Name_____________________



AP Chemistry 3: Chemical Bonding Name __________________________

A. Bonding (8.1 to 8.4)

1. why bond?

a. atoms and ions become attached (bonded) because they enter a lower energy state

b. complete valence shell = lowest energy state

1. metals lose all valence electrons, which uncovers complete valence shell (+ ion)

2. nonmetals with 5-7 valence electrons gain electrons to complete valence shell (– ion)

3 nonmetals share valence electrons with other nonmetals

a. 1-3 valence electrons share 1 for 1 ∴ doubling valence number

b. 4-7 valence electrons share to fill s and p orbitals (8 electrons = octet rule)

c. Lewis symbols

1. a system to show valence electrons

2. chemical symbol + dots for valence electrons

3. Na•, •Mg•, etc.

d. three major types of bonds

1. ionic bond: electrostatic attraction between cations and anions

2. covalent bond: shared electrons between non-metal atoms

3. metallic bond: metal atoms collectively share valence electrons

4. molecular bonds, discussed in the next section, aren't true bonds

2. ionic bonding

a. metal and nonmetal: Na(s) + ½ Cl2(g) → NaCl(s)

(oxidation-reduction reaction)

b. cations and anions: Na+(aq) + Cl-(aq) → NaCl(s) (precipitation reaction)

c. bond strength

1. lattice energy measures ΔE for bonding

2. electrostatic force measures ionic attraction

3. proportional to ionic charges

4. inversely proportional to ionic radii

3. covalent bonding

a. bonding atoms' orbitals overlap, which maximizes attraction between nuclei and bonding electrons

b. atoms share 2, 4 or 6 electrons

1. 2 (single), 4 (double), 6 (triple) bond

2. multiple bonds reduce bond distance

a. bond distance < sum of atomic radii

b. shorter bond distance = stronger bond

c. polar bond when electrons are not shared equally

1. electronegativity

a. measures atom's attraction for bonding electron pair (higher # = stronger)

b. relative scale where period 2 elements are 1.0 (Li) to 4.0 (F), with 0.5 intervals

1. noble gases are excluded

2. trend:

a. increase across period

b. decrease down groups

2. bond polarity

a. electronegativity difference between bonding atoms result in uneven sharing of electrons, which generates a partially positive charged side, δ+, and a partial negative charged side, δ-

b. notation

[pic]

c. measured as dipole moment

3. bond strength increases with polarity

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B. Lewis Structures (8.5 to 8.7)

1. shows the atoms in a molecule with their bonding and non-bonding electron pairs

a. bonding electrons (– single, = double, ≡ triple)

b. lone (non-bonding or unshared) electron pair (••)

[pic]

2. drawing Lewis structures with one central atom

|count the total number of valence electrons (subtract charge for ions) |

|CO2: 4 + 2(6) = 16 |

|IF2–: 7 + 2(7) + 1 = 22 |

|draw a skeleton structure |

|first element in formula is central, except H |

|single bonds to other atoms (max. 4) |

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|O–C–O [F–I–F]– |

|(ions are bracketed) |

|place electrons around each atom |

|8 total electrons |

|except H, Be and B or when total number of electrons is an odd number |

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|.. .. .. .. .. .. |

|:O – C – O: [:F – I – F:]– |

|.. .. .. .. .. .. |

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|count Lewis structure electrons (including bonding electrons) |

|if equal to valence electrons, stop |

|if valence e- < Lewis e-, add additional bonds to reduces # of electrons|

|by 2's |

|if valence e- > Lewis e-, add 2 or 4 electrons to central atom (3rd |

|period or higher); called expanded octet |

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|.. .. .. .... .. |

|O = C = O [:F – I – F:]– |

|.. .. .. .. .. |

|added bonds expanded octet |

3. when more than one Lewis structure is possible use formal charge to decide which is more likely

