Questions related to the final exam may be directed here and will be answered up to 24 hours before the exam.
Final Exam Friday Dec 10 9-12 AM in the GYM (All Chem 101 student will write together)
The final exam is cumulative and will cover the entire course with a weighting towards material not covered on the midterms. You must pass both the lecture and lab to pass the course.
Old copies of exams are available at ZAP in the student union.
Solved exams are being sold in Elliot (3rd floor) until Dec 2.
Pre-Exam Review Tutorials
There will be three review sessions. They are open to all Chem 101 students.
Tuesday December 7th 10:00am-noon Sci B150 (Bob Wright Centre) (run by Dr. Sirk)
Also NEW Booking! Tues Dec. 7 from 1 - 3 pm MAC A144- Chem 101 review session (Dr Sirk)
-both review sessions will be identical-
Wednesday December 8th 10:00-11:30am SSM A120 (Social Sciences & Math) (run by Dr. Briggs)
Thursday December 9th 10:00-11:30am Sci B150 (Bob Wright Centre) (run by Dr. Codding)
Friday, November 26, 2010
Tuesday, October 19, 2010
Midterm 2
All questions/comments/concerns related to the second midterm can be directed here and will be replied to within 24 hours. Questions may be submitted until 24 hours prior to the exam (until THursday at 5pm).
The location of the second midterm (held Friday Nov 19, 5-6pm) will depend on your last name:
If your last name starts with A or B, you will be writing (again)in David Lam (with Prof Briggs).
If your last name starts with C through S, you are writing in Social Sciences and Mathematics SSMA120 .
If your last name ends in T and later you will be writing in SCI B150 (with Prof. Codding).
Campus map can be found here:
http://www.uvic.ca/maps/2dmap.html
Minor changes: If you come late please come in the back, seats will be reserved to minimise disturbance.
Coverage of Second Midterm:
Chapter 8.5-8.8 Chemical Bonding, Lewis Structures, Resonance Structures, Exceptions to the Octet Rule, Bond Strengths.
Chapter 9.1-9.7 Molecular Geometry and Bonding Theories
Chapter 23.5 'Metallic Bonding', and
Chapter 12.2 'Electronic Structure of Materials, Band Theory'.
(as exemplified in Graded On-Line Quizzes 4 (part of it), 5, 6 and 7, and Assignment problem sets 3 and 4.)
The location of the second midterm (held Friday Nov 19, 5-6pm) will depend on your last name:
If your last name starts with A or B, you will be writing (again)in David Lam (with Prof Briggs).
If your last name starts with C through S, you are writing in Social Sciences and Mathematics SSMA120 .
If your last name ends in T and later you will be writing in SCI B150 (with Prof. Codding).
Campus map can be found here:
http://www.uvic.ca/maps/2dmap.html
Minor changes: If you come late please come in the back, seats will be reserved to minimise disturbance.
Coverage of Second Midterm:
Chapter 8.5-8.8 Chemical Bonding, Lewis Structures, Resonance Structures, Exceptions to the Octet Rule, Bond Strengths.
Chapter 9.1-9.7 Molecular Geometry and Bonding Theories
Chapter 23.5 'Metallic Bonding', and
Chapter 12.2 'Electronic Structure of Materials, Band Theory'.
(as exemplified in Graded On-Line Quizzes 4 (part of it), 5, 6 and 7, and Assignment problem sets 3 and 4.)
Thursday, October 7, 2010
Midterm 1
All midterm questions (content etc..) can be directed to this post. Any comments recieved before 5pm Oct 14 (24 hours before the exam) will be replied to.
The midterm will be in ECS 123 from 5-6 pm Friday Oct 15.
All students whose surnames begin with A or B should write their test in Mac A144 (David Lam Auditorium) with the other section.
The Chem 101 webpage has the official midterm information.
http://web.uvic.ca/~chem101/midterm.html
Sample Midterm : http://web.uvic.ca/~chem101/C101Midterm1A2009.print.pdf
Sample Midterm Key: http://web.uvic.ca/~chem101/C101Midterm1A2009.key.pdf
Things you need
-student ID
-soft pencil
-pen
-Sharp EL-510R calculator
Answers to the Last Comments
Anonymous said...
