1

Covalent Bonding: Orbitals

Chapter 09

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The four bonds around C are of equal length and Energy

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Can you explain this based on your knowledge of electron energy levels?

6C1s2 2s2 2p2

Bonding from s isDifferent from bondingFrom p. In addition theAngles should be within900!

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How to generate four equal orbitals?

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Hint:

A key to wave mechanics is superposition

Which is creating new waves from interference of old ones

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Let’s do some mixing

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Cross section of an sp3 orbital.

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An energy-level diagram showing the formation of four sp3 orbitals.

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Ground state of C 1s2 2s2 2p2

Promote electron at n=2 2s12p3

Hybridize at n=2 sp3 sp3 sp3 sp3

s px py pz

s px py pz

Hybridization

sp3 sp3 sp3 sp3

4sp3 orbitals of equal length, energy and in tetrahedral shape

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Hybridization

The mixing of atomic orbitals to form special orbitals for bonding.

The atoms are responding as needed to give the minimum energy for the molecule.

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Valence Bond Theory and NH3

N – 1s22s22p3

3 H – 1s1

If use the3 2p orbitalspredict 900

Actual H-N-H bond angle is 107.30

How do you explain this?

2s 2px 2py 2pz

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2s 2px 2py 2pzOriginal

Mix 1s and 3pAnd generate four

Equivalent sp3

Hybridized orbitals

sp3 sp3 sp3 sp3

3 bonding orbitalsWhich can accommodateThe 1s1 electron fromhydrogen

1 sp3

lonepair

Consider the n=2 for N

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The nitrogen atom in ammonia is sp3 hybridized.

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An orbital energy-level diagram for sp2 hybridization. Note that one p orbital remains unchanged.

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When an s and two p orbitals are mixed to form a set of three sp2 orbitals, one p orbital remains unchanged

and is perpendicular to the plane of the hybrid orbitals.

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Figure 9.13: (a) The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.

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24Sigma bond () – electron density between the 2 atomsPi bond () – electron density above and below plane of nuclei

of the bonding atoms

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The s bonds in ethylene. Note that for each bond the shared electron pair occupies the region directly between the atoms.

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A sigma () bond centers along the internuclear axis.

A pi () bond occupies the space above and below the internuclear axis.

CCH H

HH

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(a)The orbitals used to form the bonds in ethylene.

(b) The Lewis structure for ethylene.

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The orbital energy-level diagram for the formation of sp hybrid orbitals on carbon.

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When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at

180 degrees results.

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The orbitals of an sp hybridized carbon atom.

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The orbital arrangement for an sp2 hybridized oxygen atom to form CO2.

8O 1s22s22p4

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The hybrid orbitals in the CO2 molecule.

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(a) The orbitals used to form the bonds in carbon dioxide. Note that the carbon-oxygen double bonds each consist of one s bond

and one p bond. (b) The Lewis structure for carbon dioxide.

2sp orbitals from C to form two double bonds

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(a) An sp hybridized nitrogen atom. (b) The s bond in the N2 molecule. (c) The two p bonds in N2 are formed when electron pairs are shared between two sets of parallel p orbitals. (d) The

total bonding picture for N2.

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Sigma () and Pi Bonds ()

Single bond 1 sigma bond

Double bond 1 sigma bond and 1 pi bond

Triple bond 1 sigma bond and 2 pi bonds

How many and bonds are in the acetic acid(vinegar) molecule CH3COOH?

C

H

H

CH

O

O H bonds = 6 + 1 = 7

bonds = 1

10.5

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How to generate more than 4 bonds?

Remember the PCl5, SF6, etc…

The s and 3p can generate 4 orbitals

Include d orbitals to generate more!

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s p p p d d d d d

Hybridize 1s and 3p and 1d

Result in 5 dsp3 orbitals

dsp3 dsp3 dsp3 dsp3 dsp3

1 2 3 4 5

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A set of dsp3 hybrid orbitals on a phosphorus atom. Note that the set of five dsp3 orbitals has a trigonal

bipyramidal arrangement. (Each dsp3 orbital also has a small lobe that is not shown in this diagram.)

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(a) The PCl5 molecule. (b) The orbitals used to form the bonds in PCl5. The phosphorus uses a set of five dsp3 orbitals to share electron pairs with sp3 orbitals on the five chlorine atoms. The other sp3 orbitals on each chlorine atom hold lone pairs.

