• 沒有找到結果。

# Bond Theory, and

N/A
N/A
Protected

Share "Bond Theory, and"

Copied!
115
0
0

(1)

(2)

(3)

### Solution

The molecular geometry of NO3 is determined by the number of electron groups around the central atom (N).

Begin by drawing a Lewis structure of NO3.

NO3 has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three resonance structures:

The hybrid structure is intermediate between these three and has three equivalent bonds.

Use any one of the resonance structures to determine the number of electron groups around the central atom.

Determine the molecular geometry of NO3.

### Example 10.1 VSEPR Theory and the Basic Shapes

(4)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

The electron geometry that minimizes the repulsions between three electron groups is trigonal planar.

Since the three bonds are equivalent (because of the resonance structures), they each exert the same repulsion on the other two and the molecule has three equal bond angles of 120°.

### For Practice 10.1

Determine the molecular geometry of CCl4. Continued

(5)

### Procedure For…

Predicting Molecular Geometries

### Solution

Step 1 Draw the Lewis structure for the molecule.

PCl3 has 26 valence electrons.

Step 2 Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, triple bonds, and single electrons each count as one group.

The central atom (P) has four electron groups.

Predict the geometry and bond angles of PCl3.

### Example 10.2 Predicting Molecular Geometries

(6)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Step 3 Determine the number of bonding groups and the number of lone pairs around the central atom.

These should sum to your result from step 2. Bonding groups include single bonds, double bonds, and triple bonds.

Three of the four electron groups around P are bonding groups and one is a lone pair.

Step 4 Refer to Table 10.1 to determine the electron geometry and molecular geometry. If no lone pairs are present around the central atom, the bond angles will be that of the ideal geometry. If lone pairs are present, the bond angles may be smaller than the ideal geometry.

Continued

(7)

Continued

### Example 10.2 Predicting Molecular Geometries

(8)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

The electron geometry is tetrahedral (four electron groups) and the molecular geometry—the shape of the molecule—is trigonal pyramidal (three bonding groups and one lone pair). Because of the presence of a lone pair, the bond angles are less than 109.5°.

### For Practice 10.2

Predict the molecular geometry and bond angle of ClNO.

Continued

(9)

### Procedure For…

Predicting Molecular Geometries

### Solution

Step 1 Draw the Lewis structure for the molecule.

ICl4 has 36 valence electrons.

Step 2 Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, triple bonds, and single electrons each count as one group.

The central atom (I) has six electron groups.

Predict the geometry and bond angles of ICl4.

### Example 10.3 Predicting Molecular Geometries

(10)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Step 3 Determine the number of bonding groups and the number of lone pairs around the central atom.

These should sum to your result from step 2. Bonding groups include single bonds, double bonds, and triple bonds.

Four of the six electron groups around I are bonding groups and two are lone pairs.

Step 4 Refer to Table 10.1 to determine the electron geometry and molecular geometry. If no lone pairs are present around the central atom, the bond angles will be that of the ideal geometry. If lone pairs are present, the bond angles may be smaller than the ideal geometry.

Continued

(11)

Continued

### Example 10.3 Predicting Molecular Geometries

(12)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

The electron geometry is octahedral (six electron groups) and the molecular geometry—the shape of the molecule—is square planar (four bonding groups and two lone pairs). Even though lonepairs are present, the bond angles are 90° because the lone pairs are symmetrically arranged and do not compress the I Cl bond angles.

### For Practice 10.3

Predict the molecular geometry of I3. Continued

(13)

### Solution

Begin by drawing the Lewis structure of CH3OH. CH3OH contains two interior atoms: one carbon atom and one oxygen atom. To determine the shape of methanol, determine the geometry about each interior atom as follows:

Using the geometries of each of these, draw a three-dimensional sketch of the molecule as shown here:

Predict the geometry about each interior atom in methanol (CH3OH) and make a sketch of the molecule.

### Example 10.4 Predicting the Shape of Larger Molecules

(14)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

### For Practice 10.4

Predict the geometry about each interior atom in acetic acid and make a sketch of the molecule.

Continued

(15)

### Solution

Draw the Lewis structure for the molecule and determine its molecular geometry.

The Lewis structure has three bonding groups and one lone pair about the central atom. Therefore the molecular geometry is trigonal pyramidal.

Determine if the molecule contains polar bonds. Sketch the molecule and superimpose a vector for each polar bond. The relative length of each vector should be proportional to the electronegativity difference between the atoms forming each bond. The vector should point in the direction of the more electronegative atom.

The electronegativities of nitrogen and hydrogen are 3.0 and 2.1, respectively. Therefore, the bonds are polar.

Determine if NH3 is polar.

### Example 10.5 Determining if a Molecule Is Polar

(16)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Determine if the polar bonds add together to form a net dipole moment. Examine the symmetry of the vectors (representing dipole moments) and determine if they cancel each other or sum to a net dipole moment.

The three dipole moments sum to a net dipole moment. The molecule is polar.

### For Practice 10.5

Determine if CF4 is polar.

Continued

(17)

### Procedure For…

Hybridization and Bonding Scheme

### Solution

Step 1 Write the Lewis structure for the molecule.

BrF3 has 28 valence electrons and the following Lewis structure:

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The bromine atom has five electron groups and therefore has a trigonal bipyramidal electron geometry.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron geometry (see Table 10.3).

A trigonal bipyramidal electron geometry corresponds to sp3d hybridization.

Write a hybridization and bonding scheme for bromine trifluoride, BrF3.

### Example 10.6 Hybridization and Bonding Scheme

(18)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Continued

### Example 10.6 Hybridization and Bonding Scheme

(19)

Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Continued

### Example 10.6 Hybridization and Bonding Scheme

(20)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

### For Practice 10.6

Write a hybridization and bonding scheme for XeF4. Continued

(21)

### Procedure For…

Hybridization and Bonding Scheme

### Solution

Step 1 Write the Lewis structure for the molecule.

Acetaldehyde has 18 valence electrons and the following Lewis structure:

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The leftmost carbon atom has four electron groups and a tetrahedral electron geometry. The rightmost carbon atom has three electron groups and a trigonal planar geometry.

Write a hybridization and bonding scheme for acetaldehyde,

### Example 10.7 Hybridization and Bonding Scheme

(22)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Continued

### Example 10.7 Hybridization and Bonding Scheme

(23)

Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Continued

### Example 10.7 Hybridization and Bonding Scheme

(24)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

### For Practice 10.7

Write a hybridization and bonding scheme for HCN.

Continued

(25)

### Procedure For…

Hybridization and Bonding Scheme

### Solution

Step 1 Write the Lewis structure for the molecule.

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The molecule has two interior atoms. Since each atom has three electron groups (one double bond and two single bonds), the electron geometry about each atom is trigonal planar.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron geometry (see Table 10.3).

A trigonal planar geometry corresponds to sp2 hybridization.

Use valence bond theory to write a hybridization and bonding scheme for ethene, H2C＝CH2

### Example 10.8 Hybridization and Bonding Scheme

(26)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Continued

### Example 10.8 Hybridization and Bonding Scheme

(27)

Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Continued

### Example 10.8 Hybridization and Bonding Scheme

(28)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

### For Practice 10.8

Use valence bond theory to write a hybridization and bonding scheme for CO2.

### For More Practice 10.8

What is the hybridization of the central iodine atom in I3? Continued

(29)

### Solution

The H2 ion has three electrons. Assign the three electrons to the molecular orbitals, filling lower energy orbitals first and proceeding to higher energy orbitals.

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing the result by two.

Use molecular orbital theory to predict the bond order in H2. Is the H2 bond a stronger or weaker bond than the H2 bond?

### Example 10.9 Bond Order

(30)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Since the bond order is positive, H2 should be stable. However, the bond order of H2 is lower than the bond order of H2 (which is 1); therefore, the bond in H2 is weaker than in H2.

### For Practice 10.9

Use molecular orbital theory to predict the bond order in H2+. Is the H2+ bond a stronger or weaker bond than the H2 bond?

Continued

(31)

### Solution

Write an energy level diagram for the molecular orbitals in N2. Use the energy ordering for N2.

Draw an MO energy diagram and determine the bond order for the N2 ion. Do you expect the bond to be stronger or weaker than in the N2 molecule? Is N2 diamagnetic or paramagnetic?

### Example 10.10 Molecular Orbital Theory

(32)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

The N2 ion has 11 valence electrons (5 for each nitrogen atom plus 1 for the negative charge). Assign the electrons to the molecular orbitals beginning with the lowest energy orbitals and following Hund’s rule.

