# Energetics of Organic Compunds help

1. Oct 9, 2004

### parwana

Energetics of Organic Compunds!!!!!!!!!!help

The energy of interaction between a Mg cation and a Cl anion is -4.000 x 10 -19 J
What is the distance in Angstroms between these two ions?

Enter a numerical answer only, do not enter units.

i tried and tried and cant get this right

also Lewis Structure of CNH

Choose those statements that are true about the two best resonance structures for the molecule CNH

a) One resonance structure has eight valence electrons and the other has ten valence electrons

b) The best resonance structure has a single bond between the carbon and nitrogen atoms

c) Both resonance structures have a single bond between the nitrogen and hydrogen atom.

d) The best resonance structure has a triple bond between the carbon and nitrogen atoms

e) The best resonance structure has a negative charge on nitrogen

it cant be a, c is one of the answers, and the rest are troubling me.

2. Oct 9, 2004

### chem_tr

Dear friend,

Your first question involves a mathematical equation, and I don't know it. But it seems that if this distance increases, the interaction energy decreases. However, don't forget that there is a limit of coming near to each other; this is called "repulsion". So just find the necessary equation, and solve it...

About the second question, I think the likely resonance contributors are like the ones I sent in the attachment. Here, the neutral molecule HCN is not considered; the most stable structure is this. I exactly followed your query as "CNH".

In CNH molecule, there should be a triple bond between carbon and nitrogen; and one bond is present, of course, between nitrogen and hydrogen. The nitrogen is positively charged since it used its lone pair; besides the carbon is negatively charged, due to the absence of one bond.

In another conformation, there is a normal three bond pattern for nitrogen, double for carbon and single for hydrogen with a positive charge on nitrogen. But the carbon has to be 2- charged, this is not a stable conformation indeed, but possible as an alternative.

The last one is three minus on carbon and two plus on nitrogen. This is the least stable conformation; if the total number of positive and negative charges increase in a conformer, then we accept it to be highly reactive, i.e., not stable.

You may wonder why I put all the negative charges on carbon instead of nitrogen. This has to be like that, since carbon without hydrogen is a very strong base and this one must be neutralized before nitrogen. In this form, carbanion has a more electronegativity and Lewis basicity than nitrogen. But only one hydrogen changes the pattern radically; in this case, nitrogen becomes the predominant base, you are right, GCT, about this.

#### Attached Files:

• ###### rezonans_CNH.zip
File size:
2.7 KB
Views:
46
Last edited: Oct 10, 2004
3. Oct 9, 2004

### parwana

so would it be choices c and d from the asnwers i posted???

4. Oct 9, 2004

### chem_tr

Yes, it seems to be c and d... The others are not likely to be correct as you stated already.

5. Oct 9, 2004

### parwana

i got it right, now for the first one i know the formula as

Energy= k x q1 x q2 / (r)

r is the distance needed, and q1 is the -1 for an electron of Cl, and +2 for Mg, also k is 1.389 X 10^5 kj pm/mol, when distance is in picometers. But the energy they give u is in Joules, not KJ/mol.

6. Oct 9, 2004

### chem_tr

Okay, if they give us the energy in joules, then you may divide it to 1000 to convert it into kilojoules. But I suspect that the energy should be expressed as Joule/mol; two atoms of Mg and Cl cannot produce that high energy, so let me correct this first.

$$-4.000*10^{-16}~{kJ/mol} =\frac {1.389*10^{5}~{kJ.pm/mol}*q1*q2} {r~pm}$$

You can find the rest from there I think. Just change the places of energy and distance, and the result is your answer.

Note that if the energy is correctly given, you'll use avogadro's number to convert it into mole and thus express it by kilojoules over moles.

Last edited: Oct 9, 2004
7. Oct 9, 2004

### GCT

Also related to the first question is the topic of effective nuclear charge, try finding through the index of your text.......perhaps it would be of interest.

