Saturday, November 16, 2013

Midterm practice

We went over some old midterm questions in class on Thursday/Friday. Here is a copy of the 2011 second midterm (and here is the answer key). I highly recommend you try these out, along with the practice midterm posted on the course website.

99 comments:

Anonymous said...

will the midterm be all multiple choice again?

Scott McIndoe said...

Yes

Anonymous said...

I'm not sure if the links are supposed to the be the same or not but the 2011 midterm link seems to be the same as the old midterm questions we had gone over in class.

Scott McIndoe said...

Sorry, cut and paste error. Thanks for letting me know. Link should now be fixed.

Anonymous said...

how much of the midterm will have questions from chapters 1,2 and 3?

Anonymous said...

will we need to know anything from chapter 6?

Scott McIndoe said...

243: the midterm presupposes you know all the material from Chapters 1, 2 and 3, but will not test explicitly on it.
316: No; it's quite possible you will see organic examples to which you can apply your knowledge from chapters 4 and 5 to, but you won't be asked to systematically name an alkane, for example.

Anonymous said...

will the key be posted for this second set of practice midterms?

Scott McIndoe said...

yes, but later in the week

Anonymous said...

Question 25 of the 2012 Fall Midterm.

We have learned about boiling points in class, but not melting points. As far as boiling points, I would have thought that the answer would be phenol > toluene > benzene. Phenol as the highest melting point makes sense due to hydrogen bonding. I would put toluene 2nd and benzene 3rd as the shapes are the same, with an added CH3 group adding extra molecular weight to toluene.

Does a compounds melting point have additional factors, and do we need to know how to determine melting points if so?

Anonymous said...

benzene is symmetrical so it has a higher boiling point. page 166 (inoic liquids) of the textbook explains everything

Anonymous said...

when will the practice midterm from 2011 be posted?

Anonymous said...

Why is CF2Cl2 always polar when it perhaps could (my wrong thought) be symmetrical with the flourines on one axis and the chlorines on the other?

Anonymous said...

the lewis structure for SO2 has formal charges of 0 but 10 electrons on the center atom. the structure with one double bond and one single bond doesn't all have formal charges of 0 but it satisfies the octet rule. Which one is correct?

Scott McIndoe said...

302: Yeah, this was a tricky one. This exact example is straight out of the (later) notes, though. The b.p. of toluene is higher than benzene because of greater London dispersion forces, but the m.p of benzene is higher due to very efficient packing of benzene.
558: not quite, see above.
600: it already has been, on this blog.
610: CF2Cl2 is tetrahedral, so the two fluorines are always at 109.5 deg to each other and produce a larger dipole pointing in the opposite direction to the two chlorines.
613: In cases like this (where different rules point to different answers), both Lewis structures can be considered as contributing to the overall picture of the bonding.

Anonymous said...

are there three resonance structures for SO2 then?

Anonymous said...

why does benzene have a higher melting point then toluene ?

Scott McIndoe said...

915: no, just 2
921: see my comment at 856

Anonymous said...

so the resonance structures both have one single and one double bonds?

Scott McIndoe said...

Oh, yes, sorry, 3.
::O=S:=O::
:::O-S:=O:
::O=S:-O:::

Anonymous said...

Resonance structures seem ambigious. If we can draw a single dominant structure then do we assume theres only 1 structure?

Does so3 have 1 resonance structure or 4? What wiuld we put on a test

Anonymous said...

on the practice midterm, number 15, how does lattice energies work?

Anonymous said...

Hello! I was was wondering if we would be able to have scrap paper for the midterm, because the papers are double sided with small margins.

Thanks! :)

Anonymous said...

Are non-bonding electrons included in an atom's hybridization? Why? or why not?

Scott McIndoe said...

308: Yes. SO3 has even more than 4, actually, since by expanding the octet you can have 1, 2 or 3 double bonds.
412: Lattice energy is proportional to the product of the absolute charges divided by the internuclear distance.
728: No. Use the back of the data sheet.
744: Yes, they have to go somewhere!

