Bonding in Molecules Covalent Bonding Chem 59-651 Valence electrons and valence shell orbitals

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Chem 59-651

Bonding in Molecules

Covalent Bonding

The term covalent implies sharing of electrons between atoms.

Valence electrons and valence shell orbitals

- Only valence electrons are used for bonding: n s, n p, n d

- “Core” electrons are held too tightly (too low in energy)

- Filled n d orbitals are considered core electrons

Valence state electron configurations and Promotion Energies

- The promotion energy is the energy required to promote electrons from the ground state to a “valence state”, which is one type of excited state configuration that is used for bonding.

E.g. C

C*

2s 2 2p 2 ground state

2s 1 2p 3 valence state

Chem 59-651

Localized Bonding Models

Localized

implies that electrons are confined to a particular bond or atom.

The Lewis approach to bonding

Pairs of electrons are localized in bonds or as non-bonding “lone pairs” on atoms. Each bond is formed by a pair of electrons shared by two atoms.

G.N. Lewis

Octet rule

: most main group atoms will tend to end up with an n s 2 n p 6 electron configuration.

n s n p

This is mostly true for the molecules of organic chemistry not necessarily for inorganic compounds.

Chem 59-651 Rules for drawing Lewis diagrams a.

Pick the central atom.

- Atoms that are present only once in the formula, especially heavy elements and metals, tend to be at the center of the structure.

- Oxygen is often terminal and hydrogen almost always is.

- Often the formula is written with the central atom first.

(Sometimes there may be more than one central atom.) b.

Write out the valence shell electron configurations for the neutral central atom and the "terminal" atoms in their ground states. c.

If there is a negative charge distribute it among the terminal atoms in the first instance. Bear in mind that all the terminal atoms must make at least one covalent bond with the central atom, so do not create any noble gas configurations on them. Positive charge is best initially assigned by removing electrons from the central atom.

d.

The total number of unpaired electrons on the terminal atoms will have to match the number of unpaired electrons on the central atom to account for the bonds and leave no unpaired electrons. If this is not the case, once the first three steps have been carried out, there are two strategies available: e.

Move electrons between the central atom and the terminal atoms as necessary. Make sure you keep track of the formal charges because you must be specific about their location. Enclosing a Lewis structure in brackets with the charge outside is not acceptable .

f.

If and only if the central atom comes from the second period or below (Na onwards, n

=3 and up), electrons can be placed into the nd subshell. (Whether the d orbitals play a significant role in bonding in main group compounds is debatable, but they do help to predict correct structure without invoking canonical structures with unreasonable charge separations.)

N

Chem 59-651 Typical Lewis structural types:

Molecules that conform to the “Octet Rule”: saturated molecules

NH

3

CH

4

2s 2p

2s 2p

C ground state

C* valence state

3 H

4 H

1s 1s 1s

1s 1s 1s 1s

H N

H

H H

H C

H

H

These are typical of the molecules of organic chemistry.

N

Chem 59-651

Molecules that conform to the “Octet Rule”: unsaturated molecules.

2s

ClNO

2p

2s

NO

3

-

2p

N

N +

Cl

3s

Cl

3p

N O

O

2s 2p

O

2s

O -

2p

2s 2p

O -

2s

O

O N

O

2p

Chem 59-651 Resonance

Resonance implies that there is more than one possible way to distribute the valence electrons in a Lewis structure. For an adequate description, each “canonical” structure must be drawn.

O O

O N

O

O

O N

O

O N

O

If different equivalent resonance structures are possible, the molecule tends to be more stable than one would otherwise expect. This is a quantum mechanical effect that we will talk about later.

O

O N

O

Less favourable canonical structure

I expect you to be able to:

Draw Lewis structures (including resonance structures when necessary), determine bond orders , determine and place formal charges.

Chem 59-651

Molecules that don’t conform to the “Octet Rule”:

Electron-deficient molecules

BH

3

2s 2p

B

Expanded valence shell molecules

ClF

3

3s 3p

Cl

3d

Cl*

B*

F

3 H

2s

F

2p

1s 1s 1s

H

H B

H

“Lewis acids”

F

F

Cl

F

2s

F

2p

2s 2p

“Hypervalent molecules”

Chem 59-651

V alence S hell E lectron P air R epulsion Theory

A basic geometry can be assigned to each non-terminal atom based on the number of “objects” attached to it. Objects include bonded atoms

(single, double, triple, partial bonds) and “lone pairs” of electrons.

VSEPR theory lets us predict the shape of a molecule based on the electron configurations of the constituent atoms. It is based on maximizing the distance between points on a spherical surface.

