Chapter 10 Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals

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Chapter 10

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Chemical Bonding II:Molecular Geometry and

Hybridization of Atomic Orbitals

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Valence shell electron pair repulsion (VSEPR) model:

Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs.

AB2 2 0

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

linear linear

B B

3

Cl ClBe

2 atoms bonded to central atom

0 lone pairs on central atom

4

AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB4 4 0 tetrahedral tetrahedral

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB6 6 0 octahedraloctahedral

13bonding-pair vs. bonding-

pair repulsionlone-pair vs. lone-pair

repulsionlone-pair vs. bonding-

pair repulsion> >

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB2E 2 1trigonal planar

bent

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3E 3 1


tetrahedraltrigonal

pyramidal

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB3E 3 1 tetrahedraltrigonal

pyramidal

AB2E2 2 2 tetrahedral bent

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0 trigonalbipyramidal

trigonalbipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

AB2E3 2 3 trigonalbipyramidal

linear

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB5E 5 1 octahedral square pyramidal

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Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB5E 5 1 octahedral square pyramidal

AB4E2 4 2 octahedral square planar

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Predicting Molecular Geometry1. Draw Lewis structure for molecule.

2. Count number of lone pairs on the central atom and number of atoms bonded to the central atom.

3. Use VSEPR to predict the geometry of the molecule.

Example 10.1

Use the VSEPR model to predict the geometry of the following molecules and ions:

(a)AsH3

(b)OF2

(c)

(d)

(e)C2H4

Example 10.1

Strategy The sequence of steps in determining molecular geometry is as follows:

Solution

(a)The Lewis structure of AsH3 is

There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral (see Table 10.1).

Example 10.1

Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the As and H atoms). Thus, removing the lone pair leaves us with three bonding pairs and a trigonal pyramidal geometry, like NH3. We cannot predict the HAsH angle accurately, but we know that it is less than 109.5° because the repulsion of the bonding electron pairs in the As—H bonds by the lone pair on As is greater than the repulsion between the bonding pairs.

(b) The Lewis structure of OF2 is

There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral.

Example 10.1

Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the O and F atoms). Thus, removing the two lone pairs leaves us with two bonding pairs and a bent geometry, like H2O. We cannot predict the FOF angle accurately, but we know that it must be less than 109.5° because the repulsion of the bonding electron pairs in the O−F bonds by the lone pairs on O is greater than the repulsion between the bonding pairs.

(c) The Lewis structure of is

Example 10.1

There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral. Because there are no lone pairs present, the arrangement of the bonding pairs is the same as the electron pair arrangement. Therefore, has a tetrahedral geometry and the ClAlCl angles are all 109.5°.

(d) The Lewis structure of is

There are five electron pairs around the central I atom; therefore, the electron pair arrangement is trigonal bipyramidal. Of the five electron pairs, three are lone pairs and two are bonding pairs.

Example 10.1

Recall that the lone pairs preferentially occupy the equatorial positions in a trigonal bipyramid (see Table 10.2). Thus, removing the lone pairs leaves us with a linear geometry for , that is, all three I atoms lie in a straight line.

(e) The Lewis structure of C2H4 is

The C=C bond is treated as though it were a single bond in the VSEPR model. Because there are three electron pairs around each C atom and there are no lone pairs present, the arrangement around each C atom has a trigonal planar shape like BF3, discussed earlier.

Example 10.1

Thus, the predicted bond angles in C2H4 are all 120°.

Comment (1) The ion is one of the few structures for which the bond

angle (180°) can be predicted accurately even though the central atom contains lone pairs.

(2) In C2H4, all six atoms lie in the same plane. The overall planar geometry is not predicted by the VSEPR model, but we will see why the molecule prefers to be planar later. In reality, the angles are close, but not equal, to 120° because the bonds are not all equivalent.

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Dipole Moments and Polar Molecules

H F

electron richregion

electron poorregion

δ+ δ−

µ = Q x rQ is the charge

r is the distance between charges

1 D = 3.36 x 10-30 C m

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Behavior of Polar Molecules

field off field on

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Bond moments and resultant dipole moments in NH3 and NF3.

Example 10.2

Predict whether each of the following molecules has a dipole moment:

(a)BrCl

(b)BF3 (trigonal planar)

(c)CH2Cl2 (tetrahedral)

Example 10.2

Strategy

Keep in mind that the dipole moment of a molecule depends on both the difference in electronegativities of the elements present and its geometry.

A molecule can have polar bonds (if the bonded atoms have different electronegativities), but it may not possess a dipole moment if it has a highly symmetrical geometry.

Example 10.2

Solution

(a) Because bromine chloride is diatomic, it has a linear geometry. Chlorine is more electronegative than bromine (see Figure 9.5), so BrCl is polar with chlorine at the negative end

Thus, the molecule does have a dipole moment. In fact, all diatomic molecules containing different elements possess a dipole moment.