Have we covered all the information necessary to complete quiz 4?
October 14, 2010 10:51 AM
Yes
Anonymous said...
Hi, for the 2008 midterm question, the answer gives that the maximum number of electron in an atom with a principle quantum number of 4 is 32. How is this possible? Wouldn't the answer be 36 because that is how many electrons are in Kr?
Thank you.
October 14, 2010 10:55 AM
Kr has electrons in the 1s, 2s, 3s, 3p, 4s, 3d and 4p orbitals.
The electrons with a principle quantum number of 4 are in the 4s,4p,4d,4f
Steven said...
Hi Dr. Sirk,
I was wondering if a completely filled d-subshell is still considered part of an atom's valence electrons? For instance, if Zinc is 1s2,2s2,2p6,3s2,3p6,4s2,3d10....are the valence electrons 4s2,3d10?
Nope, a completely filled d orbital counts as core, so the valence electrons are 4s2
Along the same lines, how do we know which electrons of an atom are considered valence electrons? Do we just assume all electrons added after the preceding noble gas are valence?
Yes, except for a completely filled d orbital
How do we know if K+ or Ar has a larger atomic radius?
They are isoelectronic, so the one with the greatest number of protons is smaller
Finally, are we required to calculate lattice energies on the midterm? I don't remember this ever being covered in class.
No direct calculations, but you should be aware of the formula (coulomb’s law) and how charge/radius would affect the lattice energy (E=kQ1Q2/d) Q1=charge on cation, Q2=charge on anion, d=radius of anion+radius of cation
Thanks in advance!
You’re welcome
October 14, 2010 11:02 AM
Anonymous said...
2009 midterm mc question 12. How do you determine the lattice energies of the RbCl, RbBr, and RbF, which are not given on the data sheet? thanks!
You just need to determine the largest lattice energy. They all have the same charge (+1)(-1) and the same cation (Rb+) so the one will the SMALLEST radius will have the greatest lattice energy (see above question)October 14, 2010 11:47 AM
Anonymous said...
3. Which of the following electron configurations is CORRECT for the ground state of the given atom?
A. S: [Ne] 2s2 2p5
B. P: [Ne] 3s2 3d3
C. Br:[Ar] 4d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 4s2 4p1
how do you know do you know that the answer is D?
The correct electron configurations are:
A. S: [Ne] 2s2 2p6
B. P: [Ne] 3s2 3p3
C. Br:[Ar] 3d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 3d10 4s2 4p1
So only Rb is correct
October 14, 2010 3:54 PM
Anonymous said...
Question 2c on midterm 2009:
What are the n values for the transition (labeled or unlabeled) that corresponds to the ionization energy
for the hydrogen atom.
answer:From n = 1 to n = ∞
What does that mean? Thanks
October 14, 2010 4:12 PM
In ionisation the electron is completely removed from the atom. Complete removal corresponds to the n=infinity energy level. The n=infinity corresponds to no attraction to the nucleus.(Note the electron is removed from the highest ground state energy level. For H this is 1. For Li this would be 2s)
Anonymous said...
a) A red laser pointer emits light at a wavelength of 650nm. What is the frequency of this light? b) what is the energy of one mole of these photons? c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to their excited state. When the electrons return to the ground state they lose the excess energy in the form of 650nm photons. What is the energy gap between the ground state and the excited state in the laser material?
How do I calculate part c) ?
October 14, 2010 4:15 PM
E=hv or E=hc/lambda with lambda equal to 650 nm (650*10-9m)
Anonymous said...
In the 2008 midterm, question number 3 asks for correct order of Zeff. The correct answer is
Na > Be > Li
However, I thought that Zeff, for the purpose of this course, was = # of electrons - the # of core electrons, which would, in effect, make the answer Na (1) > Be (2) > Li (1)
Does this mean that we are to assume Na has a much larger Zeff than is indicated by the formula above?
October 14, 2010 4:16 PM
Good question: If we want to estimate the value of Zeff , then Zeff = # of protons - the # of core electrons is reasonable to get at a number. But that is a very rough estimate and note that Li, Na, K, Rb Cs would all have the same value (1). Recall the trends in Zeff. It increases a lot across the row and slightly down a column. So you need to use the trends for this question.
