using molecular models - cal state la€¦ · 1 © 2008 brooks/cole 1 chapter 9: molecular...

1

© 2008 Brooks/Cole 1

Chapter 9: Molecular Structures


Molecular Structures

dimethyl ether

H – C – O – C – H

H |

| H

H |

| H

..

..

ethanol

H – C – C – O – H

H |

| H

H |

| H

..

..

Two C2H6O structural isomers:

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

m.p./ °C -114.1 -141.5 b.p./ °C 78.3 -24.8


Physical models of 3D-structures:

ball and stick space filling

Computer versions:

Using Molecular Models


Hand-drawn molecules:

H

C H H

H In the plane of

the screen

Going back into the screen

Coming out of the screen

Using Molecular Models


1.! e- pairs stay as far apart as possible to minimize repulsions.

2.! The shape of a molecule is governed by the 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.

Predicting Molecular Shapes: VSEPR



Linear Triangular planar Tetrahedral

Triangular bipyramidal Octahedral

2


Basic shapes that minimize repulsions:

If the molecule contains: •! 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.

linear triangular planar

tetrahedral triangular bipyramidal

octahedral



A molecule may be described by its: •! electron-pair (e-pair) geometry •! molecular geometry

These two geometries may be different.

•! Atoms can be “seen”, lone pairs are invisible.



3 e-groups bond lone pairs pairs

2 and 3 e-group central atoms

.. ..

.. ..

.. ..

2 e-groups bond lone pairs pairs 2 0 linear

.. .. 1 1 linear

3 0 triangular planar

2 1 angular (bent)

1 2

linear

Triangular planar e-pair geometry

molecular geometry

Linear e-pair geometry

molecular geometry



2 bonds, 0 lone pairs on Be. Linear.

180.0°

180.0° “2” bonds, 0 lone pairs on C. (treat double bonds as 1 bond) Linear.

OC O

Cl Be Cl

Each C has 2 e-groups. Each H-C-C unit is linear. HC CH

180.0°

180.0°

Predicting Molecular Shapes: VSEPR 2 e-groups:


3 e-groups:

B has 3 bonds (0 lone pairs). Triangular planar.

Each C has 3 e-groups. Each C is triangular planar.

Cl B Cl

Cl

CC H HH H

120°



4 e-groups = tetrahedral e-pair geometry: bond lone pairs pairs 4 0 tetrahedral

.. ..

.. ..

1 bond, 3 lone pairs?

All molecules with only 1 bond are linear!

3 1

triangular pyramidal

2 2

angular

.. ..


3


109.5° 4 bonds, 0 lone pairs. All angles = tetrahedral angle

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

.. ..

107.5°

.. ..

.. ..

104.5°

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

H C HH

H

H N HH

O HH



VSEPR applies to each atom in a molecule. •! Alkanes: each C is tetrahedral.



Tetrahedral O

Lactic acid:

Tetrahedral C

Triangular planar C

Tetrahedral C

H

C C H

H

C

O

O

O

H

H H

..

..

.. ..

.. ..

Tetrahedral O



bond lone pairs pairs Shape

5 0 Triangular bipyramidal

Expanded octet atoms:

Remember •! lone pairs

repel the most. •! they get as far

apart as possible.

4 1 Seesaw 3 2 T-shaped 2 3 Linear

6 0 Octahedral 5 1 Square pyramidal 4 2 Square planar 3 3 T-shaped



120°

90°

Triangular bipyramidal

..

Seesaw

..

..

T-shaped

..

.. .. ..

Linear



.. ..

PF5 SF4 ClF3

F PF F

F F F SF

F

F F Cl F

F

F Xe F

..

..

.. ..

XeF2

..


4


Six e-groups = octahedral e-pair geometry

Octahedral

90°

..

Square pyramid

..

..

Square planar

Equivalent atoms



F SF F

F F

F F Br F

F

F F

F Xe F

F

F

SF6

90°

BrF5 XeF4



How do atomic orbitals (s, p, d …) produce these shapes?

Orbitals Consistent with Molecular Shapes


VB theory: bonds occur when atomic orbitals overlap.

H2 – H(1s) overlaps H(1s)

74 pm

HF – H(1s) overlaps F(2p)

109 pm



Valence Bond Theory 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, …

•! Why does C form 4 bonds at tetrahedral angles? !! C (1s2 2s2 2p2) !! 2px

1 2py1 Two bonds?

