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CHAPTER 9: MOLECULAR

GEOMETRY AND BONDING

THEORIES

SF6

Prof. Dr. Nizam M. El-Ashgar

MOLECULAR SHAPES

� Lewis do not indicate the shapes of molecules; they

simply show the number and types of bonds between

atoms.

� The overall shape of a molecule is determined by its

bond angles, the angles made by the lines joining the

nuclei of the atoms in the molecule.

PF5 NH4+ HClO4

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Molecular Shapes

Electron pairs, whether they be bonding ornonbonding (valence electrons) repel each other.

By assuming the electron pairs are placed as far aspossible from each other (to minimize repulsions),we can predict the shape of the molecule (3D).

We call this process:(VSEPR) theory

Valence Shell Electron Pair Repulsion Theor

VSEPR THEORY

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ELECTRON DOMAINS

�Electron domains:They referred to theelectron pairs as.

�Each pair of electronscount as an electrondomain, whether theyare in:

� Alone pair.

�A single bond

�A double bond

�A triple bond.

This molecule has

four electron

domains.

• Electron domain geometry: When consideringthe geometry about the central atom, we consider allelectrons (lone pairs and bonding pairs).

• When naming the molecular geometry (shape) wefocus only on the positions of the atoms (nonbondingelectrons are dropped).

• Note:

Electron domain geometry → Molecular geometry

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MOLECULAR SHAPES

ABn molecule

A = the central atom.

B = the bonded atoms

n = the number of bonded atoms.

The shape of any particular ABn can

usually be derived from one of five basic

geometric structures:

THE 5 BASIC GEOMETRIES

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• To determine the electron pair (electron domain)geometry:

•Draw the Lewis structure.

•Count the total number of electron pairs around the central atom.

•Arrange the electron pairs in one of the above geometries tominimize e−- e− repulsion, and count multiple bonds as onebonding pair.

•But then we have to account for the shape of the molecule (basedon the bonded atoms only).

An AB: one electron domain (Two atoms).

The molecule has a linear shape.

Example: HCl, O2, N3

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MOLECULAR SHAPES

1) Two electron domains; 3 atoms (AB2): The

molecule has a linear shape with 180o bond

angles.

Examples:

BeH2 CO2

2) Three electron domains; 4 atoms (AB2): The

molecule has a trigonal planar shape with 120o

bond angles.

x

xx

x

Examples:

BCl3 H2CO

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AB2 Derivative Geometries: (one derivative).

AB2E:

Elect. Dom. Geometry Molecular Shape

Trigonal planar bent; V-shape;

nonlinear; angular

Example: NO2-

3) Four electron domains; 5 atoms (AB4): The

molecule has a tetrahedral shape with 109.5o

bond angles.

Example: Methane

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AB4 Derivative Geometries: (two derivatives).

I) AB3E:

Elect. Dom. Geometry Molecular Shape

Tetrahedral Trigonal Pyramidal

Example: NH3

II) AB2E2:

Elect. Dom. Geometry Molecular Shape

Tetrahedral Bent

Example: H2O

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4) Five electron domains; 6 atoms (AB5): The molecule

has a trigonal bipyramidal shape with 90o and 120o

bond angles.

Example:

PC5

AB5 Derivative Geometries: (Three derivatives).

I) AB4E:

Elect. Dom. Geometry Molecular Shape

Tetrahedral Trigonal Pyramidal

Example: NH3

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AB5 Derivative Geometries: (Three derivatives).

I) AB4E:

There are two different positions on this molecule: axial

and equatorial.

- The axial position: is at a 90o angle to its neighbors.

- The equatorial position: is at 120o or 90o angles to its

neighbors.

AXIAL VS. EQUATORIAL

The equatorial position experiences less repulsion, so

nonbonding pairs will always occupy the equatorial

positions.

For AB3E: The lone pair exists in the equatorial position

X √

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Example: SF4

Elect. Dom. Geom.

Trigonalpipyramidal

Mol. Shape: Seesaw

II) AB3E2:

Example: BrF3

Elect. Dom. Geom. Mol. Shape

Trigonalpipyramidal T-Shape

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III) AB2E3:

Example: XeF2

Elect. Dom. Geom. Mol. Shape

Trigonalpipyramidal Linear

5) Six electron domains; 7 atoms (AB6):

The molecule has an octahedral shape

with 90o all bond angles.

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AB6 Derivative Geometries: (Two derivatives).

