Chem 1000A Final Examination - Spring 2004 - SOLUTIONS
April 20th, 2004: 9:00 to 12.00 am
Your name _______________
Instructor: Dr. M. Gerken
Student ID _______________
Time: 3h
No. of pages: 3 + 2
Total achievable marks: 116.5
Report all your answers using the correct number of significant figures. Show all units and their
conversions throughout your calculations. Use SI units for your calculations, unless noted otherwise.
WRITE ALL YOUR ANSWERS INTO THE EXAMINATION BOOKLET!
Question 1 (5 Marks)
Complete the following table.
27
Symbol
Al3+
Number of electrons
_10________
Number of neutrons
_14________
Number of protons
_13________
Overall charge
3+
_112Cd2+____
46
64
_48________
2+
Question 2 (4 Marks)
What is the maximum number of electrons that can be associated with the following combinations of
quantum numbers? Zero is a possible answer. Explain your answer briefly.
(a) n = 3, l =-1, ml = -1
This set of q.n. is not allowed, since l can’t be a negative number: zero electrons
(b) n = 2, l = 1, ml = 2, ms = ½
This set of q.n. is not allowed, since ml must not be larger than l: zero electrons
(c) n = 3
This specifies the third shell. There are 3s, 3p, and 3d orbitals present in the third shell: 18 electrons.
(d) n = 3, l = 2
This set of q.n.’s specify the 3d orbitals: 10 electrons.
Question 3 (4 Marks)
Draw a 3dxy and a 3dz2 orbital.
x
z
z
x
Question 4 (11 Marks)
You combine 2.51 g of NO and 1.32 g of F2 inside a 1.12 L Nickel can at -196 °C. You allow the reaction
mixture to warm to room temperature (25.5 °C). (a) What are the partial pressures of NO(g) and F2(g) and
what is the total pressure that you would expect if you would assume no reaction was taking place? (b) In
reality, NO and F2 react already at low temperature to form gaseous NOF. Write the balanced reaction
equation. (c) Which reactant is the limiting reagent? (d) What is the final pressure inside the Nickel can at
25.5 °C after complete reaction?
1
(a) M(NO) = 30.0061 g/mol, M(F2) = 37.9968 g/mol
n(NO) = 2.51 g/(30.0061 g/mol) = 0.0836 mol
n(F2) = 1.32 g/(37.9968 g/mol) = 0.0347 mol
pV = nRT, T = (25.5 + 273.15)K = 298.7 K; V = 1.12 L × 1 m3/1000 L = 0.00112 m3
p(NO) = nRT/V = 0.0836 mol × 8.3145 J K-1 mol-1 × 298.7 K/0.00112 m3 = 185000 J/m3
= 185000 kg m2 s-2 m-3 = 185000 kg m-1 s-2= 185000 kg m-1 s-2= 185000 Pa
p(F2) = nRT/V = 0.0347 mol × 8.3145 J K-1 mol-1 × 298.7 K/0.00112 m3 = 76800 Pa
ptot = 185000 Pa + 76800 Pa = 262000 Pa
(b) 2 NO(g) + F2(g) → 2 NOF(g)
(c) n(NO): n(F2) = 0.0836 mol : 0.0347 mol = 2.41 : 1
required ratio: 2:1. Hence: excess NO and deficiency of fluorine.
F2 is the limiting reagent.
(d) For every fluorine molecule, two NOF molecules are generated.
p(NOF) = 2×76800 Pa = 153600 Pa
p(residual NO) = 185000 Pa - 153600 Pa = 31400 Pa
ptot = 153600 Pa + 31400 Pa = 185000 Pa
Question 5 (6.5 Marks)
Balance the following redox reaction in a basic aqueous solution and show that the final equation has the
correct electron, material, and charge balance. Give the half-reactions and show as many steps as possible.
