Review Assignment #8

Chapter 9: Bonding Theories
 

28.
 

s/s overlap	pz/pz overlap	s/pz overlap

34.
 

36. (a) A double bond typically consists of one s bond and one p bond. (b) A triple bond usually consists of one s bond and 2 p bonds. (c) Since rotation about the bond axis will greatly affect the overlap of parallel p orbitals of a p bond, the molecule becomes more rigid, that is, rotation is hindered, to preserve the maximum orbital overlap.
 

38. The Lewis structures for N₂ and N₂H₄ are shown below.

The hybridization on N₂ is sp and that of N₂H₄ is sp³. Atoms with sp³ hybridization typically do not participate in multiple bonding since there are no available p orbitals to form such bonds. The only exceptions to this come with atoms that can expand their octets, that is, elements in the 3^rd row or higher of the Periodic Table. Since nitrogen can not expand its octet, no sp³ hybridized nitrogen (or carbon, or oxygen) can form multiple bonds.
 

40. The CH₂ carbon of ketene is sp² hybridized; the CO carbon is sp hybridized. (b) There are 16 valence electrons in ketene. (c) There are four s bonds, so there are eight valence electrons that make up these bonds. (d) There are two p bonds, so there are four electrons that make up those bonds. (e) There are four unpaired electrons; these are all on the oxygen atom.
 

42. The bond angle will be: 1) about 120º, 2) about 120º, and 3) about 109º. The hybridization of the central atoms of these angles are: sp², sp² and sp³, respectively.
 

47. (a) The s and s* orbitals of H₂⁺ are illustrated below. Also shown is the energy diagram for the ion.
 

(b) There is only one electron in H₂⁺. The Lewis dot structure is shown below - it follows none of the rules established for "normal" Lewis structures.

(c) The electron configuration for the ion is: s_1s¹ (d) The bond order of H₂⁺ is ½. (Recall, the bond order is given by the number of bonding electrons minus the number of atibonding electrons, divided by two). (e) If the electron in the bonding molecular orbital is promoted to the antibonding orbital, the bond order would go from +½ to -½. This is an unstable arrangement and the molecule would break apart to reduce the nuclear repulsion that would exist in the absence of bonding electron interactions.
 

48. This problem is exactly like 47, only different. The s and s* orbitals are exactly the same, and the energy diagram is similar, but it has a total of three electrons instead of one.

(b) A Lewis structure of H₂^- cannot be drawn without exceeding the "octet" rule for hydrogen (the "duet" rule? H can only have two electrons under normal conditions). (c) The electron configuration for the ion is: s_1s² s^*_1s¹ (d) The bond order of H₂^- is ½. (e) As explained above, an excited H₂^- would have a bond order of -½ and would be expected to decompose.
 

52. Using the following energy diagram, the electron configurations and bond orders for Ne₂ and Ne₂⁺ can be determined by filling in the orbitals according to the Aufbau Principle.
 

Since the bond order of Ne₂ is zero, it is not expected to be stable under any conditions. Ne₂⁺, on the other hand, should exist under some conditions since the bonding interactions are stronger than the anti-bonding interactions.
 

54. Diamagnetism is the magnetic behavior that results from a species having no unpaired electrons. Diamagnetic materials are slightly repelled by applied magnetic fields. Using the above energy diagram, the electron configurations, and hence the dia- or paramagnetism of species can be predicted
 
 
 

56. Using the energy diagram above, the electron configuration and bond orders for N₂^- and N₂ can be determined.
 

Since the bond order for N₂^- is less than that of N₂, it is expected to have a weaker and shorter bond.

66.
 

	formula	Lewis structure	No. of s bonds	No. of p bonds
a	CO2		2	2
b	NCS-		2	2
c	H2CO		3	1
d	HCOOH		4	1

Chapter 15: Introduction to Chemical Equilibria

8.
 

a	N2 (g) + O2 (g) « 2 NO (g)
b	Ti (s) + 2 Cl2 (g) « TiCl4 (l)
c	2 C2H4 (g) + 2 H2O (g) « 2 C2H6 (g) + O2 (g)
d	FeO (s) + H2 (g) « Fe (s) + H2O (g)
e	4 HCl (g) + O2 (g) « 2 H2O (l) + 2 Cl2 (g)

15.     SO₂ (g) + ½ O₂ (g) « SO₃ (g)

a	SO3 (g) « SO2 (g) + ½ O2 (g)
b	2 SO2 (g) + O2 (g) « 2 SO3 (g)
c	SO3 (g) « SO2 (g) + ½ O2 (g)

19.     2 HI (g) « H₂ (g) + I₂ (g)

23.     2 NO (g) + 2 H₂ (g) « N₂ (g) + 2 H₂O (g)

To find the equilibrium concentrations, we need to set up a table of changes, as shown below.

