Memorial University of Newfoundland
Midterm Exam 2 Blue
St. John’s, Newfoundland and Labrador
Chemistry 1050 Fall 2012
October 31st, 2012
Name:_________________________
Time: 50 Minutes
MUN #:_________________________
Dr. Peter Warburton
READ THE FOLLOWING CAREFULLY!
1.
This exam has 8 pages with 2 sections. SECTION A is comprised of short answer
questions, SECTION B is comprised of longer-answer questions. Ensure that this
examination paper is complete, i.e. that all pages are present.
2.
Failure to submit this paper in its entirety at the end of the examination may result
in disqualification.
3.
A Periodic Table and physical constants are provided on the last page of this
paper and may be detached for use during the examination.
4.
Answer each question in the space provided. Should you require more space, use
the back of the page and indicate clearly where this has been done.
5.
When answering questions show all relevant formulae, all calculations, and
justify all simplifying assumptions.
6.
Correct units should be maintained in all steps of a calculation.
7.
Numerical answers should be reported with the correct significant digits.
Do not write in the enclosed area below.
Question
Value
Section A
28
B1
4
B2
4
B3
4
Total
40
Mark
Page 1 of 8
Chemistry 1050 Midterm Exam 2
Section A
Short Answer
(28 Marks)
The value of each part of a question is given in the square brackets in the left
margin beside each question.
[3]
A1.
The wavelength of helium nuclei travelling at ninety percent of the speed of light
is 0.000370 pm. Calculate the mass of a helium nucleus.
This question is an evaluation of the de Broglie wavelength of a particle
λ=
ℎ
=
ℎ
=
λ
6.62610 1 1
10 0.9002.99810 0.000370 1 = 6.64 x 10-27 kg
[4]
A2.
Using the periodic table, give symbols for the element which fits the description:
a) smallest group 16 atom :
O
b) smallest period 2 atom :
Ne
c) most metallic atom :
Fr
d) lowest first ionization energy in group 14 :
[2]
A3.
Circle the two species that are diamagnetic:
N
[1]
A4.
Pb
Ba
S
Zn
I
Write the chemical reaction that corresponds to the second electron affinity of
oxygen.
O- (g) + e- O2- (g)
[2]
A5.
The quantum numbers listed in the table are for two electrons in the same atom.
Complete the table. For type of orbital, use format 2p, 5d, etc
Quantum numbers
n = 4, l = 0, ml = 0, ms = ½
n = 3, l = 2, ml = 1, ms = -½
Type of orbital # of radial nodes # of angular nodes
4s
3
0
3d
0
2
Page 2 of 8
Chemistry 1050 Midterm Exam 2
[1]
A6.
What is the difference between an angular node and a radial node?
A radial node means that an electron cannot be found at a certain distance from the
nucleus, while an angular node means an electron cannot be found in a certain direction.
[5]
[2]
A7.
A8.
Write shorthand ground state electron configurations for the atoms/ions listed:
Cl
[Ne] 3s23p5
Zr
[Kr] 4d25s2
N3-
[He] 2s22p6
In3+
[Kr] 4d10
Cu
[Ar] 3d104s1
NOTE: this is one of the exceptions to orbital filling
Explain:
a) Hund’s rule
When orbitals of identical energy (degenerate orbitals) are available, electrons initially
occupy these orbitals singly and with parallel spins.
b) the Pauli exclusion principle
No two electrons in an atom can have exactly the same set of the four quantum numbers.
[2]
[1]
A9.
A10.
Portions of orbital diagrams representing the ground-state electron configurations
of certain elements are shown here. Which of them violate the Pauli exclusion
principle? Which violate Hund’s rule?
Violate Pauli exclusion principle
(a) and (f)
Violate Hund’s rule
(b), (d) and (e)
What is the difference between an angular node and a radial node?
This was a repeat of Question A6, and so it was
ignored. The exam was marked out of 39 instead of
40.
Page 3 of 8
Chemistry 1050 Midterm Exam 2
[2]
A11.
What is the difference between Ψ2 and 4πr2Ψ2?
Ψ2 is the probability of finding an electron at a given point in space, while 4πr2Ψ2 is the
probability of finding an electron at a given distance r from the nucleus.
[2]
A12.
Label the 1s, 2s and 2p orbitals on the radial probability distribution below:
[1]
A13.
Rank the following ions in order of decreasing ionic size: S2-, Ca2+, P3-
These three ions are all isoelectronic, in that they all have 18 electrons. Isoelectronic
ions decrease in size as the atomic number increases. The order of decreasing size takes
us from largest to smallest, so
P3- > S2- > Ca2+4.0 g mol-1
Page 4 of 8
Chemistry 1050 Midterm Exam 2
Section B
Longer Answer Questions
(12 Marks)
The value of each part of a question is given in the square brackets in the left
margin beside each question.
[4]
B1.
Calculate the energy, frequency and wavelength of the line in the Lyman series of
hydrogen that corresponds to the transition from n = 3 to n = 1.
This is an application of the Bohr model of the hydrogen atom (Z = 1), where the
difference in energy between two allowed orbits of the electron is
1
1
− *
&' &)
1
1
∆! = 1 2.17910 % − *
3
1
1
1
∆! = 2.17910 − 9 1
8
∆! = 2.17910 − 9
∆! = −1.93710 ∆! = " #$ %
The negative sign tells us the atom is losing energy. That is, the atom is emitting a
photon with energy 1.93710 For a photon, the relationship between energy and frequency is E = hν, so ν = E/h
ν=
1.93710 !
=
= 2.92310+ ℎ 6.62610 For a photon, the relationship between wavelength and frequency is c = λν, so λ = c/ν
, 2.99810 λ= =
= 1.02610- = 102.6&
ν
2.92310+ For 1 bonus mark: Is this line in the visible spectrum? If it is, what color is this light?
If it isn’t, then what name do we use to describe this part of the electromagnetic
spectrum?
Note: no bonus mark will be given if there are no numerical answers for the frequency
AND wavelength in the problem above.
A wavelength of 102.6 nm is not in the visible spectrum. It is a photon in the ultraviolet
region.
Page 5 of 8
Chemistry 1050 Midterm Exam 2
[4]
B2.
A photon of wavelength 242 nm has just enough energy to ionize a sodium atom.
Calculate the ionization energy of 1 mole of sodium atoms
For a photon, the relationship between energy and frequency is E = hν, and the
relationship between wavelength and frequency is c = λν, so ν = c/λ, so overall
hc 6.62610 2.99810 !=
=
101 λ
242& 1&
E = 8.209 x 10-19 J
This, of course, is the energy required to ionize ONE sodium atom. If we want to know
the ionization energy for a mole of sodium atoms, we must multiply this energy by
Avogadro’s number
E = (8.209 x 10-19 J)(6.022 x 1023 mol-1) = 4.943 x 105 J mol-1 = 494.3 kJ mol-1
Only has vinyl chloride, where it is 1 mole of product. Overall reaction needs 2 moles
vinyl chloride as a product, so we multiply by a factor of two to get reaction which
has enthalpy change 2x∆H. Only has ethylene, where it is one mole of product.
Overall reaction needs 2 moles ethylene as a reactant, so we multiply by a factor of
two and reverse the reaction to get to get reaction which has enthalpy change 2x∆H. Only has HCl, where it is 2 moles of reactant. Overall reaction needs 2
moles HCl as a reactant, so reaction is the same as . Reaction includes Cl2O and
ClO2 which are not part of the overall equation, so we must eliminate these using
reaction . Since 2 moles Cl2O are reactants in , we need 2 moles Cl2O as products in
our last reaction. We have the two moles in , but as reactants, so we need to reverse
the reaction to get and take the negative of the enthalpy to get -1x∆H.
4 C (s, graphite) + 3 H2 (g) + Cl2 (g) → 2 H2CCHCl (g)
2 H2CCH2 (g) → 4 C (s, graphite) + 4 H2 (g)
2 ClO2 (g) + H2 (g) + 2 HCl (g) → Cl2O (g) + 2 H2O (g) + ½ O2 (g) + Cl2 (g)
Cl2O (g) + 322 O2 (g) → 2 ClO2 (g)
∆H = 70.0 kJ
∆H = -104.6 kJ
∆H = -423.7 kJ
∆H = 124.7 kJ
Adding the reactions together gives
4 C (s, graphite) + 3 H2 (g) + Cl2 (g) + 2 H2CCH2 (g) + 2 ClO2 (g) + H2 (g) + 2 HCl (g)+ Cl2O (g) + 3/2 1 O2 (g)→
2 H2CCHCl (g) + 4 C (s, graphite) + 4 H2 (g)+ Cl2O (g) + 2 H2O (g) + ½ O2 (g) + Cl2 (g)+ 2 ClO2 (g)
After cancellation we see we have
2 H2CCH2 (g) + 2 HCl (g) + O2 (g)→ 2 H2CCHCl (g) + 2 H2O (g)
which is our overall reaction. The enthalpy change for the overall reaction will be the
sum of the enthalpy changes for , which give ∆H = -333.6 kJ
Page 6 of 8
Chemistry 1050 Midterm Exam 2
[4]
B3.
Calculate the zero-point energy of an electron (me = 9.109 x 10-31 kg) in a one
dimensional box of length L = 185.0 nm.
The zero-point energy of an electron in a 1D box is the lowest possible energy the
electron can have, which is NOT zero. In other words, it is in the ground state where n =
1:
& ℎ ! = =
83 4
1 6.62610 89.10910 5185.0& ! = 1.76010 101 7
1&
Right now it looks as if the units don’t work out to energy.
If we recall that 1 J = 1 kg m2 s-2and express ONE of our J units in this way
! = 1.76010 Now we see the units DO actually work out to energy units of J. Because the unit
cancellation is not obvious, the units in this question were a larger part of the evaluation
mark for this question than would normally be seen on my exams.
The zero-point energy of the electron in this box is 1.760 x 10-24 J.
The End
Page 7 of 8