Wednesday, October 12, 2011

Chapter 4

Any questions on Chapter 4 material? Ask them here.

58 comments:

Anonymous said...

When/how will we know our marks on the midterm?

Scott McIndoe said...

Later this week or early next.

Anonymous said...

Where will the midterm marks be posted?

Thank you!

Scott McIndoe said...

You'll pick up the marked exams from the front of class.

Anonymous said...

http://www.youtube.com/watch?v=b4wveY2-lCo&feature=channel_video_title

Anonymous said...

Does anyone know the actual concentration of the standardised sodium thiosuphate solution in Lab 4?

Anonymous said...

Why is the orbital hybridization around an N atom in NCl3 = 4sp3? doesn't an electron jump up into the 3s orbital, leaving electrons in the 2s 2p and 3s orbitals? Also, how do you recognize hybridization in the central atom of pictures of molecules? Thanks

Scott McIndoe said...

The hybridization is sp3 because there are four electron domains. N has 2 electrons in 2s, and 3 in 2p. To make the orbitals point in the right direction (corners of a tetrahedron), you hybridize making all four orbitals degenerate. One orbital has a pair of electrons in it (the nonbonding, lone pair) and the other three are singly occupied and set up to share electrons with, in this case, chlorine. As for recognizing hybridization: just count the electron domains in the Lewis structure!

Anonymous said...

What is "axial position"?

Scott McIndoe said...

Think of a globe: it spins on an axis, and has an equator. In the same way, a trigonal pyramid can be arranged so the bonds at 180° form the axis, and the other three lie on the equator. Hence axial and equatorial positions.

Anonymous said...

how do we find the polarity of molecules??

Harman said...

What would be the hybridization and shape of carbon monoxide?

Scott McIndoe said...

Re: polarity - from the relative electronegativity of the component atoms. Add the individual dipoles as vector quantities to get the overall dipole of the molecule.

Re: CO - well, 2 atoms has to be linear! It's sp, and there is a triple bond between C and O.

Anonymous said...

so does that mean that CO has two sigma bonds and one pi bond?

Scott McIndoe said...

No, you can't have two sigma bonds between the same two atoms. It has one sigma and two pi bonds.

Anonymous said...

then how would the sp hybridization be justified?

Anonymous said...

I mean doesn't it need two sigma bonds for sp hybridization?

Anonymous said...

Can we get a T-shaped molecule if an octahedral has 3 non-bonding electron domains? I got it wrong on the practice assignment.

Anonymous said...

When we estimate polarity of a molecule, do we consider polarity due to lone pairs?

Scott McIndoe said...

Re: CO - one of the sp hybrids forms the sigma bond, the other one holds the non-bonding pair of electrons.

Re: octahedral with 3 lone pairs - what was the compound in question?

Re: polarity - well, the lone pairs contribute to the SHAPE of the molecule, which affects the polarity. Just add together the vectors you get from considering the relative electronegativities.

Anonymous said...

Re: For the octahedral question, we had to choose the electron domain geometry for the given molecular geometry.(NOT involving any compound)

Scott McIndoe said...

Ah, OK. I asked because there are no examples of such compounds (3 lone pairs and three bonding pairs would require 9 valence electrons!). T-shaped molecules come about from molecules with five electron domains, two of which are non-bonding pairs of electrons.

Anonymous said...

I encountered this question in the same assignment: "What is the molecular geometry of a molecule with 3 outer atoms and 3 lone pairs on the central atom?
Enter the molecular geometry of the molecule." and the answer was T-shaped.

Anonymous said...

regarding bond angles, in what order do a non banding pair, a double bond, or a triple bond exert greater repulsive forces on adjacent electron domains ?

Scott McIndoe said...

Re: 3 outer atoms, 3 lone pairs - I guess if such a species existed, VSEPR would predict it to be T-shaped (the other option, trigonal pyramidal, would place the 3 bulky lone pairs closer together than would the t-shaped option).

Re: relative size of double/triple/non-bonding electrons: you can't tell from what you've learned, and it rarely comes into play anyway. In a bent structure with sp2 hybridization, you could look at the bond angle and if it is <120 degrees, that would suggest that the lone pair is bulkier than the double bond. That is the case for ozone, for example, but we wouldn't expect you to know that.

Anonymous said...

Just wondering why the average bond energies on our data sheet are positive numbers (ie: H-H average bond energy is positive 432), and for the graph on page 126 of our text the energy between the H-H bond is -436. I'm not sure I understand what the positives and negatives insinuate.

Anonymous said...

When determining the overall dipole moment of a molecule, does the presense of a lone pair effect this at all? And how do we know, when asked to draw certain molecules, whether to include the lone pairs or not?

Anonymous said...

in question 9 of chapter 4 questions, how do you know that the single hydrogen atom connected to the nitrogen atom (that has the lone pair) is "coming out of the page" (as indicated in the answer)
?

Anonymous said...

On mastering chemistry this question:
Which of the following molecules or ions will exhibit delocalized bonding?
SO2,SO3, SO3 2- b)SO3 2- c) SO2 and SO3 d) SO3 and SO3 2- e) None of the above

The answer key says A but why?

Scott McIndoe said...

