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Chapter 8 Basic Concepts of Chemical Bonding An Important Principle in Chemistry The microscopic structure defines the properties of matter at our mesoscopic level. Ex. Graphite and Diamond (both are pure C). 8.1 Lewis Symbols and the Octet Rule Valence Electrons – e-’s found in the outermost shell; they are involved in chemical bonding. The #Ve-’s for representative elements = group number. Lewis symbols: The element’s symbol plus a dot for each valence electron. (See Table 8.1) Problem 1 Draw the Lewis symbol for carbon and chlorine. The octet rule Atoms tend to gain, lose or share electrons to obtain the configuration of a noble gas (the one closest to them). (Why?) Na → Na+ 1s2 2s2 2p6 3s1 → 1s2 2s2 2p6 O → O2- 1s2 2s2 2p4 → 1s2 2s2 2p6 8.2 Ionic Bonding The electrostatic attraction between + and – ions. It can be the result of an electron transfer. Energies of Ionic Bond Formation The heat of formation of ionic substances is quite negative (exothermic). Which factors are involved? a. Losing e-’s requires energy b. Gaining e-’s releases energy c. Electrostatic attraction releases energy Favorable overall (releases energy). Lattice Energy is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. (It’s a measure of the stability of an ionic compound.) LE = ΔH for:

MX(s) M+(g) + X-

(g) Always (+) Electrostatic Interaction The energy associated with electrostatic interactions is governed by Coulomb’s law: Lattice Energy (Table 8.2) LE increases with the charge on the ions. It also increases with decreasing size of ions. Usually high LE leads to high melting points.

𝐸𝑒𝑙 = 𝑘𝑄1𝑄2𝑑

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C XX

X

X

C

N XX

X

N

O XXO

Cl XCl

Problem 2 Arrange in order of increasing lattice energy: NaF, CsI, CaO Lattice Energy It cannot be determined directly by experiment, but we can use the Hess Law (Born-Haber cycle). LE is generally large enough to compensate for the loss of up to three electrons from atoms. Electron Configurations of Ions of the s- and p- Block Elements In view of the octet rule, we expect metals from representative groups 1A, 2A and 3A to form 1+, 2+ and 3+ cations, respectively; whereas, nonmetals of groups 5A, 6A and 7A to form -3, -2, -1 anions, respectively. Transition Metal Ions In forming ions, transition metals lose the valence-shell s electrons first, then as many d electrons as required to reach the charge of the ion. Ex. Fe: [Ar]3d64s2 Fe2+: [Ar]3d6 Fe3+: [Ar]3d5 8.3 Covalent Bonding A covalent bond refers to a pair of e-’s shared between two atoms (2 nonmetals usually). The electron density concentrates between the nuclei. Lewis Structures Covalent bonds can be represented by Lewis Symbols. The e-’s in the bond count for both atoms so each fluorine has an octet of e-’s. (H needs just 2. Why?) Usual Number of Bonds Group Number of Number of e−’s Tends to Valence e−’s needed for octet form 4A 4 4 4 bonds (no lone pairs) 5A 5 3 3 bonds (1 lone pair) 6A 6 2 2 bonds (2 lone pairs) 7A 7 1 1 bond (3 lone pairs)

H H+ H H H H = H2Electronsto share

A shared pairof electrons

A Covalentbond

A hydrogenmolecule

F F+ F F F F = F2

Electronsto share

A lone pair A bonding pair

A shared pairof electrons

A Covalentbond

A f luorinemolecule

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Multiple Bonds In many molecules, atoms attain complete octets by sharing more than one pair of e-’s. Multiple bonds are shorter and stronger (in general) than single bonds. 8.4 Bond Polarity and Electronegativity In a polar bond the e-’s are shared unequally (or unevenly). F2 vs HF Electronegativity A measure of the attraction an atom has for shared electrons. Electronegativity Increases

– …from left to right across a row. – …from the bottom to the top of a column.

Electronegativity and Bond Polarity The greater the difference in electronegativity, the more polar is the bond. Dipole Moments When two atoms share electrons unequally, a bond dipole results.

The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:

= Qr It is measured in debyes (D). Dipole Moment Calculations 1 Debye (D) = 3.34 x 10-30 coulomb-meters (C-m). For molecules the unit of charge is the electron: e = 1.60 x 10-19 C Distances are normally given in angstroms: 1Å = 1 x 10-10m Types of Bonds Covalent, Polar Covalent and Ionic. Differentiating Ionic and Covalent Bonding Larger Δen leads to more ionic bonds. (We do not have a well defined separation.) Rule of Thumb If Δen < 0.4, consider the bond as covalent If 0.4 < Δen < 1.6, consider a polar covalent bond If 1.6 < Δen, consider an ionic bond. Problem 3 Without using en values, rank the following bonds in order of decreasing polarity: N-P, N-Na, N-Al. Problem 4 Using electronegitivity values, determine the type of bond between Sn-Cl. What type of compound is SnCl4? (Sn = 1.8; Cl = 3.0)

O O

N N

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8.5 Drawing Lewis Structures 1. Find the sum of valence electrons of all atoms in the polyatomic ion or molecule.

– If it is an anion, add one electron for each negative charge. – If it is a cation, subtract one electron for each positive charge.