Anonymous said...
what sort of unit conversions will we need to know?
nanometers, micrometers? what else?
October 14, 2010 4:19 PM
Any of the values in Table 1.5 are testable.
Anonymous said...
can you please explain the answer to number 6 on the 2008 midterm?: What is the maximum number of electrons in an atom that can have the principal quantum number n = 4? thank you
See above
October 14, 2010 4:36 PM
First, a question asks: Arrange the bonds in order of increasing
polarity ( C--S, B--F, N--O). I know that the larger difference in
electronegativity causes a stronger polarity. For this reason, I had
ordered it as N--O, C--S, B--F. However, the answer provided explained
that because S is down a row and to the right, it has almost an
identical electronegativity to C... are we going to be expected to
know this subtle difference for the exam?
Already answered in the blog questions
Similar to this, the practice exam asks: Of the compounds below,
______ has the largest lattice energy. A) KF B) KCl C) RbCl D) RbBr E)
RbF
Would the answer to this be RbF, as it is both less electronegative
than K, and F has the highest electronegativity?
Ionic compounds, therefore you should be aware of the formula (coulomb’s law) and how charge/radius would affect the lattice energy (E=kQ1Q2/d) Q1=charge on cation, Q2=charge on anion, d=radius of anion+radius of cation. Rest explained above.
Finally, in the Prep 101 session, it was mentioned that the intensity
of a photon affects the number of electrons ejected from a surface
during the photoelectric effect, but the frequency affects the kinetic
energy of the released electron. Do we need to know the part about
intensity, as I do not recall it being mentioned in the lectures.
It was in the notes under a subtitle:observations (also in the text)
The midterm will be in ECS 123 from 5-6 pm Friday Oct 15.
All students whose surnames begin with A or B should write their test in Mac A144 (David Lam Auditorium) with the other section.
The Chem 101 webpage has the official midterm information.
http://web.uvic.ca/~chem101/midterm.html
Sample Midterm : http://web.uvic.ca/~chem101/C101Midterm1A2009.print.pdf
Sample Midterm Key: http://web.uvic.ca/~chem101/C101Midterm1A2009.key.pdf
Things you need
-student ID
-soft pencil
-pen
-Sharp EL-510R calculator
Answers to the Last Comments
Anonymous said...
Have we covered all the information necessary to complete quiz 4?
October 14, 2010 10:51 AM
Yes
Anonymous said...
Hi, for the 2008 midterm question, the answer gives that the maximum number of electron in an atom with a principle quantum number of 4 is 32. How is this possible? Wouldn't the answer be 36 because that is how many electrons are in Kr?
Thank you.
October 14, 2010 10:55 AM
Kr has electrons in the 1s, 2s, 3s, 3p, 4s, 3d and 4p orbitals.
The electrons with a principle quantum number of 4 are in the 4s,4p,4d,4f
Steven said...
Hi Dr. Sirk,
I was wondering if a completely filled d-subshell is still considered part of an atom's valence electrons? For instance, if Zinc is 1s2,2s2,2p6,3s2,3p6,4s2,3d10....are the valence electrons 4s2,3d10?
Nope, a completely filled d orbital counts as core, so the valence electrons are 4s2
Along the same lines, how do we know which electrons of an atom are considered valence electrons? Do we just assume all electrons added after the preceding noble gas are valence?
Yes, except for a completely filled d orbital
How do we know if K+ or Ar has a larger atomic radius?
They are isoelectronic, so the one with the greatest number of protons is smaller
Finally, are we required to calculate lattice energies on the midterm? I don't remember this ever being covered in class.
No direct calculations, but you should be aware of the formula (coulomb’s law) and how charge/radius would affect the lattice energy (E=kQ1Q2/d) Q1=charge on cation, Q2=charge on anion, d=radius of anion+radius of cation
Thanks in advance!
You’re welcome
October 14, 2010 11:02 AM
Anonymous said...
2009 midterm mc question 12. How do you determine the lattice energies of the RbCl, RbBr, and RbF, which are not given on the data sheet? thanks!
You just need to determine the largest lattice energy. They all have the same charge (+1)(-1) and the same cation (Rb+) so the one will the SMALLEST radius will have the greatest lattice energy (see above question)October 14, 2010 11:47 AM
Anonymous said...