a. each atom is assigned its lone electrons plus half the bonding electrons

b. formal charge = valence e- – assigned e-

c. preferred structure

1. atoms have formal charges closest to zero

2. negative formal charge reside on the more electronegative atom (upper right most on the periodic table)

d. example: NCS-

| |[:::N–C≡S:]- | [::N=C=S::]- | [:N≡C–S:::]- |

|valence e- |5 4 6 |5 4 6 |5 4 6 |

|assigned e- |7 4 5 |6 4 6 |5 4 7 |

|formal |-2 0 +1 |-1 0 0 |0 0 -1 |

∴ [::N=C=S::]- is preferred because formal charges are closest to zero and negative charge is on the nitrogen (higher electronegativity)

e. technique can produce erroneous structures (experiments are required to determine actual structure)

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C. VSEPR Model (9.1 to 9.3)

1. rules

a. maximum separation between electron pairs

b. atom positions define molecular geometry

c. lone electron pairs squeeze bond angle

(actual angle < ideal angle)

|Electron |Domain Geometry |– |: |Molecular Geometry |Bond |

|Domains | | | | |Angle |

|2 | |2 |0 | |180o |

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|3 | | | | |120o |

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| | |3 |0 | | |

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| | |2 |1 | | |

|4 | | | | |109.5o |

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| | |4 |0 | | |

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| | |3 |1 | | |

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| | |2 |2 | | |

|5 | | | | |90o |

| | | | | |120o |

| | |5 |0 | | |

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| | |4 |1 | | |

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| | |3 |2 | | |

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| | |2 |3 | | |

|6 | | | | |90o |

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| | |6 |0 | | |

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| | |5 |1 | | |

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| | |4 |2 | | |

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2. polar molecules

a. lone electron pairs distort symmetry except for sp3d-linear and sp3d2-square planar

b. different perimeter atoms

c. polar interactions increase water solubility, increase melting and boiling temperatures, decrease evaporation (volatility)

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D. Valence-Bond Theory (9.4 to 9.5)

1. explains electron domain geometries in terms of electron orbitals

2. atomic orbitals from bonding atoms merge, which allows single electrons from each atomic orbital to occupy overlapping area and simultaneously attract both nuclei (i.e. H–H: overlap of 1s orbitals)

3. more complex molecules require a fusion of s and p orbitals into equivalent (hybrid) orbitals

a. explains why covalent bonds around an atom are all the same even if electrons were originally in different shaped (s, p and d) atomic orbitals

b. 1 s + 1 p form 2 sp hybrids (2 electron domains)

ground state excited state hybridized state

|↑↓ | | | | |→ |↑ | |↑ | |

e. expanded octet hybridization

1. 1 s + 3 p + 1 d = 5 sp3d hybrids (5 domains)

2. 1 s + 3 p + 2 d = 6 sp3d2 hybrids (6 domains)

4. not all valence electrons enter hybrid orbitals

a. one electron pair per bond enters a hybrid orbital

1. sigma bond (σ)

2. electrons located between bonding atoms

b. lone pairs of electrons enter hybrid orbital

c. remaining bonding pairs of electrons from multiple bonds remain in pure p orbitals

1. pi bond (π)

2. electrons located above/below bonding atoms

d. example: ::O=C=O::

p p

sp2 p p sp2

sp2 sp sp sp2

p p sp2

sp2 p p

e. π bond electrons can spread out across entire molecule (delocalized)

1. NO3- has one π bond, which is shared evenly and simultaneously between 3 O's

2. multiple Lewis structures show all possible locations for π bonds = resonance forms

[pic]

3. bond order = sigma bond + share of π bonds

(each N–O bond has bond order = 1 1/3)

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E. Naming Binary Molecules (2.8)

1. two types of nonmetal atoms covalently bonded

2. lower electronegative atom is written first and named as the element

3. second element is given –ide ending

4. prefix used to indicate number of atoms

a. 1—mono, 2—di, 3—tri, 4—tetra, 5—penta, etc.

