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 CO 2 , H 2 CO, and CH 4 – 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 Theory Three regions of high electron density

VSEPR Theory Four regions of high electron density

VSEPR Theory Five regions of high electron density

VSEPR Theory Six 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 H 2 O

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/bp As a result, the bond angles around a center atoms can be distorted (reduced) from the predicted values CH 4 and H 2 O 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.

Molecular Geometry and Polarity • A dipole moment (bond dipole) has _____ and _____. Both must be considered when determines if a molecule is polar.

– CO 2 and H 2 O. 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 CO 2 , H 2 O, and O 3

Molecular Geometry and Polarity

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.

Valence Bond(VB) Theory • 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

Valence Bond(VB) Theory • 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

Valence Bond(VB) Theory

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 AB 3 U represents three atoms bonded to a central atom with one lone pair. An example would be NH 3

Molecular Shapes and Bonding Discussion sequence • Experimental facts and Lewis formula • VSEPR – Electronic geometry – Molecular geometry – Polarity • Valence bond theory

AB 2 Molecules - No Lone Pairs on A - Linear Molecules • The BeCl 2 molecule is linear and has melting point of 405  C.

– Draw BeCl 2 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 Be 2+ • The electron cloud on the halide is distorted by the high charge density – BeBr 2 and BeI 2 also have linear geometries

AB 2 Molecules - No Lone Pairs 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.

AB 2 Molecules - No Lone Pairs on A - Linear Molecules • Electronic Structures Lewis Formulas Be 1s  2s  2p Be · · Cl [Ne] 3s    3p · · ·· 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.

AB 2 Molecules - No Lone Pairs on A - Linear Molecules • Valence Bond Theory (Hybridization) 1s Be  2s  2p  1s sp hyb 2p Cl [Ne] 3s    3p 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.

AB 2 Molecules - No Lone Pairs 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

AB 3 Molecules - No Lone Pairs 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 -107  C – Does the molecular and electronic geometry differ?

– The data indicates that this molecule is nonpolar (no net dipole).

AB 3 Molecules - No Lone Pairs 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 BCl 2 H?

Cl Cl B  Cl

AB 3 Molecules - No Lone Pairs on A - Trigonal Planar Molecules • Electronic Structures Lewis Formulas 1s B   2s 2p .

B : 3s Cl [Ne]    3p .

Cl : 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.

AB 3 Molecules - No Lone Pairs on A - Trigonal Planar Molecules • Valence Bond Theory (Hybridization) B 1s   2s 2p   1s sp 2 hybrid Cl [Ne] 3s    3p The 2s and 2p atomic orbitals on B hybridize to produce three sp orbitals (sp 2 hybrid). Notice that chlorine has a half-filled 3p orbital that can overlap with the sp 2 hybrid orbitals of Be.

AB 3 Molecules - No Lone Pairs on A - Trigonal Planar Molecules Three regions of electron density around the central atom Illustrate bonding with the Cl atoms on the hybridized B

AB 4 Molecules - No Lone Pairs 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 • CH 4 , methane, possesses a tetrahedral geometry and has a melting point of -182  C – Would the molecular and electronic geometry differ?

– The data indicate that the molecule is nonpolar.

AB 4 Molecules - No Lone Pairs on A - Tetrahedral Molecules • Lewis structure predicts tetrahedral geometry • There are four small bond dipoles which cancel – The molecule is nonpolar • What about CCl 3 H and CH 3 Cl?

– When the symmetry lowers, the molecule becomes polar.

• Other molecules?

H H C CH 4 H H

AB 4 Molecules - No Lone Pairs on A - Tetrahedral Molecules • Electronic Structures Lewis Formulas C [He] 2s  2p .

.

C : 1s H 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.

AB 4 Molecules - No Lone Pairs on A - Tetrahedral Molecules • Valence Bond C [He]  2s 2p four sp 3  C [He] hybrid orbitals 1s H The 2s and 2p atomic orbitals on C hybridize to produce four sp orbitals (sp 3 hybrid). Notice that hyrogen has a half filled 1s orbital that can overlap with the sp 3 hybrid orbitals of C.

