Chapter 10Shapes of Molecules

The Three Dimensional Reality of Molecular Shapes

The Three Dimensional Reality of Molecular Shapes

Each atom, bonding e- pair, and lone e- pair has its own position in 3D space.

Position determined by the attractive and repulsive forces that govern all matter.

Molecular Shape is Crucial to Life Processes:

Molecular Shape is Important in Nanotechnology:Nanotechnology: “Building tiny machines at the molecular level.”

Buckyball (fullerenes) – a C60 structure discoveredin soot in 1985

Many, many uses: One use may be to “sneak” medicinesinto cells or through the blood brain barrier, where some substances (medicines) may not normally enter.

Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures2D before 3D

(1) Element symbol – nucleus + inner electronsEx: The element lithium has an element symbol Li

(2) Surrounding dots – valence electrons (outer most shell)

Lewis dot structures consist of two parts:

Li

Different elements can have the same number of dots

Be

Mg

Same Group(Column)

Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures

Electron types: Bonding Pairs & Lone Pairs

Single, Double, TripleBond types:

S

O O

O

If atoms have same groups number, place atom with higher period (row) number inthe center.

Example: SO3

Steps for writing Lewis Structures for molecules that have only single bonds:

(1) Place the atoms relative to each other, with the atom with the lower group (column)number in the center. Often, center atom is also atom with lower E.N.

Example: NF3 N

F F

F

Steps for writing Lewis Structures for molecules that have only single bonds:

(2) Determine the total number of valence e- available

Example: NF3 N 5 e- F 7 e- (x 3) = 21 e- Total = 26 e-

Continued

(3) Draw a single bond from each surrounding atom to the central atom, and subtract2 valence e-’s for each bond to find the # of e- remaining

Example: NF3 3 single bonds x 2 e- = 6 e-

e- remaining = total – bonded = 26 – 6 = 20 e- N

F F

F

(4) Distribute the remaining e- in pairs so that each atom has 8 e- (except H, has 2 e-)1st: Place lone e- pairs on surrounding (more EN) atoms to give each full valence.2nd: If any e- remain, place them on the central atom.

Example: NF3

(5) Check that each has a full valence shell. DONE.

Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures

Lewis Dot Structures are not 3D, so several different depictions are correct.

NF F

F

N

F F

FNF F

F

Etc…….

This method works for singly bonded compounds where C, N, and O, andelements in higher periods (rows), are the central atom.

In nearly all compounds:• H atoms form 1 bond• C forms 4 bonds• N forms 3 bonds• O forms 2 bonds• Halogens (F, Cl, Br, I,…) form 1 bond

Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures

In cases of multiple bonds (double, triple), there is an additional step:

(6) If a central atom still does not have an octet (after Step 4), make amultiple bond by changing a lone pair from one of the surroundingatoms into a bonding pair to the central atom.

Examples: CH4 N2

In cases where there is a polyatomic ion, Lewis Dot Structure is shownin square brackets:

Section 10.1: Resonance Structures

Have the same relative placement of atoms, but different locations of bonding and lone e- pairs.

Example: Nitrite (NO2-)

Double bonds change location.Lone pair on O atom changes location.

Neither structure depicts nitrite accurately: The two O,N bonds in this compoundhave bond lengths and bond energies that lie somewhere between the O–N and the O=N bond.

Section 10.1: Resonance Hybrids

Resonance structures are not real bonding depictions:

Structure I Structure II

In reality, Structure I and Structure II do not switch back and forth from oneinstant to the next. The actual nitritemolecule is an average of the two.

The e-’s are delocalized over the entire molecule.(Just as e-’s in a metallic bond are delocalized around the entire sea of electrons.)

Section 10.1: Resonance Hybrids

Sometimes implied (no indication).

Sometimes indicated by dotted lines.

Example: Ozone (O3)

Example: Benzene (C6H6)

Draw a Lewis Dot Structure for:

CCl4CCl2F2

CH4ONH3O

C2H6O (no O–H bonds)

10.610.8

Resonance Structures:

10.1010.12

Section 10.1: Which is the more important resonance structure?

All resonance forms contribute equally when central atom has surrounding atoms thatare all the same.

When atoms surrounding the central atom are not the same: One resonance form may“weight” the average (In other words, it “counts more” than the other forms.)

Determining the most important resonance form: Determine each atom’sformal charge – the charge it would have if the bonding electrons were equally shared,

Section 10.1: Which is the more important resonance structure?

