Quantum numbers, Orbitals, Orbitals, Orbital Representations
1. Read chapter 6 (287-291)
2. OWLs due Sunday
Oct 21
19. Wednesday, October 21, 2009
Chem. 111 with Prof. Knapp
MWF 1:25 – 2:15 in ISB 135
Quantum numbers, Orbitals, Orbitals, Orbital Representations
1. Read chapter 6 (287-291)
2. OWLs due Sunday
Oct 21
how many elements are present along the width of the block. For example, how many elements does it take to completely fill a p-subshell? That is the width of the p-block.
I was wondering what the width would be when it asks for width of periodic block.
Is anyone having trouble with getting the right answers in 6.4b MAS? I don’t understand what I am doing wrong, but I am putting the numbers into the calculator exactly as they should appear (even as they appear in the hints after getting it wrong)….
Compare the deBroglie wavelength of a bullet moving at 700 miles per hour (313 m/s) to that of an electron moving at 1.30E+7 miles per hour (5.81E+6 m/s) and a proton with a speed of 1.30E+7 miles per hour (5.81E+6 m/s).
I’ll just do the first, shouldn’t it be (in the calculator)…
6.626e-34 / (1.90e-3) * (313)
that’s h/mv… Do it out, you get 1.09e-28
However, it says the right answer is 1.11E-33.
It gives this as a solution
bullet (6.626E-34 Kg m2/s) / (1.90E-3 Kg)(313 m/s) 1.11E-33 m
Anyone know what I might be doing wrong?
Chris and Justine-
The d-orbitals (l = 2) have 2 planar nodes (# planar nodes = l). This leads to the clover-leave shape which you recognize as a d-orbital, with alternating phase in alternating lobes. Each d-orbital has a specific orientation in space, which is given by the m-sub-l quantum number, and note that each orbital has a unique orientation. Mathematically, each orbital is non-overlapping with other orbitals (remember, these are waves!).
The dz2 orbital is a weird one. But the planar nodes here take the form of two cones with their points at the nucleus. This leads to the funny donut shape.
To what temperature will a 36.0 g block of aluminum initially at 26.0*C rise if 425 J of heat is added? Specific heat capacity of aluminum is 0.897 J/(g . K)
The answer is 26+(425/(0.897*36) = 39.16 *C
What is the most likely electron configuration for a sodium ion in its ground state?
Justine, the 3d shaped orbitals has 5 different types. If it looks like it is on the x and y axis (lying flat down) it is going to be either 3dxy or 3dx2-y2 (both look the same but the other one is rotated 90 degrees). If its a on the x and z axis it will be 3dxz. X and y will be 3dyz. And if its like a doughnut around the origin with a balloon above and below it, it will be 3dz2. To match it with a picture look at page 290.
Alyson and Corey-
try reading the book section, and then return to do the OWLs. The book is really good…
I too am having the same problem as Alyson. I don’t understand that the format that owl wants us to submit the problem in.
I’m having difficulty with OWL where it asks for the specific designation for this orbital. I was wondering if anyone could help me i dont know what it is asking for and the hints aren’t really working to help.
I have the same question as Dylan, I have put in the answer 3dx2-y2 and it was wrong. I have tried everything and I am still getting it wrong, if someone could give me the answer to question 3 in 6.6-Shapes of Atomic Orbitals it would be appreciated!
whew, alright that was really helpful. i was so stuck with that. thanks!
Jessica,
The number of nodes is equal to n-1. And the number of planar nodes is equal to the quantum number l. So a 2p orbital has 1 node – as each p type orbital has a planar node, that is the only node present (l = 1). A 3p orbital must have 2 nodes – it is easy to find the planar node (agian, l = 1), as that gives the orbital its characteristic shape. But the other node is hidden – it is that ring that Aviva mentioned.
Page 289 of your book has a nice illustration of this type of phenomenon for s orbitals. The 1s orbital has zero nodes, and it has a spherical boundary surface – In figure A, the the wavefunction is always positive (red line). The 2s orbital has one node, but zero planar nodes (l = 0). So the one node has to be an internal surface, generally called a “spherical node.” Figure C shows the boundary surface for a 2s orbital (from Psi-squared); Figure A shows the wavefunction (blue line). Note that Psi is positive for small values of r, but negative for large values of r. The spherical node occurs at the point marked “2” on the x-axis.
