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.)
What is the average percentage of Mid 1, thanks
ReplyDeleteFor section A02, it is 72.4%.
ReplyDeleteI can't find the 3d diagrams for the molecular shapes. Do you think you could provide a link? Thanks.
ReplyDeleteFor question 8.88c in the 11th edition, it asks you to calculate the formal charge on N in NO2. How do you know which atom the odd electron goes on?
ReplyDeleteAll tutorials
ReplyDeletehttp://web.uvic.ca/~pcodding/info101.htm#tutorials
3-d images
http://www.wwnorton.com/college/chemistry/gilbert2/tutorials/chapter_09/vsepr/
(click on name of shape)
NO2
ReplyDeleteStart with 17 electrons
Take away 4 for the bonds leaving 13
Fill the octet of the outermost atoms first (O). Each O already has two, therefore that should use another 12.
Bonus electron goes on N.
Add in double bond.
(This will violate octet rule on N)
Hi,
ReplyDeleteCan you please explain why N3- is tetrahedral and bent? Thanks!
It isn't. Is this from question 9.21?. The answer key for that question is entirely wrong. I put the actual answers in the overview page(within the list of assignment questions).
ReplyDeletehttp://web.uvic.ca/~asirk/overview.htm
is ther anyway that i could get the sheet off your site that has all the shapes . i lost mine!
ReplyDeleteOh, that's good to know (about question 9.21) - was getting a bit worried when all my answers were wrong! Thanks!!
ReplyDeleteHi there,
ReplyDeleteCould you please explain why the compound XeF2 is linear and not bent? Thank you!
Shape sheet link
ReplyDeletehttp://web.uvic.ca/~asirk/hybrid.pdf
XeF2
ReplyDelete8+(2*7)=22 electrons
F-Xe-F
22-4(from bonds)=18 electrons
12 electrons around F,leaving 6 for Xe
Therefore Xe has 5 domains (3 non bonding pairs and 2 bonds) and is triganol bipyramidal.
The 3 non bonding pairs occupy the equatorial spaces because non bonding pairs take up more space. Recall equatorial spaces are 120 degrees apart.
The two bonds occupy the axial positions.
Therefore F-Xe-F is linear.
reposted: Hi, for question 5 on quiz 5 it says "In the nitrite ion (NO2-), __________." and you have to fill the blank. There are two possible correct answers for this question(which are both given as possible answers) but it doesn't tell you whether it is talking about overall bonds or lewis structures of the nitrite ion! The two possible CORRECT answers are A) one bond is a double bond and the other is a single bond and D) both bonds are the same. How are you supposed to know which answer to choose?
ReplyDeleteThis is a good question as it clarifies the difference between the molecule and the models used to describe it.
ReplyDeleteThe correct answer is D) Both bonds are the same because that is what the molecule actually looks like. It is incorrect to say that one bond is a double bond and one is a single bond.
If the question were about Lewis structures in general the answer would still be D, as the molecule cannot be correctly described using a single Lewis structure. It requires two Lewis structures to describe the resonance structure.
The only way A) could be the correct answer would be if the question were worded in such a way that only one Lewis structure were requested. ie: More than one Lewis structure is required to describe the nitrate ion. Please describe the bonds in one of these structures.
Does midterm 2 include the material that was on midterm 1, or only the material we have done since then?
ReplyDeleteMainly the material since Midterm 1. We won't be testing directly on anything from the first section, but of course some material has a direct spillover (shape of s and p orbitals when considering hydridisation, Lewis structures when considering electronic geometry etc..)
ReplyDeleteCan you please explain question 12.7 in the textbook?
ReplyDelete12.7: Classify each of the following materials as metal, semiconductor, or insulator: a) GaN, b) B, c) ZnO, d) Pb
I thought that GaN and ZnO would be insulators because they are ionic and Pb is a metal, but I have no idea about B. The answer says that both GaN and ZnO are semiconductors, though. Please explain!
thanks
We were working on the posted practice midterm, and are confused about
ReplyDeletequestion 9. We were wondering why this molecule could not be derived from
an octahedral electron domain geometry as well as a trigonal bipyramidal
geometry. Are the angles between domains in an octahedral geometry not
also 90?