b. mono never used for first element

5. exceptions

a. common names: NH3 (ammonia),

H2O2 (hydrogen peroxide) and H2O (water)

b. molecules that begin with H (except H2O)

1. no prefix for H ∴ H2S(g) is hydrogen sulfide

2. water solutions are acidic ∴ H2S(aq) is hydrosulfuric acid

c. organic molecules (discussed next)

F. Simple Organic Molecules—Hydrocarbons (25.1 to 25.6)

1. general properties

a. contain C and H

b. nonpolar, flammable (fuels)

2. formulas and names

a. number of carbons in parent chain

|1 |2 |3 |4 |5 |

|meth |eth |prop |but |pent |

|6 |7 |8 |9 |10 |

|hex |hept |oct |non |dec |

b. bond between carbons

1. alkanes (all single bonds) end in “ane”

2. alkenes (1 or more double bonds)

1 double end in “ene”, 2 double end in "diene"

3. alkynes (1 or more triple bonds) end in “yne”

4. cyclical

a. 3 to 6 carbon ring with single bonds between carbons: prefix "cyclo"

b. 6 carbon ring with 3 shared π bonds: benzene (called aromatic hydrocarbon)

c. branches

1. C-branches—“yl”

2. benzene branch—"phenyl"

3. location of branches

a. number of the parent carbon

b. lowest number possible

c. dash: # – word, comma: #, #

d. number of branches (2—di, 3—tri, etc.)

3. condensed structural formula

a. hydrogens are written after the carbon

b. branches are in parentheses after hydrogens

c. example: 4-ethyl-2-methyl-1-hexene

CH2C(CH3)CH2CH(C2H5)CH2CH3

d. semi-condensed (shows branches and bonds)

CH3 C2H5

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CH2=C–CH2–CH–CH2–CH3

4. functional groups

a. dramatically modify properties of hydrocarbon

b. haloalkanes: halogen replaces one or more H

1. reduces reactivity (flammability)

2. named as a branch with an “o” ending

c. oxygen containing groups

1. hydroxyl group (C–OH)

a. water soluble

b. alcohols (antiseptic, solvent, fuel)

c. acids (release H+ in solution)

2. carbonyl group (C=O)

a. aldehydes (preservative, flavors)

b. ketones (solvent)

c. esters (pleasant odors, polymers)

3. ethers have C–O–C (first anesthesia, solvent)

4. increases polarity: C–OH > C=O > C–O–C

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d. amines

1. replace H in ammonia with hydrocarbon group = amine (CH3NH2 = methylamine)

2. when NH2 branches off hydrocarbon = amino

CH3CH(NH2)CH2CH3 (2-aminobutane)

3. weak bases (neutralize acids—absorb H+)

e. summary chart

[pic]

5. isomerism

a. structural isomers: same molecular formula, different structure and name

1. move double/triple bond position

2. move branch

3. form cycloalkane from alkene

b. geometric isomers: same molecular formula, same relative position of carbons and bonds, but different spatial arrangement

1. >C=C< carbons can't rotate

a. x>C=CC=C CaO |

|> BaO. |

2. There are several oxides of nitrogen; among the more common are N2O, NO2 and NO3-.

a. Draw the Lewis structures of these molecules.

|N2O |::N = N = O:: |

|NO2 | . |

| |::O = N – O::: |

|NO3- |[:::O – N – O:::]- |

| ||| |

| |:O: |

b. Which of these molecules "violate" the octet rule? Explain.

|NO2; there is an odd number of total electrons so nitrogen must have an |

|odd number. |

c. Draw resonance structures of N2O.

|: N ≡ N – O:: Δ ::N = N = O:: Δ :::N – N ≡ O: |

d. For each resonance structure from question 2c, calculate the formal charge and evaluate which structure is most likely. Explain.

