For students taking Chemistry 101 at the University of Victoria in the Fall 2013 term.
I am doing the suggested problems from the textbook and I am wondering why for ions V3+, Hg2+, Ni2+, Mn3+, Ru2+ and Au+ there are no s orbitals in their electron configurations and how we know which ones are supposed to have s orbitals and which ones do not.
Electrons are always lost first from the subshell with the greatest value of n, so for transition metals, s electrons are always lost first, then d. Transition metal IONS never have any s electrons, so their electron configurations are always [noble gas] nd^x^, where n is the appropriate principal quantum number and x = (number of valence electrons - charge).So for Au+, n = 5 and x = 11-1 = 10, so its electronic configuration is [Xe] 5d^10^.
Sorry not chap 7 related but I just wanted to confirm... we don't have a lab exam? Are we only graded on the stuff we do in lab an our reports? And if so will the final or midterm have any lab material on it?Thanks
No lab exam: "The laboratory mark will be based on pre-lab assignments, lab quizzes, quality of technical performance, and laboratory reports". There will be no material from the labs in the midterm or final.
I just wanted to say that it is really hard to focus on both the lab and the midterm exam for students that are doing the lab this week. I am a very organized person but i still like to review concepts 2-3 days before the exam. Some of us have the lab report due exactly at the same time as the midterm. In my opinion, something has to be done to better organize the lab and avoid this conflict( eg. extension or no labs during the same week as other deparments do)Thanks
Thanks for your comment; I will pass your concern along to the course coordinator.
I am looking over my notes, and i don't understand when we are drawing lewis diagrams how we know how many electrons there are...for example a chlorine bond Cl-Cl...this has a single bond between the two chlorines and 6 dots around each one... but the hydrogen bond has no dots around each hydrogen it is just H-H so how do we know when we fill up the octet?
Yes, fair question: hydrogen is an exception because its equivalent of an octet is just two electrons - the filled shell configuration that corresponds to the nearest noble gas, in this case He. So H (and He of course) needs 2 e, all other elements need 8.
Just a question related to the online quizzes. For Molybdenum the electron configuration was [Kr]4s^14d^5. How do we know what elements do not fill their s orbitals before moving on to the d orbitals?
They're the ones in which if you DON'T fill the s orbital, you get a half- or completely-filled d-shell (these have special stability). So Mo is [Kr] 5s1 4d5 instead of [Kr] 5s2 4d4. Similarly, you'd expect Ag to be [Kr] 5s1 4d10 instead of [Kr] 5s2 4d9.
sorry not about chap 7 but related to all chapters. Are the clicker questions posted online as well, so we can review them? tank you
The clicker questions will be posted - the ones from Chapter 6 are already up and the rest will be there in due course.
Can u briefly explain what ionic seperation is? It was in the 3rd quiz and I cannot find it in the nots or text book.
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).
When writing an electron (condensed) configuration for a noble gas do we simply write [Ne] or would we write [He] 2s...... Also, is it always ok to write the condensed configuration unless specifically asked for the full configuration?
Just write [Ne]. And yes, always write the condensed electron configuration unless specifically asked for the full electron configuration.
Why is Neon omitted from the isoelectronic series given in the text (pg. 269). It has 10 e-'s. For a series do we always omit the noble gas if it would be contained in the series?Thank you for all the timely responses!!!
I think it has been omitted in this context because it is not an ion. It is, of course, isoelectronic with the ions listed.
A visitor to my office today pointed out that Figure 7.4 in the text is confusing - it looks as if electrons in the 2p orbital are spending more of their time closer to the nucleus than the 2s orbitals. I agree, it is confusing. Perhaps a better impression is given by looking at the change in electron density with r for the 2s and 2p orbitals (links take you to the Orbitron). These depictions give a better idea of the importance of the electron density close to the nucleus - the 2s electrons have a high likelihood of being found in this crucial region and as such are screened poorly from the increasing nuclear charge (in multi-electron atoms).The radial electron density depiction for a hydrogen atom is not a terribly illustrative one in this context - a better picture would show the radial functions for 2s and 2p orbitals in a multi-electron atom.
Is there any way to calculate the energy levels in a many-electron atom? I guess we can disregard the Rydberg Equation since it deals with shell and not subshell energies....Thanks!
Yes, the Rydberg equation only works for H and other one-electron systems (He+, Li2+, etc). There is no simple way of calculating energy levels in a many-electron system.
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