9. The molecular geometry illustrated below can only be derived from which
electron domain (VSEPR) geometry?
T-shaped geometry is drawn
A. trigonal bipyramidal B. tetrahedral C. octahedral D. see saw E.
trigonal planar
Good question, If there a compound with 3 lone pairs and 3 substituents, it would be T-shaped. However, such a compound does not exist and therefore the tables on octahedral geometry stop with 2 lone pairs.
ReplyDeleteThe associated readings say we only need to do section 12.2, but the assignment questions (namely 12.11 b), which is about doping semiconductors) are related to 12.3... do we need to do 12.3?
ReplyDeleteNo, the readings are only for 12.2. There are a couple questions on semiconductors that cover some material that is not testable, but will give you a better understanding of the concepts.
ReplyDeleteAre bond energy and bond enthalpy the same thing?
ReplyDeleteIn terms of what you know so far, yes energy and enthalpy are the same thing.
ReplyDelete(In terms of what you will learn later, no, they are different. Enthalpy and Entropy (almost always a smaller value) make up the total Energy.)
Hi,
ReplyDeleteI was just wondering how we determine the hybridization of atoms within a molecule. Do we look at the number of electron domains? (Ex. If there are three bonding domains it will be sp2?).
Also, there is a question on the practice midterm:
13. Place the following compounds in order of decreasing strength of intermolecular forces.
I. hexane II. 2,2-dimethylpropane III. 2,2-dimethylbutane
Have we learned enough in class to do this? Are we expected to know how to construct the molecules?
Thanks!
Hi Professor Sirk,
ReplyDeleteA question asks:
14. Band theory is the best way to describe the physical properties of transition metals. Which of the
following sets of transition metals would you expect to have filled bonding molecular orbitals and empty
antibonding orbitals, therefore the highest melting points?
A. Ti, Zr, Hf
B. Cu, Ag, Au
C. Co, Rh, Ir
D. Cr, Mo, W
E. Zn, Cd, Hg
How do we go about solving this? We only learned about the bonding/antibonding orbitals of H and He.
Thanks!
Hi Professor Sirk,
ReplyDeleteCould you please explain Q. 11 in the practice midterm??
It ask which displays the most hydrogen bonding, but all of these are molecules. I thought that hydrogen bonding only occurred between molecules...?
Thanks!
Hybridization: Yes, 2 electron domains=linear electronic geometry and sp
ReplyDelete3 electron domains=triginol planar electronic geometry and sp2
etc...
Intermolecular forces: Will not be covered. The coverage ends with metallic bonding.
(For the final, we will have reviewed organic chemistry and basic naming so you will be able to contruct the molecules)
Q 11 of Midterm: Yes-Hydrogen bonding exists between molecules, so it is assumed that you have a large number of each molecule shown.
ReplyDeleteHi Professor Sirk,
ReplyDeleteCould you explain how you define a solid to be an insulator, a metallic conductor or a semiconductor? Such as TiO2 or Ge?
thanks!
From that list, you should be able to determine all compounds except TiO2 and SiC. (For interest, those are semiconductors and insulators respectively)
ReplyDeletePure elements are metallic (if they are metals or alloys of metals), semiconductors (if they are metalloids) and insulators (if they are non metals)
Hi, about the Q11 on the practice midterm.. why is that the provided answer the answer? how can you determine that this molecule has the most hydrogen bonding going on?
ReplyDeleteAlso, Q14 on the practice midterm.. are we supposed to know this, or are we just covering anti-bonding for elements such as He and H?
Thank you!
Q11- The correct answer (D) is also the ONLY molecule has hydrogen bonding. In order to have hydrogen honding, a H must be attached to either O,F and N. (The hydrogen bonding is between the molecules)
ReplyDeleteQ14- Not covered in this version of the course (note this may be a slight difference in textbook editions as well)
Hi,
ReplyDeleteAre boiling points covered on Midterm 2?
Thanks!
Hi, is anything from chapter 11 going to be on midterm because there are many questions on the practice midterm?
ReplyDeleteThanks!
Those are both the same question.