Anonymous said...

For question 14 on the 2012 midterm,
Would the best LS for SO3 be: S with each O connected to S by double bonds because this would make the formal charge O for each of the atoms and S can expand its octet because it's in the 3rd row of the periodic table? Wouldn't this mean that there isn't resonance because this is the best LS? I'm just a bit confused by this.

Thanks!

Anonymous said...

Are there going to be questions on enthalpy or lattice energy, since these were covered in Midterm 1?

Anonymous said...

On the upcoming midterm, could we ask for blank papers from the invigilator? It would be a lot easier if we can draw lewis structures on a separate piece of paper rather than on the sides of our testpaper.

Scott McIndoe said...

1008: Fair enough; SO3 is a complicated case. The 3 Lewis structures with just one double bond are good in that they obey the octet rule, but are not so good in that formal charge is high. The Lewis structure with 3 double bonds is good because formal charge = 0, but you have to expand the octet. In such cases, it is difficult to know which is best, so it's safest to assume that they're all contributing to the overall bonding picture. Note that the other four options don't allow you to draw any other Lewis structures, so you can eliminate them from consideration.
1009: probably not.
1219: see my answer at 1010, above: no, use the back of the data sheet.

Anonymous said...

Professor McIndoe, this might not be the right place to ask this, but is it all right if I come in during your office hours and ask you some questions not pertaining to the Chem 101 course material?

Specifically, the questions have to do with spectrophotometry. Are you the right person to ask, or should I find someone else to help me?

Scott McIndoe said...

I'd avoid visiting my office hours this week with that question (everyone else there will have questions about the midterm), but in general, sure. My research expertise does not lie in spectrophotometry, but once I find out a bit more I can at least point you in the right direction.

Anonymous said...

Question 2 of the 2011 Midterm

I understand why C isn't a valid resonance structure. There should only be one double bond with N and an O.

However, why is D a valid resonance/lewis structure for ClO3-. It has 26 valence electrons, which means it would have a lone pair on Cl. With the addition of a double bond, wouldn't there then be 10 electrons on Cl, which wouldn't be valid?

Anonymous said...

on Question 23 of the 2012 Midterm 2 it says that HOCH2CH2OH has a much higher boiling point than CF4. I understand that in our notes it says the more polar the molecule the higher the boiling point however it says in the textbook that "when the molecules of 2 substances differ widely in molecular weights intermolecular attractions are generally higher in the substance with higher molecular weight"
I am confused by the contradiction if you could help?

Scott McIndoe said...

238: Cl can have an expanded octet.
522: HOCH2CH2OH can form 2 hydrogen bonds per molecule. CF4 has only London dispersion forces. So while CF4 has a higher molecular weight, the stronger hydrogen bonds more than make up for it.

Anonymous said...

I don't know if you're allowed to answer this question, but for the chemistry midterm are we expected to go off memorizing specific names of stuff?

Like could you give us a question on "neo-pentane," or "acetic acid" without giving us the lewis structure?

Like, would we need to know what the neo stands for? because it does briefly go over that in the text

Scott McIndoe said...

I try not ask questions that rely on you memorizing some random factoid. The stuff I talk about the most is generally the most important; if I mention it only in passing, it's probably only peripheral info that you're unlikely to be examined on.

Anonymous said...

When determining polarity by the symmetry of the molecule do you refer to electron domain geometry or the molecular geometry?

Anonymous said...

Why is the bond angle of y in question 5 of the 2011 midterm 109 degrees?

Scott McIndoe said...

648: Molecular geometry.
758: That oxygen atom has 4 electron domains (2 bonding, 2 non-bonding), and is therefore based on a tetrahedron.

Anonymous said...

this is such a stupid question but how do you rank the boiling points of elements (ex. from mastering chemistry: Pb, Ni, and S)

Scott McIndoe said...