Formal charge – Determined for each atom in a compound

# valence e- - (# unshared valance e- + ½ # shared valence e-)

Example: Ozone (O3)

Formal charge for OA = 6 – (4 – ½*4) = 0

Formal charge for OB = 6 – (2 – ½*6) = +1

Formal charge for OC = 6 – (6 – ½*2) = -1

In the case of ozone, both resonance structures (I and II) have the same formalcharges (but on different O atoms) so they contribute equally to the resonancehybrid.

Formal charges must sum to the total charge on the chemical species.

Section 10.1: Which is the more important resonance structure?

Formal charge – Determined for each atom in a compound

Recall – In nearly all compounds:• H atoms form 1 bond• C forms 4 bonds• N forms 3 bonds• O forms 2 bonds• Halogens (F, Cl, Br, I,…) form 1 bond

Ozone Example:

Formal charge for OA = 6 – (4 – ½*4) = 0

Formal charge for OB = 6 – (2 – ½*6) = +1

Formal charge for OC = 6 – (6 – ½*2) = -1

When formal charge is 0, an atom has its usual # of bonds.

Section 10.1: Which is the more important resonance structure?

What about a case where the resonance structures do not contribute equally?

Criteria for choosing the more important resonance structure: (1) Smaller formal charges (+ or -) are preferable to larger ones.

Resonance form I is out.

(2) The same nonzero formal charges on adjacent atoms are not preferred.

Not applicable to this example.

(3) A more negative formal charge should reside on a more electronegative atom.

O is more EN than N, so Resonance form III is the winner.

Example: Cyanate ion, NCO-

Occurs when there are different atoms around the central atom.

Section 10.1: Formal charge is not the same as the ON

What is the difference?

Formal charge -

Oxidation number -

Bonding e-’s are assigned equally to the atoms, so that each atom has ½

Bonding e-’s are assigned completely to the more EN atom

Formal charge = # valence e- - (# lone pair e- + ½ # bonding e-)

O.N. = # valence e- - (# lone pair e- + # bonding e-)

Section 10.1: Exceptions (Limitations) to the Octet and Formal Charge Rules

(1) e- deficient molecules – gaseous molecules containing Be or B as central atom

Be: 4 e- B: 6 e-

(2) Odd e- molecules– central atom has odd # of valence e-

free radicals – very reactive (b/c very unstable): often react with each other to pairup their lone e- (make you age)

Halogens much more EN than Be or B =Formal charge rules make sharing ofextra lone pairs by halogens unlikely

(3) Expanded valence shells – more than 8 valence e-; occurs only where d orbitalsare available Row 3 or higher (Review of orbital types: p289 – 295; s – 2 e-, p – 6 e-, d – 10 e-, f – 14 e-)

Resonance Structures:

10.14a10.16b

Other suggested problems:10.1510.1710.1910.24

Criteria for choosing the more important resonance structure:

(1) Smaller formal charges (+ or -) are preferable to larger ones.

Resonance form I is out.

(2) The same nonzero formal charges on adjacent atoms are not preferred.

Not applicable to this example.

(3) A more negative formal charge should reside on a more electronegative atom.

O is more EN than N, so Resonance form III is the winner.

Formal charge = # valence e- - (# unshared valance e- + ½ # shared valence e-)

Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory

Molecular shape is important in many, many scientific disciplines.

Medicine: Receptors

Nanotechnology: Membrane Transport

Ecology: Talking trees

Jack Schultz, Chemical Ecologist

Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory

Molecular shape is important in many, many scientific disciplines.

Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory

Lewis Dot Structures, 2D (Blueprint) VSEPR, 3D (House)

Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory

Each group of valence electrons around a central atom is located as far away as possible from the others in order to minimize repulsions.

e- group can be: a single bond, double bond, triple bond, lone pair, lone e-

Section 10.3: Molecular Shape and Molecular Polarity

In molecules with more than 2 atoms:

Shape and bond polarity determine molecular polarity.

Molecules with only 2 atoms.

Relative electronegativities of the twoatoms determine polarity.

Section 10.3: Molecular Shape and Molecular Polarity

In molecules with more than 2 atoms: Dipole moments

Behavior of Molecules With and Without Dipole Moments

Electric field: Polar molecules (which have a dipole moment) orient with partial charges towards oppositely charged electric plates.

