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Chapter 3: Bonding

Re: Chapter 3: Bonding

Postby goldstanda3269 » Mon Aug 24, 2015 4:17 am

ilbe wrote:I have watched hybridisation orbital in the video. Dr described ethane, ethein but not ethyne.
I have a question with regards to hybridisation orbital in ethyne.
Ethyne is triple bond thus,
C-C is sp hybridised orbital & sigma bond
other C=C are two p orbitals & pi bonds
Is this right?


Yes, that is correct.
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Re: Chapter 3: Bonding

Postby laura.cp897117 » Tue Oct 27, 2015 2:33 pm

I'm having trouble with Q2:
Which of the following representations of nitromethane have incorrect valency? (Me = methyl)

For structure A, my calculation of formal charge of N is:
# valence e = 5
# non-bonding e = 2
1/2(# bonding e) = 1/2(8) = 4
Therefore 5 - 2 - 4 = -1. The diagram shows N as having a + formal charge. What have I done wrong here?
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Fri Oct 30, 2015 2:20 am

laura.cp897117 wrote:I'm having trouble with Q2:
Which of the following representations of nitromethane have incorrect valency? (Me = methyl)

For structure A, my calculation of formal charge of N is:
# valence e = 5
# non-bonding e = 2
1/2(# bonding e) = 1/2(8) = 4
Therefore 5 - 2 - 4 = -1. The diagram shows N as having a + formal charge. What have I done wrong here?


The error is this line "# non-bonding e = 2" because in the molecule provided (nitromethane), you can see that the # non-bonding e = 0 (i.e. all the electrons are involved in bonding).

If you want to see it in slow motion (!!):

https://www.youtube.com/watch?v=FtvzYrjFXUA (nitromethane, VSEPR)

FYI: You don't need to memorise the rules for VSEPR because if it appears as a GAMSAT problem, ACER would re-list the rules which you would then need to apply correctly to one or several problems.
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Re: Chapter 3: Bonding

Postby laura.cp897117 » Fri Oct 30, 2015 9:42 am

Thanks for the video! That clarified a lot of things!
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Sun Dec 20, 2015 10:11 am

You are expected to know that H and He cannot possibly follow the octet rule (they can only have 2 electrons) but other than that, if there are any exceptions to the octet rule, the passage will suggest it is so.
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Re: Chapter 3: Bonding

Postby adrianpavone » Wed Jan 20, 2016 6:25 pm

Hi GS,

I believe there might be an issue with question 16 of Bonding: Q&A - 1. The question has a CH2 molecule and asks for the Carbon's formal charge.

The formula listed in the Gen Chem chapter 3 states that the formal charge is total valance e-'s in free atom (4) - total non-binding e-'s (2) - 1/2 bonding e-'s (1/2 4 = 2). 4 - 2 - 2 = 0. However the question states that the correct answer is A which is +2.

I will preface the above by saying that it coming out to 0 doesn't make sense to me anyway: CH4 is neutral, CH2 leaves the carbon with access to only 6 electrons and so I would expect the formal charge to be -2 following the octet rule. Looking at an example online CH3 is listed as having -1 as a formal charge (http://www.masterorganicchemistry.com/2 ... al-charge/) and CH2 is listed here as having a neutral formal charge (https://groups.chem.ubc.ca/chem330/formCharges.pdf).

I suspect that the issue is that the image shows Carbon with two bonds to Hydrogen and NO NON-BINDING ELECTRONS shown. In that particular case the formal charge would in fact be +2 as listed. Realistically though, is there ever a case where this would happen? The only case I can think of is if somehow the Carbon was somehow stripped of 2 e-'s through extreme radiation or something?
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Wed Jan 20, 2016 11:04 pm

Good analysis, including your conclusion in the final paragraph.

Had the image shown CH3 with no free electrons, you almost certainly would have chosen +1 as an answer without hesitation, even though it is a very unlikely chemical species (also note that having a -1 charge on CH3, as shown on your http://www.masterorganicchemistry.com link, is even more unlikely, but it was a correct academic exercise). See the following summary (not meant to be memorised): http://wps.prenhall.com/wps/media/objec ... _12-10.JPG There will always be some GAMSAT questions presenting familiar concepts in an unfamiliar way to be sure that you continue to use the correct principles.

Here is CH2 having a lone pair of electrons: https://upload.wikimedia.org/wikipedia/ ... arbene.png ('carbene': not meant to be memorised)

However, the molecule in question 16 is definitely not carbene. As you can see, the answer is different if a lone pair of electrons is added to the molecule shown. There are only 2 ways for you to have such an exam question: either they tell you the charge and then ask how many lone pairs must be there, or they show you the number of lone pairs and ask you for the charge. In the latter case, if you see no lone pairs then it means that there are definitely no lone pairs.

btw, to make it fair for NSB students, normally a question like this would be part of a passage which contains the basic rules to assess formal charge (which you have applied).
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Re: Chapter 3: Bonding

Postby Sheri » Sat Mar 05, 2016 7:46 pm

Hi,

For Q&A qn 8, i chose B because "Lone pairs are held closest to the nucleus of the central atom and so cause a reduction on the H-X-H angle through increased repulsion of the bonding pairs." Hence I conclude that the more lone pairs the molecule has (H2O), the smaller the H-X-H bond.
Please correct me if I'm wrong. Thank you!
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Sat Mar 05, 2016 10:04 pm

Sheri wrote:Hi,

For Q&A qn 8, i chose B because "Lone pairs are held closest to the nucleus of the central atom and so cause a reduction on the H-X-H angle through increased repulsion of the bonding pairs." Hence I conclude that the more lone pairs the molecule has (H2O), the smaller the H-X-H bond.
Please correct me if I'm wrong. Thank you!


