Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Ice-cream on April 08, 2005, 07:27:49 AM
-
hey guys, i have a few problems which i'm not too sure about...hopefully u guys can help me out!
1. A hydrogen atom in its ground state absorbs a photon of wavelength 121.5nm. To which higher energy state will it be excited?
a) n=2 b) n=3 c) n=4 d) n=5
(i used E = hc/wavelength to find te energy, which then i made it equal to deltaE = -2.18 x 10^-18 [1/(n final squared) - 1/(n initial squared)] where n initial is 0 because it is in ground state. The answer i got was n = 1.15...which doesn't match any of the choices...can any1 give me some help?)
2. Which of the following molecules has a nonlinear structure?
a) XeF2 b) BeCl2 c) O3 d) CO2 e) N20 (central atom is N)
(i'm confused about N20 becuase i think theres 2 double bonds; 1 between N-N and 1 between N-O...but i don't think that gives the lowest formal charges...any1 got any other suggestions?)
3. A ruby laser produces radiation of wavelength 633nm in pulses of 1.00e-9 s duration. If the laser produces 1.68e-16 J of energy per pulse, how many photons are produced in each pulse?
(i used E = hc/wavelength to find the energy of 1 photon. Then i divided the 1.68e-16J by that energy and got 535.468 photons...do u guys think it's right? coz i didn't use the duration time in my calculation so i'm not sure if its right)
Thanx
-
I'm no good at spectroscopy so I can't help you at 1. and 3., but as far as I remember, O3 has a nonlinear molecule.
Edit: Ok, your calculations in 3. look correct.
-
you can't do (1) using ryberg's equation. you must use bohr's equation that relates n directly to delta E for the electronic transition. bohr's equation is valid for single proton single electron system. ryberg's equation is used for generating the spectra lines for paschen, palm and other series in the hydrogen absorption spectrum.
regarding (2), i think the structure of N2O is N=N->O. You can't have N=N=O because it suggests that the central N atom contain a negative charge. The negative would be more stable if its localised on a more electronegative element.
for (3), you don't need the time for calculation unless the question ask you for ther rate of photon emmision for each laser pulse.
-
n=1 for hydrogen.
-
NNO has got delocalised electrons, just like an ozone molecule. NNO is linear, O3 is not..
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.in.gov%2Fidem%2Fair%2Fairreport%2Fimages%2FImage35.gif&hash=332e666f10d79571b2e913d610ccc00c7ed03122)
-
ok so i get why N2O is nonlinear...but what about XeF2? I got a single bond for both Xe-F and 2 lone pairs on Xe. Does that mean XeF2 is shaped liked H2O coz they both have the 2 lone pairs?
so do u guys reckon that the answer for the 1st Q is n= 1 for hydrogen? (it's just that that wasn't 1 of the choices so i just want to make sure before i go about saying the answer in the textbook is wrong!)
-
Wrong. Xe has got 8 electrons, right? So, 2 electrons are used for bonds, 6 electrons make electron pairs. So that gives us 3 pairs. So it's got like 5 electron pairs around (3 free pairs, 2 from bonds). That gives us a bipiramid looking like this.. (the internet is a great thing when my inorganic chemistry books are at my girlfriend)
XeF2 is linear.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.cbu.edu%2F%7Emcondren%2Fxef2.jpg&hash=3bb854ff1664ca882bdd4f81acff56d342aef3ff)
-
Ahh, ic....i left out 2 electrons! well, does any1 know about the 1st question about n= ? for excited hydrogen? i kept checking my answer and i still got n= 1.15 so roughly n=1. any suggestions???
-
Look what Mitch wrote.
-
i agree! why did the other ppl say that XeF2(sub two) has 4 lone pairs and such, the drawing is correct in every form, you can check it with the formal charge, its accurate. therefore, Xe(the central atom) has 3 nonbonding dommains so its a trigonal bipyramid in electron-domain-geometry but has a linear shape for the molecular geometry base on the VSEPR model.
-
Hello Geodome,
why can't he use Rydberg's formula?
Bohr's:
delta E =(h)*(R)*(1/n^2 -1/n^2); Delta E = (h)*(frequency)
upon substitution, you obtain Rydberg's
Rydberg:
frequency = (R)*(1/n^2 -1/n^2)
Just curious,
-
I got n=2 in Q.1 using Rydberg's formula.
To Ice-cream:
You shouldn't set n=0 to ground state, but n=1.