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s p p p d d d d d

Hybridize 1s and 3p and 2d

Result in 6 d2sp3 orbitals

d2sp3 ds2p3 d2sp3 d2sp3 d2sp3 d2sp3

How to form 6 orbitals?

1 2 3 4 5 6

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The relationship of the number of effective pairs, their spatial arrangement, and the hybrid orbital set required.

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The Localized Electron Model

Draw the Lewis structure(s)

Determine the arrangement of electron pairs (VSEPR model).

Specify the necessary hybrid orbitals.

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Molecular Orbitals (MO)

Analagous to atomic orbitals for atoms, MOs are the quantum mechanical solutions to the organization of valence electrons in molecules. Remember bonds are waves and wave may be arranged in constructive or destructive interference.

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Types of MOs

bonding: lower in energy than the atomic orbitals from which it is composed.

antibonding: higher in energy (unstable) than the atomic orbitals from which it is composed.

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The combination of hydrogen 1s atomic orbitals to form molecular orbitals.

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Bonding and antibonding molecular orbitals (MOs).

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(a) The molecular orbital energy-level diagram for the H2 molecule.

(b) The shapes of the molecular orbitals are obtained by squaring the wave functions for MO1 and MO2.

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A molecular orbital energy-level diagram for the H2 molecule.

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The molecular orbital energy-level diagram for the H2

- ion.

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The molecular orbital energy-level diagram for the He2 molecule.

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The molecular orbital energy-level diagram for the Li2 molecule.

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The expected molecular orbital energy-level diagram resulting from the combination of the 2p

orbitals on two boron atoms.

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The expected molecular orbital energy-level diagram for the B2 molecule.

However, B2 is found to be Paramagnetic!

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The correct molecular orbital energy-level diagram for the B2 molecule. When p-s mixing is allowed, the energies of the 2p and π2p orbitals are reversed. The two electrons from the B 2p orbitals now occupy separate, degenerate π2p molecular orbitals and thus have parallel spins. Therefore, this diagram explains the observed paramagnetism of B2.

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(a) The three mutually perpendicular 2p orbitals on two adjacent boron atoms. Two pairs of parallel p orbitals can overlap as shown in (b) and (c), and the third pair

can overlap head-on as shown in (d).

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(a) The two p orbitals on the boron atom that overlap head-on produce two s molecular orbitals, one bonding and one antibonding. (b) Two p orbitals that lie parallel

overlap to produce two p molecular orbitals, one bonding and one antibonding.

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The molecular orbital energy-level diagrams, bond orders, bond energies, and bond lengths for the diatomic molecules B2 through F2. Note that for O2 and F2 the 2p orbital is lower in energy than the π2p orbitals.

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Paramagnetic Diamagnetic

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Note that O2 is paramagnetic

When liquid oxygen is poured into the space between the poles of a strong magnet, it remains there until it boils away. This attraction of liquid oxygen for the magnetic field demonstrates the paramagnetism of the O2 molecule.

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Paramagnetism

unpaired electrons

attracted to induced magnetic field

much stronger than diamagnetism

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Bond Order (BO)

Difference between the number of bonding electrons and number of antibonding electrons divided by two.

BO = # bonding electrons # antibonding electons

2

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bond order = 12

Number of electrons in bonding MOs

Number of electrons in antibonding MOs

( - )

bond order

½ 1 0½

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Molecular Orbital (MO) Configurations

1. The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.

2. The more stable the bonding MO, the less stable the corresponding antibonding MO.

3. The filling of MOs proceeds from low to high energies.

4. Each MO can accommodate up to two electrons.

5. Use Hund’s rule when adding electrons to MOs of the same energy.

6. The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms.

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Outcomes of MO Model

1. As bond order increases, bond energy increases and bond length decreases.

2. Bond order is not absolutely associated with a particular bond energy.

3. N2 has a triple bond, and a correspondingly high bond energy.

4. O2 is paramagnetic. This is predicted by the MO model, not by the LE model,

which predicts diamagnetism.

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The molecular orbital energy-level diagram for the NO molecule. We assume that orbital order is the same as that for N2. The bond order is 2.5.

Heteronuclear Molecules and similar?

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Figure 9.42: The molecular orbital energy-level diagram for both the NO+ and CN- ions.

