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### Skills to develop

• Practical problem solving.
• ### Problems Require Written Answers

Here are some questions from old examinations. These questions require written answers from you.

You should provide an organized, nicely laied out, and easy to understand answer for your marker in order to earn high marks.

Problems require written answers are usually long, and we have separate these problems into several short problems in order to provide them in the DIALOGUE.

Do this problems yourself on paper.

### Example Written Questions

• ```Explain how Dalton's Law of partial                   !^^!
pressures applies when one collects                   !  ----- gas
a gas over water as shown in this       gas ----\     !~~!   collected
diagram.                                     |   \\   !  !   |
|~~~~~\\~!||!~~~|
|       \\||!   ---- water
Enter a letter to get some hint.             |_________\|____|
```
Hint...
As the gas bubbles through the water, it becomes saturated with H2O. The partial pressure (vapor pressure) of H2O depends on the temperature of the gas. The total pressure of collected gas P(total) = P(H2O) + P(gas). When the water levels are made equal, P(H2O) + P(gas) = P(barometric). The vapor pressure of H2O increases as T increases. The two gases H2O and collected gas behaves independently.

• A1. Draw and label the molecular orbital energy level diagram for NO(g) and insert the electrons in the appropriate orbitals for the ground state of this molecule. Hint...
For NO, it has 5+6 = 11 valence electrons (sig2s)2 (sig*2s)2 (piy)2 (piz)2 (sig2pz)2 (piy)1 [OR (pix)1] You should provide an AO-MO energy level diagram.

• A2. Write the ground state molecular orbital configuration for each of the following three species. NO, N2, O2(-). Hint...
```NO:  (sig2s)2 (sig*2s)2 (piy)2 (piz)2 (sig2pz)2 (piy)1 [OR (pix)1]
N2 has 7+7 = 14 electrons
(sig2s)2 (sig*2s)2 (piy)2 (pix)2 (sig2pz)2
O2(-) has 8+8+1 = 17 electrons (including inner shell electrons)
(sig1s)2 (sig*1s)2 (sig2s)2 (sig*2s)2 (piy)2 (piz)2 (sig2px)2 (piy)2 (piz)1
```

• A3. Compare the bond lengths (BL), bond energies (BE), bond orders (BO) and magnetism (mag) of the three species in above by filling the following table.
```   Note:  For bond lengths (BL) and bond energies (BE),
use the symbols:  S - shortest or smallest, I - intermediate
L - longest or largest
For bond orders (BO), use numerical values (1, 1.5, 2 etc)
For magnetism (mag), Use: dia - for diamagnetic para- for paramagnetic
```
Hint...
```A nice way to do this is to use a table:
|Species   |   BL   |    BE   |   BO    |   mag    |
|----------|--------|---------|---------|----------|
| NO       |   I    |    I    |  2.5    |   para   |
| N2       |   S    |    L    |  3      |   dia    |
| O2(-)    |   L    |    S    |  1.5    |   para   |
```

• For TWO PAIRS chosen from the following three, deduce and compare the shapes of the two species in each pair. Explain your reasoning. The central atom is underlined in each case. [For example, if you choose to consider the first pair, compare the shape of CO3(2-) with H3O(+), etc].
```  CO3(2-) and H3O(+)          XeF4 and SF4           H2S and ICl2(-)
-             -             --       -               -     -
```
Hint...
```CO3(2-): trigonal planar (sp2 hybridization used for C
H3O(+):  trigonal pyramidal (sp3 hybridization use for O,
4th position of tetrahedral is a lone pair of electrons.
You do the other two pairs

B1. For each of these three species, sketch the geometry, showing all
valence electrons and explain HOW you deduced each geometry.
I3(-)                    PCl3                    BrF5

Hint:...
PCl3: trigonal pyramidal; same shape as NH3 (draw the picture)
sp3 hybrid AO of P used, 4th position occupied by lone electron pairs.
You do the other two + SF4, PCl5, ClF3, XeF2, NH3, CH4, SO2, etc.

