Molecular Structure and Covalent Bonding Theories Chapter 8.

Molecular Structure and Covalent Bonding Theories

Chapter 8

The Valence Shell Electrons

• Valence shell electrons– These electrons are largely responsible for____– Electrons not present in the preceding ___ ___

• Ignore filled sets of d and f orbitals

– Used to determine the Lewis structure of a compound containing covalent bonds

• Works well for molecules containing atoms from the ____ ____ elements

Models to Describe Covalent Bonding

• Valence shell electron pair repulsion (VSEPR) model – predicts the _____ ______ of atoms in a molecule– This will be related to a physical property

called ______

• Valence bond theory – predicts how bonding will take place by ______ of atomic orbitals

VSEPR Theory• Valence shell electron are present as either ___ ___ or

____ ____. – Regions of high electron density are created.

– These regions arrange themselves to be as far away as possible form on another. As a result specific geometries are created around atoms in the molecule

• Single, double, and triple bonds are counted as one region of electron density

• Unshared pairs of valence electrons are also counted as one region of electron density

Drawing the Lewis structure accurately will reveal the number of electron density regions around the center atoms

VSEPR Theory

• Draw the Lewis structures for CO2, H2CO, and CH4

– How will these regions of electron density arrange themselves to be as far away as possible from one another?

• There are five basic shapes based on the number of electron density regions around a center atom(s)– Illustration of models with next few slides

VSEPR Theory

Two regions of high electron density

VSEPR TheoryThree regions of high electron density

VSEPR TheoryFour regions of high electron density

VSEPR TheoryFive regions of high electron density

VSEPR TheorySix regions of high electron density

VSEPR Theory

• Encountered geometries– Electronic geometry – determined by the

location of “___” the regions of electron density around the center atom(s)

– Molecular geometry – determined by the arrangement of _____ only around the center atom(s)

• The does not include lone electron pairs. The molecular shape differs from the electron shape if lone pairs are present

An example is H2O

VSEPR Theory• Lone pairs of electrons occupy more space

than bonding pairs. As a consequence, there is an order of the magnitude of repulsions – lp/lp > lp/bp > bp/bpAs a result, the bond angles around a center

atoms can be distorted (reduced) from the predicted values

CH4 and H2O What are the H-C-H and H-O-H bond angles. If a change is observed, why?

Molecular Geometry and Polarity

• The polarity can be determine once the geometry is known

• A polar bond is created if the atoms sharing the electron pair have different electronegativities– HCl and the associated dipole moment. This molecule

is polar. For diatomics, determination of polarity is easy. What if the molecule has two or more atoms? All the dipole have to be summed. If the sum equals zero, the molecule has no dipole.


• A dipole moment (bond dipole) has _____ and _____. Both must be considered when determines if a molecule is polar.– CO2 and H2O. Do these molecules have net dipoles?

• Conditions for polarity– There must be at least one polar bond or lone pair on a

central atom– The bond dipoles must not cancel or if there are two or

more lone pairs on the central atom, they must not be arranged so that their polarities cancel

CO2, H2O, and O3

Valence Bond(VB) Theory

• VB theory describes how bonding occurs• Describes how the atomic orbitals overlap to

produce the bonding geometry predicted by VSEPR– Go back and review atomic orbitals if necessary

• Electrons are arranged in atomic orbitals according to energy. The set of atomic orbitals, however, may not be of lowest possible energy upon bonding covalently to neighboring atoms.


• The valence shell orbitals (atomic orbitals) commonly combine to change their character in order to obtain a lower energy ‘mixed’ orbital set for bonding in a particular geometry– Which atomic orbitals would participate in bonding in H,

O, and C? These atomic orbitals can form a new set of hybrid orbitals upon bonding.

• Hybrization – process by which ____ ____ combine to form a set of ‘mixed’ orbitals of lower energy when bonding covalently– The ‘mixed’ orbitals are called hybrid orbitals


• Hybrid orbitals on a center atom align themselves with the bonding orbitals on the neighboring atoms– A ‘good overlap’ is necessary for sharing electrons in a

bond.