I) AB5E:

Example: IF5

Elect. Dom. Geom. Mol. Shape

Octahedral Square pyramidal

II) AB4E2

Elect. Dom. Arr. Mol. Shape

Octahedral Square planar

Example: XeF4

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SAMPLE EXERCISE 9.1

�Use the VSEPR model to predict the

molecular geometry of

a) O3

LS : Three electron domains

AB2E

The Elec. Dom. Geom. is:

Trigonal planar

The molecular shape (geometry) is Bent

b) SnCl3-

LS : Three electron domains

AB3E

The Elec. Dom. Geom. is:

Tetrahedral

The molecular shape (geometry) is Trigonal pyramidal

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PRACTICE EXERCISE

Predict the electron-domain geometry and the

molecular geometry for:

SeCl2

CO32-

SAMPLE EXERCISE 9.2

Use the VSEPR model to predict the molecular geometry of

a. SF4

LS : Five electron domains

AB4E

The Elec. Dom. Geom. is:

Trigonal bipyramidal

The molecular shape (geometry) is Seesaw

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b. IF5

LS : six electron domains

AB5E

The Elec. Dom. Geom. Is:

Octahedral

� The molecular shape (geometry) is Seesaw

BONDING VS. NONBONDING

A bonding pair of electrons: are located between

the nuclei of the two atoms that are bonded

together.

A nonbonding pair (lone pair): of electrons are

located on one atom.

Both pairs represent an electron domain.

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NONBONDING PAIRS AND BOND ANGLE

�Nonbonding pairs are physically

larger than bonding pairs.

�Therefore, their repulsions are

greater; this tends to decrease

bond angles in a molecule.

Therefore, the bond angle decreases as

the number of lone pairs increase.

104.5O107O

NHH

HC

H

HHH

109.5O

OHH

MULTIPLE BONDS AND BOND ANGLES

�Double and triple bonds

place greater electron

density on one side of

the central atom than

do single bonds.

�Therefore, they also

affect bond angles.

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SAMPLE EXERCISE

• Predict the electron-domain geometry and molecular

geometry of:

a. ClF3 Answer: (a) trigonal bipyramidal, T-shaped;

b. ICl4- Answer: (b) octahedral, square planar

SHAPES OF LARGE MOLECULES

�Larger molecules can be broken into

sections to see the shapes involved.

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SAMPLE EXERCISE 9.3

Eyedrops used for dry eyes usually contain the water-soluble

polymer called poly(vinylalcohol), which is based on the

unstable organic molecule called vinyl alcohol:

Predict the approximate values for the H – O – C and the

O – C – C bond angles.

<109.5 o

>120 o

PRACTICE EXERCISE

Predict the H – C – H ad C – C – C bond angles in the

following molecule, called propyne:

Answer: 109.5o , 180o

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MOLECULAR SHAPE AND MOLECULAR POLARITY

Bond polarity: is a measure of how equally the

electrons in a bond are shared between the two

atoms of the bond.

When there is a difference in electronegativity

between two atoms, then the bond between

them is polar.

1) For Diatomic Molecules:

• Same Atoms: non polar (H-H)

• Different Atoms: Polar (H-F)

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2) For Polyatomic Molecules:

It is possible for a molecule to contain polar

bonds, but not be polar.

For example: the bond dipoles in CO2 cancel each

other because CO2 is linear.

The dipole moment depends on:

- Polarities of the individual bonds.

- The geometry of the molecule.

Bond dipoles and dipole moments are vector

quantities – magnitude and direction.

The overall dipole moment is the vector sum of its bond dipoles.

In CO2: The dipoles are of equal magnitude but in opposite

directions, then they cancel out making the molecule nonpolar.

In H2O: the molecule is not linear and the bond dipoles do not

cancel each other (polar molecule).

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�If the dipoles do not cancel out, then the

molecule is polar. Be sure to consider shape

before deciding polarity.

SAMPLE EXERCISE 9.4

Predict whether the following molecules are polar or

nonpolar:

a. BrCl

(polar); µ = 0.57 D

b. SO2

(polar); µ = 1.63 D

c. SF6

(nonpolar); µ = 0 D

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PRACTICE EXERCISE

Determine whether the following molecules

are polar or nonpolar:

a. NF3

Answer: (a) polar because polar bonds are arranged in a

trigonal-pyramidal geometry

b. BCl3Answer: (b) nonpolar because polar bonds are arranged in a

trigonal-planar geometry

COVALENT BONDING AND ORBITAL OVERLAP

In the Lewis theory: covalent bonding occurs when

atoms share electrons. Lewis structures and VSEPR do

not explain why a bond forms.