S2- + I2 → SO42- + IS-II 2- + I02 →
S+VIO-II42- + I-I -
oxidation half-reaction (unbalanced): S2- → SO42reduction half-reaction (unbalanced): I2 → Iprinciple element balance:
oxidation half-reaction (unbalanced): S2- → SO42reduction half-reaction (unbalanced): I2 → 2Ioxygen balance:
oxidation half-reaction (unbalanced): 4 H2O + S2- → SO42reduction half-reaction (unbalanced): I2 → 2Ihydrogen balance (basic solution):
oxidation half-reaction (balanced): 4 H2O + S2- + 8OH- → SO42-+ 8 H2O
reduction half-reaction (balanced): I2 → 2IInserting electrons:
oxidation half-reaction (balanced): 4 H2O + S-II 2- + 8OH- → S+VIO42-+ 8 H2O + 8ereduction half-reaction (balanced): I02 + 2e- → 2I-I electron balanced
oxidation half-reaction (balanced): 4 H2O + S-II 2- + 8OH- → S+VIO42-+ 8 H2O + 8ereduction half-reaction (balanced): 4 I02 + 8e- → 8I-I combination:
4 H2O + S2- + 8OH-+ 4 I2 + 8e- → SO42-+ 8 H2O + 8e- + 8I –
simplification:
8OH- + S2- + 4 I2 → SO42-+ 4 H2O + 8I –
material balance:
8H, 8O, S, 8 I|8H, 8O, S, 8I
charge balance:
(8-)+(2-) = 10- |(2-)+(8-)=10correct electron, material, and charge balance!
2
|×4
Question 6(8 Marks)
Nitrogen triiodide, NI3, is unstable and decomposes into the elements.
(a) Write a balanced reaction equation for the decomposition.
(b) If you decompose 1.25 g of NI3 at exactly 1 atm of pressure and 25.5 °C, how much iodine (in g) and
how much nitrogen (in L) is produced.
(c) What are the oxidation states of N and I in NI3?
(a) 2 NI3 → N2 + 3I2
(b) M(NI3) = 394.7217 g/mol; M(N2) = 28.0134 g/mol; M(I2) = 253.81 g/mol;
n(NI3) = 1.25 g/(394.7217 g/mol) = 3.17 mmol
n(N2) = 3.17 mmol × ½ = 1.58 mmol
n(I2) = 3.17 mmol × 3/2 = 4.76 mmol
m(I2) = 4.76 × 10-3 mol ×253.81 g/mol = 1.21 g
pV = nRT
V = nRT/p = 1.58 × 10-3 mol × 8.3145 J K-1 mol-1 × 298.7 K/ 101325 Pa = 3.87 × 10-5 J/Pa
= 3.87 × 10-5 (kg m2 s-2) (kg-1 m1 s2) = 3.87 × 10-5 m3 = 0.0387 L
(c) N: -III ; I: +I
Question 7(23 Marks)
(i) Draw Lewis structures for the following compounds. (Central atom is underlined) (ii) What are the
electron-pair geometries and molecular geometries according to the VSEPR model. (iii) Do these molecules
have molecular dipole moments. Indicate the dipole moment if applicable. (iv) Specify the intermolecular
forces present in these compounds.
a) BF3
b) PCl3
c) SO32- in Na2SO3
d) NOF
(a)
..
:F :
B
: ..F :
:F..:
electron pair geometry: trigonal planar
molecular geometry: trigonal planar
no molecular dipole moment
Intermolecular forces: London dispersion forces
(b)
_
..
..
.. P
Cl
.. :
:Cl
..
:Cl
.. :
electron pair geometry: tetrahedral
molecular geometry: trigonal pyramidal
molecular dipole moment
3
Intermolecular forces: dipole-dipole interactions, dipole-induced dipole interactions, London dispersion
forces
(c)
_
..
S .. :O
O
.. :
..
:O
.. :-
+ two more resonance structures for the SO32- anion
2 Na+
electron pair geometry: tetrahedral
molecular geometry: trigonal pyramidal
molecular dipole moment
Intermolecular forces: ion-ion interactions, ion-dipole interactions, ion-induced dipole interactions, dipoledipole interactions, dipole-induced dipole interactions, London dispersion forces
(d)
_
..
..
N O
..