The stupid question is not from you but from MasteringChemistry - the short answer is, I'd look it up. Trends in boiling points are indicative of the strength of bonding forces, but that's not an easy question to answer when considering elements from the p block vs. the d block. Not a question we're likely to ask (though we *might* give you a selection of s and d block metals and ask you to order them, because the trends here are quite clear).

Anonymous said...

Question 11 from the old midterm questions we did in class concerns why Mercury is a liquid at room temperature. By looking at how mercury is at the end of the 'd' group ( for its row), wouldn't that mean that it had only antibonding electrons?
The answer says that it has an equal amount of bonding and antibonding electrons which does not make sense to me.

Scott McIndoe said...

No. It has as many bonding electrons as tungsten does... but the effect of those electrons is cancelled out by the antibonding electrons.

Anonymous said...

On question 4 of the 2011 Midterm referring to a resonance structure of nitrous oxide being the most significant contributor to the bonding in the neutral molecule. Could you maybe explain this or refer me to a place in the textbook where I could look it up?

RobinT said...

For Nitrous oxide, the structure that is the largest contributor is the one with the most negative charge on the most electronegative atom, and that atom is oxygen. So the structure with a triple bond between the nitrogens and a single bond to the oxygen.

Anonymous said...

I have tried #2 on the 2011 multiple times and i cant seem to understand why the answer is C Thanks

Anonymous said...

How in-depth do we need to know how to do MOs, do we need to know how to do them for the 2nd row of the periodic table, ie. C, N, O, F? Or just the 1st row H, and He, as were shown in class and in the previous midterms?

Thanks!

Anonymous said...

FOR TESTING PURPOSES:

#17 on the midterm says that there are resonance structures for SO2, but this is only when the octet rule is obeyed. SO2 doesnt need to obey the octet rule.

If we were writing a midterm test, and were asked to determine whether or not SO2 displayed resonance do we say yes or no?

IF we can draw a single good lewis structure (even if it breaks the octet rule) do we say the molecule doesnt display resonance?

Thanks

Miko H said...

I apologize for adding this so late but I don't quite understand infrared absorptions and molecular vibrations. (question 8, 2011 exam version A).

If there is a symmetric stretch on a linear molecule (CO2) does this create no infrared absorption because the stretches sort of 'cancel each other'?

Scott McIndoe said...

558: You can't expand the octet of N (but you can for Cl).
611: Just combinations of 1s orbitals
755: If you can draw multiple good Lewis structures, and some of them have multiple resonance structures, then say yes.
757: Yes, that's a really thinky question. You're exactly right; when the CO2 molecule stretches symmetrically, the dipole remains 0 at all times. Not true for all the other examples - the dipole changes during the vibration.

Anonymous said...

re:775: no, but im saying i can draw 1 best lewis structure. What exactly defines a "good" lewis structure? Isn't it ambigious if you say that worse lewis structures play a part? It completely goes against what we've learned in class.

On a test, if you were asked if SO2 had resonance what would the correct answer be? The reason you gave everyone a quiz mark for #1 was exactly for this ambiguity. Im wondering what we would put on a midterm>

Scott McIndoe said...

Let's take SO2. The ambiguity arises because it is possible to expand the octet of 3rd row (and beyond) elements. We gave everyone a mark for quiz question number one because it presupposed the primacy of the octet rule over all else - not something we taught. If you assume that the octet *can* be expanded, you are now faced with a dilemma. Is the one with the expanded octet and no formal charge better, or the one with an octet and elevated formal charge (as for ozone, which suffers from no ambiguity since you can't expand the octet)? Hard to choose, so it such cases it is better to assume that multiple Lewis structures contribute to the bonding picture. SO3 vs. SO3^2- is illustrative; if you assume the primacy of the octet rule, in both cases you should have a bond order of 1 and 1/3. But SO3 has shorter S-O bonds than SO3^2-, suggesting that the expanded octet Lewis structure is contributing to the bonding picture.

Anonymous said...