ReplyDeleteChapter 11 is NOT covered in this midterm (so my comments on hydrogen bonding can be reread for the final)
Q from email: Hi, Dr.Asirk,
ReplyDeletewhat is the delocalized bonding and why SO3 2- is exhibit delocalized bonding?
Thanks.
Deloalised bonding means that more than one Lewis structure is required to represent the molecule. These are resonance structures as the actual molecule is a hybrid of all those molecules.
ReplyDeleteDelocalised means the bond is spread over several bonds. These bonds are pi bonds and the total bond strength is fractional (ie 1.3, 1.5, 2.5 etc..)
To continue, a delocalised bond means the electron density of a pair of electrons is spread over multiple bonds. This is very consistant with electron probability theory. In this case the electron density is focused in the area of the bonds.
ReplyDeleteHi Dr. Sirk,
ReplyDeleteCan you explain why period 3 can possess vacant d orbitals, but period 2 and 1 cannot? Also -does the electron need to be excited in order to enter the d orbital, and if so why would it not fill the 4s(or 5s, etc) orbital first?
Thank you!
Hi Dr. Sirk,
ReplyDeleteI was just wondering what happened to the extra electrons in the p-orbitals during hybridization. For example, an oxygen may be sp3 hybridized, which means there is one 2s electron, and then each of the sp3 orbitals has one electron in it, yet there are no extra p-orbitals.. there are four p-orbital electrons on oxygen, but with sp3, the diagrams show three half-filled p-orbitals that then have a fourth orbital added with the 2s electron to form the sp3... I am confused.. I am not sure if my question makes much sense..
I am basically wondering how the orbital diagrams would look for constructing the sp, sp2, and sp3 orbitals for atoms with more electrons in the p-orbitals than shown in the hybridized model, since the hybridization only shows half-filled orbitals.
Thanks for your time!
I'm doing questions in the textbook and one question I came upon was:
ReplyDeleteIf the valence atomic orbitals of an atom are sp hybridized, how many hybridized p orbitals remain in the valence shell? How many pi bonds can the atom form?
Thanks!
Will the number of valence electrons in a molecule always equal the number of valence electrons in sigma and pi bonds together?
ReplyDeleteThanks
Hi Dr. Sirk,
ReplyDeleteCan you explain why Boron is an insulator? also - I thought ionic compounds formed insulators. Why are ZnO and GaN semiconductors?
Thank you!
Re vacant d orbitals:
ReplyDeleteThe order of the orbitals are 1s, 2s, 2p, 3s,3p, 4s,3d, 4p. Note that you don't get d orbitals until a principle quantum number of 3 is reached.
As the orbitals get higher in energy, they also get much closer together. So the energy gap from a 2s/2p to a 3d orbital is fairly significant. But for a 3s/3p to a 3d, the energy gap is less significant and the d orbitals can be accessed to expand the valence.
Hybridization:
ReplyDeleteShort answer: For O (which has 6 valence electrons) when it is sp3 hybridised, it has 1 election is two of the orbital and 2 electrons in each of the others.
So for H2O, the two orbitals with one electron are the bonding electron and the two orbitals with 2 electrons are the lone pairs.
Long Answer: I'll answer this in class, as it really needs diagrams to demonstrate.
Re: If the valence atomic orbitals of an atom are sp hybridized, how many hybridized p orbitals remain in the valence shell? How many pi bonds can the atom form?
ReplyDeleteAnswer: An sp hybridized atom will have 2 p orbitals remaining in the valence shell (but these are unhybridized). It can form two Pi bonds.
Re:Will the number of valence electrons in a molecule always equal the number of valence electrons in sigma and pi bonds together?
ReplyDeleteNo-the number of valence electrons will equal the number of electrons in the sigma and pi bonds together PLUS the number of lone pairs. (consider HCl, 2 electrons in the bond and 6 as lone pairs around Cl)
Re:Can you explain why Boron is an insulator? also - I thought ionic compounds formed insulators. Why are ZnO and GaN semiconductors?
ReplyDeleteBoron is a non metal, therefore insulator.