Molecules with out a dipole moment will not orient themselves in anyparticular direction. *Molecules with no dipole moment can be polar.

Dipole moments – a measure of molecular polarity

magnitude of partial charges on ends of a molecule (in coulombs) x

distance between them

Section 10.3: Molecular Shape and Molecular Polarity

Dipole moments: When molecular shape influences polarity

Large ∆EN between C (EN = 2.5) and O (EN = 3.5) C = O bonds are polar

CO2 molecule is linear Two identical bond polarities are counterbalanced (in otherwords, they cancel each other out)

As a result, CO2 has no net dipole moment.

Section 10.3: Molecular Shape and Molecular Polarity

Dipole moments: When molecular shape influences polarity

H2O (like CO2) also has two identical molecules bonded to the central atom.

However, H2O (unlike CO2) has a dipole moment.

Bond polarities are not canceled out because of the shape of the water molecule:

V-shaped rather than linear.

The O end of the molecule is more negative than the H ends

Section 10.3: Molecular Shape and Molecular Polarity

When different molecules have the same shape, the nature of the atomsSurrounding the central atom can have a major effect on polarity.

CCl4 – does not have a dipole

CHCl3 – has a dipole

Effect of Molecular Polarity on Behavior

Example: Boiling point of NH3 versus PH3

Why is NH3 boiling point higher?

Also determines reaction behavior: NH3 + H+ NH4+ (p392)

A closer look at molecular shapes: double bonds and lone pairs

Bond Angles:

Idealized

vs.

Actual

A closer look at molecular shapes: double bonds

Effect of double bonds on bond angle when surrounding atoms are different.

Rule: The double bond, with itsgreater e- density, repels the

two single bonds more stronglythan they repel each other.

A closer look at molecular shapes: lone pairs

Effect of lone pairs on bond angle.

Rule: Lone pairs repel bonding pairsmore strongly than bonding pairs

repel each other.

A closer look at molecular shapes: a few more details

Bond angles for:Equatorial groups = 90ºAxial groups = 120º

General Rule: The greater the bond angle, the weaker the repulsion.

In this case: Equatorial-equatorialrepulsions are weaker than axial-rquatorial repulsions.

Implication: Lone pairs, which exert stronger repulsions, will tend to occupy equatorial positions.

A closer look at molecular shapes: a few more details

General Rule: The greater the bond angle, the weaker the repulsion.

Implication: If two lone pairs arepresent, they will always occupy opposite vertices (furthest apart)

More VSEPR Practice

GeH2

PCl5SF4

ClF3

XeF2

SF6

BrF5

XeF4

NH4+

SO4-

(1) Lewis dot structure (dominant resonance form – calculate formal charge if needed)

(2) 3-D geometry (Table 10.9)

(3) Molecular polarity Bond dipoles cancel? Lone pairs present? Different surrounding atoms?

Additional Optional Homework Problems:10.30, 10.35, 10.38, 10.39, 10.51, 10.53, 10.54, 10.97

Chemical Reaction vs. Physical Interactions (Chapter 12)

Boiling point of NH3 versus PH3: Why is NH3 boiling point higher?

The N – H bonds between NH3 molecules matter,not the N – H bonds within the NH3 molecule.

EN is inversely proportion to atomic radius.

N has greater EN than P: Why?

Why is EN is inversely proportion to atomic radius?

e- shielding

If a molecule is polar, it will:(1) have a net dipole moment(2) orient itself in an electric field

If a molecule is nonpolar, it will:(1) not have a net dipole moment(2) will move about randomly in an electric field

Summary So Far: Overall Polarity of a Molecule

Steps used to determine molecular polarity:

(1) Draw the 2-D Lewis dot structure to determine thenumber and types (single, double, triple) of bondspresent, and any lone pairs present.

*When dealing with several options (resonance structures),determine the dominant resonant structure by:

1. calculating formal charge for the atoms of each molecular possibility that you are evaluating2. using the three criteria for selecting the dominant resonance structure based on formal charge

(2) Determine the 3-D shape of the molecule(3) Determine the overall polarity of the molecule

In addition to the wavelength of energy interaction with the moleculeThe symmetry of the molecule (Lewis dot structure + VSEPR!!!)

will also determine whether a photon can be absorbed.

*Symmetry is NOT a net dipole moment:

• Symmetrical = mirror image

• Asymmetrical = not mirror image

Is HCl symmetrical?

Is CO2 symmetrical?