VSEPR Predicting Molecular Shapes: VSEPR

Molecular Structures

Two C2H6O structural isomers:

H H H H | | .. | .. | H – C – C – O – H H – C – O – C – H | | .. | .. | H H H H ethanol dimethyl ether

Chapter 9: Molecular Structures m.p./ °C -114.1 -141.5 b.p./ °C 78.3 -24.8

Molecular shape is important! Small structural changes cause large changes in physical (and chemical) properties.

Using Molecular Models Using Molecular Models Physical models of 3D-structures: Hand-drawn molecules: Going back into the screen ball and stick space filling H In the plane of the screen C Computer versions: H H H Coming out of the screen

Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR

1. e- pairs stay as far apart as possible to minimize repulsions. 2. The shape of a molecule is governed by the Linear Triangular planar Tetrahedral number of bonds and lone pairs present. 3. Treat a multiple bond like a single bond when determining a shape. Each is a single e-group. 4. Lone pairs occupy more volume than bonds.

Triangular bipyramidal Octahedral

1 Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR

Basic shapes that minimize repulsions: A molecule may be described by its: • electron-pair (e-pair) geometry • molecular geometry

linear triangular tetrahedral triangular octahedral planar bipyramidal These two geometries may be different. If the molecule contains: • Atoms can be “seen”, lone pairs are invisible. • only bonding pairs – the angles shown are correct. • lone pair/bond mixtures – the angles change a little. . lone pair/lone pair repulsions are largest. . lone pair/bond pair are intermediate in strength. . bond/bond interactions are the smallest.

Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR 2 and 3 e-group central atoms 2 e-groups: Linear e-pair Triangular planar e-pair 180.0° geometry geometry Cl Be Cl 2 bonds, 0 lone pairs on Be. Linear. 2 e-groups 3 e-groups bond lone bond lone pairs pairs pairs pairs 180.0° 2 0 3 0 “2” bonds, 0 lone pairs on C. linear triangular planar O C O (treat double bonds as 1 bond) Linear. .. .. 1 1 2 1 ..

linear angular (bent)

Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR 3 e-groups: 4 e-groups = tetrahedral e-pair geometry:

bond lone 120° pairs pairs 4 0 Cl B Cl B has 3 bonds (0 lone pairs). Triangular planar. tetrahedral

C .. .. l ..

3 1 triangular pyramidal

Each C has 3 e-groups.

.. .. H C C H .. Each C is triangular planar.

H H 2 2 1 bond, 3 lone pairs?

angular .. All molecules with only 1 bond are linear!

2 Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR VSEPR applies to each atom in a molecule. H 109.5° 4 bonds, 0 lone pairs. • Alkanes: each C is tetrahedral. H C H All angles = tetrahedral angle

...... 3 bonds, 1 lone pair. H N H Lone-pair/bond > bond/bond repulsion H-N-H angle is reduced. H

107.5°

...... 2 bonds, 2 lone pairs. H O Two lone pairs .. H-O-H angle even smaller. H 104.5°

Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR Lactic acid: Expanded octet atoms: bond lone Tetrahedral O pairs pairs Shape H Remember Triangular planar C 5 0 Triangular bipyramidal .. O .. • lone pairs H 4 1 Seesaw .. repel the most. C C C O H .. 3 2 T-shaped • they get as far .. .. H H O 2 3 Linear apart as possible. Tetrahedral C H Tetrahedral C 6 0 Octahedral 5 1 Square pyramidal Tetrahedral O 4 2 Square planar 3 3 T-shaped

Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR

F F F F Cl F F Xe F 90° F P F F S F F ...... F F ...... 120°

......

Triangular bipyramidal Seesaw T-shaped Linear ......

PF5 SF4 ClF3 XeF2

3 Predicting Molecular Shapes: VSEPR Predicting Molecular Shapes: VSEPR Six e-groups = octahedral e-pair geometry

F F F F F

90° F S F .. F Br F F Xe F F F F F

90° ......

Octahedral Square pyramid Square planar

Equivalent atoms

SF6 BrF5 XeF4

Orbitals Consistent with Molecular Shapes Orbitals Consistent with Molecular Shapes VB theory: bonds occur when atomic orbitals overlap.

H2 – H(1s) overlaps H(1s) HF – H(1s) overlaps F(2p) How do atomic orbitals (s, p, d …) produce these shapes?

109 pm 74 pm

Valence Bond Theory Orbitals Consistent with Molecular Shapes

This works for H2 and HF, but… • Why does Be form compounds? . Be (1s2 2s2) . No unpaired e- to share.

. Experiments show: linear BeH2, BeCl2, … One s orbital + one p orbital → two sp hybrids.