So then for all molecules that can be created by breaking and not breaking the octet rule (3rd row) we assume there is resonance.

On a midterm we are not capable of googling bond length and so this assumption must be made or else we are basically guessing whether or not there is resonance.

Also, for s03, do we say there are 3 resonance structures or 4? If 4, how do you describe the delocalization as one of them breaks the octet and one doesnt?

Anonymous said...

are the last few questions on the midterm on the chemistry page organic chem questions?

Anonymous said...

how do you explain question 24 in the midterm posted on the Chemistry main page

Anonymous said...

How do you know to find the bond angles if the diagram does not show non-bonding electrons?

Scott McIndoe said...

932: You could even argue there are *7* Lewis structure, since you can also draw structures in which there are two double bonds and formal charges of +1 and -1. The overall bonding can be thought of as an average of all the good Lewis structure, weighted towards the ones thought to be the best. If you can decide which these are - such as the SO3 case - just assume they're all weighted equally.
941: Yes
954: C, D and E all have the same number of electrons (more than A and B), but different surface areas. Pick the one with the highest surface area (the one that looks least like a sphere).
1002: Complete the octet!

Anonymous said...

Why does the metal in 6B group have the highest melting point but not the one with 9 valence electron,which is half of the number of the total electrons in s,p,d band(18)?Does it mean the antibonding appear just after 6B?

I know in the electron band structure,s,p,d bands have some part of energy overlapping.But how can I see how many electron belongs to bonding MO and how many are not?

For W 6s25d4,how can I know 6 electrons are all bonding electrons and s-d MO band are half-filled?I tried to fill the electron into the band diagram by moving a ruler upward on page 144 .If the ruler arrives at the middle of the s band,s band is occupied by 1 electron.Meanwhile d band will also contain some electrons.If I keep moving to the middle of the d band,now d band is occupied by 5 electron,but the horizontal line is above the middle of s band.Is that part empty?How does the half-filled s-d come from? Or I should move down a little bit and that's where the anti-bonding MO appear?

Mercury has 9 electrons,it has as many anti-bonding electrons as bonding electrons?

I'm quite confused about the band theory though...Thanks

Scott McIndoe said...

The s and d electrons are of similar energy, and the s and d bands overlap completely. So they can be considered as one continuous band that can accommodate 12 electrons, as shown in your lecture book. This band is half-filled at W (6 e) and completely filled at Hg (12 e). This is of course a simplification of the real picture, but it does a pretty good job of predicting the properties of the metals in the s and d block.

Anonymous said...

What's the bond order for He2- ? I'm a bit confused by this one.

Thanks!

Scott McIndoe said...

That's a pretty wild species you've dreamt up there. I can assure you that it is not remotely stable!

Anonymous said...

Okay thanks! It was one of the options on the Fall 2012 midterm, question 13 and I couldn't figure it out. Thanks!

Anonymous said...

Why is the melting point of Chromium higher then the others?

Anonymous said...

Question 21 on the Fall Midterm 2012 ^^

Scott McIndoe said...

See my answer at 747. Cr has a half-filled s-d band, so bonding is maximized.

Anonymous said...

how much larger must a polar molecule be for the dispersion forces to dominate over the dipole dipole interactions?

Anonymous said...

why are CH4,Br2,Al increasing in boiling point?

Anonymous said...

how many multiple choice? 25 as well? because the given sample exams had more than that

Anonymous said...

do we have to know the band structure of a metal? like 4.39 in the textbook

Anonymous said...

i was told there was a error on the pink exam but not until after i was already finished and had left so i couldnt go back in will i now automatically just get that question wrong?

Trey Larue said...

I just had a quick question about number 10 and 11 on the midterm. The two questions are asking about the hybrid orbitals in propene. Two of the carbons have 3 electron domains (meaning there are two sp2 orbitals) and one of the carbons has 4 electron domains (meaning there is one sp3 orbital). That was my logic on solving these two problems but then number 11 didn't offer one sp3 orbital as an option. So, where did I go wrong?

