Chemical Structures (Part 2)

Chapter 11

Some images Copyright © The McGraw-Hill Companies, Inc.

Brad Collins, DTCC

Organic Structures• Organic compounds do not follow the AXn general

formula, but: • DO generally follow the octet rule.

• Hydrogen and halogens are always outer atoms • In neutral atoms, N and P will retain their lone

pair; O and S will retain their 2 lone pairs • Rarely see expanded octets

11.1

Organic Compounds• Use structural formulas to depict organic

compounds • Isomer - compounds with the same molecular

formula, but different structural formulas.

11.1

• Ethanol: • Molecular: C2H6O • Structural: • Condensed structural: CH3CH2OH

• Dimethyl ether: • Molecular: C2H6O • Structural: • Condensed structural: CH3OCH3

Organic Compounds• Condensed structural formulas

• Hydrogens follow the atom they are bonded to. • CH3, not H3C

• Single bonds are omitted • CH3CH3OH, not CH3–CH3–O–H

• Multiple bonds ARE shown • CH2=CH2 or CH CH

• Side chains are often shown: • but may be written in parenthesis: CH3CH(OH)CH3

11.1

Organic Compounds• Condensed structural formulas

• Cyclic structures a drawn with the geometric figure they resemble: • Cyclohexane, C6H12

11.1

PracticeDraw the Lewis structure for diethyl ether, CH3CH2OCH2CH3

11.1

Bond Enthalpy• The enthalpy required to break a particular bond in

1 mole of gaseous molecules

H2 (g) H (g) + H (g) ΔH0 = 436.4 kJCl2 (g) Cl (g)+ Cl (g) ΔH0 = 242.7 kJHCl (g) H (g) + Cl (g) ΔH0 = 431.9 kJ

N2 (g) N (g) + N (g) ΔH0 = 941.4 kJ N N

Bond Enthalpy

Bond Energies Single bond < Double bond < Triple bond

O2 (g) O (g) + O (g) ΔH0 = 498.7 kJ O O

11.2

Bond Enthalpy• In polyatomic molecules (e.g., H2O) the average

bond enthalpy is listed

H2O (g) H (g) + OH (g) ΔH0 = 502 kJ

OH (g) H (g) + O (g) ΔH0 = 427 kJ

Average OH bond energy = 502 + 4272

= 464 kJ

11.2

Bond Enthalpy

Note: Forming the listed bond, releases the tabled amount of energy • Breaking bonds is

endothermic • Forming bonds is

exothermic

11.2

Bond Enthalpy (BE) and Enthalpy changes in reactions

ΔH0 = total energy input – total energy released

= ΣBE(reactants) – ΣBE(products)

Imagine a reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

11.2

Bond Enthalpy and ∆Hrxn

H2(g) + F2 (g) —> 2HF(g) !

∆Hrxn = –543.2 kJ

11.2

PracticeUse bond energies to calculate the enthalpy change for:

H2 (g) + F2 (g) 2HF (g)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

ΔH0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ11.2

Shapes of Molecules• VSEPR Theory

• Valence Shell Electron Pair Repulsion • Molecules will form shapes that minimize the

repulsive force between the electron pairs (bonding pairs and lone pairs) • Arrangement describes how the electron pairs

arrange themselves around an atom • Shape describes how the molecule actually

looks in 3-dimensional space

11.3

Atomic Arrangements Two Items Around a Central Atom

• Classify molecules according to how many atoms (B) are bonded to the central atom and how many lone pairs (E) are on the central atom.

• Two outer atoms, no lone pairs (AB2) • Electron pairs align opposite the central atom, A

AB2 2 0

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

linear

B B

11.3

linear

B B

Cl ClBe

2 atoms bonded to central atom0 lone pairs on central atom 10.1

Beryllium Chloride

Atomic Arrangements Three Items Around a Central Atom

• Three outer atoms, no lone pairs (AB3) • Electron pairs align at 0, 120, and 360º around

the central atom, A

AB3 3 0

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

11.3

trigonal planar

trigonal planar

120º

10.1

Trigonal Planar

Boron Trifluoride

• 2 outer atoms, one lone pair (AB2E) • Electron pairs align at 0, 120, and 360º around

the central atom, A

AB2E 2 1

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

11.3

bent

Atomic Arrangements Three Items Around a Central Atom

trigonal planar

B B

Atomic Arrangements Four Items Around a Central Atom

• Four outer atoms, no lone pairs (AB4) • Electron pairs align at ~109.5º angles around the

central atom, A

AB4 4 0

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

11.3

tetrahedraltetrahedral

B

B

B

B

Methane

11.3

Atomic Arrangements Four Items Around a Central Atom

• Three outer atoms, one lone pair (AB3E) • Electron pairs align at ~109.5º angles around the

central atom, A

AB3E 3 1

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

tetrahedral trigonal pyramidal

11.3

Atomic Arrangements Four Items Around a Central Atom

• Two outer atoms, two lone pairs (AB2E2) • Electron pairs align at ~109.5º angles around the

central atom, A

AB2E2 2 2

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

11.3

tetrahedral bent

Effect of Lone Pairs

bonding-pair vs. bonding

pair repulsion

lone-pair vs. lone pair repulsion

lone-pair vs. bonding

pair repulsion< <

11.3

Atomic Arrangements Five Items Around a Central Atom

• Five outer atoms, no lone pairs (AB5) • Electron pairs align at ~109.5º angles around the

central atom, A

AB5 5 0

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

Trigonal bipyramidal

B

B

B

BB

Trigonal bipyramidal

11.3

11.3

Phosphorus Pentachloride

Atomic Arrangements Six Items Around a Central Atom

• Six outer atoms, no lone pairs (AB6) • Electron pairs align at ~109.5º angles around the

central atom, A

AB6 6 0

Class

# of atoms bonded to

central atom

# lone pairs on

central atomArrangement of electron pairs

Molecular Shape

11.3

Octahedral

B

B

B

B

B

B

Octahedral

11.3

11.3

Sulfur Hexafluoride

Atomic Arrangement

Summary

11.3

Polarity• Definition: Unequal sharing of electrons in covalent

bonds • Electrons spend more time with one atom in the bond • Bond has negative and positive end

H F

electron rich regionelectron poor

region

δ+ δ−

11.4

Polar Molecules• To be polar, molecules must meet 2 criteria:

• Must have a polar bond (dipole) • Must have a molecular shape that allows a dipole moment to exist.

• Dipoles are vectors (have magnitude and direction) • Dipole moment is the vector sum of all the dipoles present.

• Examples: HF, NH3, and NF3

H F

electron rich regionelectron poor

region

δ+ δ−

HF

11.4

11.4

NH3 NF3

Which of the following molecules have a dipole moment? H2O, CO2, SO2, and CH4

O HH

dipole moment polar molecule

SO

O

CO O

no dipole moment nonpolar molecule

dipole moment polar molecule

C

H

H

HH

no dipole moment nonpolar molecule

11.4

Does CH2Cl2 have a dipole moment?

11.4

11.4

Dipole Moments of Polar Molecules

Trends in Polarity• Molecules with a linear, trigonal planar, or

tetrahedral arrangement are predicted to be nonpolar if: • There are no lone pairs on the central atom • All the outer atoms are the same

• If one or more of these aren’t true, the molecule is usually polar.

• Does NOT work for trigonal bipyramidal or octahedral arrangements.

11.4

Practice: Polar MoleculesPredict the polarity of the following molecules:

Chloroform, CHCl3

Phosphorus trichloride, PCl3

Bromine pentafluoride, BrF5

11.4