• Table 8-2 (refer to it)– The label given to a set of hybridized orbitals reflects

the number and type of atomic orbitals used to produce the set.

• Indicates the electronic geometry in agreement with VSEPR

Molecular Shapes and Bonding

• Simples structures will be analyzed based on geometry type.

• Experimentally determined findings will be discussed in light of these models.

• Terminology– A – central atom– B – atoms bonded to A– U – lone pairs of electrons around A

AB3U represents three atoms bonded to a central atom with one lone pair. An example would be NH3

Molecular Shapes and Bonding

Discussion sequence

• Experimental facts and Lewis formula

• VSEPR– Electronic geometry– Molecular geometry– Polarity

• Valence bond theory

AB2 Molecules - No Lone Pairs

on A - Linear Molecules• The BeCl2 molecule is linear and has melting point

of 405C.– Draw BeCl2 and discuss electronic geometry

• Does the molecular geometry differ?

– The molecule does not satisfy the octet rule

– The compound bonds covalently due to the high charge density on Be2+

• The electron cloud on the halide is distorted by the high charge density

– BeBr2 and BeI2 also have linear geometries


on A - Linear Molecules• The molecule possesses

two polar bonds (Be-Cl) EN = 1.5

• The molecule, however, has no net dipole because the two bond dipoles are equal but in opposite directions.

Cl-Be-Cl ::

::

::

Bond dipoles cancel. This is a nonpolar molecule.


on A - Linear Molecules• Electronic Structures Lewis Formulas

1s 2s 2p

Be 3s 3p

Cl [Ne] ··

Be ··

Cl··

··

.

The 2s orbital is full indicating that it will not bond. How will the Be atom make these electrons available for bonding? What happens in this molecule? Experimental data indicates that the Be-Cl bonds are identical.


on A - Linear Molecules• Valence Bond Theory (Hybridization)

1s 2s 2p 1s sp hyb 2p

Be 3s 3p

Cl [Ne] The two atomic orbitals on Be hybridize to produce two sp

hybrid orbitals that have properties between the s and p atomic orbitals. Notice that chlorine has a half-filled 3p orbital that can overlap with the sp hybrid orbitals of Be.


on A - Linear Molecules

Two regions of electron density around the central atom

Illustrate how the sp orbitals overlap with the 3p orbitals on Cl


on A - Trigonal Planar Molecules• Group IIIA elements that form covalent

compounds by bonding to three other atoms– Octet rule is not satisfied but no big deal

• Boron trichloride is a trigonal molecule with a melting point of -107C– Does the molecular and electronic geometry

differ?– The data indicates that this molecule is nonpolar

(no net dipole).


on A - Trigonal Planar Molecules• Lewis structure predicts

trigonal planar geometry• There are three bond dipoles

of equal length but different direction. – The bond dipoles cancel each

other

The molecule has no net dipole

How about BCl2H?

BCl

Cl

Cl


on A - Trigonal Planar Molecules• Electronic Structures Lewis Formulas

1s 2s 2p

B 3s 3p

Cl [Ne] Suppose that an electron in the 2s atomic orbital is promoted

to an empty 2p atomic orbital allowing for 3 unfilled atomic orbitals for bonding. This would produce, however, unequal energies for the three B-Cl bonds.

B:

:

.

Cl:

:.


on A - Trigonal Planar Molecules• Valence Bond Theory (Hybridization)

1s 2s 2p 1s sp2 hybridB

3s 3pCl [Ne]

The 2s and 2p atomic orbitals on B hybridize to produce three sp orbitals (sp2 hybrid). Notice that chlorine has a half-filled 3p orbital that can overlap with the sp2 hybrid orbitals of Be.


on A - Trigonal Planar Molecules

Three regions of electron density around the central atom

Illustrate bonding with the Cl atoms on the hybridized B


on A - Tetrahedral Molecules• Group IVA elements that form covalent

compounds by bonding to four other atoms– Four electrons are shared and the octet rule is generally

satisfied

• CH4, methane, possesses a tetrahedral geometry and has a melting point of -182C– Would the molecular and electronic geometry differ?