Re: hydrogen bond energy. Bond dissociation energies are +ve, because this is an endothermic process. On the other hand, combining two H atoms produces energy, hence the value is -ve.

Re: dipole moment. The lone pairs contribute to the shape of the molecule, but they are part of the atom to which they are attached and we consider only differences in electronegativity between different atoms when looking at dipoles. It's safe to always draw the lone pair, just make sure you ignore them when it comes to describing the resulting shape (i.e. NH3 is trigonal pyramidal, NOT tetrahedral).

Re: Q9, ch4 - it could either be coming out of the page or going into it, both are correct. However, the key point is that it doesn't lie in the same plane as the C-H bonds.

Re: SO3 etc... I assume you drew SO3 with an expanded octet, three double bonds, and 0 formal charge on all atoms, and they drew it with two single bonds, an octet on S, and formal charges of +2 on S and and -1 on two of the O's. This is a poor question really, because it's a question of which is "more correct" - keep to an octet, or minimize formal charge? We'll rty not to ask ambiguous questions like this one!

Anonymous said...

On Sunday's quiz question 3 got me - "In comparing the same two atoms bonded together, the __________ the bond order, the __________ the bond length, and the __________ the bond energy." I answered greater-shorter-greater but the key says just the opposite, smaller-longer-smaller. Help?!

Scott McIndoe said...

They're both right... did you get marked wrong?

Anonymous said...

I was marked wrong on that as well

Anonymous said...

So which one for number 1 would be more correct? A few more people also said they had SO3 2- as an answer.

Will we be credited back for number 1 and 3 then?

Scott McIndoe said...

I've made the course coordinator aware of this issue. I'll let you know.

Anonymous said...

In this quiz question, "In comparing the same two atoms bonded together, the __________ the bond order, the __________ the bond length, and the __________ the bond energy."

Why is "greater, shorter, greater" wrong and " smaller, longer, smaller " correct ?

Scott McIndoe said...

see above

Anonymous said...

How to molecular orbital diagram for atoms involving p orbitals?...I couldn't find any example.

Anonymous said...

^ draw*

Scott McIndoe said...

You don't need to be able to draw an MO diagram for anything more complicated than the combination of 1s orbitals.

Anonymous said...

does that imply to all the sections? I remember doing some examples involving p orbitals in Dr.Burford's class.

Scott McIndoe said...

Even if you covered second row diatomics in Dr Burford's section, it won't be examinable. I assume he covered it just for fun (the O2 MO diagram shows nicely how O2 is paramagnetic, for example).

Anonymous said...

when two atoms are bonded by a triple bond, what is the hybridization of the orbitals that make up the sigma bond component of the bond?

would it be sp?

Scott McIndoe said...

yes, sp

Anonymous said...

Draw the Lewis structure for NO2(Question 5 on the lecture notebook in chapter 3)

I drew it with Nitrogen, with a single electron, having a double bond with an oxygen and single bond with another oxygen.

I know nitrogen cannot possible have an expanded octet because it is not period 3 or beyond.

Then why does the answer key show nitrogen with two double bonds to each oxygen with the lone electron on the nitrogen?

Anonymous said...

What is the flexibility of the markers on the grading of molecular sketches?...there are a few ways to draw them (going into the page, different angles)

Scott McIndoe said...

You can draw it from whatever angle, but it needs to be correct and clear.

Scott McIndoe said...

Re: NO2; yes, that answer is incorrect. You drew it correctly.

Anonymous said...

For Question 2 on Chapter 5 lecture notebook(predict the trend)

Would we need to know precisely that SbH3 has a greater boiling point than NH3?

I predicted nh3 would be highest since it would have dipole-dipole, hydrogen bonding and dispersion forces. whereas others have dispersion.(The answer key said Sbh3 has highest bp)

Also wouldn't all the molecules in that group be polar?(tetrahedral geometry with one lone pair)

Scott McIndoe said...

Yes, they're all polar, but only NH3 has hydrogen bonding. I wouldn't want to guess off the top of my head which of SbH3 and NH3 has the highest bp, so no, not a question I'd ask.

Anonymous said...

In molecules with an odd number of valence electrons where you have to put a single electron on an atom, is the single electron counted as an electron domain?

Thanks!

Anonymous said...

I don't quite get the answer given to this question: The ion I3- is known but the ion F3- is not. explain
why F3- does not form..... I tried to think of Molecular Orbital Bond order to explain it but they would have the same bond order...why is my approach wrong?

Thank You!

Anonymous said...

A question in the text about metallic bonding asks which would be more ductile, Zn or Si? How would you explain this in relation to metallic bonding? I understand that the more antibonding electrons there are, the less strong a metal is, but I thought Si was not a metal. The answer in the text says it would be Zn.

Scott McIndoe said...

Re: single electron - yes, it counts as a domain (the electron has to go in an orbital, it's just that that orbital is only half-filled).

Re: The central I of I3- is able to expand its octet, but F can't do that.

Re: Si vs Zn - we'll cover this more in Chapter 7, when we look at different types of solid.

Anonymous said...

For a molecule like(He2)- what would be the bond order?


This would have 5 electrons so would this question be examinable?

If so, how would you solve this?

Scott McIndoe said...

He2- is not examinable (and it's certainly not a stable species!)

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