For students taking Chemistry 101 at the University of Victoria in the Fall 2013 term.
I was just checking the scores for the online quizzes and I could find 4 and 5 but quiz 6 seems to have disappeared? How can I found out how I did?thankscheers anonymous
Yes, apparently a few people are reporting this problem - Dr. Briggs assures me that the system has recorded the grades and that it seems to be working just fine, but that he is not sure why some students can't see their grades. Hopefully it is just a minor glitch that will sort itself out. In the meantime, the answer keys for quizzes 456 have been posted.
I was just wondering about pi and sigma bonds. In the notes and text it says that single bonds are always sigma bonds but then it also says that sigma bonds are stronger than pi bonds. But, arent double and triple bonds stronger than single bonds? if this is the case, then why are sigma (single) bonds not weaker than pi bonds?
"Sigma bonds are stronger than pi bonds" is not an equivalent statement to "single bonds are stronger than double bonds".A double bond is made up of 1 sigma + 1 pi bond, and a triple bond of 1 sigma + 2 pi. So:1 sigma + 2 pi > 1 sigma + 1 pi > 1 sigma (where > means stronger than, not longer than!).
In the practice midterm, you are asked about NO2. It has 17 e-, so would that lead to 2 reasonance structures? and would the shape of NO2 be linear? except that it is apparently polar. Could you maybe shed some light?
for section 9.7 (molecular bonding theory) there are no recommended assignment questions. I am just wondering what kind of related questions we could expect on an exam, and how we might prepare for these questions.
for the previous post, i meant that there are no suggested problems for the ninth edition text
NO2 is polar; the unpaired electron has to reside in an orbital on N, so this means that you need 3 electron domains around the N, and the molecule is bent.Just try some of the questions in the "molecular orbitals" section; if the question involves elements other than H or He, don't bother with them.
just wondering why in a molecule such as water the oxygen must hybridize if it already has two p orbitals with unpaired electrons...can't these electrons just bond with the hydrogen and the other filled orbitals act as the lone pairs? or is there another reason to hybridization appart from the promotion of electrons to make orbitals with only one electron?
Good question. The reason is that if the two p orbitals were used to bond to the H, this would make the two H atoms 90° to one another (unfavourable due to crowding)and leave the non-bonding electrons in an s and a p orbital. It turns out that this is a higher energy arrangement, and that electron pair repulsion is mimimised by sp3 hybridisation.
This may be an obvious question, but I was reading chapter 9 in the text and noticed that there was another section (9.8) on MOs However, our notes and review do not cover this section. Does this mean that we do not need to know 9.8?
No. 9.8 covers MO theory for second row diatomics; we haven't covered it and it is not examinable.
For #5 in version B of the second midterm, how is it possible for oxygen to have a formal charge of zero if it is singly bonded to the central P and has a full octet? Wouldn't the formal charge be 6-7=-1?? Thanks a lot :)
The best Lewis structure is O=PCl3 (expanded octet on P, formal charge of 0 on all atoms).
i'm guessing atoms that are not central in a molecule don't need to hybridize since this requires energy...but are there cases in which they do hybridize? like the double-bonded oxygen in formaldehyde...is there a specific "type" of atom (like oxygen) that due to its valence distribution must hybridize? thanks.
The central atom is the most important (because it determines the shape of the molecule), but you do need to consider the outer atoms. Formaldehyde is a good example - the O is sp2 hybridised, maximising both sigma and pi overlap.
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