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Chemistry: Post your doubts here!

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You are on the right track by first counting the number of carbons. Actually, both C AND D start with seven carbons, so we eliminate both C and D.
Therefore, we focus on A and B.
Step 1: Replace the Br atom with CN
Step 2: hydrolyze the CN to COOH




Generally, the smaller the atoms orbitals that overlap, the stronger the bonds. This trend is seem by the bond strength decreasing from Cl-Cl>Br-Br>I-I.
H-F being stronger than H-Cl can also be explained in that manner.

However although F atoms are really small and thus expected to have the most overlap it is actually an exception to the general rule. This is because the two nuclei (each 9 protons) will repel each other if there are too close together (lone pair of electrons will also repel each other). Thus the bond length ends up longer than expected.

So how do we decide between A and B?

But why is that only with F-F? I mean H-F bond also consists of smaller atomic orbitals that overlap,no?
 
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So how do we decide between A and B?

Try drawing out the resulting products from A and B. You should be able to see that B gives us the desired product.

But why is that only with F-F? I mean H-F bond also consists of smaller atomic orbitals that overlap,no?

Yes.
But F-F has three lone pairs on each F atom , H-F has no lone pairs on the H atom.
Thus, the overlapping orbitals effect is much more significant than the repulsion effect of lone pair electrons in H-F.
 
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But isn't the second and third I.E's difference the greatest? and so, the 3rd electron can be said to be in a new shell not the 2nd one?

You are actually, right. The 3rd electron is in the inner shell. So if forms X2+.

Sometimes the values in the table are "inspired" from the data booklet.

So it referring to the data booklet can also confirm the answer.

we can see that it "resembles" the element of Calcium

Screen Shot 2015-01-18 at 11.09.50 AM.png
 
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Try drawing out the resulting products from A and B. You should be able to see that B gives us the desired product.



Yes.
But F-F has three lone pairs on each F atom , H-F has no lone pairs on the H atom.
Thus, the overlapping orbitals effect is much more significant than the repulsion effect of lone pair electrons in H-F.

Your last line seems confusing.. Do you mean, thus, the overlapping orbitals of F-F experience more lone pair lone pair repulsions than those in H-F ?
 
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Yes, you can simplify it in that way.



Answer is B.
thankyou
Q: whats the difference between methylbenzene and benzene reactions? For As level only
btw, see A and B give these products and none of them seems to be like the desired one to me:(
 

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thankyou
Q: whats the difference between methylbenzene and benzene reactions? For As level only
btw, see A and B give these products and none of them seems to be like the desired one to me:(

There are errors in your drawings (which is why it is important for me to see them rather then provide the answer straightaway).
Screen Shot 2015-01-18 at 4.38.12 PM.png

This will explain why B is the answer.
Screen Shot 2015-01-18 at 4.55.08 PM.png
 
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Form an equation between the Kc and the respective concentrations,then solve for the concentration of HI
View attachment 50295

Thank you again!! Can you help me with another one please?

May june 2014 P12 Q4
Hydrogen and carbon dioxide gases are mixed in equal molar amounts at 800 K. A reversible

reaction takes place.

H2(g) + CO2(g) H2O(g) + CO(g)

At equilibrium, the partial pressures of H2 and CO2 are both 10.0 kPa. Kp is 0.288 at 800 K.

What is the partial pressure of CO in the equilibrium mixture?

A 5.37 kPa B 18.6 kPa C 28.8 kPa D 347 kPa

5
 
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Thank you again!! Can you help me with another one please?

May june 2014 P12 Q4
Hydrogen and carbon dioxide gases are mixed in equal molar amounts at 800 K. A reversible

reaction takes place.

H2(g) + CO2(g) H2O(g) + CO(g)

At equilibrium, the partial pressures of H2 and CO2 are both 10.0 kPa. Kp is 0.288 at 800 K.

What is the partial pressure of CO in the equilibrium mixture?

A 5.37 kPa B 18.6 kPa C 28.8 kPa D 347 kPa

5

Try using the method shown in the previous question and express the Kp in terms of the equilibrium concentrations. Let the unknown pressure of CO be x.
Post what you have after that and I'll guide you from there if you're still stuck.

Edited: Changed "moles of CO" to "pressure of CO"
 
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Try using the method shown in the previous question and express the Kp in terms of the equilibrium concentrations. Let the unknown moles of CO be x.
Post what you have after that and I'll guide you from there if you're still stuck.
So
20/0.288 = p(H20) + p (CO)
and then i don't understand, it doesn't give me nothing for h20 and is the p(co) 1/4 x Tp? what is the Tp??
 
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So
20/0.288 = p(H20) + p (CO)
and then i don't understand, it doesn't give me nothing for h20 and is the p(co) 1/4 x Tp? what is the Tp??

I made a typo in the previous message, the unknown x is for partial pressure (not moles) of CO.

Screen Shot 2015-01-18 at 11.15.18 PM.png

solve for x...
 
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Hi,
Can anyone help me out with the following questions:)The marking scheme is also given.


Thank you,
 

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