# ChemSiddiqui

Senior Members

378

## Community Reputation

79 Excellent

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Atom
• Birthday 07/27/1989

## Profile Information

• Location
UK
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no interesting hobbies!
• College Major/Degree
Chemistry degree coming up in 2011
• Favorite Area of Science
Chemistry
• Biography
I am an average person
• Occupation
Student
1. Hello everyone, I am trying to teach myself theoretical chemsitry and have done well so far. I am using this book called Modern Quantum Chemistry by Szabo. I was reading about Couloumb and Exchange opertaor and I have a question. I understand why Coloumb and exchange operator integrals arise, due to electrostatic potential and antisymmetry of the slater determinant. Then there was this exercise which made me pause. It asks to prove that the coulomb operator denoted by Jii and the exchange operator denoted by Kii are equal. Now in Chemists notation it the couloub integral Jab is given as Jij = (ii|jj) = <ij|ij> and Kab is defined as Kii = (ij|ji) = <ij|ji>. Now the following to prove is; 1).Jii = Kii My attempt at it is Jii = (ii|ii) and also Kii = (ii|ii) so they are equal 2). Jij*= Jij My attempt at solution: Jij = (ii|jj) = (ij|ij) = Jij* other to prove are: 3). Kij* = Kij 4). Jij = Jji 5). Kij = Kji I dont want to consider (3-5) yet unless I know I am on the right direction on proving them and so I ask if you think the above solutions are reasonable? any suggestion you can give me to maybe do these sort of proofs? Any help will be most appreciated. Thanks.
2. Hi everyone, I am a beginner in the field of laser science and I have been reading a few things about techniques like velocity mapping and charged particle imaging. I understand most of it, but there are certain terms/words that needs clarification. I am going to point them out and explain what I understand from then if any of you can explain if i am correct then please say so and if i am not, please correct me. Ok so first: " In the case of neutral atoms, radicals or molecules, a quantum-state-specific resonance enhanced multi-photon ionisation (REMPI) scheme is generally used to produce positively charged species with unchanged velocities; the recoil of a very light electron typically causes a negligible change to the velocity of the neutral species when it is ionised " (Taken from literature) Ok so the first term resonance. How I understand it is this. If a molecule/atom absorbs a photon at a frequency that corresponds to the Delta E and hence satisfies the Bohr frequency condition the molecule makes a transition to a high energy state (lets call this the excited state) and such a process is called (resonant) absorption. Ok so that would mean that for the molecule to be resonantly ionised the photon must have energy that correspond to/higher than the ionisation potential for that molecule? And in the case of REMPI do we mean resonance enchaned because we go through a (resonant) intermediate state before ionisation? What I understand from recoil electron is this: the motion of electron( here ejection) initiated by collision. In the same text at other places, it talks about the recoil velocity or ion recoil. Does that mean in the same context? any clarification will be most appreciated.
3. Hey everyone, OK i am a bit stuck. I am doing some excercises on graphs and here is this table that contains data to plot a graph to determine the activation energy T/K---------------------------- $10^{13} k / cm^3 molecule^{-1} s^{-1}$ 300----------------------------1.04 350----------------------------1.87 400----------------------------3.04 450----------------------------4.60 500----------------------------6.58 I have constructed an arrhenius plot( ln(k) vs 1/T) from this data using excel. I am getting a slope of -1379. using ln(k) =ln(A)-Ea/R *(1/T) so slope = -Ea/R and Ea = -slope*R this means that Ea is $(1379*8.3144)/10^{13} = 1.14 X 10^{-9} molecule^{-1} cm^3 s^{-1} K^{-1}$. I want to get it into kJ mol-1. So I multiply it by Na and I get $6.90 X 10^{14} mol^{-1} cm^3 s^{-1} K^{-1}$ how do I get rid of the other terms to get joules? Do I need to multiply the whole number by 10-6 and then take the square to make it $mol^{-1} m^{3} s^{-2} K^{-1}$ but I still would need to get rid of K? How do I do that? sorry i know this is a bit messy but could anyone help. It will be most apprecitated. If any of you did the plot what value in kJ mol-1 are you getting? Thanks for your help
4. Hey everyone, I am a little stuck with some bit of thermodynamics. Basically, I am doing some questions and the first was what expression is equivalent to the partial derivative $(dV/dT)p$. I took the cross-derivative of the equation for free gibs energy ($dG =VdP -SdT$) and got $-(dS/dP)T$. Now this is where I am stuck. the next question says to work out an expression for deltaS as a function of pressure using the expression that I just found. I dont know what $-(dS/dP)T$ is equal to, so my question is that can I intergrate a partial derivate on it own between two limits for pressure to get deltaS? I think not; I think that I must first find what it equals(eg. equals to Volume, enthalphy etc),separate variables and then perform intergration. Any advise will be great and will be much appreciated.
5. well when you say lattice strucutre, what you actually mean is a very ordered and arranged structue. spin-lattice relaxation (T1) is in NMR, and i think here 'lattice' refers to the liquid molecules in the solution that are tumbing together. spin is an intrinsic property of electron due to them having spin angualr momentum (denoted by ms), its not really spin spin, but electrons behave as if they are spinning. Environment is what is around the molecule in question. Hope I have helped. Feel free to correct me if I am wrong anywhere.
6. Hi there, try group theory. A good book for this is in introduction to molecular symmetry and group theory by A.vincent. It should answer most of your questions above, or simply punch group theory into google. good luck.
7. Hello Everyone, I was just wondering if it is possible to hypnotise somebody? Has anyone here ever been hypnotised? share you experiences, opinions here....
8. Hi everyone, I was just wondering, because I did this experiment in which we had to oxidise 2-methoxyphenol to get vanillin. Use use nitrobenzene rather than any other common oxidising agent like potassium permanganate etc. I was wondering what that was? Could it be that oxidising agents like potassium permangante need to be acidified and this can dehydrate the hydroxy group on the 2-methoxyphenol and so we dont get the desired product? I am just curiuos!. Your suggestions are appreciated!
9. for future reference grignard reagents usually are used if a carbon-carbon bond formation is involved, but of course the conditions are important in choice of reagents!
10. ## How do you find electronegativity of an element?