In the valence-bond theory:

- Bonds form when orbitals on atoms overlap.

- The overlap of the orbitals allows two electrons of

opposite spin to share the common space between the

nuclei, forming a covalent bond.

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BOND LENGTH

The bond length of a covalent bond corresponds

to the minimum of the potential energy curve.

• As two nuclei approach each other their atomic orbitals

overlap.

• As the amount of overlap increases, the energy of the

interaction decreases.

• At some distance the minimum energy is reached.

• The minimum energy corresponds to the bonding distance

(or bond length).

• As the two atoms get closer, their nuclei begin to repel and

the energy increases.

• At the bonding distance: the attractive forces between nuclei

and electrons just balance the repulsive forces (nucleus-

nucleus, electron-electron).

COVALENT BONDING AND ORBITAL OVERLAP

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HYBRID ORBITALS

�To explain geometries: we assume that the

atomic orbitals on an atom mix to form new

orbitals called hybrid orbitals.

�Hybridization: The process of mixing

atomic orbitals.

�The number of hybrid orbitals must equal

the number of atomic orbitals.

SP HYBRID ORBITALS

• BeF2

• Be: Z = 4 The ground state configuration:

• Promotion of an electron gives: An excited

state configuration:

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� Be can now form two bonds with the two F atoms since F

atom has an unpaired electron the p orbital (2p5).

� However, this arrangement would not make the two Be –

F bonds equal because one would involve and s orbital

and one would involve a p orbital.

� We can form 2 hybrid sp orbitals which have 2 lobes like a

p orbital, but 1 lobe is much larger than the other

� These two degenerate orbitals would align themselves

180° from each other ( linear Geometry).

s + p sp

�The hybrid orbitals and bonding in BeF2 is

linear in geometry.

�So BeF2 is Linear

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HYBRID ORBITALS NOTES

�Whenever we mix a certain number of

atomic orbitals, we get the same number of

hybrid orbitals.

�Each of the hybrid orbitals is equivalent to

the others but points in a different

direction.

�Thus mixing one 2s and one 2p orbital

yields two equivalent sp hybrid orbitals

that point in opposite directions.

SP2 HYBRIDIZATION

Using a similar model for Boron:

B: (Z = 5) → 1S22S22p3

s + 2 p → sp2

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�The three degenerate orbitals would align

themselves 120° from each other (trigonal planar)

geometry.

� The hybrid orbitals and bonding in BF3 is trigonal

planar.

SP3 HYBRIDIZATION

Using the model for Carbon:

C: (Z = 6) → 1S22S22p4

s + 3 p → sp3

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The four degenerate orbitals would align themselves

109.5° from each other (tetrahedral) geometry.

� The hybrid orbitals and bonding in CH4 is

tetrahedral.

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HYBRID ORBITAL CHART

Hybridization Involving d Orbitals

• Since there are only three p-orbitals, trigonalbipyramidal and octahedral electron domain geometriesmust involve d-orbitals.

• Trigonal bipyramidal electron domain geometriesrequire sp3d hybridization.

• Octahedral electron domain geometries requiresp3d2 hybridization.

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sp3d Hybridization:

For geometries involving expanded octets on

the central atom, we must use d orbitals in our

hybrids.

sp3d2 Hybridization:

s + 3p + 3d → sp3d2

Octahedral

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1. Draw the Lewis structure.

2. Determine the electron domain geometry with VSEPR.

3. Specify the hybrid orbitals required for the electron pairs

based on the electron domain geometry.

HYBRIDIZATION SUMMARY

5 pairs = sp3d2

Electron pairs: electron domains (bonding and nonbonding)

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SAMPLE EXERCISE 9.5

� Indicate the hybridization of orbitals employed by

the central atom in NH2-.

The Lewis structure of NH2– is:

o There are four electron domains around N.

o The electron-domain geometry is tetrahedral.

o The hybridization that gives a tetrahedral electron-

domain geometry is sp3

PRACTICE EXERCISE

�Predict the electron-domain geometry and the

hybridization of the central atom in

(a) SO32-

Answer: (a) tetrahedral, sp3

(b) SF6

b) octahedral, sp3d2

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Examples – Determine the hybridization on

the central atom of each:

1. NCl3 sp3

2. CO2 sp

3. H2O sp3

4. SF4 sp3d

5. BF3 sp2

6. XeF4 sp3d2

Examples – Determine the hybridization on EACH

atom:

O

C C C HH

sp2 sp

C C N H

HH

O

H

H

sp3

sp2

sp3

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• σ-Bonds: electron density lies on the axis

between the nuclei.