!! p orbitals are at 90° to each other !! Experiments show: tetrahedral CH4, CCl4, …


One s orbital + one p orbital ! two sp hybrids.


5


sp Hybrid Orbitals Be compounds (BeH2, BeF2 …):

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

2p 2p 2p 2p 2p 2p

2s Isolated Be atom

2s

Promotion Orbital hybridization

Ene

rgy,

E Two unhybridized

p orbitals

Two sp hybrid orbitals on Be in BeF2


sp2 Hybrid Orbitals B forms three sp2 hybrid orbitals:

!!One s orbital mixes with two p orbitals. !!One p orbital is unmixed.


sp2 Hybrid Orbitals B compounds (BH3, BF3 …):

Each sp2 hybrid (120° apart) holds one e-. Three equivalent covalent bonds form.

2p 2p 2p 2p 2p 2p

2s Isolated B atom

2s

Promotion Orbital hybridization

Ene

rgy,

E One unhybridized

and vacant p orbital

Three sp2 hybrid orbitals of B in BF3


sp3 Hybrid Orbitals C forms four sp3 hybrid orbitals:

!!One s orbital mixes with three p orbitals. !!All p orbitals are mixed.

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


sp3 Hybrid Orbitals N and O compounds (NH3, H2O…) have more e-:


sp3 Hybrid Orbitals “Octet rule” molecules have tetrahedral e-pair shape. •! sp3 hybridized (CH4, NH3, H2O, H2S, PH3, …)

H

C

H H

H

" bond

6


Summary:

Mixed Hybrids (#) Remaining Geometry s+p sp (2) p+p Linear s+p+p sp2 (3) p Triangular planar s+p+p+p sp3 (4) Tetrahedral

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 Expanded Octets


A carbon atom can have a: •! tetrahedral center (CH4, CHF3 , C2H6…) = sp3 •! triangular-planar center (H2CO, C2H4 …) = sp2

CC H HH H

Hybridization in Molecules with Multiple Bonds


C (sp2) + C (sp2) overlap (" bond): C C

H

H H

H

Unhybridized C p orbitals each contain one e-.

C C

H H

H " bond

C C

H H

H

overlap




Formaldehyde is similar:


A third type of C center is seen: !! linear center (C2H2, acetylene) = sp hybridized

CC H H



" bond: C (sp) + C (sp) overlap: C C H H

C C H H overlap C C H H


7


# bonds prevent bond rotation:

Non-rotating double bonds allow cis-trans isomerism to occur.



•!The dipoles cancel because of CO2’s shape. •! the bond dipoles have equal size but point in opposite

directions.

the arrow points to $-, the + shows $+ O = C = O!

$- $- 2$+

Molecular Polarity


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


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

Observed dipole, µ = 1.85 D!

.. ..

H H O

+

Net dipole

Molecular Polarity


Molecular Polarity

8


Molecular Polarity


Molecular Polarity

F F

F

C

F

CF4 is non polar

No net dipole

F F

H

C

F

CHF3 is polar

Net dipole

+


PF5

PF4Cl

PF3Cl2

PF3Cl2

+

+

Molecular Polarity


Molecules are sticky and attract each other.

Noncovalent Interactions


London Forces

$+ $- $+ $-

•! Strength (0.05 % 40 kJ/mol): Small molecule = few e- = weak attraction.

Large molecule = many e- = stronger attraction.

•! The only force between nonpolar molecules.


Atom Molecule # of e- bp (°C) He 2 &269 Ne 10 &246 Ar 18 &186 Kr 36 &152

More e- = larger attraction = greater stickiness = higher b.p.

F2 18 &188 Cl2 34 &34 Br2 70 +59 I2 106 +184 CH4 10 &161 C2H6 18 &88 C3H8 26 &42 C4H10 34 0

London Forces

9


Polar molecules attract each other.

Strength = 5 % 25 kJ/mol.

Dipole-Dipole Attractions


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

With equal number of e- (and same shape): dipole/dipole > London

Dipole-Dipole Attractions


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

F, O & N are small with large electronegativities.

!! results in large $+ and $- values.

H-bonds are usually drawn as dotted lines.

Hydrogen Bonds


H on one molecule interacts with O on another molecule.

Hydrogen Bonds


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

Hydrogen Bonds


Phospholipids form lipid bilayers:

Noncovalent Forces in Living Cells

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

10


Biomolecules: DNA and Molecular Structure

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





Complementary base pairs:



using molecular models - cal state la€¦ · 1 © 2008 brooks/cole 1 chapter 9: molecular...

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