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Chemistry Paper 1 problems...

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ayesha.:
m/j/03 q 27,28,34

Vin:
*Buzz* People need help though it's too late. I'm helpless I've not done the portion. :\

Good job tmisterr and CHEMMASTER6000! +REP! :)

Deadly_king:

--- Quote from: ayesha. on June 09, 2010, 08:06:01 pm ---m/j/03 q 27,28,34

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27. Normally the strength of halogen bonds get weaker as we move down the group. In other words C-F bond is much stronger than C-Cl bond. So in order for a radical to be formed, homolytic bond fission should undergo across a C-Cl bond.
A : CHFClC*FCl..........it shows that a C-F bond has been broken as one F atom has been lost. This cannot be the case since fluorine forms very stable molecules which are unlikely to break.
B : Same explanation as for A.
D : A C-H bond is also much stronger than a C-Cl bond. It is said that it is stronger tahn even a C-F bond. So its not likely to occur.  

Therefore the answer is C.

28. Sodium reacts with acids or alcohols to release hydrogen gas. However one mole of an acid or alcohol releases only half mole of hydrogen. Therefore X should either contain either 2 acidic groups. 2 alcohol groups or 1 acidic and 1 alcohol group.

Only compound D corresponds to that. One mole of all the other compounds will release only half mole of hydrogen gas.

34. All of these are compounds are seen to be ionic ones and are all across the same period in the periodic table. Therefore they form ions with the same number of electrons.
As we move along the period 3 from sodium to argon the covalent character of the compounds increases. Sodium(the first element in the period) is a metal while chlorine is a gas and exists as diatomic molecules with simple covalent structure.
But its electonegativity difference does not increase since fluorine has a much higher electronegativity value than any other element.

As we move from sodium to argon the electronegativity value increases.

Therefore answer is C since only 2 and 3 are correct.

Reference : http://www.chemguide.co.uk/atoms/bonding/electroneg.html

Arthur Bon Zavi:

--- Quote from: Deadly_king on September 05, 2010, 09:13:27 am ---27. Normally the strength of halogen bonds get weaker as we move down the group. In other words C-F bond is much stronger than C-Cl bond. So in order for a radical to be formed, homolytic bond fission should undergo across a C-Cl bond.
A : CHFClC*FCl..........it shows that a C-F bond has been broken as one F atom has been lost. This cannot be the case since fluorine forms very stable molecules which are unlikely to break.
B : Same explanation as for A.
D : A C-H bond is also much stronger than a C-Cl bond. It is said that it is stronger tahn even a C-F bond. So its not likely to occur.   

Therefore the answer is C.

28. Sodium reacts with acids or alcohols to release hydrogen gas. However one mole of an acid or alcohol releases only half mole of hydrogen. Therefore X should either contain either 2 acidic groups. 2 alcohol groups or 1 acidic and 1 alcohol group.

Only compound D corresponds to that. One mole of all the other compounds will release only half mole of hydrogen gas.

34. All of these are compounds are seen to be ionic ones and are all across the same period in the periodic table. Therefore they form ions with the same number of electrons. as we move along the period 3 from sodium to argon the covalent character of the compounds increases. But its electonegativity difference does not increase since fluorine has a much higher electronegativity value than any other element.

As we move from sodium to argon the electronegativity value increases.

Therefore answer is C since only 2 and 3 are correct.

Reference : http://www.chemguide.co.uk/atoms/bonding/electroneg.html

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+rep...dead brottha!!

Deadly_king:

--- Quote from: ruby92 on June 09, 2010, 08:02:02 pm ---can someone explain o/n 2006 q 9 and 21
o/n 2008
22 and how is 3 posibble for 32?

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Nov 06 No 9 and 21.

9. Standard enthalpy change of reaction = -286 + 44 - (-283) = 41 KJmol-1
First the hydrogen gas needs to be converted to water in liquid form and then into gas. This whole process requires (44 - 286) KJmol-1.
Then Carbon dioxide must be converted to carbon monoxide which will require 283 KJmol-1.
addition of all these energies leads to answer C.

21.
When n=1 ---------> C2H5Cl (1 type only) Chloroethane

When n=2 ---------> C2H4Cl2 (2 types) 1,2 dichloroethane and dichloroethane

When n=3 ---------> C2H3Cl3 (2 types) 1,2 trichloroethane and trichloroethane.

When n=4 ---------> C2H2Cl4 (2types) tetrachloroethane and 1tricholro 2 chloroethane.

In all there are 7 different choroethanes that can be formed. Hen answer is C.
It is easier to understand when you draw their displayed formula. Unfortunately i dont know how to draw them here.

Nov 08 No 22 and 32

22. The hydrocarbon has to undergo a monochloro-substitution which means that this reaction will not affect the carbon-carbon double bond. One hydrogen atom needs to be replaced by a chlorine atom. Chirality refers to  one carbon atom being attached to four different group of atoms.

A is rejected since it will not show cis-trans isomerism even with the substitution of chlorine as the same carbon atom carries two hydrogen atoms. Moreover a double bond is not indicative of a chiral centre.

B and C are also rejected as one carbon atom contains two methyl groups. Again no cis-trans isomerism and no chiral centres even upon substitution.

Answer is D.
Upon substitution of a hydrogen atom on the third carbon atom by a chlorine atom, the carbon atom becomes a chiral centre being attached to four different group of atoms namely methyl group, hydrogen atom, cholrine atom and propene.

Drawing the respective compounds will allow you to eliminate those not showing cis-trans isomerism.


32. A Brosted-Lowry acid is one which can donate an H+ molecule that is a proton donor.

1. H30+ <-------> H+ + H20

2. NH4+ <-------> NH3 + H+

3. H2O <--------> H+ + OH-
Water is an acid as well as an alkali according to Bronsted-Lowry. It can as well give H+ to form OH- and also accepts a proton to form H30+

Since all these reactions are reversible (<-------->) all three of them can donate a proton. Hence answer is A.


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