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CHE 390F Physical and Inorganic Chemistry
Fall 2006
Mid-Term Examination KEY
Student Name
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#1
#2
#3
#4
#5
#6
#7
Total
/10
/20
/10
/10
/10
/20
/20
/100
The following defines the honour code expections of the Faculty of Applied Science and Engineering.
Please write your name above and adhere to the high standards of the Faculty.
The University and its members have a responsibility to ensure that a climate that might encourage, or
conditions that might enable cheating, misrepresentation or unfairness not be tolerated. To this end, all
must acknowledge that seeking credit or other advantages by fraud or misrepresentation, or seeking to
disadvantage others by disruptive behaviour is unacceptable, as is any dishonesty or unfairness in
dealing with the work or record of a student. It shall be an offence for a student knowingly: (a) to forge
or in any other way alter or falsify any document or evidence required for admission to the University,
or to utter, circulate or make use of any such forged, altered or falsified document, whether the record be
in print or electronic form; (b) to use or possess an unauthorized aid or aids or obtain unauthorized
assistance in any academic examination or term test or in connection with any other form of academic
work; (c) to personate another person, or to have another person personate, at any academic examination
or term test or in connection with any other form of academic work; (d) to represent as one's own any
idea or expression of an idea or work of another in any academic examination or term test or in
connection with any other form of academic work, i.e. to commit plagiarism; (e) to submit, without the
knowledge and approval of the instructor to whom it is submitted, any academic work for which credit
has previously been obtained or is being sought in another course or program of study in the University
or elsewhere; (f) to submit any academic work containing a purported statement of fact or reference to a
source which has been concocted.
Read each question carefully. The exam lasts 1 hr.
If you have any questions - ASK!!!! DO NOT TEAR OFF PAGES - DRAW A BOX AROUND YOUR ANSWER….
Possibly useful information:
R = 8.314 J/mol-K
Avogadro’s number: 6.02 x 1023
Boltzmann’s constant: 1.38 x 10-23 J/K
Planck’s constant: 6.625 x 10-34 J-s
Mass of a proton: 1.67 x 10-24 g
Page 1 of 13
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1. Circle True or False. (Do not guess. +1 point for a correct answer. 0 points if no answer is provided,
-1 point for incorrect answer). (10 points possible)
(a)
True
I wrote my name and student number on the cover sheet.
(b)
True
False
The addition of an electron to an atom of a halogen is an exothermic process
(c)
True
False
It is easier to ionize magnesium (Mg) than to ionize barium (Ba).
(d)
True
False
Nitrogen has lower electron affinity than carbon – both answers acceptable
since magnitude and sign need to be also considered
(e)
True
False
Cl- is bigger than Ca2+.
(f)
True
False
Anions are always bigger than the atoms from which they are formed.
(g)
True
False
The fluoride anion, F--, is diamagnetic
(h)
True
False
An expanded octet may occur in any row except for period 1 or 2.
(i)
True
False
A dipole moment is the product of the number of atoms and the charge
(j)
True
False
The formal charge on P in PCl5 is 0
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2. (a) Sketch the approximate molecular orbital diagram for LiF (10 points)
For full credit, all of the following must have been included
• All axes labelled, and an energy axis shown
• Bonding, non-bonding orbitals labelled
• Fluorine must be clearly shown as being lower in energy than Li since it is the more
electronegative species
• All orbitals must be labelled (σ , σ *, non-bonding, s, px, py, pz)
• Core levels not necessary but could have been shown for completeness (i.e. 1s levels
from the Li and the F)
• Correct number of electrons must have been shown for all levels
• Ordering of bonding and non-bonding orbitals must be correct
• Number of molecular orbitals must equal the number of atomic orbitals
(b) Phosphoric acid is H3PO4. Draw the Lewis structure of phosphate PO43- in which not all of the
P-O bonds are the same to account for the number of hydrogens in the acid. (3 points) Calculate the
bond order based on this structure. (5 points)
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-3
O
-1
P
O
O
-1
O
-1
Bond order: P=0 : 2; P-O- : 1 – aggregate value since resonance form: 5/4
For full credit, all electrons must be shown, formal charges assigned, and overall ion charge
shown as -3. Double and single bonds should be shown to illustrate resonance structures (i.e. one
double bond and three single bonds distributed). Again note that the question stated that you were
to show the phosphate ion structure such that it would reflect the number of H’s in the acid
(H3PO4). This meant that the structure should show 3 equivalent single bonds and one nonequivalent double bond.
(c) From the following list, which is expected to have the largest ionization energy and which will have
the smallest ionization energy and why? Fe–, Fe, Fe+, Fe2+, Fe3+ (5 points)
Lowest : Fe- ; Highest : Fe3+ It always takes more energy to remove an electron from a positively
charged ion than from a neutral atom. Also, depending on what the electronic configuration is of the
cation in question, you may be faced with the situation of trying to remove an electron from a filled
shell, which would require an enormous amount of energy. Full credit requires an explanation
regarding the effect of the nuclear charge on these differently charged species and a comment on the
Zeff value in each case.
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3. Consider the following sets of quantum numbers for an electron in a hydrogen atom:
I. n = 3 l = +2
ml = + 1/2
ms=–1/2
II. n = 3 l = 0
ml = –1
ms=–1/2
III. n = 3 l = +1
ml= –1
ms=–1/2
(a) Which of the three sets (I., II., III.) is an acceptable set of quantum numbers? (2 points)
III
(b) For each of the unacceptable sets, give a concise explanation of why they are unacceptable.
Please specify which explanation belongs to which unacceptable set. (8 points)
I is not an acceptable set. The acceptable range of values are
n: 1, 2, 3…
l: 0, 1, 2, … (n-1)
ml : 0, ± 1, ± 2, ± 3, … ± l
ms : ± 1/2
The magnetic quantum number, which refers to the angular momentum of the electron in the
z-direction, or the orientation of the electron, can only have integer values .
II is not acceptable since range for ml is from 0 to ± l, and l was already defined as 0.
4. Write the electron configuration for the hydride ion H– . Write the electron configuration for a He
atom. Which species is more reactive H– or He? Can you explain why using the electron
configurations? (10 points)
1s
-
H :
1s
He :
1s
H:
We know that He is an inert gas as its 1s orbital is already fully populated. In the case of H,
the 1s orbital is half-filled; however, in its hydride (anionic form, and thus negatively charged state),
it can readily react with positively charged species i.e. Li+ to form ionic species such as Li+H-. This is
less likely to happen in the case of He which would require the addition of an electron to an
unpopulated 2s orbital, which is energetically less favourable, in comparison to the addition of an
electron to the half-filled 1s orbital of H. The main consideration is the balance with the nuclear
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charge. In the case of H, there is only one proton so the stabilization of that second electron is much
lower than in the case of He, which has two protons in its nucleus. The key for this is the negative
charge associated with the hydride, even though there is a filled 1s orbital, and the difference in the
nuclear charge and how it affects the stability of the electrons.
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5.
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Consider the molecule XeF4. (10 points)
(a) Draw the Lewis structure for this molecule
F
F
Xe
F
F
xenon tetrafluoride
Various perturbations on this form were accepted.
The necessary components include formal charge labels on all the atoms.
(b) What is the arrangement of electron density around the Xe atom (in words)?
This is an AB4E2 case (2 lone pairs on the Xe, 4 ligands). The molecular geometry is
square planar with the lone pairs oriented above and below the plane defined by the
four Xe-F. The electrons, both lone pair and bonding, are arranged in an octahedral
configuration – electron density is octahedrally arranged. It is important to recognize
the difference between molecular geometry and electron density arrangement.
F
F
Xe
F
F
(c) What is the hybridization of Xe in the molecule XeF4?
sp3d2: For full credit, a clear explanation as to how this hybridization strategy was
determined was required. This would include an illustration of the pre-hybridization
distribution of electrons in Xe, and the distribution upon hybridization to clearly show
how an sp3d2 hybrid is formed.
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Question 6
(a)
Determine the point group of diborane (10 points). Show all your work and label (or depict)
all symmetry elements on the figure. For clarity, it might be helpful to use multiple sketches
to depict the symmetry elements
H
H
H
H
B
B
H
H
D2h : E, 3 x C2 (x,y,z), i, σ xy, σ xz, σ yz
It is important to realise that identifying the point group does not necessarily identify all the symmetry
elements. It simply allows you to determine the specific point group through effectively a minimum
path decision tree. You need to also clearly show all the existing symmetry elements in the molecule.
In this case, you should have clearly illustrated or shown where they are located and appropriate
labelling must have been used. Specifically, you should have defined the coordinate axes and
appropriately labelled all mirror planes, rotation axes, and inversion centers. The identify element, E,
must have been explicitly identified. It was not sufficient to only state the point group. As the question
indicated, “all symmetry elements” were to be shown and labelled appropriately.
(b)
Reduce the following representations to irreducible representations: (10 points)
C3v
Γ1
E
2C3
3σ v
6
3
2
Γ2
5
-1
-1
Here is the corresponding C3v character table.
3σ v
C3v
E
2C3
A1
1
1
1
A2
1
1
-1
E
2
-1
0
Γ 1 = 3A1 + A2 + E
A1:
A2:
(1/6)[(6)(1)(1) + (3)(1)(2) + (2)(1)(3)] = 3
(1/6)[(6)(1)(1) + (3)(1)(2) + (2)(-1)(3)] = 1
E:
(1/6)[(6)(2)(1) + (3)(-1)(2) + (2)(0)(3)] = 1
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Γ 2: A2 + E
A1:
(1/6)[(5)(1)(1) + (-1)(1)(2) + (-1)(1)(3)] = 0
A2:
(1/6)[(5)(1)(1) + (-1)(1)(2) + (-1)(-1)(3)] = 1
E:
(1/6)[(5)(2)(1) + (-1)(-1)(2) + (-1)(0)(3)] = 2
For this, all work must be shown and the appropriate answer shown in its final form. This
includes appropriate and correct labelling of the irreducible representations.
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Question 7.
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In a recent paper by Yun et al., Inorg. Chem., 42 (7), 2253 -2260, 2003. A New
Quasi-One-Dimensional Ternary Chalcogenide: Synthesis, Crystal Structure, and Electronic Structure
of Nb1+xV1-xS5 (x = 0.18), a novel compound was prepared, the structure of which is shown below in two
different orientations.
View of Nb1+xV1-xS5 down the c-axis (needle axis)
showing an individual layer and the coordination
around the Nb and M atoms. Large filled circles
View of Nb1+xV1-xS5 down the b-axis (needle axis)
are Nb atoms, small filled circles are M atoms, and
showing the stacking of the layers. The M site can
open circles are S atoms.
be occupied by either Nb or V. The approximate
ratio of V to Nb occupancy in the M site is 5:1.
(a) How many S atoms surround a Nb atom? (2 points)
From the diagram, it is clear that there are 6 S atoms arranged around a given Nb
atom and that there are 2 additional S atoms participating from the adjacent offset
rows
(b) How many S atoms surround an M site? Note that the M site can be occupied 18% of the
time by Nb, and 82% of the time by V. (2 points)
Again, in this case, one needs to inspect carefully the two diagrams where it will be
clear that there are 6 S atoms arranged around the M site. The arrangement of the S
atoms around the M central resembles that of a distorted octahedron.
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Question 7 (cont’d)
(c) The authors describe this structure as being composed to two set of chains. Identify these
chains and describe how they are oriented and interconnected (6 points)
Chain 1 Chain 2
Chain 1: two Nb chains linked by sharing S atoms in a chain arrangement
(arrangement information was not necessary for full credit.) Nb atoms are arranged in
a zig-zag motif.
Chain 2: two M chains linked by sharing S atoms in a octahedral chain arrangement
(arrangement information was not necessary for full credit.) M atoms are arranged in
a zig-zag motif
The two chains (Chain 1 and Chain 2) are interlinked by two S atoms (per repeat)
(d) The authors conducted detailed calculations of the band structure for the two sets of
chains.(10 points)
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The MO diagrams of monomeric NbS6 (a) and NbS8
(b). The band structures for a single [NbS5] chain (c)
and a double [Nb2S8] chain (d).
MO diagram of monomeric VS6 (a). The band
structures for a single [VS4] chain (b) and a
double [V2S6] chain (c)
• For the NbS5 chain and the double [Nb2S8] chain, which orbitals contribute the most to
the electronic conductivity of the chain. Justify your answer. (4 points)
The diagrams clearly show that a band gap exists between the z2 level and xy level, both for the
monomeric NbS6 and NbS8, and the NbS5 and Nb2S8 chains. This suggests that the z2 level
defines the top of the valence band, and thus is the orbital that contributes the most to the
conductivity of the chain (i.e. it would be considered the highest occupied molecular orbital
(HOMO) and the xy (or xz) orbitals would be the lowest unoccupied molecular orbital
(LUMO) An analogy can be drawn by inspecting the diagram illustrating the molecular
orbital energies of the VS6 chain. Again in this case, we see that there is a clear gap between
the z2 level and the x2-y2 and yz orbitals.
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Question 7 (cont’d)
• As we saw in our studies of MO theory, the original energy levels of the atomic orbitals
change when molecular orbitals are formed. What can you say about the difference
between the monomeric NbS6 and NbS8 molecular orbitals? What happens when the
additional 2 S atoms are added? (6 points)
From the diagrams it is clear that the the orbital energies of the xy and x2-y2 orbitals are
significantly affected by the addition of the other 2 S atoms. All the other energy levels remain
the same suggesting that the atomic orbitals of these two additional S atoms are positioned in
such a way that upon binding, they alter the energy of the x2-y2 and xy orbitals only. This
provides a clue as to the symmetry and location of the binding sites on the Nb for these two
additional S atoms since it is only these two orbitals that are affected. Note that there appears
to be some additional splitting of the energy levels in the NbS8 case but these are not labelled
so we cannot comment other than to say that whatever those molecular orbitals levels are,
they also have some spatial orientation that is congruent with that of the incoming S atoms so
that they also undergo a redistribution of energies (both up and down). It is important to note
that you should have been able to draw analogies from our class discussions on how the
molecular orbitals are formed from the blending of different (and appropriate) atomic orbitals
to how the interaction of the S atoms with the existing MO of the NbS6 chain would have been
affected (qualitatively speaking).
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