Chemistry: Post your doubts here!

[Turki AbdulAziz]

Look at the groups in which Nitrogen and Sulfur reside.

Nitrogen is in group 5 and Sulfur in group 6.
That means Nitrogen has 5 electrons in its outer-shell, and Sulfur, 6.

The number of bonds an atom can form is,
No of bonds = Full valence shell - Number of valence electrons

For example, nitrogen already has 5 electrons, it needs 3 more to achieve a noble gas structure. So it will form 3 bonds.
The above formula works the same way,
No of bonds in N = 8 - 5 = 3

So with that in mind, Nitrogen can form 3 bonds and Sulfur can form 2.

The next thing to note, is that each bond will have two electrons.
Since nitrogen needs to form 3 bonds, it will need 6 electrons.
Sulfur needs to form 2 bonds, so it will need 4 electrons.

Out of the 8 electrons that can exist in the outer-shell,

-Nitrogen needs 6, which leaves 2 electrons not participating in any bond.
-Sulfur needs 4, which leaves 4 not participating in a bond.

A lone pair = 2 electrons not participating in a bond.

Nitrogen therefore has 1 lone pair and Sulfur has 2.

Hope that helped you!

This is true, but this is under the circumstance where Nitrogen forms 3 bonds and Sulfur forms 2 bonds. Sulfur does form two bonds with nitrogen in the structure displayed in question and hence it does have two lone pairs of electrons like you said. But if you look at the Nitrogen it only forms a total of two single bonds (one with each atom of sulfur) and not 3 bonds like how it is supposed to. So out of the 5 valence electrons in nitrogen 2 are used for bonding leaving 3 electrons free. Now this is where I'm really confused as we now have 1 lone pair and one electron in the outer shell ??

Look at the structure displayed in question: http://freeexampapers.com/A-Level/Chemistry/CIE/2012-Jun/9701_s12_qp_22.pdf

 

[The Sarcastic Retard]

This is true, but this is under the circumstance where Nitrogen forms 3 bonds and Sulfur forms 2 bonds. Sulfur does form two bonds with nitrogen in the structure displayed in question and hence it does have two lone pairs of electrons like you said. But if you look at the Nitrogen it only forms a total of two single bonds (one with each atom of sulfur) and not 3 bonds like how it is supposed to. So out of the 5 valence electrons in nitrogen 2 are used for bonding leaving 3 electrons free. Now this is where I'm really confused as we now have 1 lone pair and one electron in the outer shell ??

Look at the structure displayed in question: http://freeexampapers.com/A-Level/Chemistry/CIE/2012-Jun/9701_s12_qp_22.pdf

That means 1 lone pair and 1 free radical. We are not asked for free radical so ignore that. Hence 1 lone pair on nitrogen.

 

[psychiatrist]

Firstly, I've deleted my original explanation for Q18, I realized it was nonsensical after I reread it. Sorry about that.

Q6. The chart shows that for every 1 isotope that has 10 nucleons, there exist 4 isotopes with 11 nucleons.
This is the crucial interpretation you need to understand on the chart. The calculation is to find the average of each isotope.

If I were to reword the scenario as: for every 1 person who have $10 , there are 4 people with $11, what is average amount of money for each person?
(1 x 10 + 4 x 11)/5

Q17. Use data booklet and compare to the 2nd IE to the neighbors.
A: Mg (Compare to Na and Al) Does Mg has a higher 2nd IE than its neighbors? No.
View attachment 52835

B: Al (Compare to Mg and Si) Does Al has a higher 2nd IE than its neighbors? Yes.
View attachment 52840

C: Si (Compare to Al and P) Does Si has a higher 2nd IE than its neighbors? No.
View attachment 52839

D : P (Compare to Si and S) Does P has a higher 2nd IE than its neighbors? No.
View attachment 52838


Q18. Use the data booklet to compare the 1st IE of the options and the two elements that come after it.
A. Carbon (compared to nitrogen and oxygen)
View attachment 52841

B. Nitrogen (compared to oxygen and fluorine)
View attachment 52842

C. Fluorine (compared to Neon and Sodium)
View attachment 52843

D. Sodium (compared to magnesium and aluminum)
View attachment 52844

Based on the chart in the question, the 2nd element should have the highest IE and the 3rd element should have the lowest I.E. This fits option C.

Q27. You have already figured that there are 56 neutrons and 43 protons.
Statement 1: there are 13 more neutrons than protons. Thats correct isn't it?

Thanks a lot sir.

 

[psychiatrist]

https://docs.google.com/viewerng/vi...wp-content/uploads/2012/11/9701_w14_qp_33.pdf
Q1 part f. HOW ARE WE SUPPOSED TO DRAW THE HESS CYCLE!!!?? Anyone?

I am not sure if this is correct. Wait till Metanoia checks my answer.

 

[Turki AbdulAziz]

https://docs.google.com/viewerng/vi...wp-content/uploads/2012/11/9701_w14_qp_33.pdf
Q1 part f. HOW ARE WE SUPPOSED TO DRAW THE HESS CYCLE!!!?? Anyone?

This is what the Hess Cycle would look like (Teacher showed it to me):

 

[princess Anu]

Both Mg0 and mgOh are sparingly soluble in h2o or is it that mgo is insoluble?

 

[My Name]

Both Mg0 and mgOh are sparingly soluble in h2o or is it that mgo is insoluble?

MgO is almost insoluble ( due to high L.E).

 

[The Sarcastic Retard]

MgO is almost insoluble ( due to high L.E).

