The Shapes of Molecules

Chapter 10 The Shapes of Molecules The Shapes of Molecules 10.1 Depicting Molecules and Ions with Lewis Structures 10.2 Valence-Shell Electron-Pai...
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Chapter 10

The Shapes of Molecules

The Shapes of Molecules

10.1 Depicting Molecules and Ions with Lewis Structures

10.2 Valence-Shell Electron-Pair Repulsion (VSEPR) Theory and Molecular Shape

10.3 Molecular Shape and Molecular Polarity

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Lewis Electron - Dot Symbols For main group elements The A group number gives the number of valence electrons. Place one dot per valence electron on each of the four sides of the element symbol. Pair the dots (electrons) until all of the valence electrons are used. Example:

Nitrogen, N, is in Group 5A and therefore has 5 valence electrons.

. N. .

. . N. :

. . N: .

:

. :N . .

The steps in converting a molecular formula into a Lewis structure.

Molecular formula

Step 1 Atom placement

Lewis structure – The nonmetal atoms can acquire stable noblegas structure by sharing electrons to form an electron-pair bond.

Place atom with lowest EN in center Step 2

Add A-group numbers

Sum of valence e-

Step 3 Remaining valence e-

Octet rule – the nonmetals achieve a noble-gas structure by sharing eight electrons. This generalization is referred to as Octet rule.

Draw single bonds. Subtract 2efor each bond.

Step 4

Give each atom 8e(2e for H)

Lewis structure

Duet rule – Hydrogen atoms are surrounded by a duet of electrons in molecules.

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Molecular formula

For NF3

:

: F:

: F:

:

Sum of valence e-

:

Atom placement

N

: F:

5e-

F 7e-

X 3 = 21e-

Total

26 e-

:

Remaining valence e-

N

Lewis structure Single bond – a single electron pair is shared between two bonded atoms. Double bond – two electron pairs are shared between two bonded atoms. Triple bond – three electron pairs are shared between two bonded atoms.

SAMPLE PROBLEM Writing Lewis Structures for Molecules with One Central Atom PROBLEM:

Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone.

PLAN: Follow the steps outlined. SOLUTION:

F

:

: Cl : F:

:

:

: Cl C : F: :

Make bonds and fill in remaining valence electrons placing 8e- around each atom.

F

:

Steps 2-4: C has 4 valence e-, Cl and F each have 7. The sum is 4 + 4(7) = 32 valence e-.

Cl C

:

Step 1: Carbon has the lowest EN and is the central atom. The other atoms are placed around it.

Cl

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SAMPLE PROBLEM

Writing Lewis Structure for Molecules with More than One Central Atom PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that is being used as a gasoline alternative in car engines. SOLUTION:

Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds. There are 4(1) + 4 + 6 = 14 valence e-. C has 4 bonds and O has 2. O has 2 pair of nonbonding e-.

:

H C

O

H

:

H

H

Resonance: Delocalized Electron-Pair Bonding

O3 can be drawn in 2 ways -

O O

O O

O

O

Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure. B

B O

O O

O

A

C

O

O O

O

O

A

C

Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs. is used to indicate that resonance occurs.

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Resonance (continued)

Three criteria for choosing the more important resonance structure:

Smaller formal charges (either positive or negative) are preferable to larger charges.

Avoid like charges (+ + or - - ) on adjacent atoms. A more negative formal charge should exist on an atom with a larger EN value.

SAMPLE PROBLEM PROBLEM: PLAN:

Write resonance structures for the nitrate ion, NO3-.

After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms.

SOLUTION:

O

Writing Resonance Structures

Nitrate has 1(5) + 3(6) + 1 = 24 valence e-

O

O

O

N

N

N

O

O

O

O

O

N does not have an octet; a pair of e- will move in to form a double bond.

O

O

O

O

N

N

N

O

O

O

O

O

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Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge is the charge an atom would have if the bonding electrons were shared equally. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons) B

For OC

O

For OA # valence e- = 6

O

# nonbonding e- = 4

A

# bonding e- = 4 × 1/2 = 2

# valence e- = 6

O

# nonbonding e- = 6

C

For OB

# bonding e- = 2 X 1/2 = 1

# valence e- = 6

Formal charge = -1

Formal charge = 0 # nonbonding

e-

=2

# bonding e- = 6 X 1/2 = 3 Formal charge = +1

SAMPLE PROBLEM Writing Lewis Structures for Octet Rule Exceptions PROBLEM:

PLAN:

Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2.

Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. Note that (a) contains P which is a Period-3 element and can have an expanded valence shell.

SOLUTION:

(a) H3PO4 has two resonance forms and formal charges indicate the more important form.

-1 0

O

0 H O

P

O 0 H

0

+1 O H 0

0

0 0 H O

0 O

0

P

O H 0

0 0 O H more stable 0 lower formal charges

(b) BFCl2 will have only 1 Lewis structure. F B Cl

Cl

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The orientation of polar molecules in an electric field. Negative end of HF

Polarity –is called bond polarity or chemical polarity. Polarity describes how equally bonding electrons are shared between atoms. Polar – is a result of an unsymmetrical electron distribution. (HF, H2O) Dipole – is the bond or molecule which contains positive or negative pole.

Positive pole of the electric field

Electric field ON

Electric field OFF

Nonpolar – is a symmetrical distribution of electrons which leads to the a bond or molecule without positive or negative pole.(CO2, BF3)

If the electric field is off, the polar molecule, HF, orientated randomly. If the electric field is on, the polar molecule, HF, aligns their positive ends toward negative poles of the electric field.

SAMPLE PROBLEM Predicting the Polarity of Molecules PROBLEM:

From electronegativity (EN) values and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: (a) Ammonia, NH3

(b) Boron trifluoride, BF3

PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity.

SOLUTION:

The dipoles reinforce each other, so the overall molecule is definitely polar.

(a) NH3

ENN = 3.0 H

ENH = 2.1

N H

H

H

N H

H

bond dipoles

H

N H

H

molecular dipole

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SAMPLE PROBLEM Predicting the Polarity of Molecules

continued

(b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar.

F B F

F

1200

F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar.

The Central Themes of VB Theory Basic Principle A covalent bond forms when the orbitals of two atoms overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons. The two wave functions are in phase so the amplitude increases between the nuclei.

Themes A set of overlapping orbitals has a maximum of two electrons that must have opposite spins. The greater the orbital overlap, the stronger (more stable) the bond. The valence atomic orbitals in a molecule are different from those in isolated atoms. There is a hybridization of atomic orbitals to form molecular orbitals.

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Hybrid Orbitals Key Points The number of hybrid orbitals obtained equals the number of atomic orbitals mixed. The type of hybrid orbitals obtained varies with the types of atomic orbitals mixed.

Types of Hybrid Orbitals sp

sp2

sp3

sp3d

sp3d2

VSEPR VSEPR--Valence ValenceShell ShellElectron ElectronPair PairRepulsion RepulsionTheory Theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to maximize repulsions. These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-group arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-groups are defining the object arrangement,but the molecular shape is defined by the relative positions of the atomic nuclei. Because valence electrons can be bonding or nonbonding, the same electron-group arrangement can give rise to different molecular shapes. A - central atom

X -surrounding atom

E -nonbonding valence electron-group

AXmEn integers

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Figure 10.2

Electron-group repulsions and the five basic molecular shapes.

linear

tetrahedral

trigonal planar

trigonal bipyramidal

octahedral

The sp hybrid orbitals in gaseous BeCl2.

atomic orbitals hybrid orbitals

orbital box diagrams

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The sp2 hybrid orbitals in BF3.

The sp3 hybrid orbitals in CH4.

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The sp3 hybrid orbitals in NH3.

The sp3 hybrid orbitals in H2O.

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The sp3d hybrid orbitals in PCl5.

The sp3d2 hybrid orbitals in SF6.

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SAMPLE PROBLEM PROBLEM:

Postulating Hybrid Orbitals in a Molecule

Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following: (a) Methanol, CH3OH

PLAN:

(b) Sulfur tetrafluoride, SF4

Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid orbitals. Use partial orbital box diagrams to indicate the hybrid for the central atoms.

SOLUTION:

H

(a) CH3OH

The groups around C are arranged as a tetrahedron.

C O H H H

SAMPLE PROBLEM

O also has a tetrahedral arrangement with 2 nonbonding epairs.

Postulating Hybrid Orbitals in a Molecule

continued

2p

2s

2p

sp3 single C atom

hybridized C atom

2s

sp3 hybridized O atom

single O atom

(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs. F F S F F

3d

3d

3p sp 3d 3s

S atom

hybridized S atom

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Lewis electron - dot symbols for elements in Periods 2 and 3.

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