Jessica –
The bigger orbitals have more “rings” of electrons around them. For example, in a 2p model will look like a normal dumbbell shape, while a 3p model will look like the dumbbell shape surrounded by a bigger cloud of electrons in a dumbbell shape ( with a space in between). Another example: the 1s model will look like a spherical ball of electrons, the 2s will look like that sphere with another bigger circle around it (with a space in between them), and the 3s will look like that with yet another bigger sphere around those two (also with a space in between). Its all explained very well in OWL 6.6a!!
Hope that helped!
How can you tell whether a dot model is modeling a 2p or 3p? i can tell what the subshell label is based on its shape but how do you know if its 2,3, etc.?
Hi Dylan,
I am not sure which question you are looking for. If the question was “what was the specific designation for this orbital?, then the answer would be 3dx2-y2.
If it doesn’t work, try to look up on page 290 and look at the figure 6.14. That is a good way to guide you to get answers.
Hope it is helpful.
@Alyssa,
Yes. Redox and oxidation numbers (Chapter 3.9) will be on exam II.
Hi Professor,
I was wondering if re-dox reactions will be on exam 2?
thanks!
Hey Taylor, he doesn’t really post anything for the OWL grades, but you can kind of figure it out yourself. He also needs to wait till the end of the semester to fix the 24 hour grace period problems people have done.
Thanks Michael. Also does anyone know about the extra credit with posting on the blog, can we post 4 questions and that will be our 4 points extra credit or do we have to answer 2 questions?
I’m having a lot of trouble with the tutorial 6.6e; I can’t seem to get past the third question where it asks (at least for me) to name the specific “3d” orbital. If anyone could give me and hand with this it would be greatly appreciated.
The answer to your second question, “To what temperature will a 36.0 g block of aluminum initially at 26.0*C rise if 425 J of heat is added? Specific heat capacity of aluminum is 0.897 J/(g . K)?”, is ~39.16*C
@Michael
The answer to “What is the specific heat of aluminum metal if 53.7 J of heat must be added to raise the temperature of 5.0 g of aluminum from 25.0*C to 35.0*C?” is 1.074j/(K*g)
Unless Professor Knapp has already announced that class is canceled Wednesday the 25th, which he has not, then class is still on. Classes go until 6:00pm on Wednesday before Thanksgiving. Recess begins after that.
Does anyone know if we still have class the Wednesday before Thanksgiving? My bio class is canceled and I was wondering if this class might be too.
nope! thank you for your help!
Is there a way to find out our overall grade or something for the OWLS?
Calculate the enthalpy change for the following reaction using standard molar enthalpy of formation data:
2C_6_H_10_ (g) + 17O_2_ (g) –> 12CO_2_ (g) + 10H_2_O (L)
Calculate how much heat must be added to 50.0 g of liquid water at 15.0*C to convert it to steam at 100*C. The specific heat capacity of liquid water is 4.184J/(g . *C) and that of water vapor is 1.86 J/(g . *C). The heat of vaporization of water is 2135 J/g.
To what temperature will a 36.0 g block of aluminum initially at 26.0*C rise if 425 J of heat is added? Specific heat capacity of aluminum is 0.897 J/(g . K)
What is the specific heat of aluminum metal if 53.7 J of heat must be added to raise the temperature of 5.0 g of aluminum from 25.0*C to 35.0*C?
@Alyssa
I had trouble with that one too. What you have to do is fill in all of the columns completely. For example, the one i got was an electron in a 5d orbital. this tells me that n is going to be 5, so you would fill the n column with 5’s. The fact that it is in the d orbital means L will be 2 (found that in the book), so you fill the L column with 2’s. Then m can be anything from -L to L, so in this case you would put the values of -2,-1,0,1,and 2 into the m column. If you are reading the table horizontally now, each row should read: 5, 2, -2 5, 2,-1 etc. I hope that wasn’t too confusing…
I am having trouble with OWL question 6.5 tut – Quantum Numbers. Does anyone know how to complete this tutorial?
thanks!