2. Determine the skeleton structure and connect the atoms with lines. a. The central atom is usually the least en atom, or the one that normally forms more bonds. (Never H!) Keep track of the electrons: 26 - 6 = 20 b. Oxyacids: O is bonded to the “central” atom, and H atoms bonded to O. c. If there are several options, choose the one in which atoms contain their usual # of bonds. 3. Fill the octets of the outer atoms first.

Keep track of the electrons: 26 - 6 = 20; 20 - 18 = 2

4. Put the remaining e- on the central atom, in pairs. Keep track of the electrons: 26 - 6 = 20; 20 - 18 = 2; 2 - 2 = 0 5a. If you run out of electrons before the central atom has an octet… …form multiple bonds until it does. 5b. If a central atom has more than eight electrons, it is ok if the atom is in the 3rd period or beyond. (These elements can use unfilled d orbitals for bonding.) Also: B often has 6 e-’s instead of 8. Make sure you have the right # valence e-’s. Problem 5 Draw the Lewis Structures for: CCl4 NF3 CO2 CO Formal Charge and Alternative Lewis Structures Formal Charge (FC) is the apparent charge on an atom in a Lewis Structure. To assign formal charges. For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms (# bonds). Subtract that from the number of valence electrons for that atom: the difference is its formal charge.

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Formal Charge and Alternative Lewis Structures FC is used to determine:

- The most likely (dominant) skeleton structure and Lewis structure. - The most likely distribution of electrons. - The reactivity of different sites in a molecule. - An atom with the normal # of bonds and normal # of lone pairs won’t have FC.

Warning: FC vs Oxidation Number Problem 6 Determine the dominant structure for SO2. 8.6 Resonance Structures This is the Lewis structure we would draw for ozone, O3. But… …both O-O bonds are the same length. (???) One Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures, resonance structures, to describe the molecule. Just as green is a synthesis of blue and yellow… …ozone is a synthesis of these two resonance structures. Resonance Structures or Resonance Forms are all the individual Lewis Structures in cases where two or more Structures are equally good descriptions of a single molecule. Remember: The actual structure is a hybrid of all resonance structures. The electrons that form the second C-O bond below move among the two oxygens and the carbon. They are not localized; they are delocalized (do not flip back and forth).

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Resonance in Benzene (C6H6) Benzene has two resonance structures. All C–C bonds are equivalent (all 1½ bonds) Benzene is very stable, “resonance stabilized” (spreading out e- is favorable) Drawing Lewis Structures 6. Optimize the formal charge, if possible, and show all equivalent or near-equivalent resonance structures. (central N needs and octet) Valid resonance structures for NNO (16 ve-) are: FC: -1 +1 0 -2 +1 +1 0 +1 -1 Worst! Best! Problem 7 Draw the Lewis Structures for: F2CO NO3

– ClO2–

8.7 Exceptions to the Octet Rule 1. Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons. 2. Ions or molecules with less than an octet. Consider BF3: If we try to give B an octet we get (not an accurate description): Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.

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3. Atoms in 3rd period and beyond can have more than an octet. Presumably d orbitals in these atoms participate in bonding. Which one is the better structure? Problem 8 Draw the Lewis Structures for: O2

– HClO3 XeF4 SF4 SF6 8.8 Strengths and Lengths of Covalent Bonds The strength of a bond is measured by determining how much energy is required to break the bond (This is the bond enthalpy.). The bond enthalpy for a Cl-Cl bond, D(Cl-Cl), is measured to be 242 kJ/mol. The higher the bond dissociation energy the stronger the bond. Since breaking the bonds always requires energy, then is ΔH+. (Gas Phase.) Forming bonds (being the opposite process) is exothermic ΔH–. ΔH = 436 kJ/mol Average Bond Enthalpies (Table 8.4) These are average values for different molecules. (Notice how C=C is not 2x C–C.) Bond Enthalpies and Enthalpies of Reactions

Another way to estimate Hrxn is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. In other words:

Hrxn = (bond enthalpies of bonds broken) – (bond enthalpies of bonds formed)

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Bond Enthalpies and Enthalpies of Reactions

CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g) One C-H bond and one Cl-Cl bond are broken; one C-Cl and one H-Cl bond are formed.

H = [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(H-Cl)] = [(413 kJ) + (242 kJ)] - [(328 kJ) + (431 kJ)] = 1894 kJ – 1998 kJ = -104 kJ

Keep in mind that the Hrxn obtained in this manner is an estimate since the BDE’s are average values. (The energy of a H–C bond in a CH4, will be a bit different than the C-H bond in chloroform, CHCl3.) If the reactants are not in the gas phase, then the extra energy to convert to gas needs to be included. Problem 9

Predict the Hrxn for the reaction of acetone with hydrogen to yield 2-propanol. Problem 10

Predict the Hrxn for the reaction:

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Bond Enthalpy and Bond Length (Table 8.5) We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases.

H3C C CH3

O

+ H2

(g)(g) H3C C CH3

OH

(g)

H

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Structures of Organic Molecules Atoms have their normal number of bonds and lone pairs. The condensed structural formula indicates the skeleton structure. Problem 11 Fill in any multiple bonds and lone pairs. Problem 12 Write the structure and fill in any multiple bonds and lone pairs. CHCCHCCHOCH2NHCOOH

CH3CO2H H C C O

H

H

O

H H C C O

H

H

O

H

CH3CH2NH2 H C C N

H

H

H

H

H H

CH3COCH3 H C C C

H

H

O

H

H

H

H C

H

H

C

O

O C

Br

C

H

C N