3 4 5 6 Number of

Objects

Geometry

2 linear trigonal planar tetrahedral trigonal bipyramidal*

Octahedral

Chem 59-651

The geometry around an atom is described by the general formula:

AX

m

E

n

Where X is a bonded atom, E is a lone pair and (m+n) is the number of objects (sometimes called the steric number , SN) around the central atom A.

4 5 6 Number of

Objects

Geometry

2 linear

Formula

(Shape)

AX

2

3 trigonal planar

AX

3

(trig. planar)

AX

2

E

(bent) tetrahedral

AX

4

(tetrahedral)

AX

3

E

(pyramidal)

AX

2

E

2

(bent) trigonal bipyramidal

AX

5

(t.b.p. or square pyramidal)

AX

4

E

(seesaw)

AX

3

E

2

(T-shaped)

AX

2

E

3

(linear)

Octahedral

AX

6

(octahedral)

AX

5

E

(square pyramidal)

AX

4

E

2

(square planar)

AX

3

E

3

(T-shaped)

Chem 59-651 Less common geometries

8 Number of

Objects

Geometry

7 pentagonal bipyramidal square anti-prismatic

XeF

5

NMe

4

+

Xe -

F F F F F

Xe is described as AX

5

E

2 and has a pentagonal planar shape derived from the pentagonal bipyramidal geometry.

Chem 59-651

Refinement of VSEPR theory predicted geometries

The relative steric demand of objects is different and amount of repulsion caused by the object will alter the arrangement of the atoms around the central atom.

Lone pair of electrons

Multiple bond polarized toward central atom

Normal single bond

Long single bond polarized away from central atom

109.5°

CH

4

NH

3

106.6°

OH

2

104.5°

Chem 59-651

Valence Bond Theory

Valence bond theory (VBT) is a localized quantum mechanical approach to describe the bonding in molecules. VBT provides a mathematical justification for the Lewis interpretation of electron pairs making bonds between atoms. VBT asserts that electron pairs occupy directed orbitals localized on a particular atom. The directionality of the orbitals is determined by the geometry around the atom which is obtained from the predictions of VSEPR theory.

In VBT, a bond will be formed if there is overlap of appropriate orbitals on two atoms and these orbitals are populated by a maximum of two electrons.

σ bonds : symmetric about the internuclear axis

π bonds : have a node on the inter-nuclear axis and the sign of the lobes changes across the axis.

Chem 59-651

Valence Bond Theory

Detailed valence bond theory treatment of bonding in H

2

.

VBT considers the interactions between separate atoms as they are brought together to form molecules.

H

A

1s 1

φ

A

(1)

H

B

1s 1

φ

B

(2) electron

Ψ

1

=

φ

A

(1)

φ

B

(2)

Ψ

2

=

φ

A

(2)

φ

B

(1)

Atomic wavefunction on atom B

Quantum mechanics demands that electrons can be interchangeable so we must use a linear combination of

Ψ

1 and

Ψ

2

.

Ψ

+

= N (

Ψ

1

+

Ψ

2

) (bonding, H-H)

Ψ

-

= N (

Ψ

1

-

Ψ

2

) (anti-bonding)

N is a normalizing coefficient

C is a coefficient related to the amount of ionic character

Ψ

3

=

φ

A

(1)

φ

A

(2) (ionic H H + )

Ψ

4

=

φ

B

(1)

φ

B

(2) (ionic H + H )

Ψ molecule

= N [

Ψ

1

+

Ψ

2

] + ( C [

Ψ

3

+

Ψ

4

])

Ψ molecule

= N [

Ψ covalent

+ ( C

Ψ ionic

)]

Chem 59-651

Valence Bond Theory

Valence bond theory treatment of bonding in H

2 used.

and F

2

– the way it is generally

F

2s 2p

H

A

1s 1

φ

A

α

H

B

1s 1

φ

B

β

F

2s 2p

This gives a 1s-1s

σ bond between the two H atoms.

2p z

2p z

This gives a 2p-2p the two F atoms.

σ

Z axis bond between

For VBT treatment of bonding, people generally ignore the antibonding combinations and the ionic contributions.

Chem 59-651

Valence bond theory treatment of bonding in O

2

Z axis

O

2s 2p

2p z

2p z

This gives a 2p-2p

σ bond between the two O atoms.

O

2s 2p

Z axis

O O

Lewis structure

2p y

2p y

(the choice of 2p y is arbitrary)

This gives a 2p-2p

π bond between the two O atoms. In VBT,

π bonds are predicted to be weaker than

σ bonds because there is less overlap.