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A partial molecular orbital energy-level diagram for the HF molecule.

Heteronuclear Molecules and similar?

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The electron probability distribution in the bonding molecular orbital of the HF molecule. Note the greater electron density close to the

fluorine atom.

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Combining LE and MO Models

bonds can be described as being localized.

bonding must be treated as being delocalized.

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Delocalized molecular orbitals are not confined between two adjacent bonding atoms, but actually extend over three or more atoms.

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(a) The benzene molecule consists of a ring of six carbon atoms with one hydrogen atom bound to each carbon. (b) Two of the

resonance structures for the benzene molecule.

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The bonding system in the benzene molecule.

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(a) The p molecular orbital system in benzene is formed by combining the six p orbitals from the six sp2 hybridized carbon

atoms. (b) The electrons in the resulting p molecular orbitals are delocalized over the entire ring of carbon atoms, giving six

equivalent bonds. A composite of these orbitals is represented here.

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Electron density above and below the plane of the benzene molecule.

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The resonance structures for O3 and NO3-. Note

that it is the double bond that occupies various positions in the resonance structures.

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(a) The p orbitals used to form the π bonding system in the NO3-

ion. (b) A representation of the delocalization of the electrons in the π molecular orbital system of the NO3

- ion.

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QUESTIONThe hybridization of the phosphorus atom in the cation PH2

+ is:

1) sp2.

2) sp3.

3) dsp. 4) sp. 5) none of these.

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ANSWER

1) sp2.

Section 9.1 Hybridization (p. 413) The two of the three hybridized orbitals are bonding orbitals, the third hybridized orbital contains the lone pair.

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QUESTION

Atoms which are sp2 hybridized form ____ pi

bond(s). 1) 0 2) 1 3) 2 4) 3 5) 4

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ANSWER2) 1 Section 9.1 Hybridization (p. 416) An atom with sp

2 hybridization has a p orbital

that remains unhybridized and can overlap with another p orbital on a neighboring atom to create a pi bond.

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QUESTION

The hybridization of the central atom in I3– is:

1) sp 2) sp

2

3) sp3

4) dsp3

5) d2sp

3

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ANSWER

4) dsp3

Section 9.1 Hybridization (p. 421) Three of I3

– orbitals contain lone pairs.

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QUESTION

Which of these statements about benzene is true? 1) All carbon atoms in benzene are sp

3

hybridized. 2) Benzene contains only bonds between C

atoms. 3) The bond order of each C—C bond in benzene

is 1.5. 4) Benzene is an example of a molecule that

displays ionic bonding. 5) All of these statements are false.

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ANSWER3) The bond order of each C—C bond in

benzene is 1.5. Section 9.5 Combining the Localized Electron and Molecular Orbital Models (p. 439) This unusual bond order is due to the delocalization of the bonding electrons. On average three electrons are shared between two atoms.

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QUESTIONWhich statement about N2 is false? 1) It is a gas at room temperature. 2) The oxidation state is +3 on one N and –3

on the other. 3) It has one sigma and two pi bonds between

the two atoms. 4) It can combine with H2 to form NH3. 5) It has two pairs of nonbonding electrons.

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ANSWER2) The oxidation state is +3 on one N and –3

on the other. Section 9.1 Hybridization (p. 418) For a molecular form of an element, all atoms share electrons equally and thus should have oxidation states of zero.

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QUESTIONThe hybridization of the central atom, Al, in AlBr3 is: 1) sp 2) sp

2

3) sp3

4) dsp3

5) d2sp

3

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ANSWER

2) sp2

Section 9.1 Hybridization (p. 416) Aluminum has three valence electrons, each of which forms a bond with a Br atom.

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QUESTIONThe hybridization of Br in BrF3 is: 1) sp 2) sp

2

3) sp3

4) dsp3

5) d2sp

3

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ANSWER

4) dsp3

Section 9.1 Hybridization (p. 421) Bromine has two lone pairs as well as three bonding pairs.

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QUESTIONWhat is the bond order of Ne2? 1) 0 2) 1/2 3) 1 4) 1 1/2 5) 2

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ANSWER1) 0 Section 9.2 The Molecular Orbital Model (p. 429) Ne atoms have a full octet and have no half-filled orbitals to overlap for bonding.

