For students taking Chemistry 101 at the University of Victoria in the Fall 2013 term.
Is there any other midterm apart from the one in class and the one on the website that we could maybe see? thank you
No, sorry, I don't have any. The end of the lab manual has past exam papers with relevant material in, though.
When we solved number 5 on the multiple choice, Why is P double bonded with O ? Why isn't it a single bond?
I just had a couple of questions. First, I notice on the sample midterm posted (http://web.uvic.ca/~chem101/C101Midterm2AFall2010Answers.pdf), there are no answers for some of the written questions. Will an updated version be posted before the midterm? Also, in written question four of that midterm, I was wondering how to go about calculating the bond order of the N-O bond. I know that the formula is (nonbonding - bonding)/2, but how exactly does one determine which electrons are non-bonding and bonding respectively in such a problem?
Re: P=O - because expanding the octet of P is preferable to having more formal charges on the structure.Re: missing questions - I will go over these in class tomorrow. Re: bond order can be estimated by inspecting the different resonance structures and averaging the bond orders of the important ones. So, for example, benzene has two resonance structures, and when you average them, you find the C-C bond order is 1.5.
Re: bond orderSo for the example in the sample midterm O2NCl, the third resonance structure with the double bond from N to Cl is ignored because we are only looking for the N-O bond order?Then it's just ((2+1)+(1+2))/4 to find the bond order?
No, it's ignored because the Lewis structure with a double bond to Cl is an unimportant one. Lewis structures with +ve charges on electronegative elements and lots of formal charges are not good ones.
For the sample midterm on the website, written question number one it talks about bond energies. the question is H2+C2H4 -> C2H6. and the answer is you break the H-H bond and C=C and make C-C and two C-H. but for the breaking what about the four C-H bonds and for forming what about the six C-H? Do we just ignore them?
Breaking four C-H bonds and forming four C-H bonds cancel each other out.
Hello!I have a few questions:- which forces exactly are van der Waals forces?- i'm a little confused: are dipole-dipole forces stronger or dispersion forces? The text says one thing and the diagram another.- researchers don't know which PO4^3- structure is dominant.. will you be asking us questions like that? Thank you for all your help!!!
Which of the following is most likely to exhibit liquid-crystalline behavior? how do you go about answering this question
Look for *molecules* (i.e. not ions) that are rodlike and polar.
vdW forces are those between neutral molecules, i.e. hydrogen bonding, dipole-dipole, dispersion. Dipole-dipole forces are stronger than dispersion, for sure. You will have to show me exactly what discrepancy you're talking about between text and diagram."researchers don't know which PO4^3- structure is dominant" is not a question, so no.
When drawing the molecular shape, do we draw the lone pairs in or do we only draw the actual atoms in the molecule and the shape the lone pairs pushes the atoms into but not the lone pairs? In questions 4 and 5 in the back of chapter 4 in the notebook, the tutorial draws them without the lone pairs
It's correct either way; it's easy to show the shape if you do include them, just make sure you name the shape and not the electron domain geometry!
In question 1 in the study questions at the back of chapter 5, how come ion-dipole forces are not being overcome when dissolving CsI in liquid H2O?
Would you be able to go over the O2NCl bond order part of the question in tommorow's lecture?
Because you're forming ion-dipole interactions, not breaking them.I may do ClCNO2 quickly.
For 1d on the in-class midterm...Why is it between 110-124pm for the bond length?There is a clear triple bond between n-n but there is a lone pair of electrons...how does that affect bond length?
There are 3 Lewis structures, which you can rank in importance. The one with an NN triple bond is most important, but the N=N=O one is also pretty good, so the weighted average of these two will lie between a triple and a double bond.
On the midterm there was a general consensus of people getting the first question on the written wrong. I drew an identical structure as in the answer key but I don't have the number of valence electrons but it doesn t ask for it. What is wrong with my answer? Also I provided a range of electron lengths (122-136pm) for rather than an exact numerical value.. This was done in a practice midterm..why would this not be accepted?Also for the multiple choice, the drawings for liquid crystal states are totally ambiguous ( for nematic state)
I thought that the nematic crystal question on the exam showed too much disorder to be considered a nematic.. in the diagram in the text and lecture book they are quite ordered but on the midterm there were axis that were aligned at 90degrees... is there a better, more clear definition of what a nematic crystal is that includes the picture shown in the test?
Come and see me if you'd like something remarked.Re: nematic liquid crystals. The definition you got in class was "ordered along the long axis of the molecule only". B clearly shows some order - it's not completely disordered as in A. That order consists of some alignment along the long axis, and no separation into layers, therefore = nematic.
For the final exams in the back of the lab manual.. where are the answers?
You can buy old exam papers with solutions from ChemSoc, 3rd floor Elliott, this week.
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