Double bond:

Triple bond:

σ

σ bond + bond + 2

π

π bond bond

The Lewis approach and VBT predict that O this is wrong!

2 is diamagnetic –

Chem 59-651

Directionality

The bonding in diatomic molecules is adequately described by combinations of “pure” atomic orbitals on each atom. The only direction that exists in such molecules is the inter-nuclear axis and the geometry of each atom is undefined in terms of VSEPR theory (both atoms are terminal). This is not the case with polyatomic molecules and the orientation of orbitals is important for an accurate description of the bonding and the molecular geometry.

Examine the predicted bonding in ammonia using “pure” atomic orbitals:

2s 2p

N

3 H

1s 1s 1s

H N

H

H

The 2p orbitals on N are oriented along the X, Y, and Z axes so we would predict that the angles between the 2p-1s

σ bonds in NH

3 would be 90°.

We know that this is not the case.

106.6°

Chem 59-651

Hybridization

The problem of accounting for the true geometry of molecules and the directionality of orbitals is handled using the concept of hybrid orbitals.

Hybrid orbitals are mixtures of atomic orbitals and are treated mathematically as linear combinations of the appropriate s, p and d atomic orbitals.

Linear sp hybrid orbitals

A 2s orbital superimposed on a 2p x orbital

Ψ

1

=

1

2

φ s

+

1

2

φ p

Ψ

2

=

1

2

φ s

1

2

φ p

The 1/

2 are normalization coefficients.

The two resultant sp hybrid orbitals that are directed along the X-axis

(in this case)

Chem 59-651

Orthogonality and Normalization

Two properties of acceptable orbitals (wavefunctions) that we have not yet considered are that they must be orthogonal to every other orbital and they must be normalized . These conditions are related to the probability of finding an electron in a given space.

Orthogonal means that the integral of the product of an orbital with any other orbital is equal to 0, i.e.:

Ψ Ψ

m

∂τ =

0 where n

≠ m and

δτ means that the integral is taken over “all of space” (everywhere).

Normal means that the integral of the product of an orbital with itself is equal to 1, i.e.:

Ψ Ψ

n

∂τ =

1

This means that we must find normalization coefficients that satisfy these conditions.

Note that the atomic orbitals (

φ

) we use can be considered to be both orthogonal and normal or “orthonormal”.

Chem 59-651

Example of the orthogonality of

Ψ

1 and

Ψ

2

Ψ

1

=

1

2

φ s

+

1

2

φ p

Ψ

2

=

1

2

φ s

1

2

φ p

Ψ Ψ

2

∂τ =



 1

2

φ s

+

1

2

φ p 



 1

2

φ s

1

2

∫ Ψ Ψ

2

∂τ =

2

1 ∫ ( )

2

1 ∫

( )

+

1

2

( )

1

2

( )

Ψ Ψ

2

∂τ =

1

2

( ) ( ) ( ) ( )

2 2 2

1

Ψ Ψ

2

∂τ

2

1

2

0

Thus our hybrid sp orbitals are orthogonal to each other, as required.

Chem 59-651

Hybridization

Valence bond theory treatment of a linear molecule: the bonding in BeH

2

BeH

2

Be

2s 2p

The promotion energy can be considered a part of the energy required to form hybrid orbitals.

Be* sp 2p

Be* (sp)

2 H

1s 1s

H Be H

The overlap of the hybrid orbitals on Be with the 1s orbitals on the H atoms gives two Be-H (sp)-1s

σ bonds oriented 180° from each other.

This agrees with the VSEPR theory prediction.

Chem 59-651

Valence bond theory treatment of a trigonal planar molecule: the bonding in BH

3

B

B*

2s sp 2

B* (sp 2 )

2p

2p

H

H B

H

This gives three sp 2 orbitals that are oriented 120° apart in the xy plane – be careful: the choice of axes in this example determines the set of coefficients.

Ψ

1

=

Ψ

2

=

Ψ

3

=

1

3

φ s

1

3

φ s

1

3

φ s

+

1

6

φ p x

+

1

6

φ p x

2

6

φ p x

1

2

φ p y

1

2

φ p y

Chem 59-651

Valence bond theory treatment of a trigonal planar molecule: the bonding in BH

3 sp 2 2p

B*

3 H

1s 1s 1s

H B

H

H

The overlap of the sp 2 hybrid orbitals on B with the 1s orbitals on the H atoms gives three B-H (sp 2 )-1s

σ bonds oriented 120° from each other.

This agrees with the VSEPR theory prediction.