– The data indicate that the molecule is nonpolar.


on A - Tetrahedral Molecules• Lewis structure predicts

tetrahedral geometry

• There are four small bond dipoles which cancel– The molecule is nonpolar

• What about CCl3H and CH3Cl?– When the symmetry lowers, the

molecule becomes polar.

• Other molecules?

CH

H

HH

CH4


on A - Tetrahedral Molecules

• Electronic Structures Lewis Formulas

2s 2p

C [He] 1s

H Suppose that an electron in the 2s atomic orbital is promoted

to an empty 2p atomic orbital allowing for 4 unfilled atomic orbitals for bonding. This would produce, however, unequal energies for the four C-H bonds.

C:..

H .


on A - Tetrahedral Molecules • Valence Bond

2s 2p four sp3 hybrid orbitals C [He] C [He] 1s H The 2s and 2p atomic orbitals on C hybridize to produce four sp

orbitals (sp3 hybrid). Notice that hyrogen has a half-filled 1s orbital that can overlap with the sp3 hybrid orbitals of C.

Many AB4 type molecules have this hybridization

NH4+ is an AB4 type polyatomic ion


on A - Tetrahedral Molecules

Four regions of electron density around the central atom

Illustrate how the hydrogen atomic orbitals bond to the hybridized carbon sp3 orbitals

Alkanes CnH2n+2• alkanes are saturated hydrocarbons• have the general formula CnH2n+2.

CH4 - methane

C2H6 or (H3C-CH3) - ethane

C3H8 or (H3C-CH2-CH3) - propane

• C atoms are located at the center of a tetrahedroneach alkane is a chain of interlocking tetrahedraC atom at the center of each tetrahedronenough H to form a total of four bonds for each C

AB3U Molecules - One Lone Pair

- Pyramidal Molecules• Group VA elements (e.g. N) have five electrons in

the valence and commonly bond to three atoms leaving a lone pair.– The octet rule is satisfied

• The most common molecule is NH3.– How many regions of electron density around nitrogen?

The bong angle is in this molecule is ~107. Why?

• Other common molecules are NF3, PF3, and the polyatomic ion SO3

2-.


- Pyramidal Molecules• The Lewis structure predicts

tetrahedral electronic geometry.– Is the molecular geometry

different?

• There are three bond dipoles? Detail.– Is the molecule polar?– How about NF3? How do the

polarities of the two molecules compare (later)?

NH

HH

:


- Pyramidal MoleculesElectronic Structures Lewis Formulas

2s 2pN [He]

2s 2pF [He] 1s H There are three half-filled atomic orbitals on the nitrogen

(2p). The data suggests, however, that there are four nearly equivalent orbitals (not three). Three orbitals are for bonding and one for a lone pair.

AB3U Molecules - One Lone Pair - Pyramidal Molecules

• Valence Bond

2s 2p four sp3 hybrids

N [He]

The 2s and 2p atomic orbitals hybridize to form four sp3 hybrid orbitals. This hybridization is also necessary to produce the correct geometry for bonding.

Illustrate bonding with hydrogen 1s atomic orbital.

Once again there are four regions of electron density around the center atom


• Let’s compare NH3 with NF3.• The geometry of the both molecules is

already known– Electronic geometry is _________– Molecular geometry is _________

• How does the lone pair influence polarity? It’s contribution has to be included to determine polarity of a molecule.


• The bond dipoles go opposite directions on NH3 and NF3

– For NH3, the net dipole is enhanced by the lone pair.