 
Since the equilibrium concentration of NO is given as 0.062 M, x can be calculated as follows:

0.062 = 0.100 - 2x
x = 0.019

So, the equilibrium concentrations are:

[H₂]_eq = 0.050 - 2(0.019) = 0.012 M
[N₂]_eq = 0.019 M
[H₂O]_eq = 0.10 + 2(0.019) = 0.138 M

Substituting into the definition of K yields:

 

Since we are solving for [Cl₂], we can rearrange the equation to:

 

To rearrange for [Br]:

The concentration of Br₂ is:

Now, [Br] can be calculated using the above equation,

So, the mass of Br is:

b.     2 HI (g) « H₂ (g) + I₂ (g)

The equilibrium pressures of H₂ and I₂ are:

 

Rearranging the equilibrium expression for P_HI yields:

So, the mass of HI is:

35.     2 NO (g) « N₂ (g) + O₂ (g)

To find the equilibrium concentrations, we need to set up a table of concentrations as shown below:

Substituting into the equilibrium expression,

x² = (2.4´ 10³) (0.0400 - 0.800x + 4x²)

Rearranging yields,

0 = 9,599x² -1,920x + 96

So, for the quadratic equation:  a = 9,599; b = -1,920; c = 96

Solving the quadratic equation for x yields: x = 0.101 and 0.0990. Only the second solution makes physical sense as the first would give an equilibrium concentration of NO that is negative.

So, x = 0.0990

The equilibrium concentrations are then:

[NO] = 0.200 - 2(0.0990) = 0.00200 M
[O₂] = 0.0990 M
[N₂] = 0.0990 M

39a.     PH₃BCl₃ (s) « PH₃ (g) + BCl₃ (g)

The table of concentrations that is relevant is:

Substituting into the equilibrium expression gives:

x = 0.0432

So the equilibrium concentrations are:

[PH₃]_eq = 0.0432 atm
[BCl₃]_eq = 0.0432 atm

41.     I₂ (g) + Br₂ (g) « 2 IBr (g)

 

0.250 - 2x + 4x² = 280x²

0 = 276x² + 2x - 0.250

So, solving the quadratic equation when a = 276, b = 2 and c = -0.250, yields x = 0.0267 and -0.0339. Only the positive value of x makes physical sense. So, x = 0.0267, and the equilibrium concentrations are:

[I₂] = [Br₂] = 0.0267 M
[IBr] = 0.500 - 2(0.0267) = 0.447 M

44.

a. an increase in the partial pressure of a reactant will shift the reaction to the right, that is, more product will be formed.

b. since the reaction is endothermic, heat can be viewed as a reactant; therefore increasing the temperature has the same effect as increasing the concentration of reactants and will shift the equilibrium to the right.

c. removing CO₂, a reactant, will shift the equilibrium to the left.

d. decreasing the total pressure will have no effect on the equilibrium position since the number of gaseous molecules is the same for the reactants and products.

e. removing some of the product will shift the equilibrium to the right.

f. adding a catalyst will have no effect on the equilibrium position.

52.     H₂ (g) + S (s) « H₂S (g)

The equilibrium concentrations are:

 

The amount of sulfur can be ignored because it is a solid material and, as such, its concentration is independent of the actual amount present.

54.     A (g) « 2 B (g)

A table of pressure changes can be set up as follows:

Since the partial pressure of A is 0.36 atm at equilibrium, we can write:

0.36 = 0.55 - x
x = 0.19

So, P_B = 2x = 0.38 atm

P_total = P_A + P_B = 0.36 atm + 0.38 atm = 0.74 atm

 

b) Since the equilibrium shifts to the left at higher temperature, the reaction as written must be exothermic.

c) By LeChâtelier's Principle, the equilibrium position shifts to the product side as products are removed from a system. In the case of the exhaust system, as the Ni(CO)₄ is formed in the pipes, it is swept out with the waste gases. Thus, equilibrium with CO is never really achieved and more Ni(CO)₄ is continually formed because its concentration never increases to its equilibrium value.