Tuesday, September 11, 2007

2007 Chapter 6

Questions on the material covered in Chapter 6? Leave a comment here.

47 comments:

Anonymous said...

Are there microscopes powerful enough to see atoms? If so, what do electrons look like?

Scott McIndoe said...

Yes, but they are not optical microscopes, which at best can resolve down to about a micron. Atomic force microscopes provide a three-dimensional image with sub-atomic resolution, and electron microscopes can also image atoms. However, there are good fundamental reasons why we'll never see a picture of an electron - in Friday's class, we'll see how Heisenberg's uncertainty principle means that we can know the location of an electron only in terms of a probability.

Anonymous said...

Just wondering if you could tell us what kind of questions we should expect on midterms. Would they be multiple choice, fill in the blanks, short answer or a mix/other? Just so we can start to study accordingly. Thanks alot:)

Scott McIndoe said...

Check out the assessment page of the course website - there is a sample midterm exam posted there. Basically, they're a mix of multiple-choice and short-answer questions.

Anonymous said...

For the suggested practice questions for this chapter, are we only supposed to do the black questions from the assignment list or all of the questions on the list?

Scott McIndoe said...

All of them - but you'll only have the answers for the RED ones unless you've purchased the solutions manual.

Anonymous said...

Ok i was doing a practice problem and noticed one the units was a "pm" and the answer has to be meters. I checked in the back of the text for unit conversions but didnt find it...so what is a pm????

Scott McIndoe said...

A picometre = 10^-12^ m.

Anonymous said...

I want to get on the companion website for the text but, it requires a code that I don't think I have becasue I bought my book used. Is there anyway to get one?
Thanks

Anonymous said...

No question - just wanted to say I'm really loving the way you've structured the lectures, especially the clicker questions! Very unique & fun.

Anonymous said...

Are the online quizzes open-book, so we can look at all our notes and our textbooks as we do them?

Scott McIndoe said...

Re: companion website - here's Prentice Hall's advice on getting an Access Code. The benefit is probably marginal, though, so I suggest you just spend your time with the textbook - the online stuff will just present some of the same things in a slightly different way.

Re: clickers; glad you like them, I do too.

Re: online quizzes; absolutely, go ahead and use your notes and textbook.

Anonymous said...

Do we need to know the names of the scientists that contributed the theories/principles? (ex: Heisenberg, de Broglie, Schrodinger...)

Anonymous said...

Many of the textbook Qs refer to the table of the electromagnetic spectrum; for example they ask what kind of radiation (visible, UV, microwave...) is a wave with __wavelength. Do we need to memorize the wavelengths of each kind of light?

Anonymous said...

I don't have a question. I just wanted to tell you that I love your accent!

Scott McIndoe said...

Re: wavelength ranges. While it is possible that we can ask you about anything you've been taught, we tend to not ask questions that rely solely on you having rote memorized a table of data. I do think it is reasonable that you know the wavelengths of visible light (400-750 nm) and the general order i.e. radio > microwave > IR > visible > UV > x-ray > gamma.

Re: famous scientists; see answer above. We're not going to ask you "who came up with the uncertainty principle?", but we might just ask "what is the significance of Heisenberg's uncertainty principle [in the following instance]?". The names are provided partly for interest's sake, but also because people often find things easier to remember when they're associated with a person. The ideas associated with the 3 scientists you mentioned as well as Planck, Hund and Pauli are all important concepts from Chapter 6.

Re: accent; thanks.

Anonymous said...

So at this point, do we also know everything we need to tackle quiz #2?

Anonymous said...

Just wondering if there is a study group organized yet? or is there some sort of message board where students can discuss questions together?

Scott McIndoe said...

Re: quiz #2; yes.

Re: message board - no, but it is a good idea. What I will do is post an invitation on the blog for students to ask other students questions, and I will refrain from commenting under that post. It might be a good place to ask your question about study groups.

Anonymous said...

Could you possibly provide us with a sample of the "info sheet" we will have for the test just so when we are working on questions etc, we will know what we will be givin during the test and what we will have to memorize. Thanks!

Also...is there a differnt midterm for each chem 101 section or do we all take the same test?
Thanks again!

Scott McIndoe said...

I sure can - there's a data sheet at the back of your lab manual.
The midterm is the same for all four sections.

Anonymous said...

During high school chemistry I was always taught that the elements Al, Ga, In, Sn, Tl, Pb, Bi, and Po were also transition metals, and that Ge and Sb were metalloids. In this course, because of the electron orbitals, should we consider these to be nonmetals and only elements up to those in column 12 to be transition metals?

Scott McIndoe said...