Also due to no H2 bond in MgO.

 

[ashcull14]

 

[The Sarcastic Retard]

View attachment 52875

Here: https://www.xtremepapers.com/commun...st-your-doubts-here.9859/page-619#post-908062

 

[nehaoscar]

http://freeexampapers.com/A-Level/Chemistry/CIE/2012-Jun/9701_s12_qp_22.pdf

Question 4 b part i and ii

For part i, to find moles oh H2 you do
n = v/24 = 0.08/24
= 6.67 x 10^-3 mol

Now it asks you to find the moles for hydrogen ATOMS and not H2 ...
so that means a single atom of Hydrogen right?

So then wouldn't you divide 6.67 x 10^-3 mol by 2 to get the moles rather than multiply...
The mark scheme says multiply by 2 but then by doing that aren't you finding the moles for 2H2 instead of H?

Please explain why it's multiplied by 2?!?!?

 

[Xaptor16]

okay so for question 12, i get that we use their molar masses and whatever, but that way we would be using 184 for NO2 and 32 for O2, question is why do we take 32 for oxygen? or do we not use the molar masses?

 

[Xaptor16]

http://freeexampapers.com/A-Level/Chemistry/CIE/2012-Jun/9701_s12_qp_22.pdf

Question 4 b part i and ii

For part i, to find moles oh H2 you do
n = v/24 = 0.08/24
= 6.67 x 10^-3 mol

Now it asks you to find the moles for hydrogen ATOMS and not H2 ...
so that means a single atom of Hydrogen right?

So then wouldn't you divide 6.67 x 10^-3 mol by 2 to get the moles rather than multiply...
The mark scheme says multiply by 2 but then by doing that aren't you finding the moles for 2H2 instead of H?

Please explain why it's multiplied by 2?!?!?

a molecule of H2 has two hydrogen atoms so:
hydrogen molecule : hydrogen atom
1 : 2
6.67 x 10^-3 : x

x = ( 6.67 x 10^-3 ) x 2

 

[qwertypoiu]

View attachment 52877


okay so for question 12, i get that we use their molar masses and whatever, but that way we would be using 184 for NO2 and 32 for O2, question is why do we take 32 for oxygen? or do we not use the molar masses?

2Mg(NO3)2 ----> 2MgO + 4NO2 + O2

The equation shows us that for every 4 moles of NO2 released, 1 mole of O2 is released.
Mass of NO2: (14 + 16x2) x 4 = 46 x 4 = 184 grams
Mass of O2: (16x2)x1 = 32 grams.
184/32 = 5.75 = 1/0.174 = A!

 

[Xaptor16]

2Mg(NO3)2 ----> 2MgO + 4NO2 + O2

The equation shows us that for every 4 moles of NO2 released, 1 mole of O2 is released.
Mass of NO2: (14 + 16x2) x 4 = 46 x 4 = 184 grams
Mass of O2: (16x2)x1 = 32 grams.
184/32 = 5.75 = 1/0.174 = A!

thankyou!

 

[Metanoia]

http://freeexampapers.com/A-Level/Chemistry/CIE/2012-Jun/9701_s12_qp_22.pdf

Question 4 b part i and ii

For part i, to find moles oh H2 you do
n = v/24 = 0.08/24
= 6.67 x 10^-3 mol

Now it asks you to find the moles for hydrogen ATOMS and not H2 ...
so that means a single atom of Hydrogen right?

So then wouldn't you divide 6.67 x 10^-3 mol by 2 to get the moles rather than multiply...
The mark scheme says multiply by 2 but then by doing that aren't you finding the moles for 2H2 instead of H?

Please explain why it's multiplied by 2?!?!?

1 mole of H2 = ? mol of H atoms

 

[nehaoscar]

a molecule of H2 has two hydrogen atoms so:
hydrogen molecule : hydrogen atom
1 : 2
6.67 x 10^-3 : x

x = ( 6.67 x 10^-3 ) x 2

Oh i see! Thankyou!

 

[Lola_sweet]

https://docs.google.com/viewerng/vi...11/9701_May-June-2011-All-Question-Papers.pdf

Qp 22
Qs)2a) hw do we calculate the bond energy

 

[asadalam]

Is So3 connected via PD PD forces or Van Der waals?Shouldnt it be PD PD Due to difference in e-vity of S and O?Or is SO3 symmetrical?.
Also are large scale uses of halogenoalkanes,naoh and other such compounds included.What are included.

Also when HCL is dissolved in water is the bonding changed to form ions of H30 and Cl or due to hydrogen bonds being formed?

Metanoia
The Sarcastic Retard

 

[Xylferion]

https://docs.google.com/viewerng/vi...11/9701_May-June-2011-All-Question-Papers.pdf

Qp 22
Qs)2a) hw do we calculate the bond energy

Delta H = Bonds broken - Bonds formed

You're already told the value of Delta H, so use that as a starting point. Now work out the total energy for bonds broken and the total energy for bonds formed.
Bonds broken:-

1 x C=C
4 x C-H
1 x H-F

Bonds formed :-

1 x C-F
5 x C-H
1 x C-C

Bonds broken = 610 + 4(410) + 562 = 2812 kJmol-1
Bonds formed = X + 5(410) + 350 = 2400 + X kJmol-1

X is the bond energy of C-F which we are going to find.

-73 kJmol-1 = 2812 - ( 2400 + X )
-73 = 2812 - 2400 - X
X = 2812 - 2400 + 73
X = 485 kJmol-1

Hope that helped!