B2.  Briefly explain why PCl5 exists, but NCl5 does not exist.
Hint...
The 3d orbitals are available for P to form sp3d hybridized orbitals,
but the d orbitals are not available for N. ....

C1.  Define allotrope.
Hint...
Allotrope: two solids or liquids of the same elements that have different
properties.  Examples: graphite and diamond; red and white phosphorus.
A common terminology you should know.

C2.  Assign formal charges and oxidation numbers to each of the
atoms in both structures.
:F::
/
::O = Xe::            [ ::O = C = N:: ] (-)
\
:F::

Hint...
Formal charge:  F   O  Xe   ||  O  C  N
0   0  0    ||  0  0  -1
oxidation no.:  F   O  Xe   ||  O  C  N
0   0  0    ||  0  0  -1

C3.  Balance the following redox equation.  Show your work.
V + H2O  ---> HV2O7 (3-) + H2
```
Hint...
You will have more fun doing this one than reading answers on my screen. Write down the problem and practice it. The following few questions give hints to this problem.

• C3a. Balance the following redox equation. Show your work.
V + H2O ---> HV2O7 (3-) + H2
(a) What is oxidized?

Consider...
V -> HV2O7(3-) Oxidation state of V change from 0 to +5 <-- oxidized
H2O -> H2 Oxidation state of H change from +1 to 0 <-- reduced

Excellent...
H is reduced

• C3b. Balance the following redox equation. Show your work.
V + H2O ---> HV2O7 (3-) + H2
(b) To balance 2V --> HV2O7 (3-)
How many electrons should be added to the right hand side?

Consider...
To balance charges: 2V --> HV2O7 (3-) + 10 e-

Excellent...
You should realize that we put in 2 V on the left side to balance V

• C3c. Balance the following redox equation. Show your work.
V + H2O ---> HV2O7 (3-) + H2
(c) How many electrons should be added to the left side to balance
?e + H2O --> H2

Consider...
2 e- + H2O --> H2

Excellent...
2 H from -1 to 0

• C3d. Balance the following redox equation. Show your work.
V + H2O ---> HV2O7 (3-) + H2
(d) What factor should you multiply equation (I) in order to cancel the e- when you add the following two equations to get an overall equation?
(I) 2 e- + H2O --> H2
(II) 2V --> HV2O7(3-) + 10 e-

Consider...
The overall equation is 2V + 5H2O --> HV2O7(3-) + 5H2

Excellent...
The 10 e- came from V

• C3e. Balance the following redox equation. Show your work.
V + H2O --> HV2O7(3-) + H2
(e) To balance the charge, how many OH- ions should be added in
2V + 5H2O + ? OH(-) --> HV2O7(3-) + 5H2 ?

Consider...
2V + 5H2O + 3 OH(-) --> HV2O7(3-) + 5H2 ?

Excellent...
Because of HV2O7(3-)

• C3f. Balance the following redox equation. Show your work.
V + H2O --> HV2O7(3-) + H2
(f) What do you add to balance the rest of the atoms
2V + 5H2O + 3 OH(-) --> HV2O7(3-) + 5H2 ?

Consider...
2V + 5H2O + 3OH(-) --> HV2O7 (3-) + 5H2 + H2O
or
2V + 4H2O + 3OH(-) --> HV2O7 (3-) + 5H2
will balance it.

Excellent...
H2O on the right hand side or reduce the H2O on the left hand

• C3g. Balance the following redox equation. Show your work.
V + H2O --> HV2O7(3-) + H2
(g) When the equation is balanced, what is the coefficient for H2?

Excellent...
You can get this number without balance the whole equation. But it is a risky practice!

• D1. Use the bond energy data to estimate the enthalpy of formation of HOCl(g)
```   [Bond energies: H2:436.0kJ/mol; O2:498.2kJ/mol;  Cl2:242.6kJ/mol;
HO:463.0kJ/mol; ClO:220.0kJ/mol; HCl:431.8kJ/mol]
```

Excellent...