You are not wrong, in fact, you stated the correct reasoning. However, the question stem stated "Put the following molecules in the order of increasing H-X-H bond angle" meaning that you must begin with the smallest angle as described which is indeed H2O.
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Re: Chapter 3: Bonding

Postby ElleRT » Wed Jun 08, 2016 9:12 pm

For Q&A set 1:

Q12. Which of the following structures could not be considered as planar?

I don't understand how B and C could be considered planar.

Thanks
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Sat Jun 11, 2016 6:21 am

ElleRT wrote:For Q&A set 1:

Q12. Which of the following structures could not be considered as planar?

I don't understand how B and C could be considered planar.

Thanks


C is definitely NOT planar. The 2 molecules that you should instantly recognise are: D which must be linear and therefore planar; and C which, like NH3 is a 3 dimensional tetrahedron and therefore cannot be planar (see the table in CHM 3.5 or the videos in this chapter).
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Re: Chapter 3: Bonding

Postby gamsat773618 » Sun Nov 27, 2016 6:48 am

Hello,
Q3 and Q4 , if we have similar questions in the real exam, do we expect them to provide us with the periodic table? or we have to memorise the atomic no. and relevant info? many thanks
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Mon Nov 28, 2016 12:10 am

gamsat773618 wrote:Hello,
Q3 and Q4 , if we have similar questions in the real exam, do we expect them to provide us with the periodic table? or we have to memorise the atomic no. and relevant info? many thanks


I think you are referring to: Q3 and Q4 in Q & A - 1

It is not so much that they expect that you have memorised the atomic number nor the atomic mass, however, it is clear that they expect that you are aware of the positions of the first 20 elements of the periodic table (so you know the relative trends and valencies) and a sense of the most famous transition elements. When you complete ACER's Red Booklet in particular, but some of their other materials, this will become very clear. Reviewing our videos on the Periodic Table provides a good summary.
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Re: Chapter 3: Bonding

Postby lilsleepy » Thu Feb 08, 2018 8:36 pm

Hi Gold Standard,
On page CHM-36 2018 edition of the book, it says 'actual formal charge of the oxygen in c=o is 1/3 = -2/3' in CO3^-2 . Why does 1/3= -2/3? Does that mean the overall charge can be both 1/3 & -2/3?

In regards to -2/3, I would like to clarify the -2 & 3. When considering formal charges, do you have to consider both the single bonded oxygens? Is that why the charge is -2 instead of -1? Does the 3 in the denominator represent the 3 resonance structures of CO3^-2?


Another question regarding electrons around S on page CHM-37.
Why does the surrounding electrons in Sulfur 12? I can't see as to why there are 12 electrons surrounding Sulfur instead of the usual 8.

Thank you :)
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Re: Chapter 3: Bonding

Postby goldstanda3269 » Fri Feb 09, 2018 9:03 pm

lilsleepy wrote:Hi Gold Standard,
On page CHM-36 2018 edition of the book, it says 'actual formal charge of the oxygen in c=o is 1/3 = -2/3' in CO3^-2 . Why does 1/3= -2/3? Does that mean the overall charge can be both 1/3 & -2/3?


Your interpretation does not match the text. The overall charge is -2 and does not change. The -2 charge can be considered to be one -1 charge on each of 2 oxygens, or the better interpretation is -2/3 charge on each of 3 oxygens (equal charges) which adds up to -2.

Please look again at the images on page CHM-35.


lilsleepy wrote:Another question regarding electrons around S on page CHM-37.
Why does the surrounding electrons in Sulfur 12? I can't see as to why there are 12 electrons surrounding Sulfur instead of the usual 8.


I know that this won't seem satisfying, but your question is interesting but it has nothing to do with the GAMSAT. As soon as possible, please use real ACER practice materials to gauge your preparation and you will see that they would never ask such questions except when they actually provide detailed guidance. Some students spend most of their time reading content and little time completing GAMSAT-level practice questions so when they see the real exam, they are surprised at the style of questions (very little assumed knowledge and extensive reasoning). In-depth studying of details often feels like time lost.

OK, but just for fun!

Main group elements in the third period and below form compounds that deviate from the octet rule by having more than 8 valence electrons due to d-orbitals. Sulfur, phosphorus, silicon, and chlorine are common examples of elements that can form an expanded octet.
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