As insane_alien said one can calculate electronegativity. there are scales of electronegativity and of them all (3 i think) paulings scale is most commly used by chemists but it is just qualitative, and the numbers are supposed from a scale of 1-4. But allens, mulikkens etc have mathematical expressions with which it is possible to calculate the electronegativity and they are in good agreement to the paulings scale of electronegtivities.
11. Thanks a lot timo (didnt you use to have a different username before? how can one change it?). I used the right units this timeand got a value of n=4 and I can work out the states myself now. good day!
12. hmm....Ok!. I just did the calculation and the result seems a little absurd. Here goes; $E= 5.448 X 10^-(19) J; n=?; R=1.0967 X 10^7 m^-1; Z=1 for H atom$ $n^2 = \frac{RZ^2}{E} => n = 4.48 X 10^12.$ what have I done wrong here? can you let me know? Thanks
13. do you mean $En = -R \frac{Z^2}{n^2}$? but then the question gives a positive value of ionisation energy and with this expression we cant get the value of n as the underoot will be complex. I think if we used the first equation in my previous post such that to set ny = infinity and then from there we can work out nx? what you say to that? thanks for your help though!
14. Hey everyone, I was going through some past exam papers and this question came up and I am as the title says stuck. here goes the problem; what states of hydrogen atom have an ionisation energy of 5.448 x 10^(-19) J? This question requires students to do some mathematical calculations. I was wondering if can I use the formulea $Enx-Eny = hcR (\frac{1}{nx^2} - \frac {1}{ny^2})$ and then apply $En = - \frac {hcR}{n^2}$ (where R is the rydberg constant) to get n and work out the quantum numbers from there? any help most appreciated. thanks
15. hi there everyone, I just drew a mo diagram of c2(C=C) and see that the bond between the two carbons is a double bond and that is due to the 2p pi orbitals. so where does the lone pair come from in the dicarbon then? I reckon they come from 2s bonding and antibonding orbitals. frstly could anyone confirm that please. given that, the dinitrogen has a similar mo diagram. comparing the two, can we say that the lone pair in dinitrogen have the same energy as in dicarbon as they too come from the 2s bonding and antibonding mo'? any comment or help appreciated.
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