From the information below, identify element X.
a. The wavelength of the radio waves sent by an FM station
broadcasting at 97.1 MHz is 30.0 million (3.00 e7) times
greater than the wavelength corresponding to the energy difference
between a particular excited state of the hydrogen
atom and the ground state.
b. Let V represent the principal quantum number for the valence
shell of element X. If an electron in the hydrogen atom
falls from shell V to the inner shell corresponding to the excited
state mentioned above in part a, the wavelength of light
emitted is the same as the wavelength of an electron moving
at a speed of 570. m/s.
c. The number of unpaired electrons for element X in the ground
state is the same as the maximum number of electrons in an
atom that can have the quantum number designations n = 2,
mL = -1, and mS = -1/ 2.
d. Let A equal the charge of the stable ion that would form when
the undiscovered element 120 forms ionic compounds. This
value of A also represents the angular momentum quantum
number for the subshell containing the unpaired electron(s)
for element X.
posted before, but still need help. I did a and b already with help.
C. All i know for c is that n=2 means 2 possible subshells, mL=-1 means that L must ne 1 so it's a P subshell, so there are 6 electrons max so far, right?, but what about the ms=-1/2, what would that do?
yes i completely agree. i went through the entire part c and d myself and came up wit h platinum as my final answer.. i am not sure though. so c is just that there is only on unpaired electron
now part d says something about element 120 so i figured that was the atomic number, which means that the 120 element would be in group II which means it would have 2 valence electrons and a charge of +2 as an ion making the angular momentum (m)of element X = 2
Here is the way I interpret c. Check my thinking.
This part c is JUST for the determination of the number of unpaired electrons. If n=2, then L can be 0 or 1. If L = 0, theh m can't be -1 so L = 0 is out and L must be = 1 in this part. Given that m = -1 and s = -1/2 (m is m_{l} and s is m_{s}) and L = 0, I read that to mean that there can be ONLY the one electron. The ONLY other configuration that will give N= 2, L = 1, m_{l} = -1 is for m_{s} to be +1/2 and that is specifically excluded by the problem stating that m_{s}= -1/2. So I conclude that the number of unpaired electrons is 1.
(This says nothing about any other part of the element EXCEPT the number of unpaired electrons; that is, nothing about the valence shell--we already know N=5) or the number of electrons, either total or in any of the subshells. Don't be afraid to disagree. Check my thinking carefully.
is that remotely correct
I agree with the element #120 being in group II AND with the ion being +2 which also means that the angular momentum is 2. That means, although you don't say explicitly, that L = 2. So we are looking for an element with the valence shell of N=5, 1 unpaired electron in the L=2 subshell, a 5s2 configuration for that N=5 valence shell. Pt has a 5d9 6s1 so I don't think that is right because the unpaired electron is an s electron AND the outside valence shell is N=6 and not N=5. I think the problem states that the unpaired electron is an L = 2 or a d electron and we must have N=5. So now what do you think?
I didn't intend to bold all of that but I still think what I wrote makes sense, even with the bolding which I meant to cut off after the word subshell in line 4.
because of the m=2 I thought that L could be either 2 or 3 (d or f) and i thought that m=2 meant that the unpaired electron had to be in the d shell
i do understand what you are saying about platinum and 5s2 though
I am confused about your statement of m=2. m_{l} = -1 in part c is only for the purpose of determining the number of unpaired electrons. We know now that is 1 so that section is history. Part d states the angular momentum of the unpaired electron is 2 so L must be 2 for the subshell in which the unpaired electron is found and that makes it a d electron.
no, I wasn't talking about part c. I was saying that I thought that if the angular momentum was 2 that L could be 2 OR 3 because if L is 3 then the angular momentum can also be 2. But regardless I thought that the unpaired electron was located in the d shell.
The angular momentum of the atom is determined by the azimuthal quantum number, which in this problem, according to part d, is 2. So L = 2. L is the azimuthal quantum number. In every day English, it determines the ellipticity of the orbitals.
ok, I understand now. So where is the unpaired electron located? was I right about the d shell?
I see what you are saying. I think, however, that if it says the angular momentum is 2 then it must be 2 and we can't assume it COULD be 2 OR 3. For that matter, if it COULD be 3 then it might also be 1, or zero, since an atom with n=5 can have values for L of 0, 1, 2, or 3. I believe this whole part d is for the distinct purpose of telling us that L = 2 AND that the unpaired electron is in that particular subshell.
ok, so if it has a valence configuration of 5s2 and an unpaired electron in d then the only element I see with that configuration is Y, yttrium, which has a configuration of 4d1 5s2
I think it is Y. But I don't want to talk you into anything you don't believe. This is the first time I've seen this problem. However, I believe our reasoning is correct. This may be the most difficult railroad problem I have seen. And I've seen many.
I agree with it being Y! it's the only thing that make sense. Thanks a bunch for your help!! oh I redid the other problem about the A,B,C and I got 0.537 atm as the total pressure.
That's what I obtained also for the total P. That problem wasn't a piece of cake, either, but it was easier than this radio problem. The radio problem was stated so confusingly AND it jumped around from pillar to post. I had trouble understand exactly what the sentence said. Anyway, thanks for using Jiskha and I'm glad I was able to help. Good luck. Come back anytime.
Thank you very very much!!!!!!!!