You were probably taught that they were metals, as opposed to transition metals, and that is quite true. The s- and f-blocks are also metals. In class, I called the p-block "main group elements" without discriminating between metals, metalloids and non-metals; we don't cover this subject in 101 because, as you pointed out, you already know it!

Anonymous said...

your notes say that as the principal quantum number 'n' in the bohr model approaches infinity, energy = zero, but a higher orbit means a higher energy state, are these contradictory?

thank you!

Scott McIndoe said...

It sounds contradictory, but it's not - the sign of the energy is negative. Read "the energy states of the hydrogen atom" from section 6.3 of your textbook for a more detailed explanation.

Anonymous said...

I don't know if this is more of a physics question or a chem question, but:

Since work is transfer of energy and is equal to force times displacement, and since energy is quantized. Does this mean that both force and displacement would be quantized as well?

Anonymous said...

when doing electron configurations...when is it that you have to jump to the 4f and 5f orbitals?? we were confused as to when these are applicable??? andddd why is it that no two electrons have the same 4 quantum numbers...we have a general idea but want a real answer!! thanks....

Scott McIndoe said...

Re: quantization –yes, but I’m not sure how much sense it makes to think about it in classical terms. At the kind of scale at which it might become important, we can’t locate particles with confidence as they become more wave-like in character (see Heisenberg uncertainty principle).

Re: electron configurations, a “real answer”... In the solution to the Schrodinger equation, 3 quantum #’s (n, l, ml) arise from the space geometry of the solution and the 4th arises from electron spin. No 2 electrons can have an identical set of quantum #’s according to Pauli, so the quantum #’s set limits on the # of electrons which can occupy a given state and therefore give insight into the building up of the periodic table of the elements. As for the Pauli exclusion principle – it’s part of one of our most basic observations of nature: particles of half-integer spin must have antisymmetric wavefunctions, and particles of integer spin must have symmetric wavefunctions. If we write a mathematical description of the wavefunction for a two electron system, it will vanish identically if the states of the electrons are the same, implying that it is impossible for both electrons to occupy the same state.

When do you fill the 4f and 5f orbitals? Look carefully at the periodic table, and you’ll notice that the atomic number increases in a jump between Ba (56) and Lu (71) – in between are the 14 lanthanide elements (the 4f orbitals are being filled). Similarly, between Ra (88) and Lr (103), the 14 actinide elements appear (the 5f orbitals are being filled. So: for any condensed electron configuration – first find the element, then go one row up and to the far right to find the noble gas corresponding to the core electrons, e.g. for thallium (Tl, element 81), this would be [Xe] (element 54). So you need to account for 27 valence electrons. These are 6s2 4f14 5d10 6p1 – because you have to cross the s, f, d and enter the p block in traversing the periodic table from the left hand side to get to Tl.

Edmond Li said...

Dr. McIndoe:

I was doing the second online quiz and I encountered questions 06-mc-027 and 06-mc-036. For the choices they only had the letters a to e but there was nothing beside these letters. I looked at the question above and below the two questions to see if the letters correspond (i.e. to see if the question uses the same answers as the previous/next question) but they don't. Is this some sort of mistake or did I miss something?

I probably shouldn't post my student # online but I'll come to talk to you at a lecture if you want.

Scott McIndoe said...

These questions have images under each of the options; I'm not sure why your browser failed to render them. Contact Dr. Briggs (briggsag@uvic.ca), the course coordinator, and he will be able to discount these questions for you.

Anonymous said...

Hello. I am just wondering about an inconsistency in the periodic tables we have been shown. The table that was put up during the lectures show the lanthanides group from atomic number 56 - 70, but on the back of our lab manuals it shows the lanthanides as atomic numbers 57 - 71. The same problem can be said for the actinides. I am just wondering which one is correct. By the way, I think you are doing a great job in the course; I am really enjoying the lectures.

Scott McIndoe said...

The one that was put up in lectures is correct, and it is the one you'll get in the midterms and final. There IS a copy of the new one somewhere in your lab manual, at the start of the old exams section I believe. The new one is consistent with the textbook.
Just why so many PTs are printed incorrectly is actually due to a long-standing argument about, you guessed it, electron configurations. The argument centres around whether La or Lu is the Group 3 element. Even though the argument was cleared up over 50 years ago (it's Lu), this minor error has persisted.

Anonymous said...

Hi Dr. McIndoe,
I just finished the third lab quiz and on it I encountered a question regarding ionic separation. I can't remember this concept or find it in the text (it isn't even on wikipedia!). Just wondering if you could refresh my memory or point me in the right direction..
Thanks! I'm really digging your class!

Scott McIndoe said...

It's just asking for the distance between cation and anion, so it is a question about ionic size (Section 7.3 of your notes).
The Wikipedia is an amazing resource, great for background and (surprisingly?) accurate for scientific material, but it can rapidly drown you in information...