All single bonds are σ-bonds.

• π-Bonds: electron density lies above and below

the plane of the nuclei.

• A double bond consists of one σ-bond and one π-

bond.

• A triple bond has one σ-bond and two π-bonds.

• Often, the p-orbitals involved in π-bonding come

from unhybridized orbitals.

MULTIPLE BONDS

SIGMA (σ) BONDS

�Sigma bonds are characterized by

� Head-to-head overlap.

� Cylindrical symmetry of electron density about

the internuclear axis.

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MULTIPLE BONDS

�A pi (π):

characterized by

�Side-to-side overlap:

Overlap of the orbitals is

perpendicular to the

internuclear axis.

�Electron density above and

below the internuclear axis.

PI VS. SIGMA

� Unlike a sigma bond, in a pi bond there is no probability of

finding the electron on the internuclear axis.

� Since the p orbitals in a pi bond overlap sideways rather

than directly facing each other, the total overlap in a pi

bond tends to be less than in a sigma bond.

� This means that pi bonds tend to be weaker than sigma

bonds.

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TYPE OF BOND

� Single bonds are sigma bonds.

� A double bond consists of 1 sigma bond and 1 pi bond.

TYPE OF BOND

�A triple bond consists of 1 sigma bond and 2 pibonds.

�Since pi bonds occur using unhybridized porbitals, it can only occur with sp and sp2

hybridization.

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SAMPLE EXERCISE 9.6

� Formaldehyde has the Lewis structure:

Describe how the bonds in formaldehyde are formed in terms of

overlaps of appropriate hybridized and unhybridized orbitals.

Explanation:

Structure: Trigonalplanar geometry with bond angles of

about 120. It is sp2 hybrid orbitals on C .These hybrids are used to make the two C—H and one C—O σ bonds to C. There remains an unhybridized 2p orbitalon carbon, perpendicular to the plane of the three sp2

hybrids.The O atom also has three electron domains around it : sp2

hybridization as well.One of these hybrids participates in the C—O σ bond, whilethe other two hybrids hold the two nonbonding electronpairs of the O atom. The remain unhybridized 2p orbitalform π bond with the unhybridized 2p orbital with carbon

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PRACTICE EXERCISE

� Consider the acetonitrile molecule:

a) Predict the bond angles around each carbon atom.

Answer: (a) approximately 109around the left C and 180on the right

C

b. Describe the hybridization at each carbon atom.

Answer: (b) sp3, sp

c. Determine the total number of σ and π bonds in the

molecule.

Answer: ( (c) five σbonds and two bonds

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DELOCALIZED ELECTRONS: RESONANCE

When writing Lewis structures for species like the nitrate

ion, we draw resonance structures to more accurately

reflect the structure of the molecule or ion.

LOCALIZED AND DELOCALIZED ELECTRONS

�Localized electrons:

belong to the two

atoms that form the

bond.

�When resonance

structures exist for a

molecule, the

electrons move

around and are called

delocalized.

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DELOCALIZED Π BONDING IN BENZENE

PRACTICE EXERCISE

Which of the following molecules or ions will exhibit

delocalized bonding: SO3, SO32-, H2CO, O3, NH4

+?

Answer: SO3 and O3, as indicated by the presence of two or more

resonance structures involving bonding for each of these molecules

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SAMPLE INTEGRATIVE EXERCISE

� Elemental sulfur is a yellow solid that consists of S8

molecules. The structure of the S8 molecule is an eight-

membered ring. Heating elemental sulfur to high

temperatures produces gaseous S2 molecules:

S8(s) � 4S2(g)

a) With respect to electronic structure, which element in

the second row of the periodic table is most similar to

sulfur?

Sulfur: [Ne]3s23p4 electron configuration.

It is expected to be most similar electronically to oxygen

(electron configuration, [He]2s22p4).

b. Use the VSEPR model to predict the S – S – S bond

angles in S8, and the hybridization at S in S8.

d. Use average bond enthalpies (Table 8.4) to estimate the

enthalpy change for the reaction just described. Is the

reaction exothermic or endothermic?

The reaction is endothermic

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HW EXERCISES:

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44 46 51 55 56 61 66 69

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