Chemtr. where did you get your zip file? I'm not quite sure, but it seems that one of the resonance structures is wrong, the one on the right, the lone pair of the nitrogen comes out of nowhere (the nitrogen on the left strucutre should not be neutral). Also, try drawing an alternate route of delocalization, from the right structure, move the electron toward the nitrogen, in this case, nitrogen will be neutral and carbon will have a positive charge (this structure is relatively stable, it's a carbocation). This is just a consideration, perhaps I'm wrong. Two negative charges on carbon? Extremely unstable.

I need to review this subject a bit more, hwoever from what I remember, C and D seem to be correct, also consider the structure I proposed above, although carbon in this structure does not have a full octet (a carbocation), this would probably be the other stable structure (since two is asked for this problem). Although C triple bond N-H has a charge separation that makes it unstable, I believe it would be more stable than the other, carbocation.

Last edited: Oct 9, 2004
8. Oct 10, 2004

### chem_tr

Hello GCT,

I think that carbon without hydrogen is a very powerful base than nitrogen; that's why it behaves as a base before nitrogen does. Please review the relative pKb values of amines and carbanions; it will solve the problem.

I updated my post where you've seen my attachment; I forgot to increase the resonance conformers in which only one bond is present among C, N, and H. This is the lest stable conformer, but I think it goes on like that. Only one hydrogen changes the case drastically, I agree with you about that. In HCN, nitrogen is more basic than carbon, as more basic carbon already catched one molecule of acid. Please review the attachment in this thread; with this attachment, you can produce the molecular orbital pattern for cyanide anion by putting enough electrons. Please remember that the atomic orbital on the left must be of carbon, since nitrogen is mole electronegative and must be of lower energy (like the one on the right). You'll see that the sigman non-bonding molecular orbital is completely owned by carbon, and it is HOMO for bonding orbitals, so this must be used before the ones on nitrogen.

Please send your comments, GCT, I may also be wrong; maybe the one you're thinking is correct. I'd like to know your ideas.

#### Attached Files:

• ###### asimetrik_MO_diyagram.zip
File size:
6 KB
Views:
85
9. Oct 10, 2004

### parwana

i got 6.9X10^20 , but its wrong, also it says to give the answer in Angstroms??
You are looking for the interaction between a pair of ions,
not a mole of pairs of ions.

Mg 's charge is +2, and Cl is -1, so i have everything set up, but it is still not right

10. Oct 10, 2004

### chem_tr

Dear parwana,

If you are certain about the dimension, then you'll have to multiply the value with 10-16 with avogadro's number, roughly 6*1023, to convert it into kJ/mol.

The result you're finding is very probably in picometer form, a picometer is 10-12 meters. An angstrom is one tenth of a nanometer, i.e., 10-10 meters. So I think you'll divide your result by 100 and it will be over.

Last edited: Oct 10, 2004
11. Oct 10, 2004

### parwana

6.9 X 10^18 then??

can u try to solve it and see if iam right??

12. Oct 10, 2004

### parwana

also this is the hint it gave me

You are looking for the interaction between a pair of ions,
not a mole of pairs of ions.

13. Oct 10, 2004

### parwana

forget it i got it right, thanks to chem tr

14. Oct 10, 2004

### GCT

Did you acknowledge the other isomer that I suggested? It seems that you are showing delocalization in one direction, how about delocalizing the pi electron to the nitrogen from the leftmost conformation in your structure?

15. Oct 10, 2004

### GCT

The nitrogen will be neutral, the carbon will become a carbocation, everything else in the molecule will be neutral (near neutral).

16. Oct 11, 2004

### chem_tr

Dear GCT,

If you are talking about hydrogen cyanide, HCN, then you are right to delocalizing the electrons towards nitrogen side. But I think the molecule we are talking about is methylidineammonium, if we write it as HCNH with additional hydrogen. CNH is the anionic form of it, and it eventually converts to HCN. So you are right after it converts to HCN. I investigated the question without thinking of the other alternatives, just studied CNH. It is really a weird structure due to the position of the proton. I suppose it is better to handle this compound as HCN in order not to fail.

Thanks, chem_tr

Last edited: Oct 11, 2004