Glad you like the class, I appreciate the feedback.

An aside: I've had more questions about the data sheet; it's on the assessment page or just click here.

Anonymous said...

Hi - I have a question about using Heisenberg's uncertainty formula.
In your class example you multiplied the 1% uncertainty with the velocity. In the questions in the textbook (6.45, 6.46) they seem to put just the uncertainty (in the solution they have 0.01x10^5 m/s in the formula for a given velocity of 3.00x10^5 m/s +-1%).
So what exactly do you put as the value for delta v?
Thank you

Scott McIndoe said...

If the uncertainty is given as e.g. 3.00 +/- 0.01 × 10^5^, use the uncertainty directly (in this case, 0.01 +/- 10^5^). However, if it is given as a percentage, e.g. 3.00 +/- 1% × 10^5, you have to work out the absolute uncertainty by multiplying the percentage error (1% or 0.01) by the original quantity (i.e. 0.01 × 3.00 × 10^5^ = 0.03 × 10^5^).

Anonymous said...

Hello:
What is the difference between Zeff and the attraction of an atom on its electrons? Because as you go down a column the attraction on outer electrons decreases but Zeff increases slightly.
Also, does Zeff vary depending on which subshell of electrons you are considering for a particular atom? Thanks :)

Scott McIndoe said...

You're asking essentially the difference between the effective nuclear charge (how many protons are the electrons in a given subshell experiencing attraction to) and the first ionization energy (how easy is it to remove the outermost - highest energy - electron). Zeff increases slightly as you go down a group because the approximation that the core electrons are "perfect shielders" is not, well, perfect. The first ionization energy decreases as we go down a group because the electrons are further away from the nucleus (in higher subshells) and can therefore be more easily removed.

Yes, Zeff does depend on what subshell you're considering! The 1s electrons of gold, for example, experience as Zeff of 79, because they have no shielding. The 2s electrons of gold have a Zeff of 79 - 2 = 77, because the 1s electrons screen some of the charge. The 3s have 79 - 2 - 8 = 69, and so on.

Anonymous said...

Hello, I just checked the scores on my first three quizzes and had a question. For the one about the correct electron configuration of Mo, you say it's one of the ones following exception where it only has one s electron so it can get a half-full d subshell. Are we supposed to guess this from where it is? Because you said in the lectures that the only ones we need to know are Cu and Cr.

Anonymous said...

Ohh, is that what ionic separation meant? I thought you meant lattice energy, or something like that - like how much it takes to separate ions. Oops.

Scott McIndoe said...

Re: Mo electron configuration; yes, Mo is in the same group as Cr so has the same electron configuration. I am mildly surprised that this question was set - W is the next element down and it has a 6s2 5d4 configuration, and there are a LOT of other exceptions among the heavier transition metals that you couldn't be expected to predict (e.g. Ru, Rh, Pd). However, the quizzes are open book and your textbook has a complete PT with electron configurations of all elements in it (near the end of Ch. 6). If you're ever unsure about a quiz answer: consult your textbook.

Anonymous said...

Just wanted to open by saying that I'm really enjoying your lectures. You have a pace that's easy to follow and make things really clear. I was just reading up on electron spin and wondered if an electron can ever "change" it's direction of spin. If it's striped to another atom in an ionic bond will it retain its original polarity/spin direction, and pair only with a complimentary atom, or can it change direction? I hope the question is clear enough.

Thanks.

Scott McIndoe said...

Thanks. Yes, electrons can change their spin, and there is an entire type of spectroscopy (electron spin resonance, ESR) that relies on this phenomenon. It can't happen of course when electrons are paired in an orbital (if one of a set of paired electrons flipped its spin, it would violate the Pauli exclusion principle), so ESR is applicable only to paramagnetic compounds (those with at least one unpaired electron).

Jacob33 said...

Which of the magnetic quantum numbers correspond to which plane?

Thank you.

Scott McIndoe said...

We can't tell. The main thing is the respective orientations of the orbitals with respect to one another, e.g. that the 3 p orbitals are all perpendicular to each other.

Anonymous said...

In quiz 2 there was a question that asked what the electron configuration was of a ground state molybdenum. Apparently it is one of the exceptions to the rule, but I do not think that we learned about it in class; as you said the only anomalous elements were copper and chromium. ?.

Scott McIndoe said...

See my answer above re: Mo. I didn't say Cr and Cu were the only ones that were anomalous, just that they were the ones you should learn. As you go further down the periodic table, anomalous electronic configurations become increasingly common, due to the similarity in energy of the outermost orbitals, and there is very little point in memorizing these exceptions.