|: N ≡ N – O:: |::N = N = O:: |:::N – N ≡ O: |

|5 5 6 |5 5 6 |5 5 6 |

|5 4 7 |6 4 6 |7 4 5 |

|0 1 -1 |-1 1 0 |-2 1 1 |

|N ≡ N – O is the most likely because it has the smallest absolute |

|formal number and the negative charge is on the O, which has the highest|

|electronegativity. |

e. Which side of the N–O bond is +δ? Explain.

|The N side is +δ because N has a lower electronegativity than O (listed |

|to the left of O). |

f. Rank the strength of the N–O bond in order of strongest (1) to weakest (3). Explain your answer.

|N2O |NO2 |NO3- |

|1 |2 |3 |

|The bond order in N2O = 2, in NO2 = 1.5 and in NO3- = 1.33 ∴ N2O > NO2 >|

|NO3-. |

3. Complete the chart for SeF2, SeF4, and SeF6.

| |SeF2 |SeF4 |SeF6 |

|Lewis |.. |F |F F |

|Structure |F – Se – F || |\ / |

| |.. |F – Se – F |F – Se – F |

| | |.. \ |/ \ |

| | |F |F F |

|Se- Hybridization|sp3 |sp3d |sp3d2 |

|Domain Geometry |tetrahedral |trigonal |octahedral |

| | |bipyramidal | |

|Molecular |bent |seesaw |octahedral |

|Geometry | | | |

|Ideal Bond Angle |109.5o |90o, 120o |90o |

|Polarity |polar |polar |nonpolar |

4. Consider the ion SF3+.

a. Draw a Lewis structure.

|F |

|| |

|F – S – F |

|.. |

b. Identify the type of hybridization exhibited by sulfur.

|sp3 |

c. Identify the electron-domain and molecular geometries.

|Electron-domain geometry |Molecular geometry |

|Tetrahedral |Trigonal pyramid |

d. Predict whether the F-S-F bond angle is equal to, greater than or less than 109.5°. Explain

|The F-S-F bond angle in SF3+ is slightly smaller than 109.5o because the|

|non-bonding pair of electrons takes up more space. |

5. Consider the ion SF5-

a. Draw a Lewis structure.

| F |

|.. / |

|F – S – F |

|/ \ |

|F F |

b. Identify the type of hybridization exhibited by sulfur.

|sp3d2 |

c. Identify the electron-domain and molecular geometries.

|Electron domain geometry |Molecular geometry |

|Octahedral |Square pyramid |

6. Two Lewis structures can be drawn for the OPF3 molecule.

|Structure 1 |Structure 2 |

|.. | |

|: O : |: O : |

|.. | .. |.. || .. |

|: F – P – F : |: F – P – F : |

|.. | .. |.. | .. |

|: F : |: F : |

|.. |.. |

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Which Lewis structures best represents a molecule of OPF3? Justify your answer in terms of formal charge.

|F – P – O |F – P = O |

|7 4 7 |7 5 6 |

|7 5 6 |7 5 6 |

|0 -1 1 |0 0 0 |

|Structure 2 is the best representation because the formal charges on |

|each atom are closest to zero, with the negative charge (if it exists) |

|on the more electronegative element. |

7. a. Draw the condensed structural formula and name four structural isomers of C4H9Cl.

|Formula |Name |

|CH2ClCH2CH2CH3 |1-chlorobutane |

|CH3CHClCH2CH3 |2-chlorobutane |

|CH2C)CH(CH3)CH3 |1-chloro-2-methylpropane |

|CH3CCl(CH3)CH3 |2-chloro-2-methylpropane |

b. Draw the structural formula and name two geometric isomers of C2H2Cl2.

|Formula |Name |

|H Cl |trans-1,2-dichloroethene |

|\ / | |

|C = C | |

|/ \ | |

|Cl H | |

|Cl Cl |cis-1,2-dichloroethene |

|\ / | |

|C = C | |

|/ \ | |

|H H | |

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