3. Which of the following electron configurations is CORRECT for the ground state of the given atom?
A. S: [Ne] 2s2 2p5
B. P: [Ne] 3s2 3d3
C. Br:[Ar] 4d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 4s2 4p1
how do you know do you know that the answer is D?
The correct electron configurations are:
A. S: [Ne] 2s2 2p6
B. P: [Ne] 3s2 3p3
C. Br:[Ar] 3d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 3d10 4s2 4p1
So only Rb is correct
October 14, 2010 3:54 PM
Anonymous said...
Question 2c on midterm 2009:
What are the n values for the transition (labeled or unlabeled) that corresponds to the ionization energy
for the hydrogen atom.
answer:From n = 1 to n = ∞
What does that mean? Thanks
October 14, 2010 4:12 PM
In ionisation the electron is completely removed from the atom. Complete removal corresponds to the n=infinity energy level. The n=infinity corresponds to no attraction to the nucleus.(Note the electron is removed from the highest ground state energy level. For H this is 1. For Li this would be 2s)
Anonymous said...
a) A red laser pointer emits light at a wavelength of 650nm. What is the frequency of this light? b) what is the energy of one mole of these photons? c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to their excited state. When the electrons return to the ground state they lose the excess energy in the form of 650nm photons. What is the energy gap between the ground state and the excited state in the laser material?
How do I calculate part c) ?
October 14, 2010 4:15 PM
E=hv or E=hc/lambda with lambda equal to 650 nm (650*10-9m)
Anonymous said...
In the 2008 midterm, question number 3 asks for correct order of Zeff. The correct answer is
Na > Be > Li
However, I thought that Zeff, for the purpose of this course, was = # of electrons - the # of core electrons, which would, in effect, make the answer Na (1) > Be (2) > Li (1)
Does this mean that we are to assume Na has a much larger Zeff than is indicated by the formula above?
October 14, 2010 4:16 PM
Good question: If we want to estimate the value of Zeff , then Zeff = # of protons - the # of core electrons is reasonable to get at a number. But that is a very rough estimate and note that Li, Na, K, Rb Cs would all have the same value (1). Recall the trends in Zeff. It increases a lot across the row and slightly down a column. So you need to use the trends for this question.
Anonymous said...
what sort of unit conversions will we need to know?
nanometers, micrometers? what else?
October 14, 2010 4:19 PM
Any of the values in Table 1.5 are testable.
Anonymous said...
can you please explain the answer to number 6 on the 2008 midterm?: What is the maximum number of electrons in an atom that can have the principal quantum number n = 4? thank you
See above
October 14, 2010 4:36 PM
First, a question asks: Arrange the bonds in order of increasing
polarity ( C--S, B--F, N--O). I know that the larger difference in
electronegativity causes a stronger polarity. For this reason, I had
ordered it as N--O, C--S, B--F. However, the answer provided explained
that because S is down a row and to the right, it has almost an
identical electronegativity to C... are we going to be expected to
know this subtle difference for the exam?
Already answered in the blog questions
Similar to this, the practice exam asks: Of the compounds below,
______ has the largest lattice energy. A) KF B) KCl C) RbCl D) RbBr E)
RbF
Would the answer to this be RbF, as it is both less electronegative
than K, and F has the highest electronegativity?
Ionic compounds, therefore you should be aware of the formula (coulomb’s law) and how charge/radius would affect the lattice energy (E=kQ1Q2/d) Q1=charge on cation, Q2=charge on anion, d=radius of anion+radius of cation. Rest explained above.
Finally, in the Prep 101 session, it was mentioned that the intensity
of a photon affects the number of electrons ejected from a surface
during the photoelectric effect, but the frequency affects the kinetic
energy of the released electron. Do we need to know the part about
intensity, as I do not recall it being mentioned in the lectures.
It was in the notes under a subtitle:observations (also in the text)
Saturday, September 25, 2010
Introduction
This Blog is to answer general questions about the lecture section of Chem 101 (Section A02). Enter your question in the comments and I will reply within the comments as well. As no identification is required to post, the blog is moderated (to reduce spam) so comments will not appear immediately but should be posted with a reply within 24 hours. Feel free to look over other people's comments and chime in.
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