Chem 59-651

Valence bond theory treatment of a tetrahedral molecule: the bonding in CH

4

2p

C

C*

2s sp 3

C* (sp 3 )

H

H C

H

H

This gives four sp 3 orbitals that are oriented in a tetrahedral fashion.

Ψ

Ψ

Ψ

1

2

3

=

=

=

Ψ

4

=

1

4

φ s

1

4

φ s

+

+

1

4

φ s

1

4

φ s

1

4

φ p x

1

4

φ p x

+

1

4

φ p x

1

4

φ p x

+

1

4

φ p y

1

4

φ p y

+

1

4

φ p y

1

4

φ p y

+

1

4

φ p z

1

4

φ p z

1

4

φ p z

1

4

φ p z

Chem 59-651

Valence bond theory treatment of a tetrahedral molecule: the bonding in CH

4

2s 2p

C

C* sp 3

C* (sp 3 )

4 H

1s 1s 1s 1s

H

H

C

H

H

The overlap of the sp 3 hybrid orbitals on C with the 1s orbitals on the H atoms gives four C-H (sp 3 )-1s

σ bonds oriented 109.47° from each other. This provides the tetrahedral geometry predicted by VSEPR theory.

P

P*

Chem 59-651

Valence bond theory treatment of a trigonal bipyramidal molecule: the bonding in PF

5

3s 3p

P* ( sp 3 d)

3d

3d

PF

5

AX

5 has an VSEPR theory geometry so we need hybrid orbitals suitable for bonds to 5 atoms. n s and n p combinations can only provide four, so we need to use n d orbitals (if they are available) .

3s 3p z

3p y

3p x

3d z2 sp 3 d z2

The appropriate mixture to form a trigonal bipyramidal arrangement of hybrids involves all the n s and n p orbitals as well as the n d z2 orbital.

Chem 59-651

Valence bond theory treatment of a trigonal bipyramidal molecule

The orbitals are treated in two different sets.

Ψ

Ψ

2

=

Ψ

1

3

=

=

1

3

φ s

+

1

3

φ s

+

1

3

φ s

1

6

φ p x

+

1

6

φ p x

2

6

φ p x

1

2

φ p y

1

2

φ p y

These coefficients are exactly the same as the result for the trigonal planar molecules because they are derived from the same orbitals (sp 2 )

Ψ

4

=

Ψ

5

= −

1

2

φ p z

+

1

2

φ p z

+

1

2

φ d z

2

1

2

φ d z

2

These coefficients are similar to those for the sp hybrids because they are formed from a combination of two orbitals (pd).

Remember that d orbitals are more diffuse than s or p orbitals so VBT predicts that the bonds formed by hybrids involving d orbitals will be longer than those formed by s and p hybrids.

Chem 59-651

Valence bond theory treatment of a trigonal bipyramidal molecule: the bonding in PF

5

P* ( sp 3 d)

3d

F

2s 2p

F F

2s 2p 2s 2p

F

2s

F

2p

2s 2p

The overlap of the sp 3 d hybrid orbitals on P with the 2p orbitals on the F atoms gives five P-F (sp 3 d)-2p

σ bonds in two sets: the two axial bonds along the z-axis (180° from each other) and three equatorial bonds in the xy plane (120° from each other and 90° from each axial bond). This means that the 5 bonds are not equivalent!

Chem 59-651

An alternative, and maybe more reasonable, version of VBT treatment of a trigonal bipyramidal molecule:

The d orbitals are too high in energy to mix effectively with the s and p orbitals, so the trigonal bipyramidal molecule is actually composed of an equatorial set of trigonal

(sp 2 ) hybrids and the axial bonds come from an MO interaction between the two ligand orbitals and the p z orbital on the central atom.

Ψ

1

=

Ψ

2

=

1

2

φ p zFa

+

1

2

φ p zFb

1

2

φ p zFa

1

2

φ p zFb

σ =

1

2

Ψ

1

+

1

2

φ

3 p zP

σ

*

=

1

2

Ψ

1

1

2

φ

3 p zP

Chem 59-651

The square pyramidal AX

5 geometry requires mixing with a different d orbital than in the trigonal bipyramidal case.

Sb(C

6

H

5

)

5 d orbitals

You should consider what orbital(s) would be useful for such a geometry and we will see a way to figure it out unambiguously when we examine the symmetry of molecules.

S

Chem 59-651

3s

Valence bond theory treatment of an octahedral molecule: the bonding in SF

6

3p

3d

S*

S* ( sp 3 d 2 ) 3d

F F F F F F

3s 3p z

3p y

3p x

3d z2

3d x2-y2 sp 3 d 2

The overlap of the sp 3 d 2 hybrid orbitals on S with the 2p orbitals on the F atoms gives six S-F (sp 3 d 2 )-2p

σ bonds 90° from each other that are equivalent . You can figure out the normalization coefficients. As in the case of the TBP, there is also an MO approach that does not require d orbitals.

N

Chem 59-651

Valence bond theory treatment of

π

-bonding: the bonding in ClNO

2s 2p

Cl N O sp 2 2p

N*(sp 2 )

σ σ

π

There are three “objects” around N so the geometry is trigonal planar. The shape is given by AX

2

E (angular or bent).

Cl

3s 3p

O

2s 2p

Cl

N

O

A drawing of the VBT

π bond in ClNO.

The overlap of the sp 2 hybrid orbitals on N with the 3p orbital on Cl and the 2p orbital on O give the two

σ bonds and it is the overlap of the “left over” p orbital on N with the appropriate orbital on O that forms the (2p-

2p)

π bond between the two atoms.

O -

Chem 59-651

Valence bond theory treatment of

π

-bonding: the bonding in the nitrate anion

2s 2p

O

N

O N

N +

O sp 2 2p

N + *(sp 2 )

σ

σ

σ

π

There are three “objects” around N so the geometry is trigonal planar. The shape is given by AX

3

(trigonal planar).

2s 2p

N O

O -

O

O

2s 2p

O VBT gives only one of the canonical structures at a time.

2s 2p

The overlap of the sp 2 hybrid orbitals on N with the the 2p orbitals on the

O give the three (sp 2 -2p)

σ bonds and it is the overlap of the “left over” p orbital on N with the appropriate orbital on the uncharged O atom that forms the (2p-2p)

π bond.

Chem 59-651

Valence bond theory treatment of

π

-bonding: the bonding in ethene

2s 2p

H H

Each C

C C

C*(sp 2 )

Each C* sp 2

σ σ

σ

2p sp 2

π

2p

H H

There are three “objects” around each C so the geometry is trigonal planar at each carbon. The shape is given by AX

3 for each carbon.

σ

σ C*(sp 2 )

H

4 H

H

H

C C

1s 1s 1s 1s

H

The overlap of the sp 2 hybrid orbitals on C with the the 1s orbitals on each H give the four terminal (sp 2 -1s)

σ bonds. The double bond between the C atoms is formed by a (sp

π bond.

2 - sp 2 )

σ bond and the (2p-2p)

S

S*

Chem 59-651

Valence bond theory treatment of

π

-bonding: the bonding in SOCl

2

3s 3p

Cl

O S

3d

Cl

S*(sp 3 )

σ

σ sp 3

σ π

3d There are four “objects” around S so the geometry is tetrahedral and the shape is given by AX

3

E (pyramidal).

Cl Cl

O

2s 2p Cl

Cl

S

O

The overlap of the sp 3 hybrid orbitals on S with the 3p orbitals on Cl and the 2p orbital on O give the three

σ bonds and, because the lone pair is located in the final sp 3 hybrid, it is the overlap of the “left over” d orbital on S with an appropriate p orbital on O that forms the (3d-2p)

π bond in the molecule.

Cl

Chem 59-651

Valence bond theory treatment of bonding: a hypervalent molecule, ClF

3

3s 3p F

F Cl

3d

Cl* F

Cl* ( sp 3 d)

F F F

3d

There are five “objects” around

Cl so the geometry is trigonal bipyramidal and the shape is given by AX

3

E

2

(T-shaped).

Consider this: Why are such molecules T-shaped instead of pyramidal?

The overlap of the sp 3 d hybrid orbitals on Cl with the 2p orbitals on the F atoms gives three P-F (sp 3 d)-2p

σ bonds in two sets: the two axial bonds along the z-axis (less than 180° from each other because of the repulsion from the lone pairs) and the one equatorial bond halfway between the other Cl bonds. Again, the bond lengths will not be the same because there is more d contribution to the axial hybrid orbitals.

Chem 59-651

Summary of Valence Bond Theory

1.

Write an acceptable Lewis structure for the molecule.

2.

Determine the number of

VSEPR objects around all central atoms and determine the geometry around the atom.

3.

Construct hybrid orbitals suitable for the predicted bonding.

4.

Link orbitals together to make bonds.

5.

Describe the bonding. Include the names of the orbitals involved in each bond. Draw pictures of the bonds formed by the overlap of these orbitals.

H Be H

Two objects around Be, so AX

2

(linear)

Two orbitals pointing 180° from each other needed, so use two sp hybrids

Two (sp-1s) Be-H

H

σ sp

Be H bonds.

1s

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