Anonymous said...

I'm pretty positive that there would be 4 sp3 orbitals because there is 4 electron domains. There would be 3 sp2 orbitals for two of the carbons.

Scott McIndoe said...

153: Not much larger. Depends of course on the size of the dipole.
201: Because London dispersion forces are weaker than metallic bonding, and LDF increase with number of electrons.
219: yes (as you found out)
242: you don't need to be able to reproduce it, but understand it, yes
623: I can only speak for the room I was in... I recorded the ID of the one student who left before the announcement. Some rooms had no announcement, but we will sort it all out.
800: 809 is correct

Anonymous said...

There was an error on the exam?

Anonymous said...

when will the midterm marks be posted?

Anonymous said...

question 10 and 11 on the midterm seem a little dodgey....

Anonymous said...

What force is the least important in Supramolecular chemistry?
And which force is responsible for helping simple glucose dissolve in water?

Thank you!

Scott McIndoe said...

324: Yes, but it will get sorted out.
330: Later this week, I hope, but the bubble sheet processing was *slow* last time.
209: How so? Too easy?
247: The question was basically "which of these forces is the strongest" and the answer was ionic bonding.
Hydrogen bonding.

Anonymous said...

will the answers to the different versions of the midterms be posted before the grades go up?

Anonymous said...

hi there, i was just looking at the mark sheet for the second midterm from this year. for number 10 asking about sp2 hybrid orbitals in propene. the correct answer on the mark sheet says C.6 but on my exam 6 was at B. just am wondering if the scanner would correct the mistake?

Anonymous said...

Pretty much same thing as the lost comment...but for question 11 as well, where the correct answer is 4. On the answer key 4 is D, but on the test we had, it was C.

Anonymous said...

Discrepancy in version B question #10. The answer key has the correct answer as 6, but C has the answer 6, whereas on my exam 6 is choice B, which I answered.

Scott McIndoe said...

Thanks for letting me know; I've passed on your comments to Dr Briggs for checking.

Scott McIndoe said...

You are checking the right answer key, correct? There are 2, one for each version.

Anonymous said...

Is the "LINK" button for the midterm answer keys broken right now? It isn't working for me

Scott McIndoe said...

Probably. Dr Briggs is getting Version B re-scanned. Stay tuned.

Anonymous said...

For questions 2/3 on the midterm (version A), why is N3- drawn with double bonds (putting a -1 charge on the two lateral Ns, and a +1 charge on the middle N) instead of with single bonds (putting only a -1 charge on the middle N and no charge on the lateral Ns)?

Scott McIndoe said...

To complete the octet on the central N - and you should check your formal charge calculations... :)

Anonymous said...

do you think the test results will be up by tomorrow?

Anonymous said...

I notice you put up links for the solutions of the midterm, however the links are dead, could you fix that please? Thanks!

Scott McIndoe said...

708: Maybe
753: Dr Briggs has taken them down while Version B gets rescanned.

Anonymous said...

Are we going to go over the midterm in class today?

Anonymous said...

I have seen the sign on the 3rd floor of the Elliott building that you are selling old Chem 101 finals. Will you be providing us with a few copies as you have done with the midterms, or do we need to purchase them?

Anonymous said...

How many multiple choice and written questions will there be on the final? thanks

Scott McIndoe said...

829: No, I didn't! Monday.
507: Yes, we will be posting one or two.
933: Somewhere in the 65-75 range.

Anonymous said...

For the 9:33 comment I meant to ask how many of each will there be, sorry for not making that clear. thanks

Scott McIndoe said...

All MC

Anonymous said...

In one of the MC questions, it says: Give the structural formula for an aldehyde that is an isomer of acetone.

How can we solve this?

Scott McIndoe said...

Move the carbonyl group to an end C instead of the middle one, i.e. CH3CH2CHO instead of CH3COCH3.
More generally, these sorts of questions just require you to draw multiple structures that obey the octet rule and check that they're all actually different.