ZnO and GaN are compound semiconductors which (short answer) you don't need to know and (longer answer which you can ignore) Ga and Zn are slightly less metallic and can form compounds which are semiconductors. These compound semiconductors have tunable band gaps by addition of various materials and therefore can make light emitting diodes of tunable colours (such as traffic lights)
Hi Dr. Sirk,
ReplyDeleteI'm a bit confused about when we show lone pairs when drawing a 3D structure...Does electron domain geometry show the lone pairs while molecular geometry does not? Also, would we ever be asked to draw a 3D structure with the shading to show bonds going into/out of the page while showing double/triple bonds?
Thanks!
Re Lone Pairs-Yes, you have answered your own question correctly. If asked for the molecular structure you do not need to show lone pairs. For the electronic structure you do need to show them.
ReplyDeleteThere is no official way to draw the orbitals, but you should be able depict both double and triple bonds. (If you feel that your drawings are unclear, you can add a description such as "px-px overlap to give a pi bond and py-py overlap to give a pi bond" with arrows pointing to your bonds)
Two reposted questions:
ReplyDeleteAnonymous said...
Hi Dr. Sirk,
Could you please explain how to do a question like #14 on the midterm, when you are dealing with bonding molecular orbitals and anti-bonding orbitals of transition metals - are you supposed to look at the bond order?
November 12, 2010 11:45 PM
Anonymous said...
In section 23.5 of the textbook, there's a subsection called molecular orbital model for metals which talks about 4s, 4p and 3d orbitals and "Fermi level" (bottom of pg. 993-994)
Is this stuff we need to know? I don't recall any of it from our notes
November 16, 2010 6:53 PM
Quick answer to both of these-You don't need to know (Q 14 is something that was covered in the 10th edition of the text in more detail)
ReplyDeleteDo we need to know how to calculate dipole moments? u=Qr ?
ReplyDeleteNope.
ReplyDeletedo we need to know about aromatic compounds?
ReplyDeleteDr Sirk,
ReplyDeleteLooking at question 10 from the practice midterm, can you explain how polarity of a molecule effects it's boiling point? Why would a nonpolar molecule have a lower boiling point than a polar molecule?
"There is no official way to draw the orbitals, but you should be able depict both double and triple bonds. (If you feel that your drawings are unclear, you can add a description such as "px-px overlap to give a pi bond and py-py overlap to give a pi bond" with arrows pointing to your bonds)"
ReplyDeleteI don't really understand what you are saying here. To show double and triple bonds can we not simply draw two or three lines? Could you possibly give an example where we would need to show double or triple bonds like you are saying above?
Thank you!
sorry - can you explain again why boron is an insulator? it is along the metalloid line, so wouldn't that make it a semi conductor?
ReplyDeletethanks again!
Text book question: 9.43 - fill out the table:
ReplyDeleteThe last one, XeF2, shouldn't that be trigonal bipyramidal? The answer at the back says it's octahedral.. I think it's a typo. But can you clarify?
Thanks,
Hi Dr. Sirk,
ReplyDeleteI have built up a few questions.. mostly from the intro to the section 9.1-9.7 that I did not feel I grasped fully.
How does VSEPR differ from the Lewis concept of chemical bonding? (Is it that with VSEPR, molecules are drawn with their actual shape, and that Lewis structures are not?)
How does molecular orbital theory differ from VSEPR? (Is it the concept of the bonding orbitals and non-bonding orbitals?)
Why do hybrid orbitals differ from pure atomic orbitals? (Is this because pure atomic orbitals have only s,p,d,f, whereas hybrid orbitals have combinations of these?)
Would the size of a p-orbital affect the amount of overlap between orbitals (in reference to Q2 of the additional i-clicker questions)
Finally, what happens to the other 2s electron after the other 2s is promoted to the p-orbitals in hybridization? I know that the 2s orbital disappears, but the text only mentions that one electron is promoted..
Thank you in advance for helping me with all of these ideas!!
Re aromatic: Aromatic compounds are composed of multiple single bond, double bond combinations and have not been covered. Delocalised bonds and resonance structures will be covered.
ReplyDeleteRe question 10: This is from chapter 11 and therefore final exam material, not midterm material.
ReplyDeleteFor compounds of similiar molecular weight, a polar molecule will have stronger intermolecular forces. Dipole-dipole forces are stronger than london dispersion forces. See the last slide from Tuesday.
Re drawings of double and triple bonds
ReplyDelete1) For Lewis structures, yes two lines or three lines is fine.
2)If a question asks for you depict an orbital overlap that gives a double bond (or triple bond) you will need to show a 2-D or 3-D structure. ( Slide 15 of Chapter 9 shows images of these overlaps)
Another example would be question 4b of the 2009 midterm
ReplyDeleteLet's repost this entire one:
ReplyDeleteHow does VSEPR differ from the Lewis concept of chemical bonding? (Is it that with VSEPR, molecules are drawn with their actual shape, and that Lewis structures are not?)
Yes-Lewis structure counts electrons and gives the bonds and resonance structures. It says nothing about the 3-D geometry. So molecules are drawn in a simple fashion.
VSEPR uses the Lewis structures to give some idea of the geometry of the molecule (I say some idea because each atom has its specific geometry, but as single bonds can spin the overall shape can still be a mystery, especially for large molecules-see protein folding determination, impossibility of)
How does molecular orbital theory differ from VSEPR? (Is it the concept of the bonding orbitals and non-bonding orbitals?)
MO theory gives the shape of the orbitals (not of the molecule). It also gives the energy of the orbitals (very important but only covered in a very basic sense in class-ie bonding MOs are lower in energy and antibonding are higher in energy) Eventually this information can be used to predict and explain reactivity and catalysis of various chemical reactions.
Why do hybrid orbitals differ from pure atomic orbitals? (Is this because pure atomic orbitals have only s,p,d,f, whereas hybrid orbitals have combinations of these?)
Yes. Hybrid orbitals are used to explain the position of electrons in molecules and the geometry.
Would the size of a p-orbital affect the amount of overlap between orbitals (in reference to Q2 of the additional i-clicker questions)
ReplyDeleteYes. 3p orbitals are bigger than 2p orbitals. Therefore the atoms will be further apart.
Finally, what happens to the other 2s electron after the other 2s is promoted to the p-orbitals in hybridization? I know that the 2s orbital disappears, but the text only mentions that one electron is promoted..
That electron goes into the sp (or sp2 or sp3).
However, to keep track of the electrons in the orbitals one needs to form a Lewis structure and then the electrons actually end up in whatever orbitals they need to. This gets pretty complicated and doesn't tell us anything so we don't get into it (if you feel so inclined, you can track the electrons in CO)
Re Boron: Whether or not a material is a semiconductor depends on the bandgap. Generally material with a band gap (the energy gap between the valance band and the conduction band) of 1-3 eV.
ReplyDeleteNow I just looked up Boron and it does have a bandgap in that region, so it should be a semiconductor. However, on further investigation, it is very difficult to prepare pure Boron, so it isn't of practical use.
This is a long winded way of saying, I was incorrect above when I was asked why Boron was an insulator. The correct answer should have been pure Boron is a semiconductor (but I was mislead by my practical knowledge of Boron)
Hi there,
ReplyDeleteI have a question about the Midterm Practice one did in lecture.
Number 2. Why answer is E :-38
not B:38
I calculated
E(enthalpies of bonds broken)-E(bonds formed)
=(c-c)+(H-cl)-(c-cl)-(c-H)
=348j+431j-328j-413j
=38j
not -38j
Do we need to know how to classify things as semiconductor, conductor, and insulator? In 12.2, the only way they're explained is with reference to the band gap, but it doesn't tell you how to classify a substance (that's 12.3, but I do have the 11th international version and it's possible that it's slightly different). Thanks.
ReplyDeleteQuestion 2 of midterm. The correct answer is A (-44). I have corrected that on the answer key.
ReplyDeleteThe difference between that and your answer is that you break a double bond in the reactants (614) and then form a single bond (348) in the products.
You can use just a single bond in the reactants because breaking the pi part of a double bond is different than a sigma.
Insulator (large bandgap), Semiconductor (medium bandgap), Metal (small/no bandgap)is all you need to know for the classification. Non metals are insulators, metalloids are semicondutors and metals are conductors.
ReplyDeleteA few questions from long ago that got lost:
ReplyDeleteHi Professor Sirk, A question asks: 14. Band theory is the best way to describe the physical properties of transition metals. Which of the following sets of transition metals would you expect to have filled bonding molecular orbitals and empty antibonding orbitals, therefore the highest melting points? A. Ti, Zr, Hf B. Cu, Ag, Au C. Co, Rh, Ir D. Cr, Mo, W E. Zn, Cd, Hg How do we go about solving this? We only learned about the bonding/antibonding orbitals of H and He. Thanks!
By Anonymous on Midterm 2 on 11/9/10
Don't need to know (This section was taught a bit differently when a previous version of the textbook was used)
Can you please explain question 12.7 in the textbook? 12.7: Classify each of the following materials as metal, semiconductor, or insulator: a) GaN, b) B, c) ZnO, d) Pb I thought that GaN and ZnO would be insulators because they are ionic and Pb is a metal, but I have no idea about B. The answer says that both GaN and ZnO are semiconductors, though. Please explain! thanks
By Anonymous on Midterm 2 on 11/7/10
Answer: You don't need to know compound semiconductors yet. You just need to be able to discuss the elements. For my notes and corrections on Boron, see a few posts up.
Why the bond angle in BrF3 is smaller than PF5?
ReplyDeleteThey are both Trigonal bipyramidal.
Is it because there is two lone pairs in BrF3 and only one lone pair in PF5 then lone pairs take more space?
Yes.
ReplyDeleteLong answer: Both have trigonal bipyramidal electronic geometries. PF5 has no lone pairs and has a trigonal bipyramidal molecular geometry while BrF3 has two lone pairs and a T-shaped molecular geometry. The lone pairs take up more space than F and therefore some of the bond angles are less than 90 degrees.
Hi Dr. Sirk,
ReplyDeleteI noticed a discrepancy for Molecular Orbital Theory between the notes from the board and the online slide. For He2 +, I think you mentioned in class that we should bring together He and He first and then lose the electron. However, in the textbook and in the slide notes it shows He being combined with He+ right away. Which method is correct? Also, in one page of our notes there is a diagram for the MO model showing "an infinite chain of atoms" (after the "Metal Bonding" slide)...what does this mean?
Either method is correct. As long as the correct number of electrons end up in the MO (in this case, 3 electrons) it does not matter if you ionise one of the He first (which would give 2 electrons in the He AO and 1 in the He+ AO, and three in the MO) or after (which would give 2 electrons in the He AO and 2 in the He AO, ionize to lose one and 3 in the MO)
ReplyDeleteI personally prefer to bring in all the electrons for each atom (easier to double check you have correct number corresponding to the valence-->this becomes important in later courses) and ionize the entire species, but they are both correct.
An infinite chain of atoms will result in an infinite number of MO's (recall each AO gives one MO) and eventually these are called bands.
Hi, are we responsible for learning the names and structures for all of the electron pair geometry structures and molecular geometry structures for VSEPR theory.
ReplyDeleteie. do we have to know that the electron pair geometry is octahedral or more specifically that it is square planar
Hi, I was wondering if there was any way we could get the answers to the quizzes so we can review them, they show the score, if you got the question right or wrong, but doesn't show the correct answer abcd etc.
ReplyDeleteHi Dr. Sirk,
ReplyDeleteI know that when looking at the bond angles in molecules that have multiple bonds, the angles will be smaller; however, if a molecule has one double bond and two single bonds, but also has 2 other resonance structures will the angles all be 120 if they're trigonal planar?
Is BCl3 polar?
ReplyDeleteIf a question comes up asking us to determine whether a molecule is polar or not, will a double bond affect the polarity if the molecule would normally be non-polar?
ReplyDeleteAlso, if all of the molecular geometries have no non-bonding pairs,will they all be non-polar? (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral)?
Re : Are we responsible for learning the names and structures for all of the electron pair geometry structures and molecular geometry structures for VSEPR theory.
ReplyDeleteAnswer: Yes, yes and yes.
Re quiz answers
ReplyDeleteThe quizzes with answer key are posted here (bottom of page)
http://web.uvic.ca/~chem101/quizzes.html
Re: bond angles
ReplyDeleteYes all angles will be equivalent as the double bond is spread equally (delocalized) over the three resonance structures. An example is SO3. All angles are 120 degrees.
BCl3 is non polar (It is also triganol planar and breaks the octet rule)
ReplyDeleteRe: If a question comes up asking us to determine whether a molecule is polar or not, will a double bond affect the polarity if the molecule would normally be non-polar?
ReplyDeleteAnswer: If there is a double and 2 single bonds then the central atom must be
a) bonded to different species (ie H2C=O) and therefore polar
b) It is only one resonance structure and all bonds are actually equivalent (ie SO3)
Re:Also, if all of the molecular geometries have no non-bonding pairs,will they all be non-polar? (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral)?
Answer: Only if all the atoms are equivalent. Consider the following tetrahedral molecules with no lone pairs. CCl4 is non polar, CF2Cl2 is polar, CHClBrF is polar, CH4 is non polar.
the midterm answer key says that IF4- in non-polar. Does that mean that the non-bonding pairs cancel each other out? does this apply to other molecules as well?
ReplyDeleteQuestion: 1.What is the polarity of XeF2?
ReplyDeleteAnswer: XeF2 is non polar (trigonal bipyramidal electronic and linear molecular geometry)
2.What is the molecular geometry of CH3+
Trigonal pyramidal
3.How many pi bonds will diazine have?
Three
4.Are molecules with a tetrahedral molecular geometry of always non polar if all the substituents are the same?
Yes
Re lone pairs
ReplyDeleteWhen considering polarity, you can ignore lone pairs (except for their very important contribution to the molecular geometry)
So does we just draw the molecular geometry without the lone pairs and determine the polarity from that?
ReplyDeleteQ23.30: What is the difference between insulators, conductors, and semiconductors?
ReplyDeleteI understand there is a bigger band gap, but why is that happening?
Yes: The lone pairs do not affect the polarity.
ReplyDelete(Longer answer: They would, but any time the lone pairs are unequally distributed, the molecular geometry is asymmetric.
Band gap
ReplyDeleteShort answer: all you need to know is that the band gap (the energy through which an electron must be promoted to conduct electricity) changes.
Long answer: the band gap reflects the difference in energy for the bonding and antibonding MO's. This energy is also related to bond strength. Carbon as diamond has a strong bond strength and a large distance between MO's and is an insulator. Silicon has a medium bond strength and a medium distance between MO's and is an semiconductor.
I was wondering when the answer key for Midterm 2 will be posted?
ReplyDeleteThanks
The key will be posted after the exams have been graded (likely by Friday)
ReplyDeleteAnd the answer key is posted:
ReplyDeletehttp://web.uvic.ca/~chem101/midterm.html
Is hydrogen bonding a Van der Waals force? My notes say it is, but the textbook says only dipole-dipole and London dispersion forces are. Thanks!
ReplyDeleteDo we need to know the crystal structures of CsCl, ZnS, and CaF2 (shown on p. 467 of 11th ed)?
ReplyDeleteHow far are we supposed to read in chapter 25 again? I wrote down 25.5, p. 1062 but p. 1062 is in the middle of 25.3, so that can't be right....
ReplyDeleteThanks!
Can you explain the bond order question? #4c written
ReplyDeleteThanks,
Re: Van der Vaals forces.
ReplyDeleteFor the purposes of this course, Van der Waals forces mean hydrogen bonding as well as dipole-dipole and LDF. The text only lists the second two, but the class notes are quite clear. (For further confusion Van der Waals often only means LDF)
Crystal structure-Nope. You should know that different ionic solids have different crystal structures depending on charge and size. You don't need to know what they are.
ReplyDeleteReading in organic chemitry are to the end of 25.5. You don't need to know mechanisms though.
ReplyDeleteBond order 4c.
ReplyDeleteThere are three resonance structures that can be drawn. Two are more stable than the third. Therefore, the two stable ones contribute more to the final product and the bond order is a little less than 1.5.
In regards to #4 c on the second midterm, how can you determine that the third resonance structure is the least stable?
ReplyDelete
ReplyDeleteHow to Check JSC Result 2019?