```   1       1        1
- H2  + - O2  +  - Cl2   ----> HOCl      = H-O-Cl
2       2        2
delta H  = 436/2 + 498.3/2 + 242.6/2 - 463 - 220) kJ mol^-1
= -94.55 kJ mol^-1
```
Provide some comment.

• D2. Use the enthalpy of combustion data given to determine the standard enthalpy of formation of ethane, C2H6 (g).
```      C(graphite) + O2(g)   --> CO2(g)               -393.5 kJ/mol
C2H6(g) + 7/2 O2(g)   --> 2CO2(g) + 3H2O(l)    -1559.8kJ/mol
H2(g) + 1/2 O2(g)     --> H2O(l)               -295.5 kJ/mol
```

Excellent...

```     2C(graphite) + 2O2(g)   --> 2CO2(g)         2 * -393.5 kJ/mol
2CO2(g) + 3H2O(l)  --> C2H6(g) + 7/2 O2(g)      1559.8 kJ/mol
3H2(g) + 3/2 O2(g)     --> 3H2O(l)          3*  -295.5 kJ/mol
____________________________________________________________________
2C (graphite) + 3H2 (g) ---> C2H6 (g)              -83.7 kJ/mol
```

• E. Exactly 1.00 mol of pure NOBr(g) is placed in a 10.0 L vessel and heat to 350 K, at which temperature, it undergoes partial decomposition according to the reaction:
NOBr(g) --> NO(g) + 1/2 Br2(g)
If the final total pressure in the vessel is found to be 340.0 kPa, what fraction of the original NOBr molecules are dissociated?

Consider...
Calculate n by using n = PV / RT from total pressure and volume. n = 340 * 10 / (8.314 * 340) = 1.1684 mol. So, we have 0.1684 mol Br2, which corresponds to 0.3368 mol of NOBr decomposing.

Excellent...
A test of stoichiometry and gas law.

• F1. If you accidentally spilled 50 mL of battery acid (38 % by weight H2SO4, and density of 1.29 g/mL), what minimum volume of 2.5 M sodium bicarbonate would be required to neutralize the spill according to the following equation?
H2SO4 + 2NaHCO3 --> 2H2O + 2CO2 + Na2SO4

Excellent...
This is intended for you to practice.

• F2. A 0.91 g sample of sucrose (C12H22O11) was placed in a bomb calorimeter with excess amount of O2. Complete combustion of sucrose to CO2 and H2O produced a temperature change of 3.50 K. If the heat capacity of the calorimeter is 4250 J/K, calculate the standard molar enthalpy of combustion of sucrose at 25 C.

Intended...

• G1. Write the electronic configuration for the ground state of In, Ni(2+), and Cr.

Hint...
Atomic number for In, Ni, and Cr are 49, 28, and 24 for In, Ni, and Cr respectively. Try the Aufbau procedure.

• G2. Draw a labelled graph which shows how the ionization energy depends on Z for the elements in the second period (Li to Ne). Explain the general trend and any exceptions to this trend.

Hint...
A test on the variation of property of elements of the periodic Table.

• G3. Define valence electrons in an atom. How many valence electrons are there for arsenic?

Consider...
You should know the names of common elements. Arsenic is As, below N, and P.

Excellent...
Practice the first half of the question yourself.

• H. Estimate the enthalpy of sublimation of solid calcium (Ca(s)) by making use of the following information.
```    Enthalpy of dissociation of Cl2(g)           242.6 kJ/mol
Standard enthalpy of formation of CaCl2(s)  -795.0
First ionization energy of Ca(g)             590.
Second ionization energy of Ca(g)            1145.
Third ionization energy of Ca(g)            4912.
First electron affinity of Cl(g)            -348.7
First electron affinity of Ca(g)             156.
Lattice energy of CaCl2                     2253.
```

Consider...

```    Ca(2+) + 2 Cl
| 242           |
Ca(2+) + Cl2    |
|               |  348.7 * 2
| 1145       2 Cl(-) + Ca(2+)
|                  |
Ca(+) + Cl2        |
| 590              |
Ca(g) + Cl2(g)     | 2253
| ?                |
Ca(s) + Cl2(g)     |
| 795              |
CaCl2(s)------------
```