Example 10.2

(b) Because fluorine is more electronegative than boron, each B−F bond in BF3 (boron trifluoride) is polar and the three bond moments are equal. However, the symmetry of a trigonal planar shape means that the three bond moments exactly cancel one another:

An analogy is an object that is pulled in the directions shown by the three bond moments. If the forces are equal, the object will not move. Consequently, BF3 has no dipole moment; it is a nonpolar molecule.

Example 10.2

(c) The Lewis structure of CH2Cl2 (methylene chloride) is

This molecule is similar to CH4 in that it has an overall tetrahedral shape. However, because not all the bonds are identical, there are three different bond angles: HCH, HCCl, and ClCCl. These bond angles are close to, but not equal to, 109.5°.

Example 10.2

Because chlorine is more electronegative than carbon, which is more electronegative than hydrogen, the bond moments do not cancel and the molecule possesses a dipole moment:

Thus, CH2Cl2 is a polar molecule.

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Change in Potential Energy of Two Hydrogen Atomsas a Function of Their Distance of Separation

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Change in electron density as two hydrogen atoms approach each other.

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Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals

1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals.

2. Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

3. Covalent bonds are formed by:

a. Overlap of hybrid orbitals with atomic orbitals

b. Overlap of hybrid orbitals with other hybrid orbitals

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Formation of sp3 Hybrid Orbitals

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Formation of Covalent Bonds in CH4

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Predict correctbond angle

sp3-Hybridized N Atom in NH3

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Formation of sp Hybrid Orbitals

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Formation of sp2 Hybrid Orbitals

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# of Lone Pairs+

# of Bonded Atoms Hybridization Examples

2

3

4

5

6

sp

sp2

sp3

sp3d

sp3d2

BeCl2

BF3

CH4, NH3, H2O

PCl5

SF6

How do I predict the hybridization of the central atom?

1. Draw the Lewis structure of the molecule.

2. Count the number of lone pairs AND the number of atoms bonded to the central atom

Example 10.3

Determine the hybridization state of the central (underlined) atom in each of the following molecules:

(a)BeH2

(b)AlI3

(c)PF3

Describe the hybridization process and determine the molecular geometry in each case.

Example 10.3

Strategy The steps for determining the hybridization of the central atom in a molecule are:

Solution

(a) The ground-state electron configuration of Be is 1s22s2 and the Be atom has two valence electrons. The Lewis structure of BeH2 is

Example 10.3

There are two bonding pairs around Be; therefore, the electron pair arrangement is linear. We conclude that Be uses sp hybrid orbitals in bonding with H, because sp orbitals have a linear arrangement (see Table 10.4). The hybridization process can be imagined as follows. First, we draw the orbital diagram for the ground state of Be:

By promoting a 2s electron to the 2p orbital, we get the excited state:

Example 10.3

The 2s and 2p orbitals then mix to form two hybrid orbitals:

The two Be−H bonds are formed by the overlap of the Be sp orbitals with the 1s orbitals of the H atoms. Thus, BeH2 is a linear molecule.

Example 10.3

(b) The ground-state electron configuration of Al is [Ne]3s23p1. Therefore, the Al atom has three valence electrons. The Lewis structure of AlI3 is

There are three pairs of electrons around Al; therefore, the electron pair arrangement is trigonal planar. We conclude that Al uses sp2 hybrid orbitals in bonding with I because sp2 orbitals have a trigonal planar arrangement. The orbital diagram of the ground-state Al atom is

Example 10.3

By promoting a 3s electron into the 3p orbital we obtain the following excited state:

The 3s and two 3p orbitals then mix to form three sp2 hybrid orbitals:

The sp2 hybrid orbitals overlap with the 5p orbitals of I to form three covalent Al−I bonds. We predict that the AlI3 molecule is trigonal planar and all the IAlI angles are 120°.

Example 10.3

(c) The ground-state electron configuration of P is [Ne]3s23p3. Therefore, P atom has five valence electrons. The Lewis structure of PF3 is

There are four pairs of electrons around P; therefore, the electron pair arrangement is tetrahedral. We conclude that P uses sp3 hybrid orbitals in bonding to F, because sp3

orbitals have a tetrahedral arrangement. The hybridization process can be imagined to take place as follows. The orbital diagram of the ground-state P atom is

Example 10.3

By mixing the 3s and 3p orbitals, we obtain four sp3 hybrid orbitals.

As in the case of NH3, one of the sp3 hybrid orbitals is used to accommodate the lone pair on P. The other three sp3 hybrid orbitals form covalent P−F bonds with the 2p orbitals of F. We predict the geometry of the molecule to be trigonal pyramidal; the F−F angle should be somewhat less than 109.5°.

Example 10.4

Describe the hybridization state of phosphorus in phosphorus pentabromide (PBr5).

Example 10.4

Strategy Follow the same procedure shown in Example 10.3.

Solution The ground-state electron configuration of P is [Ne]3s23p3. Therefore, the P atom has five valence electrons. The Lewis structure of PBr5 is

There are five pairs of electrons around P; therefore, the electron pair arrangement is trigonal bipyramidal. We conclude that P uses sp3d hybrid orbitals in bonding to Br, because sp3d hybrid orbitals have a trigonal bipyramidal arrangement.

Example 10.4

The hybridization process can be imagined as follows. The orbital diagram of the ground-state P atom is

Promoting a 3s electron into a 3d orbital results in the following excited state:

Example 10.4

Mixing the one 3s, three 3p, and one 3d orbitals generates five sp3d hybrid orbitals:

These hybrid orbitals overlap the 4p orbitals of Br to form five covalent P−Br bonds. Because there are no lone pairs on the P atom, the geometry of PBr5 is trigonal bipyramidal.

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sp2 Hybridization of Carbon

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Unhybridized 2pz orbital (gray), which is perpendicular to the plane of the hybrid (green) orbitals.

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Sigma bond (σ) – electron density between the 2 atoms

Pi bond (π) – electron density above and below plane of nuclei of the bonding atoms

Bonding in Ethylene, C2H4

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Sigma (σ) and Pi Bonds (π)

Single bond 1 sigma bond

Double bond 1 sigma bond and 1 pi bond

Triple bond 1 sigma bond and 2 pi bonds

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Another View of π Bonding in Ethylene, C2H4

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sp Hybridization of Carbon

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Bonding in Acetylene, C2H2

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Another View of the Bonding in Acetylene, C2H2

Example 10.5

Describe the bonding in the formaldehyde molecule whose Lewis structure is

Assume that the O atom is sp2-hybridized.

Example 10.5

Strategy Follow the procedure shown in Example 10.3.

Solution There are three pairs of electrons around the C atom; therefore, the electron pair arrangement is trigonal planar. (Recall that a double bond is treated as a single bond in the VSEPR model.) We conclude that C uses sp2 hybrid orbitals in bonding, because sp2 hybrid orbitals have a trigonal planar arrangement. We can imagine the hybridization processes for C and O as follows:

Example 10.5

Carbon has one electron in each of the three sp2 orbitals, which are used to form sigma bonds with the H atoms and the O atom. There is also an electron in the 2pz orbital, which forms a pi bond with oxygen. Oxygen has two electrons in two of its sp2 hybrid orbitals. These are the lone pairs on oxygen. Its third sp2 hybrid orbital with one electron is used to form a sigma bond with carbon. The 2pz orbital (with one electron) overlaps

with the 2pz orbital of C to form a pi bond.

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Molecular orbital theory – bonds are formed from interaction of atomic orbitals to form molecular orbitals.

O

O

No unpaired e-

Should be diamagnetic

Experiments show O2 is paramagnetic

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Energy levels of bonding and antibonding molecular orbitals in hydrogen (H2).

A bonding molecular orbital has lower energy and greater stability than the atomic orbitals from which it was formed.

An antibonding molecular orbital has higher energy and lower stability than the atomic orbitals from which it was formed.

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Constructive and Destructive Interference

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Two Possible Interactions Between Two Equivalent p Orbitals

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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.

Molecular Orbital (MO) Configurations

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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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General molecular orbital energy level diagram for the second-period homonuclear diatomic molecules Li2, Be2, B2, C2, and N2.

Example 10.6

The ion can be prepared by bombarding the N2 molecule with fast-moving electrons.

Predict the following properties of :

•electron configuration

•bond order

•magnetic properties

•bond length relative to the bond length of N2 (is it longer or shorter?)

Example 10.6

Strategy From Table 10.5 we can deduce the properties of ions generated from the homonuclear molecules.

How does the stability of a molecule depend on the number of electrons in bonding and antibonding molecular orbitals?

From what molecular orbital is an electron removed to form the ion from N2?

What properties determine whether a species is diamagnetic or paramagnetic?

Example 10.6

Solution From Table 10.5 we can deduce the properties of ions generated from the homonuclear diatomic molecules.

(a)Because has one fewer electron than N2, its electron configuration is

(b) The bond order of is found by using Equation (10.2):

bond order = ½ (9 − 4) = 2.5

(c) has one unpaired electron, so it is paramagnetic.

Example 10.6

(d) Because the electrons in the bonding molecular orbitals are responsible for holding the atoms together, should have a weaker and, therefore, longer bond than N2. (In fact, the bond length of is 112 pm, compared with 110 pm for N2.)

Check Because an electron is removed from a bonding molecular orbital, we expect the bond order to decrease. The ion has an odd number of electrons (13), so it should be paramagnetic.

Chapter 10 Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals

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Transcript of Chapter 10 Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals