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CHE 390F Physical and Inorganic Chemistry
Fall 2007
Mid-Term Examination - 1
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#1
#2
#3
#4
#5
Total
/5
/5
/10
/10
/20
/50
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 8
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1. Describe the hybridization state of chlorine in chlorine trifluoride (ClF3). Be sure to fully justify your
answer (5 points)
d orbitals can be used to explain the bonding in molecules that exceed the octet rule. In this case, the
chlorine atom in ClF3 is not sp2 hybridized, but is rather dsp3 hybridized. The s, p, and d valence
orbitals of the chlorine atom all mix together to generate 5 hybrid valence orbitals. 2 of these orbitals
are "lone pairs", which contain two electrons each. The other 3 orbitals are bonding orbitals and each
contains 1 electron from the chlorine atom and 1 electron from the fluorine atom. So yes, you do have to
use all 7 electrons from the chlorine atom.
ClF3 an interhalogen compound . The central atom is Chlorine. The following relationship is applicable
to a great extent to find out the type of hybridization:
H=(V+M-C+A)
'H' is the Hybridization;
'V' valence electrons
'M' number of monovalent atoms linked to the central atom
'C' change possessed by the cation
'A' change possessed by the Anion.
In case of ClF3 H=1/2(7+3)=5. Hence '5' orbitals are involved in the hybridization. So the hybridization
is sp3d. "dsp3" is not considered, as the 'd' orbital involved in chlorine atom is that of the ultimate shell.
Only in transition metals do the penultimate 'd' orbitals participitate in the hybridization.
Since we changed the question to CIF3 – i.e. Carbon, and not chlorine as the central atom, the question
then reverts to the more conventional hybridization model of a general sp3 hybrid case. For complete
credit, your answer should include a discussion as to the relative differences in the % character of the
hybrid based on the halogen that is attached.
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2. When a He atom absorbs a photon to form an excited configuration 1s12s1 (He*), a weak bond forms
with another He atom to form the diatomic molecule HeHe*. Construct a molecular orbital
description of the bonding in this species. (5 points)
He HeHe* He*
2s
1s
1s
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3. The following table details the 1st ionization energies for Sr, Ba, and Ra.
Data obtained from: http://physics.nist.gov/PhysRefData/IonEnergy/tblNew.html
Sr
Ba
Ra
Ionization Energy (eV)
5.6949
5.2117
5.2784
• Define ionization energy (2 points)
Energy required to remove an electron from the stable atom to create a cation.
• Explain the observed trend in ionization energies for Sr, Ba, Ra. (8 points)
The decrease in ionization energy from Sr -> Ba -> Ra reflects the weakening of the interaction
between the outermost valence electron and the nucleus. The fact that the 1st ionization energy
for Ba and Ra are so close and yet Ba and Ra differ substantially in the number of electrons is
that the shielding electrons in Ra are not in fact very good at shielding the nucleus thus the
outermost electron still feels an attractive interaction with the nucleus. In this case, this
interaction is effectively almost the same as it is for Ba. This effect is attributed to the lanthanide
contraction, which refers to a decrease in the radii of the elements following the lanthanides
compared to what would be expected if there were no f-transition metals.
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4. Predict the structure of I(CF3)Cl2. Do you expect the CF3 group to be in an axial apical or equatorial
position? Why? (Ref: Minkwitz and Merkei, Inorg. Chem. 1999, 38, 5041) (5 points)
It seems unreasonable give the size of the triflate group (CF3) that it would sit in an equatorial
plane, along with a Cl. Rather it makes much more sense for the two Cl groups to sit in the
equatorial plane and have the CF3 group sit axially.
The paper by Minkwitz and Merkei is consistent with this view.
An error was made in setting up this question – I should have used the term apical rather than axial.
As such, full credit will be given to all for this question.
The correct answer is that the CF3 group would
sit in an equatorial position, not in the apical positions.
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(b) The bond dissociation enthalpy for the nitrogen-nitrogen bond in N2 is 945 kJ mol-1 and for
[N2]- it is 765 kJ mol-1. Account for this difference in terms of MO theory. Do you expect [N2]to be paramagnetic or diamagnetic? (5 points)
The addition of an electron to the stable [N2] results in a decrease in bond order, which is
reflected by the decrease in the bond dissociation energy / enthalpy. This would also result in
an single unpaired electron, so the system would be paramagnetic
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In a recent paper, a computational approach was taken to resolve and interpret the
photoelectron spectra of the anionic species FeS2-. The following two conformations have
been suggested as plausible ground state configurations of the FeS2 species
Fe
Fe
S
S
S
•
S
In order for bonding to occur, and for molecular orbitals to be created, several conditions
should be met. Describe these in detail (5 points).
Bonding only occurs (or is most likely to occur) under the following conditions when it
comes to creating molecular orbitals (MO’s) from the constituent atomic orbitals (AO’s)
• AO’s must have similar geometry / axial directionality
• AO’s must have similar symmetry
• AO’s must be of similar energy levels
•
Write out a mathematical expression that describes how molecular orbital wavefunctions
are generated from the constitutent atomic orbitals wavefunctions. (3 points)
m
" MO =
n
##C "
ij AO(ij )
j=1 i=1
or in words :
The molecular orbital is formed by a linear combination of all the atomic orbitals of the bonded atoms wherein
all the atomic orbitals in fact participate with the extent of participation defined by the coefficient C ij . Strictly
speaking the core orbitals do participate however, the corresponding coefficient would be very small and so these
are often ignored, and then emphasis is placed on the valence orbitals. Nevertheless, you need to indicate that
this is the case.
!
•
For the two conformations or arrangements shown above, suggest which orbitals of the
Fe and S atoms would be involved in bonding. Is it possible for other orbitals to
participate, and if so, which ones and why? Justify your answer. Note that for this
answer, you should focus on the most strongly participating orbitals. (7 points)
S: 1s22s22p63s23p4
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Fe: 1s22s22p63s23p63d104s2
For this case, the most obvious consideration is that you would need to include
the 3p orbitals of the S. The 3d of Fe are also likely candidates because of their spatial
distribution. This is particularly true if you look at the two conformations. In both the
bent and triangular cases, there would be a high degree of directionality to the bonding
suggesting that orbitals with the appropriate orientation would be necessary. In this case,
it is clear that the d-orbitals would participate. However, the 4s-orbital also has a close
enough energy, and spherical symmetry, to also be involved in active bonding.
•
From the computational simulations, it was suggested that the ground state of the FeS2anion could be represented as S2-Fe3+S2- and that it exists in a high-spin state. Define the
term high-spin and what it implies for the properties of the anion. Propose how the spin
state might change for the neutral FeS2 case and justify your answer. (5 points)
High spin implies that the electrons in the outermost molecular orbitals are unpaired.
This typically means a material has the potential to be paramagnetic. To address the
question of how the spin state might change for FeS2, it is necessary to consider the
distribution of the electrons within these orbitals in the case of the anion and whether it is
possible to fully pair up all the electrons in the neutral case, or whether the neutral
compound remains in a high-spin state (although with likely a lower total spin). In this
case, the anion exists with 5 unpaired electrons distributed among the predominantely 3d
orbitals (in terms of character). This suggests that removing an electron to form the FeS2
neutral species would not result in a significant restructuring of the electronic
configuration and thus FeS2 would remain in a high-spin state.
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