• Why does C form 4 bonds at tetrahedral angles? . C (1s2 2s2 2p2) 1 1 . 2px 2py Two bonds? . p orbitals are at 90° to each other

. Experiments show: tetrahedral CH4, CCl4, …

4 sp Hybrid Orbitals sp2 Hybrid Orbitals

2 Be compounds (BeH2, BeF2 …): B forms three sp hybrid orbitals: . One s orbital mixes with two p orbitals. . One p orbital is unmixed. 2p 2p 2p 2p 2p 2p Two unhybridized

E p orbitals Promotion Orbital hybridization

Energy, Energy, Two sp hybrid

orbitals on Be in BeF2 2s 2s Isolated Be atom

Each sp hybrid (180° apart) holds one e-. Two equivalent covalent bonds form.

sp2 Hybrid Orbitals sp3 Hybrid Orbitals

3 B compounds (BH3, BF3 …): C forms four sp hybrid orbitals: . One s orbital mixes with three p orbitals. . All p orbitals are mixed. 2p 2p 2p 2p 2p 2p One unhybridized

E and vacant p orbital Promotion Orbital hybridization

Energy, Energy, Three sp2 hybrid

orbitals of B in BF3 2s 2s Isolated B atom

Each sp2 hybrid (120° apart) holds one e-. In C, each sp3 hybrid (109.5° apart) holds one e-. Three equivalent covalent bonds form. Four equivalent covalent bonds form.

sp3 Hybrid Orbitals sp3 Hybrid Orbitals - N and O compounds (NH3, H2O…) have more e : “Octet rule” molecules have tetrahedral e-pair shape. 3 • sp hybridized (CH4, NH3, H2O, H2S, PH3, …)

H σ bond

5 Hybridization in Expanded Octets Hybridization in Molecules with Multiple Bonds Summary: A carbon atom can have a: • tetrahedral center (CH , CHF , C H …) = sp3 Mixed Hybrids (#) Remaining Geometry 4 3 2 6 2 s+p sp (2) p+p Linear • triangular-planar center (H2CO, C2H4 …) = sp s+p+p sp2 (3) p Triangular planar s+p+p+p sp3 (4) Tetrahedral H C C H H H d orbitals can also form hybrids: Mixed Hybrids (#) Remaining Geometry s+p+p+p+d sp3d (5) d+d+d+d Triangular bipyramid s+p+p+p+d+d sp3d2 (6) d+d+d Octahedral

Hybridization in Molecules with Multiple Bonds Hybridization in Molecules with Multiple Bonds

H H Formaldehyde is similar: C (sp2) + C (sp2) overlap (σ bond): C C H H Unhybridized C p orbitals each contain one e-.

H H σ bond C C overlap C C H H H H

Hybridization in Molecules with Multiple Bonds Hybridization in Molecules with Multiple Bonds

A third type of C center is seen: σ bond: C (sp) + C (sp) overlap: H C C H . linear center (C2H2, acetylene) = sp hybridized

H C C H

H C C H overlap H C C H

6 Hybridization in Molecules with Multiple Bonds Molecular Polarity

π bonds prevent bond rotation:

the arrow points to δ-, the + shows δ+ O = C = O δ- 2δ+ δ-

• The dipoles cancel because of CO2’s shape. Non-rotating double bonds allow cis-trans isomerism • the bond dipoles have equal size but point in opposite to occur. directions.

Molecular Polarity Molecular Polarity

Molecule µ (D)

H2 0 HF 1.78 HCl 1.07 HBr 0.79 HI 0.38

H2O 1.85

H2S 0.95

CO2 0

CH4 0

CH3Cl 1.92

CH2Cl2 1.60

CHCl3 1.04

CCl4 0

Molecular Polarity Molecular Polarity • Polar molecules: bond dipoles do not cancel • Water is polar:

.. .. O Net dipole H H +

Observed dipole, µ = 1.85 D

7 Molecular Polarity Molecular Polarity

H F + No net dipole Net C C dipole F F F F F F

CF4 is non polar CHF3 is polar

Molecular Polarity Noncovalent Interactions

Molecules are sticky and attract each other. + + PF Cl PF5 3 2

PF Cl 4 +

London Forces London Forces Atom Molecule # of e- bp (°C) He 2 −269 More e- Ne 10 −246 = larger attraction = greater stickiness Ar 18 −186 = higher b.p. Kr 36 −152

δ+ δ- δ+ δ- F2 18 −188

Cl2 34 −34 Br 70 +59 • Strength (0.05 ↔ 40 kJ/mol): 2 I 106 +184 Small molecule = few e- = weak attraction. 2 CH 10 −161 Large molecule = many e- = stronger attraction. 4 C H 18 −88 • The only force between nonpolar molecules. 2 6 C3H8 26 −42

8 Dipole-Dipole Attractions Dipole-Dipole Attractions

Polar molecules attract each other. nonpolar # of e- bp (°C) polar # of e- bp (°C)

SiH4 18 −112 PH3 18 −88

GeH4 36 −90 AsH3 36 −62

Br2 70 +59 ICl 70 +97

- Strength = 5 ↔ 25 kJ/mol. With equal number of e (and same shape): dipole/dipole > London

Hydrogen Bonds Hydrogen Bonds An especially large dipole-dipole attraction. . 10 ↔ 40 kJ/mol . Occurs when H bonds directly to F, O or N

H on one molecule F, O & N are small with large electronegativities. interacts with O on another molecule. . results in large δ+ and δ- values.

H-bonds are usually drawn as dotted lines.

Hydrogen Bonds Noncovalent Forces in Living Cells Phospholipids form lipid bilayers:

Water is a liquid at room T (not a gas).

Polar end = hydrophilic (water loving). Nonpolar end = hydrophobic (water hating).

9 Biomolecules: DNA and Molecular Structure Biomolecules: DNA and Molecular Structure

In DNA there are 4 possible bases—adenine (A), thymine (T), guanine (G), or cytosine (C)

Biomolecules: DNA and Molecular Structure Biomolecules: DNA and Molecular Structure

Complementary base pairs: