chapter 8 bonding: general concepts. copyright © houghton mifflin company. all rights reserved.crs...

Chapter 8

Bonding: Bonding: General ConceptsGeneral Concepts

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 8–2

QUESTIONThe force of attraction between two unlike ions can be calculated using Coulomb’s law. Which of the following is a valid conclusion that can be drawn, using Coulomb’s law, about ionic attractions?

1. As two positive ions approach, the energy of interactionwill become more negative.

2. Smaller ions will have less energy of interaction.3. A +3 ion interacting with a –1 ion will have a greater

energy of interaction than a –3 ion and a +1 ion.4. The distance between ions is inversely related to the energy

of interaction between the ions.


ANSWERChoice 4 applies a proper chemical application of Coulomb’s law. 30 illustrates Coulomb’s law. In general, the greater the charge density (facilitated by small volume for a large charge) the greater attractive or repulsive force an ion can develop. Since smaller ions may be able to have small separating distances between ions, their interacting force will increase. Coulomb’s law relates that the interacting force would increase as the distance between ions decreases.

Section 8.1: Types of Chemical Bonds


QUESTION


QUESTION (continued)What useful interpretation about bonding, energy, and inter-atomic distance can be drawn from this graph?

1. The maximum chemical stability for the two atoms lies tothe left of the graph where the inter-atomic distanceapproaches zero.

2. The best distance for a stable bond is not the smallest possible distance between two atoms.3. The best distance for a stable bond is the smallest possible distance between two atoms.4. None of these choices seem to be logical conclusions to

me.


ANSWERChoice 2’s interpretive statement agrees with the observations shown in the figure. The arrangement with lowest energy shows the atoms getting closer, but if they continue to get closer the energy goes up quickly. (If forced closer than the bonding distance, repulsive nuclear forces between the atoms cause an increase in the energy of the system.)

Section 8.1: Types of Chemical Bonds


QUESTIONGallium, located near the middle of the periodic table, has an intermediate electronegativity value. Of the following, which offers the more accurate explanation for the fact that as Ga combines with elements in group VIA the bonding in such combinations becomes less polar from the lighter group VI atoms to the more massive group members?

1. More massive atoms tend to have a higher electronegativity.

The difference between Gallium’s electronegativity and theatoms in group VIA becomes larger as Gallium combineswith increasingly more massive group VI atoms.

2. The decrease in polarity of the bonding between Ga andgroup VIA atoms follows the gradual decrease inelectronegativity in the group.

3. I am not yet ready to explain the trend.


ANSWERChoice 2 properly connects the drop in bond polarity to the fact that as electronegativity in a group decreases the difference between a fixed element’s electronegativity, such as Ga, and those others will decrease. See page 334 for a view of this trend.

Section 8.2: Electronegativity


QUESTIONThe bonds between carbon and oxygen in CO2 and the bonds between carbon and oxygen in acetic acid have exactly the same difference in electronegativity. Yet CO2 is a non-polar gas, whereas acetic acid is polar. Which statement provides a factual reason for this difference?


QUESTION (continued)1. CO2 has no dipole moment because the symmetry of the

C–O bonds places them 180°C apart; thus, the polarity predicted in the bonds is cancelled by molecular shape. Since acetic acid is polar, the shape must not allow for canceling the bond polarity.

2. Polarity in a bond does not always predict polarity in a molecule because the dipole moment could be increased causing the molecule to be less polar.

3. The partial negative charge of each oxygen, compared to carbon, is decreased in CO2 because it gets shared among two atoms.

4. The C–O bond in acetic acid remains polar because the other parts of the molecule form a tetrahedron.


ANSWERChoice 1 provides the connection between molecular shape and possible cancellation of bond polarity. Since CO2 has one oxygen on each side of the central C atom (180°) apart, the increased pull on electron charge of oxygen over carbon gets canceled by the same strength of pull from the other oxygen.

Section 8.3: Bond Polarity and Dipole Moments


QUESTIONOf the following, which would be isoelectronic with Rb+ and smaller in size?

1. Se2–

2. Mo6+

3. Kr4. K+


ANSWERChoice 2 provides the correct response to both conditions. Mo6+ is isoelectronic (same number of electrons, now that it has lost 6 electrons to have 36), and since it is the only one with more protons, it will have a greater nuclear attraction, which assists in decreasing ionic size.

Section 8.4: Ions: Electron Configurations and Sizes


Arrange the following, without consulting any specific listed radii, in order from smallest to largest size (radius).

S2– ; Cl–; K+; Ar

1. Ar; S2– ; Cl–; K+

2. K+; Ar; Cl–; S2– 3. S2– ; Cl–; Ar ; K+

4. S2– ; Cl–; K+; Ar

QUESTION


ANSWERChoice 2 provides the correct sequence from smallest to largest. They are all isoelectronic so the nuclear charge (atomic number) is a major factor. The higher the nuclear charge, the smaller the size.

Section 8.4: Ions: Electron Configurations and Sizes


QUESTIONWhat evidence exists that the drive for atoms to obtain an octet is not the main impetus for forming solid ionic crystals?

1. Many non-octet-based ionic compounds exist.2. When the various energy requirements for changing solid

metals and gaseous non-metals into solid ionic crystals aretaken into account, the negative electron affinity value istypically far too large.

3. Without the large crystal lattice energy value, obtaining theoctet for both ions is still endothermic.

4. These don’t seem to make sense; I thought getting an octetwas the primary impetus for ionic crystal formation.


ANSWERChoice 3 correctly points out that the largest contributor in forming a solid ionic crystal from a solid metal and gaseous non-metal is the crystal lattice energy. Check the visual representation of this in Figure 8.9 on page 344.

Section 8.5: Energy Effects in Binary Ionic Compounds


QUESTIONWhich would likely have the larger negative value for its crystal lattice energy CaO or KF and predict approximately how much larger your choice will be over the compound not selected?

1. CaO; approximately two times larger2. CaO; approximately four times larger3. KF; approximately two times larger4. KF; approximately four times larger


ANSWERChoice 2 correctly uses Coulomb’s law to make the prediction. Since the attraction between gaseous ions will be directly influenced by charge and inversely related to distance, the larger +2 and –2 charges of Ca and O respectively will combine for an effect four times larger than the +1 and –1 of K and F.

Section 8.5: Energy Effects in Binary Ionic Compounds


QUESTION


QUESTION (continued)Based on this graph, approximately what % ionic character and what electronegativity difference correlate to an ionic compound?

1. 25 %; 2.02. 50%; 1.03. 75%; 2.04. 50 %; 1.7


ANSWERChoice 4 shows both the accepted definition for bond character to be considered ionic (50%), and the electronegativity difference used to classify bonds as ionic (1.7). Remember, though, that any compound that conducts an electric current when melted will be considered ionic.

Section 8.6: Partial Ionic Character of Covalent Bonds


QUESTION


QUESTION (continued)Using tabulated average bond energies, determine the H value for the electrolysis of water:

2 H2O 2H2 + O2

1. –425 kJ2. +355 kJ3. + 941 kJ4. + 509 kJ


ANSWER

Choice 4 correctly uses stoichiometry, bond energy values and the proper sign convention. The table value for bond energy is written for one bond. Determining that 2 H2O represents four O–H bonds is very important in the calculation.

Section 8.8: Covalent Bond Energies and Chemical Reactions


QUESTIONAccording to our current bond formation theory, number of shared electron pairs, bond length, and bond energy are related. Which of the following correctly states the relationship among these three?

1. More shared e– pairs ensures a shorter bond length witha lower bond energy.

2. Fewer shared e– pairs ensures a longer bond length with a lower bond energy.

3. More shared e– pairs ensures a longer bond length, with a lower bond energy.

4. Fewer shared e– pairs ensures a shorter bond length, with a higher bond energy.


ANSWERChoice 2 correctly indicates the relationship between bond length, shared e– pairs and bond energy. Greater bond energy indicates a stronger bond, which is enhanced when more electron pairs are involved in bonding. The strength of a bond is inversely related to the bond length.

Section 8.8: Covalent Bond Energies and Chemical Reactions


QUESTIONThe formaldehyde compound, also known as methanal, has been linked to indoor air pollution and related health effects. It can be used in some cases as a disinfectant and is found in some resins and glues. The correctly drawn Lewis structure of H2CO would have how many unshared electrons?

1. Zero unshared e–

2. One unshared e–

3. Two unshared e–

4. Four unshared e–


ANSWERChoice 4 is correct if the proper Lewis structure is drawn. The carbon atom shares a double bond with oxygen in this compound (in addition to two C–H bonds). This leaves four unshared e– around oxygen.

Section 8.10: Lewis Structures


QUESTIONDinitrogen monoxide has several uses ranging from a dentistry anesthetic to automobile racing enhancement. Starting with two possible basic structures given here, diagram two different Lewis structures. If your first structure contains four unbonded electrons around oxygen and your second structure contains six unbonded electrons around oxygen, how many bonds would be between the N atoms in the first and second compounds?

N–N–O N–N–O

1. 2, 32. 1, 33. 2, 24. 1, 2


ANSWERChoice 1 provides the correct number of bonds for both NNO compounds. This is an example of resonance. By following the Lewis electron dot structure rules for both compounds, the first compound should have double bonds between N=N and N=O. The second compound would have a triple bond between the N atoms and a single bond between N and O.

Section 8.12: Resonance


QUESTIONThe N2O molecule has sometimes found use as an aerosol propellant. One structure that satisfies the Lewis electron dot rules is N=N=O. In this case, what would be the formal charge of the middle N atom?

1. Zero2. +13. –14. I am not sure how to determine the formal charge on

an atom in a compound.


ANSWERChoice 2 for this structure is correct. Add together the total unbonded e– and ½ the shared e–. Subtract this from the assigned valence number of electrons based on the group number for the element from the periodic table. For the middle N, these values would be: 5 – (0 + ½(8)) = + 1.

Section 8.12: Resonance


QUESTIONThe compound BF3 can react with silicon to help etch computer chips. After diagramming the complete Lewis structure for this compound, determine its shape and provide one of the following names for the shape you drew.

1. Trigonal planar2. Trigonal bipyramidal3. T–shaped4. Linear


ANSWER

Choice 1 correctly names the shape of BF3. This compound has three equal electron densities around the B atom. These densities will repel each other equally thus producing a compound with 120° angles for the F to B to F bonds.

Section 8.13: Molecular Structure: The VSEPR Model


QUESTIONWhat is the shape and number of unbonded electron pairs around the I atom in the ICl2

– ion?

1. Trigonal planar; 32. Bent (V shape); 23. Linear; 34. Bent (V shape); 1


ANSWERChoice 3 correctly describes the shape for ICl2

– and reports the correct number of unbonded electron pairs that help produce the shape. After providing octets for both Cl atoms and adding in the extra electron (noted as ion charge in the formula), there remains three pairs of unbonded electrons around the I atom. These arrange symmetrically around the central I atom. This allows the two Cl atoms to be 180° apart from each other, which draws them into a linear position with I in the center.

Section 8.13: Molecular Structure: The VSEPR Model

chapter 8 bonding: general concepts. copyright © houghton mifflin company. all rights reserved.crs...

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