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QUESTION

What is the bond order of C2+?

1) 0 2) 1/2 3) 1 4) 1 1/2 5) 2

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ANSWER4) 1 1/2 Section 9.2 The Molecular Orbital Model (p. 429) Remember when using MO diagrams that Hund’s Rule still applies when filling the orbitals and that forgetting this rule can give the wrong bond order.

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QUESTIONHow many electrons are involved in pi bonding in benzene, C6H6? 1) 12 2) 30 3) 3 4) 6 5) 18

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ANSWER4) 6 Section 9.5 Combining the Localized Electron and Molecular Orbital Models (p. 439) Each carbon atom in the benzene ring donates an electron to the pi bonding.

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QUESTIONWhich of the following substances contains two pi bonds? 1) C2H4 2) C3H8 3) C2H2 4) C2H6 5) CH4

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ANSWER3) C2H2 Section 9.1 Hybridization (p. 418) Each carbon in C2H2 has two unhybridized p orbitals for pi bonding.

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QUESTIONConsider the skeletal structure shown below:

N—C—C—N

Draw the Lewis structure and answer the following: How many of the atoms are sp hybridized? 1) 0 2) 1 3) 2 4) 3 5) 4

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ANSWER5) 4 Section 9.1 Hybridization (p. 418) Each nitrogen and carbon has an unhybridized p orbital for pi bonding.

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QUESTIONConsider the skeletal structure shown below:

N—C—C—N

Draw the Lewis structure and answer the following: How many pi bonds does the molecule contain? 1) 0 2) 2 3) 4 4) 6 5) 7

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ANSWER3) 4 Section 9.1 Hybridization (p. 418) The pi bonds create a triple bond between each CN pair.

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QUESTIONWhich of the following statements about the molecule BN is false? 1) It is paramagnetic. 2) Its bond order is 2. 3) The total number of electrons is 12. 4) It has two pi bonds. 5) All of these are true.

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ANSWER1) It is paramagnetic. Section 9.3 Bonding in Homonuclear Diatomic Molecules (p. 432) BN has 8 valence electrons, all of which are paired so that there is no paramagnetism. This pairing can be seen in its Lewis structure and using MO Theory.

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QUESTIONWhich of the following molecules contains the shortest C — C bond? 1) C2H2 2) C2H4 3) C2H6 4) C2Cl4 5) 2 and 4

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ANSWER1) C2H2 Section 9.3 Bonding in Homonuclear Diatomic Molecules (p. 430) Triple bonds are shorter than double bonds, which are shorter than single bonds. The strength of the bonds increases with decreasing length.

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QUESTIONIf four orbitals on one atom overlap four orbitals on a second atom, how many molecular orbitals will form? 1) 1 2) 4 3) 8 4) 16 5) None of these

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ANSWER3) 8 Section 9.1 Hybridization (p. 414) Molecular orbitals form during the combination of atomic orbitals, but orbitals cannot be created or destroyed. If eight atomic orbitals mix, eight molecular orbitals are formed.

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QUESTIONWhich of the following molecules or ions is not paramagnetic in its ground state? 1) O2 2) O2

+

3) B2 4) NO 5) F2

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ANSWER5) F2 Section 9.3 Bonding in Homonuclear Diatomic Molecules (p. 432) All species can be made paramagnetic by the promotion of an electron from a filled orbital. In all cases covered here, we will assume that the molecules and ions are in their electronic ground state.

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QUESTIONThe fact that O2 is paramagnetic can be explained by: 1) the Lewis structure of O2. 2) resonance. 3) a violation of the octet rule. 4) the molecular orbital diagram for O2. 5) hybridization of atomic orbitals in O2.

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ANSWER4) the molecular orbital diagram for O2. Section 9.3 Bonding in Homonuclear Diatomic Molecules (p. 434) Application of Hund’s Rule will leave two electrons unpaired in the two bonding orbitals.

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QUESTIONAs the bond order of a bond increases, the bond energy ______ and the bond length ______. 1) increases; increases 2) decreases; decreases 3) increases; decreases 4) decreases; increases 5) More information is needed to answer this

question.

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ANSWER3) increases; decreases Section 9.2 The Molecular Orbital Model (p. 426) Multiple bonds between atoms increases the electron density, shielding the nuclei from one another, allowing the atoms to draw closer together.