent ionic character of the bond?  (assuming discrete charges) = dipole moment calculated using the bond distance and assuming complete electron transfer = (1)(1.602 x 10 -19 C)(0.127 x 10 -9 m) 1 D 3.336 x 10 -30 C  m = 6.10 D % ionic character = 1.08 D 100% = 18% 6.10 D

40 Electronegativity and Percent Ionic Character

41 Guidelines For Drawing Lewis Structures The following general guidelines are useful in drawing Lewis structures for molecules and ions. 1) The central atom is usually the least electronegative atom (excluding hydrogen, which will never be the central atom). 2) For molecules or ions that obey the octet rule number of bonds = (# e - needed for octet rule) - (# valence e - ) 2 3) There is usually one electron from each atom making a covalent bond. 4) Common number of covalent bonds formed H - 1 bond (no exceptions)O - 2 bonds F - 1 bond (no exceptions)N - 3 bonds Cl, Br, I - 1 bondC - 4 bonds (almost no exceptions)

42 Example: Draw the Lewis structure for the following molecules a) NOF b) CH 2 O c) CH 3 CHO

43 Example: Draw the Lewis structure for the following molecules a) NOF b) CH 2 O c) CH 3 CHO NOFN 5 valence e - central atom = N (least electronegative) O 6 valence e - F 7 valence e - 18 valence e - total(8 + 8 + 8) = 24 e - needed for octet rule number of covalent bonds = [ 24 - 18 ]/2 = 6/2 = 3

44 Examples of Covalent Bonding NOFcentral atom = N # bonds = ( 24 - 18 ) = 3 2 CH 2 Ocentral atom = C # bonds = ( 20 - 12) = 4 2 CH 3 CHO # bonds = ( 32 - 18) = 7 2

45 Organic Molecules For organic molecules, we can often use the way in which the formula for the molecule is written as a guide to its Lewis structure. Example: What is the Lewis structure for diethyl ether, whose chemical formula is CH 3 CH 2 OCH 2 CH 3 ?

46 Organic Molecules For organic molecules, we can often use the way in which the formula for the molecule is written as a guide to its Lewis structure. Example: What is the Lewis structure for diethyl ether, whose chemical formula is CH 3 CH 2 OCH 2 CH 3 ?

47 Coordinate Covalent Bond In most covalent bonds each atom contributes the same number of electrons to the bond. In a coordinate covalent bond, (sometimes called a dative bond) both electrons come from the same atom. One example of a coordinate covalent bond is in the ammonium ion (NH 4 + ). Coordinate covalent bonds can form when there is an atom that has one or more available lone pairs of electrons. Note that in the ammonium ion nitrogen is making more bonds than usual, and that all four of the N-H bonds are identical.

48 Formal Charge Formal charge (FC) is a number that represents in an approximate way the electron density around a particular atom in a molecule or ion. Note that it does not represent the real charge on the atom - it is a bookkeeping device to keep track of electron density in the molecule or ion. It is similar to, but not the same, as oxidation number. Formal charge is assigned as follows: FC = (# valence e - in atom) - [ (# nonbonding e - ) + 1 / 2 (# bonding e - ) ] The sum of the formal charges must add up to the charge of the molecule or ion. Example:FC(N) = 5 - [ 0 + 1 / 2 (8) ] = +1 FC(H) = 1 - [ 0 + 1 / 2 (2) ] = 0

49 Use of Formal Charge Formal charge can be used to determine which resonance structure is most important in representing a molecule or ion for cases where the resonance structures are not equivalent to one another. The best resonance structure is the one which: 1) Makes the formal charges of all atoms as close to zero as possible. 2) If there are nonzero formal charges, places negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms.

50 Example: Which of the following structures for N 2 O is the best structure? # bonds = (24 - 16) = 4 2

51 Example: Which of the following structures for N 2 O is the best structure? # bonds = (24 - 16) = 4 2 0 +1 -1 -1 +1 0 -2 +1 +1 The structure on the right can be ruled out because it has a formal charge that is more different from zero than the other two structures. Of the two remaining structures, the one on the left places the negative formal charge on oxygen, the more electronegative atom, and so is better than the one in the middle.

52 Resonance Structures It is not always possible to represent bonding in a molecule or ion with a single Lewis structure. Example: What is the Lewis structure for O 3 ? # bonds = ( 24 - 18) = 3 2 Experimentally, the two bonds in O 3 are the same, and inter- mediate between a single and a double bond. That suggests the real structure is some combination of the above two Lewis structures. Resonance structure - Two or more valid Lewis structures for a molecule or ion. The actual structure is an “average” of the resonance structures. We use an arrow (  ) to indicate the Lewis structures that contribute to the representation of the molecule or ion.

53 Resonance structures differ only in the arrangement of electrons, and not in the arrangement of the atoms making up the molecule or ion. Example: NO 3 - (nitrate ion) # bonds = ( 32 - 24) = 4N = central atom 2 In this case there are three equivalent Lewis structures that differ only in the location of the N=O double bond. The N-O bond is equal to 1 1 / 3 of a covalent bond.

54 Formal Charge and Resonance Structures We can sometimes use formal charge to identify the most important resonance structure for a molecule or ion. Consider our previous case of the N 2 O molecule. 0 +1 -1 -1 +1 0 -2 +1 +1 The average structure will have a larger contribution from the resonance structure on the left, and a smaller contribution from the resonance structure on the right, based on the formal charges observed.