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing the result by two.

Continued

### Example 10.10 Molecular Orbital Theory

(33)

The bond order is 2.5, which is a lower bond order than in the N2 molecule (bond order = 3); therefore, the bond is weaker. The MO diagram shows that the N2 ion has one unpaired electron and is therefore paramagnetic.

### For Practice 10.10

Draw an MO energy diagram and determine the bond order for the N2+ ion. Do you expect the bond to be stronger or weaker than in the N2 molecule? Is N2+ diamagnetic or paramagnetic?

### For More Practice 10.10

Use molecular orbital theory to determine the bond order of Ne2. Continued

### Example 10.10 Molecular Orbital Theory

(34)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

### Solution

Determine the number of valence electrons in the molecule or ion.

Number of valence electrons

= 4 (from C) + 5 (from N) +

1 (from negative charge) = 10

Write an energy level diagram using Figure 10.15 as a guide. Fill the orbitals beginning with the lowest energy orbital and progressing upward until all electrons have been assigned to an orbital. Remember to allow no more an two electrons (with paired spins) per orbital and to fill degenerate orbitals with single electrons (with parallel spins) before pairing.

Use molecular orbital theory to determine the bond order of the CN ion. Is the ion paramagnetic or diamagnetic?

(35)

Continued

### Molecules and Ions

(36)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

### Solution

Calculate the bond order using the appropriate formula:

If the MO diagram has unpaired electrons, the molecule or ion is paramagnetic. If the electrons are all paired, the molecule or ion is diamagnetic.

Since the MO diagram has no unpaired electrons, the ion is diamagnetic.

### For Practice 10.11

Use molecular orbital theory to determine the bond order of NO. (Use the energy ordering of O2.) Is the molecule paramagnetic or diamagnetic?

Continued

(37)

### Taste

(38)

Chapter Ten Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

(39)

### B-A-B angle?

(40)

Chapter Ten Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

(41)

### possible, they make a

(42)

Chapter Ten Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

(43)

(44)

2

B B

(45)

### Linear Geometry

(46)

Three Electron Groups:

Trigonal Planar Electron Geometry

2

3

(47)

(48)

2

3

4

(49)

4

4+

(50)

2

3

4

5

(51)

(52)

2

3

4

5

6

(53)

(54)

(55)

(56)

3

2

(57)

3

4

(58)

4

3

2

2

(59)

5

4

(60)

5

4

3

2

(61)

5

4

3

2

2

3

(62)

6

5

(63)

6

5

4

2

(64)

(65)
(66)

2

4

2

4

(67)

(68)

-30

(69)

(70)

(71)
(72)

(73)
(74)

2

2

2

4

(75)

2

2

(76)

(77)

2

2

(78)

(79)

3

0

(80)

(81)

(82)

(83)

(84)

(85)

(86)

(87)

(88)

(89)

(90)

(91)

(92)

(93)

(94)

(95)

2

3

2

3

4

3

2

(96)

3

(97)

-

2

(98)

(99)
(100)

(101)
(102)

(103)

## Interaction of p Orbitals

(104)

(105)

(106)

(107)
(108)

(109)

2s

(110)

(111)

(112)

(113)

(114)

### 10.7

(115)

A) the approximate atomic number of each kind of atom in a molecule B) the approximate number of protons in a molecule. C) the actual number of chemical bonds in a molecule D)

You are given the wavelength and total energy of a light pulse and asked to find the number of photons it

Step 1: With reference to the purpose and the rhetorical structure of the review genre (Stage 3), design a graphic organiser for the major sections and sub-sections of your

Wang, Solving pseudomonotone variational inequalities and pseudocon- vex optimization problems using the projection neural network, IEEE Transactions on Neural Networks 17

Define instead the imaginary.. potential, magnetic field, lattice…) Dirac-BdG Hamiltonian:. with small, and matrix

Monopolies in synchronous distributed systems (Peleg 1998; Peleg

For R-K methods, the relationship between the number of (function) evaluations per step and the order of LTE is shown in the following

Microphone and 600 ohm line conduits shall be mechanically and electrically connected to receptacle boxes and electrically grounded to the audio system ground point.. Lines in