Visualizing Molecules - VSEPR, Valence Bond Theory, and Molecular Orbital Theory

Home , Formal charge, Octet rule, Valence bond theory

Nestor S. Valera

Visualizing Molecules

1. Lewis Structures and Valence-Shell Electron-Pair Repulsion Theory

2. Valence Bond Theory and Hybridization

3. Molecular Orbital Theory

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

Place atom Molecular Step 1 with lowest formula EN in center Atom Add A-group Step 2 placement numbers

Sum of Draw single bonds. Step 3 valence e- Subtract 2e- for each bond. Give each Remaining Step 4 atom 8e- valence e- (2e- for H)

Lewis structure

Molecular formula For NF3

Atom placement : : N 5e- : F: : F: : Sum of N F 7e- X 3 = 21e- valence e- - : F: Total 26e Remaining : valence e-

Lewis structure

SAMPLE PROBLEM 10.1 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 in Figure 10.1 .

SOLUTION: Cl Step 1: Carbon has the lowest EN and is the central atom. Cl C F The other atoms are placed around it. F Steps 2-4: C has 4 valence e-, Cl and F each have 7. The : : Cl : sum is 4 + 4(7) = 32 valence e-. : : :Cl C F : Make bonds and fill in remaining valence : : electrons placing 8e- around each atom. : F : :

SAMPLE PROBLEM 10.2 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 : H C O H : H

SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with Multiple Bonds. PROBLEM: Write Lewis structures for the following:

(a) Ethylene (C2H4), the most important reactant in the manufacture of polymers

(b) Nitrogen (N2), the most abundant atmospheric gas PLAN: For molecules with multiple bonds, there is a Step 5 which follows the other steps in Lewis structure construction. If a central atom does not have 8e-, an octet, then two e- (either single or nonbonded pair)can be moved in to form a multiple bond. SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom. H : H H H C C C C H H H H

- (b) N2 has 2(5) = 10 valence e . Therefore a triple bond is required to make

the octet around each N.

. . : : : : :

Here are some helpful generalizations used in constructing Lewis structures

• H atoms form one bond • Carbon atoms form four bonds • Nitrogen atoms form three bonds • Oxygen atoms form two bonds • Halogen atoms form one bond when they are “surrounding” atoms; but more than one bond when they are “central” atoms. Fluorine is always a “surrounding” atom. • The “central” atom is usually large and of low EN • The “octet” rule may be violated by almost all atoms except those that belong to the second period.

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

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.

(a) Ethylene (C2H4), the most important reactant in the manufacture of polymers

(b) Nitrogen (N2), the most abundant atmospheric gas

- SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e . H can have only one bond per atom.

H H H H

C : C C C H H H H

- (b) N2 has 2(5) = 10 valence e . Therefore a triple bond is required to make

the octet around each N.

. . : : : : :

It is possible to have more than one satisfactory Lewis structure.

O can be drawn in 2 ways - 3 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. Resonance structures are NOT capable of independent existence. B B O O O O O O O O O

A C 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. 13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Can you see the “resonance” or “canonical” figures?

- PROBLEM: Write resonance structures for the nitrate ion, NO3 .

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

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

O O O

Some resonance structures are more “important” or dominant than others.

16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The formal charge helps us selecting the best resonance structure. Formal charge is the charge an atom would have if the bonding electrons were shared equally. An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons) B

For OC For OA O # valence e- = 6 # valence e- = 6 O O # nonbonding e- = 6 # nonbonding e- = 4 A C For OB # bonding e- = 2 X 1/2 = 1 # bonding e- = 4 X 1/2 = 2 - # valence e = 6 Formal charge = -1 Formal charge = 0 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1 17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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.

EXAMPLE: NCO- has 3 possible resonance forms -

N C O N C O N C O

A B C formal charges -2 0 +1 -1 0 0 0 0 -1

N C O N C O N C O

Forms B and C have negative formal charges on N and O; this makes them more preferred than form A.

Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.

PROBLEM: Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2. Draw the Lewis structures for the molecule and determine STRATEGY: 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 +1 0 (b) BFCl will have only 1 0 0 O 2 O Lewis structure. 0 H O P O H 0 0 H O P O H 0 O 0 O 0 F 0 0 H H more stable B Cl Cl 0 0 lower formal charges 20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The VSEPR - Valence Shell Electron Pair Repulsion Theory is a good 3-D visualization tool for ions and molecules.

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

The positions of the electron-groups define the object arrangement or electron- group geometry, but the relative positions of the atomic nuclei define the molecular shape or molecular geometry.

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 atoms E -nonbonding valence electron-group

AXmEn

linear trigonal planar tetrahedral

trigonal bipyramidal octahedral

Multiple bonds and lone pairs affect actual bond angles.

Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. 1200 1220 Effect of Double Bonds H larger EN H 1200 C O 1160 C O ideal H greater H electron real density Effect of Nonbonding(Lone) Pairs

Lone pairs repel bonding pairs more strongly than bonding pairs Sn repel each other. Cl Cl

Draw the molecular shape and predict the bond

angles (relative to the ideal bond angles) of PF3

SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair

The shape is based upon the tetrahedral arrangement. F P F The F-P-F bond angles should be <109.50 due F to the repulsion of the nonbonding electron P F pair. F F The final shape is trigonal pyramidal. <109.50

The type of shape is

AX3E

(b) For COCl2, C has the lowest EN and will be the center atom.

There are 24 valence e-, 3 atoms attached to the center atom.

Cl C O C does not have an octet; a pair of nonbonding electrons will move in from the O to make a Cl double bond. Type AX The shape for an atom with three atom 3 O attachments and no nonbonding pairs on the central atom is trigonal planar. C O 0 Cl Cl The Cl-C-Cl bond angle will 124.5 C be less than 1200 due to Cl the electron density of the Cl C=O. 1110 26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of

(a) SbF5 and (b) BrF5.

- SOLUTION: (a) SbF5 - 40 valence e ; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal. F F F F Sb F Sb F F F F F

- (b) BrF5 - 42 valence e ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

F F F Br F F 27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. To predict molecular shapes when there is more than one central atom, we simply apply VSEPR principles several times.

PROBLEM: Determine the shape around each of the central atoms in

acetone, (CH3)2C=O. PLAN: Find the shape of one atom at a time after writing the Lewis structure.

SOLUTION: tetrahedral O tetrahedral H H H C C C H H H trigonal planar

O >1200 H C H H C C H HH

Ethane and ethanol have multiple tetrahedral centers.

Valence Bond Theory strengthens VSEPR principles by invoking directional properties of the orbitals.

Basic Principles:

• A covalent bond forms when the orbitals of two atoms overlap • The overlap region, which is between the nuclei, is occupied by a pair of electrons spinning in opposite directions. Mathematically, this means that the the two wave functions that describe the two orbitals are in phase so the amplitude increases between the nuclei.

Hydrogen, H2

Hydrogen fluoride, HF

“Hybrid” orbitals arise from the “mixing” of appropriate atomic 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. The use of hybrid orbitals rationalizes the equivalence of bonds. Types of Hybrid Orbitals

sp sp2 sp3 sp3d sp3d2

Gaseous BeCl2 is a linear molecule with two equivalent Be - Cl bonds

atomic orbitals

hybrid orbitals

The sp hybrid orbitals in gaseous BeCl2(continued).

orbital box diagrams with orbital contours

BF3 is a triangular molecule.

CH4 is a tetrahedral molecule.

The electron pair geometry of NH3 is tetrahedral but the molecule is pyramidal.

3 The sp hybrid orbitals in H2O.

3 The sp d hybrid orbitals in PCl5.

3 2 The sp d hybrid orbitals in SF6.

! (or “sigma”) bonds are formed from the direct or “head on” overlap of orbitals.

The ! (or “sigma”) bonds in ethane (C2H6) are formed from sp3 hybrid orbitals. both C are sp3 hybridized s-sp3 overlaps to ! bonds

sp3-sp3 overlap to form a ! bond relatively even distribution of electron density over all ! bonds

" (or “pi”) bonds are formed from the lateral or “side-ways” overlap of orbitals.

2 The ! (or “sigma”) bonds in ethylene (C2H4) are formed from sp hybrid orbitals while the unhybridized p orbitals of carbon give rise to the # bonds. overlap in one position - !

p overlap - #

The ! (or “sigma”) bonds in acetylene (C2H2) are formed from sp hybrid orbitals while the unhybridized p orbitals of carbon give rise to the # bonds. overlap in one position - !

p overlap - #

• MO’s are formed from all of the valence orbitals. • n AO’s combine to form n MO’s. • For MO’s formed from a given set of contributing AO’s, the greater the numbe of interatomic nodes, the higher the energy. • Like AO’s, MO’s obey Pauli’s exclusion principle and Hund’s rules. • Unlike AO’s, MO’s encompass the whole molecule.

• The ! label means that MO is cyclindrically symmetric about the internuclear axis. • The # label means that MO undergoes a phase change upon rotation about the internuclear axix by 180o. • The $ label means that MO undergoes a phase change upon rotation by the internuclear axix by 90o. • The * label means that there is a nodal plane between the nuclei, and this plane is orthogonal to the internuclear axis. • Bonding MO’s have lower energies than their parent AO’s • Antibonding MO’s have higher energies than their parent AO’s

MOT - Homonuclear Diatomic Molecules

Molecular orbitals are formed from the linear combination of atomic orbitals

The energies of the 2s and 2p atomic orbitals decrease going over a period

Bond distance and bond dissociation energy vary with bond order

MOT - Heteronuclear Diatomic Molecules

MO’s derived from s orbitals in a heteronuclear diatomic molecule

The MO energy level of HF show that electron density is drawn towards the more electronegative ﬂuorine

A more rigorous MO diagram for CO