Many AB 4 NH 4 + type molecules have this hybridization is an AB 4 type polyatomic ion

AB 4 Molecules - No Lone Pairs on A - Tetrahedral Molecules Four regions of electron density around the central atom Illustrate how the hydrogen atomic orbitals bond to the hybridized carbon sp 3 orbitals

Alkanes C n H 2n+2 • alkanes are saturated hydrocarbons • have the general formula C n H 2n+2 . CH 4 - methane C 2 H 6 or (H 3 C-CH 3 ) - ethane C 3 H 8 or (H 3 C-CH 2 -CH 3 ) - propane • C atoms are located at the center of a tetrahedron each alkane is a chain of interlocking tetrahedra C atom at the center of each tetrahedron enough H to form a total of four bonds for each C

AB 3 U 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 NH 3 .

– How many regions of electron density around nitrogen? The bong angle is in this molecule is ~107  . Why?

• Other common molecules are NF 3 , PF 3 , and the polyatomic ion SO 3 2 .

AB 3 U Molecules - One Lone Pair - 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 NF 3 ? How do the polarities of the two molecules compare (later)?

H H N H

AB 3 U Molecules - One Lone Pair - Pyramidal Molecules Electronic Structures Lewis Formulas 2s  2p    2p N [He] 2s F [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.

AB 3 U Molecules - One Lone Pair - Pyramidal Molecules • Valence Bond 2s N [He]  2p  four sp 3 hybrids The 2s and 2p atomic orbitals hybridize to form four sp 3 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

AB 3 U Molecules - One Lone Pair - Pyramidal Molecules • Let’s compare NH 3 with NF 3 .

• 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.

AB 3 U Molecules - One Lone Pair - Pyramidal Molecules • The bond dipoles go opposite directions on NH 3 and NF 3 – For NH 3 , the net dipole is enhanced by the lone pair.

– For NF 3 , 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 NF 3 H H N H H-N-H = 107.3

 N F F F F-N-F = 102.1



AB 2 U 2 - 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 • H 2 O 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 H 2 S and OCl 2

AB 2 U 2 - Two Lone Pairs V-Shaped Molecules • 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 H H O

AB 2 U 2 - Two Lone Pairs V-Shaped Molecules • Electronic Structures Lewis Formulas O [He] 1s 2s   2p .

H 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.

AB 2 U 2 - Two Lone Pairs V-Shaped Molecules • Valence Bond 2s O [He]   2p   four sp 3 hybrids The hybrid orbitals that are full belong to the lone pairs. The half-filled orbitals are used for bonding.

Trigonal Bipyramidal Electronic Geometry • AB 5 , AB 4 U, AB 3 U 2 , and AB 2 U 3 • Hybridization is sp 3 d.

• 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 (PF 5 )

Trigonal Bipyramidal Electronic Geometry, AB 5 • 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 AB 5 ) polar? – Show molecule with this geometry.

· · ·· F ·· · · ·· F · · As · · F ·· · · ·· F ·· ·· F ·· · · · · trigonal bipyramid

Trigonal Bipyramidal Electronic Geometry, AB 5 • Electronic Structures Lewis Formulas P [Ne] F [He] 2s   3s 3p   2p

.

·· P .

·· F ·· .

.

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

Trigonal Bipyramidal Electronic Geometry, AB 5 • 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 •

AB 4 U

, AB 3 U 2 , and AB 2 U 3 • Go through the procedure for the molecule, SF 4 .

– 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, AB 4 U

Trigonal Bipyramidal Electronic Geometry with Lone Pairs • AB 4 U,

AB 3 U 2

, and AB 2 U 3 • The ClF 3 molecule – Where are the likely locations for the lone pairs?

• The molecular geometry is termed as _______.

What is the electronic geometry?

The Molecular Geometry, AB 3 U 2

Trigonal Bipyramidal Electronic Geometry with Lone Pairs • AB 4 U, AB 3 U 2 , and

AB 2 U 3

• The I 3 species – Where are the likely locations for the lone pairs?

• The molecular geometry is termed as _______.

The Molecular Geometry, AB 2 U 3 .

Octahedral Electronic Geometry • AB 6 , AB 5 U, and AB 4 U 2 • Hybridization is sp 3 d 2 .

• Occurs for Group VIA elements below oxygen.

• What are the predicted bond angles for this geometry?

Octahedral Electronic Geometry: AB 6 • The VSEPR theory predicts octahedral for the electronic and molecular geometry.

• Is this molecule polar?

• What are the bond angles?

F F F S F F F octahedral

Octahedral Electronic Geometry: AB 6 • 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