– For NF3, the net dipole is decreased due to the lone pair

• Additionally, the H-N-H angle is greater than the F-N-F angle due to closer approach of the lone pair to nitrogen on NF3

NH

HH

:

N

FFF

:

H-N-H = 107.3

F-N-F = 102.1

AB2U2 - Two Lone Pairs - V-Shaped Molecules

• Group VIA elements (e.g. O) have six electrons in the valence and commonly bond to two atoms leaving two lone pairs.– The octet rule is generally satisfied

• H2O is the most common molecule of this type. – The molecular geometry is ______ and the electronic

geometry is _____

– Other examples of this type of molecule is H2S and OCl2


• The Lewis structure predicts that the molecule is bent in agreement with experimental data.– The actual H-O-H bond angle is

104.5 due to repulsions from two lone pairs

• There are two bond dipoles (O-H). Additionally, the net dipole is enhanced by the lone pairs.– Illustrate

OHH

:

:



2s 2p

O [He] 1s

H There are two half-filled atomic orbitals on the nitrogen (2p).

The data suggests, however, that there are four nearly equivalent orbitals (not two). Two orbitals are for bonding and two for lone pairs.

O:

:

..

H .


• Valence Bond 2s 2p four sp3 hybridsO [He]

The hybrid orbitals that are full belong to the lone pairs. The half-filled orbitals are used for bonding.

Trigonal Bipyramidal Electronic Geometry

• AB5, AB4U, AB3U2, and AB2U3

• Hybridization is sp3d.• The lone pairs (if present) will arrange

themselves to minimize repulsive forces.– lp/lp >> lp/bp > bp/bp

• This geometry is common for P, As, and Sb.– All five valence electrons are shared (PF5)

Trigonal Bipyramidal Electronic Geometry, AB5

• The VSEPR theory predicts trigonal bipyramidal for the electronic and molecular geometry.

• There are three equatorial atoms and to axial atoms. What are the bond angles? Are the individual bonds polar? Is the molecule (type AB5) polar? – Show molecule with this geometry.

AsF

F

F

F

F··

····

··

····

··

··

··

··

····

····

··

trigonal bipyramid



3s 3p

P [Ne] 2s 2p

F [He]

P··

...

F···· .··

The 3d subshell is empty and participates in the rehybridization (sp3d).


• Hybridization involves one d orbital form the empty 3d subshell and the 3s and 3d orbitals.– Illustrate from page 332

– Can also occur for n=4, 5, and 6

• There are no unshared pairs.– 5 covalent bonds

This type of hybridization does not occur for N. Why?

Trigonal Bipyramidal Electronic Geometry with Lone Pairs

• AB4U, AB3U2, and AB2U3

• Go through the procedure for the molecule, SF4.– Where is the preferred

location of the lone pair?• lp/lp>>lp/bp>bp/bp

– This molecular geometry is termed as ______.

Is the molecule polar?

The Molecular Geometry, AB4U



• The ClF3 molecule– Where are the likely

locations for the lone pairs?

• The molecular geometry is termed as _______.

What is the electronic geometry?

The Molecular Geometry, AB3U2



• The I3- species

– Where are the likely locations for the lone pairs?

• The molecular geometry is termed as _______.

The Molecular Geometry, AB2U3.

Octahedral Electronic Geometry

• AB6, AB5U, and AB4U2

• Hybridization is sp3d2.

• Occurs for Group VIA elements below oxygen.

• What are the predicted bond angles for this geometry?

Octahedral Electronic Geometry: AB6

• The VSEPR theory predicts octahedral for the electronic and molecular geometry.

• Is this molecule polar?• What are the bond

angles?

SF

F

F

F

F

F

octahedral

Octahedral Electronic Geometry: AB6

• Hybridization involves two d orbital form the empty 3d subshell and the 3s and 3p orbitals.– Illustrate from page 336

– Can also occur for n=4, 5, and 6

• There are no unshared pairs.– 6 covalent bonds

Does this type of hybridization occur for N?

Variations of Octahedral Shape

• If lone pairs are incorporated into the octahedral structure, there are two possible new shapes. – One lone pair - square pyramidal– Two lone pairs - square planar

• The resulting hybridization will be the same.

Larger Molecules

• Cyclic molecules

• Linear and branched molecules

• Containing multiple types of elements

Molecular Structure and Covalent Bonding Theories Chapter 8.

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Transcript of Molecular Structure and Covalent Bonding Theories Chapter 8.