:..F :
electron pair geometry: trigonal planar
molecular geometry: bent
molecular dipole moment
Intermolecular forces: dipole-dipole interactions, dipole-induced dipole interactions, London dispersion
forces
Question 8 (5 Marks)
Which of the following statements are correct, which ones are incorrect:
(a) Colligative properties depend exclusively on the molar mass of a compound. (incorrect)
(b) Colligative properties depend exclusively on the number of particles. (correct)
(c) Colligative properties depend on the number of particles and the average molar mass of these
particles. (incorrect)
(d) The depression of the boiling point is a colligative property. (incorrect)
(e) Dissolution of compounds increases the boiling point of the solvent in that solution. (correct)
Question 9 (4 Marks)
When 2.3 g of ethanol (C2H5OH) is dissolved in 1.00 kg of pure sulfuric acid (H2SO4) the freezing point of
this solution is 9.45 °C. Determine how many particles are formed as one molecule of ethanol is dissolved
in sulfuric acid.
The cryoscopic constant, Kf.p. for ethanol is 6.12 °C kg/mol. Pure sulfuric acid has a freezing point of 10.37
°C.
∆Tf.p. = Kf.p. × molality ; ∆Tf.p. = 10.37 °C – 9.45 °C = 0.92 °C
Molality = ∆Tf.p./ Kf.p. = 0.92 °C/(6.12 °C kg/mol) = 0.1503 mol/kg
n(dissolved particles) = 0.1503 mol/kg ×1.00 kg = 0.1503 mol
M(C2H5OH) = 46.06904 g/mol
n(C2H5OH) = 2.3 g/(46.06904 g/mol) = 0.050 mol
4
For every ethanol molecule dissolved in sulfuric acid, we find three particles in solution.
Question 10 (6.5 Marks)
Determine the specified masses for the following species.
(a) Mass (in g) of one mol of O2
31.9988 g
(b) Mass (in g) of one molecule of O2
31.9988 g mol-1/(6.022 × 1023 mol-1) = 5.314 × 10-23 g
(c) Mass (in u) of one mol of O2
31.9988 u×6.022 × 1023 mol-1 = 1.927 × 1025 u mol-1
(d) Apparent molar mass of air (80% N2 and 20% O2 – ignore the other minor components)
M(N2) = 28.0134 g/mol, M(O2) = 31.9988 g/mol
M(air) = 0.80 ×28.0134 g/mol + 0.20 ×31.9988 g/mol = 28.8105 g/mol
Question 11 (10.5 Marks)
The combustion analysis of 3.372 g of an organic carboxylic acid sample (CxHyOz) produced 7.061 g of
CO2 and 1.445 g of water. (a) Determine the empirical formula of this compound. (b) The molar mass of
this carboxylic acid was determined to be 168.2 g/mol. Determine the molecular formula. (c) Write the
balanced reaction equation of this combustion.
M(CO2) = 44.0098 g/ mol; M(H2O) = 18.01534 g/mol
n(H2O) = 1.445 g/(18.01534 g/mol) =0.08021 mol
n(CO2) = 7.061 g/(44.0098 g/ mol) = 0.1604 mol
n(H) = 0.08021 mol H2O × 2 mol H/1 mol H2O = 0.16042 mol H
n(C) = 0.1604 mol CO2× 1 mol C/1 mol CO2 =0.1604 mol C
m(H) = 0.16042 mol ×1.0079 g/mol =0.16167 g
m(C) = 0.1604 mol × 12.011 g/mol =1.927 g
m(O) = m (acid = CxHyOz) – m(H) – m(C) = 3.372 g -0.16167 g – 1.927 g = 1.283 g
n(O) = 1.283 g/(15.999 g/mol) = 0.0802 mol
n(C) : n(H) : n(O) = 0.1604 mol: 0.16042 mol : 0.0802 mol
≈2:2:1
empirical formula: C2H2O
molar mass of C2H2O: M(C2H2O) = 42.0368 g mol-1
experimental molar mass of benzoic acid: M(acid) = 168.2 g mol-1
M(acid)/ M(C2H2O) = 4
molecular formula of benzoic acid: C8H8O4
C8H8O4 +
8O2 → 8CO2 + 4H2O
5
Question 12 (3 Marks)
Write the electron configuration of Gadolinium (Gd) in the orbital box notation.
Gd:
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s
2s
2p
3s
3p
4s
3d
↑↓
5s
↑↓ ↑↓ ↑↓↑↓ ↑↓
4d
↑↓↑↓ ↑↓
5p
↑↓
6s
↑↓ ↑ ↑
4f
↑↓ ↑↓ ↑↓
4p
↑ ↑ ↑ ↑
Question 13 (5 Marks)
Write the electron configurations of Cr, Cr3+, and Cr6+ using the noble-gas notation. What are the magnetic
properties of Cr, Cr3+, and Cr6+?
Cr: [Ar]4s1 3d5
Cr3+:[Ar]4s0 3d3
Cr6+:[Ar]4s0 3d0
Both, Cr and Cr3+ are paramagnetic. Cr6+ is diamagnetic.
Question 14 (7 Marks)
Carbon dioxide (CO2) is a gas at room temperature. Describe the bonding situation in CO2 using the valence
bond theory. (a) What is the hybridization of the carbon atom? (b) Draw energy diagram indicating the
formation of the hybrid orbitals on carbon starting from the atomic orbitals on C, generating the hybrid
orbitals on C. (c) Describe the bonding (all bonds) in CO2 in terms of orbital overlap.
..
..
O
C
O
..
..
(a) C: sp hybridization
(b) Carbon:
E
↑ ↑
↑ ↑
2p
↑ ↑
px, py
sp
↓↑
2s
(c)
C=O bonds: overlap between the sp hybrid orbital on C and a 2p orbital on O, forming the σ bond and
overlap between the 2p orbital on C with a 2p orbital on O, forming the π bonds
Question 15 (7 Marks)
Carbon forms at least three oxides: Carbonmonoxide (C≡O), carbondioxide (O=C=O), and a suboxide
(O=C=C=C=O). (a) Specify the oxidation states for every oxygen and carbon in these three oxides. (b)
6
What are the hybridizations for the carbon atoms in these three oxides? Are they different or the same?
Explain your answer briefly.
(a) C+II≡O-II
O-II=C+IV=O-II
O-II=C+II=C0=C+II=O-II
(b) The hybridization of all carbon atoms in these compounds is sp hydridization. All the carbon atoms have
a linear electron-pair geometry and require two unhybridized p-orbitals for the formation of two π bonds.
Question 16 (7 Marks)
(a) What is the minimum energy (in kJ/mol) and longest wavelength (in nm) of electromagnetic radiation
that is necessary to ionize He+ to give He2+. (b) You ionize He+ with monochromatic light (light that has one
specific wavelength). You detect the kinetic energy of the photoelectron of 127.3 kJ/mol. What is the
wavelength and frequency of the light that you used?
(a)
∆E = (Efinal - Einitial) = Ry(Z2/nfinal2 – Z2/ninitial2)
For He: Z=2
∆E = En=∞ -En=1 = -Ry(4/∞ - 4/1) = 4Ry =4 × 2.1799 × 10-18 J = 8.7196 × 10-18 J
E = hc/λ
λ = hc/E = 6.626 × 10-34 Js × 2.998× 108 m s-1/ 8.7196 × 10-18 J = 2.278 × 10-8 m = 22.78 nm
(b)
kinetic energy of one photoelectron: Ekin= 127300 J/mol(6.022 × 1023 mol-1) = 2.114 × 10-19 J
energy of photon used = energy required to ionize He+ + kinetic energy of the electron
= 8.7196 × 10-18 J + 2.114 × 10-19 J = 8.931 × 10-18 J
E = hν
ν = E/h = 8.931 × 10-18 J/6.626 × 10-34 Js = 1.348 × 1016 Hz
c=ν λ
λ = c/ ν = 2.998× 108 m s-1 /(1.348 × 1016 Hz) = 2.224× 10-8 m = 22.24 nm
7