55 Exceptions to the Octet Rule While the octet rule works well for many substances, there are several important exceptions to the general rule. 1) Compounds of beryllium (Be) and boron (B). Beryllium (which often forms covalent compounds) usually makes two covalent bonds, boron usually makes three covalent bonds. Be1s 2 2s 2 B1s 2 2s 2 2p 1

56 2) Molecules or ions with an odd number of electrons. In this case, it is impossible to pair up the electrons so that every atom satisfies the octet rule. In this case, the less electronegative atom generally is the one that will be one electron short of an octet. Examples: NO and ClO. 3.0 3.5 3.0 3.5 Compounds with an odd number of electrons are usually very reactive, as they would like to acquire an additional electron to satisfy the octet rule. (16-11)/2 = 2.5 (16-13)/2 = 1.5

57 3) Elements below the second period of the periodic table. Since these elements have d orbitals in addition to their s and p orbitals, there is room in the valence shell for more than 8 electrons. The elements can form compounds with an expanded octet (more than 8 valence electrons). N1s 2 2s 2 2p 3 P1s 2 2s 2 2p 6 3s 2 3p 3 3d 0 NF 3 PF 3 PF 5 NF 5 - does not occur We usually consider Lewis structures that contain atoms with more than an octet of electrons only if no reasonable structure obeying the octet rule can be found. We also use formal charge to guide us (discussed later).

58 Example: What are the Lewis structures for SF 2, SF 4, and SF 6 ?

59

60 Formal Charge and the Octet Rule We sometimes see cases where a resonance structure that is better based on formal charge is worse based on satisfying the octet rule. In these cases, how do we decide which resonance structure is more important?

61 All the atoms in the structure at left satisfy the octet rule. Sulfur violates the octet rule in the structure at right, but the formal charges are all equal to zero. So both are likely to be important. Note that there are a number of other equivalent resonance structures for the structure at right.

62 Bond Dissociation Energy We define the bond dissociation energy (D) as the amount of energy required to break (dissociate) one mole of a specific bond in a specific molecule in the gas phase. ethane D = 376. kJ/mol CH 3 CH 3 (g)  CH 3 (g) + CH 3 (g) propane D = 356. kJ/mol CH 3 CH 2 CH 3 (g)  CH 3 CH 2 (g) + CH 3 (g) butane D = 352. kJ/mol CH 3 CH 2 CH 2 CH 3 (g)  CH 3 CH 2 CH 2 (g) + CH 3 (g)

63 Average Bond Energy If we look at the amount of energy required to break a particular type of bond in different molecules we observe that the values don’t change much from molecule to molecule. That suggests that we define an average bond energy that represent the “average” amount of energy required to break one mole of a particular type of bond. 1) Since these are average values, the actual amount of energy required to break a specific bond in a particular molecule may differ from the average. 2) The larger the value for bond energy the stronger the bond. 3) Average bond energies depend on both the identity of the two atoms bonded together and the bond order. C - C347. kJ/mol C = C620. kJ/mol C  C812. kJ/mol

64 Table of Average Bond Enthalpies

65 Example: How much energy would it take to break apart one mole of formaldehyde (CH 2 O) molecules into atoms in the gas phase?

66 CH 2 O(g)  C(g) + 2 H(g) + O(g) energy = D(C=O) + 2 D(C-H) = (745 kJ/mol) + 2 (414 kJ/mol)  1573. kJ/mol experimental value = 1511. kJ/mol

67 Estimating  H rxn From Bond Energies Consider a chemical reaction taking place with gas phase reactants and products. We can represent the reaction in two ways reactants  products  H rxn reactants  atoms  products In the second pathway we break all of the chemical bonds in the reactant molecules to form atoms, and then make the new chemical bonds required to form our final products. Since enthalpy is a state function, the change in enthalpy for both of the above processes must be the same. Therefore, we may say  H rxn  (  bond energies for reactants) – (  bond energies for products) The approximation is due to the fact that the values for bond energies are average values and so approximate.

68 Example Chloroethane may be formed by the following process C 2 H 4 (g) + HCl(g)  CH 3 CH 2 Cl(g) Estimate the value for  H rxn for the above process.

69 Chloroethane may be formed by the following process C 2 H 4 (g) + HCl(g)  CH 3 CH 2 Cl(g) Estimate the value for  H rxn for the above process. Bonds broken = 4 (C-H) + (H-Cl) + (C=C) = 4 (414) + (432) + (620) = 2708 kJ/mol Bonds formed = 5 (C-H) + (C-Cl) + (C-C) = 5 (414) + (339) + (347) = 2756 kJ/mol  H rxn  (2708 kJ/mol) - (2756 kJ/mol) = - 48 kJ/mol

70 Bonding in Metals Metals have a general tendency to give up electrons to form cations. This indicates that the valence electrons in a metallic element are only loosely attracted to the nucleus of the metal. A simple model for bonding in metals is to treat the valence electrons as forming a “sea” of electrons which can easily move about in the metal. Because of this, this suggests that it should be easy to move electrons through the volume of the metal. In fact, metals are good conductors of electricity because these loosely bound electrons can easily move through the metal. The above also accounts for the malleability and ductility of metals. Because there are usually no strong localized bonds in metals, it becomes easy to alter their shape (hammer them into thin sheets or draw them into wires).

71 End of Chapter 8 “The underlying physical laws necessary for the mathematical theory of a large part of physics and the whole of chemistry are thus completely known, and the difficulty is only that the application of these laws leads to equations much too complicated to be soluble.” - P. A. M. Dirac “The great importance in Lewis’ theory is that it provided chemists with a valuable way of visualizing the electronic structures of atoms and molecules, and for practical purposes his ideas are still used today.” - Keith J. Laidler “My name is Bond - Covalent Bond.” - anonymous


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