Chemistry: Post your doubts here!

[Namehere]

Thank you so much, it is great that you elaborate so well so the concept seems simple!

Can you help me with the following questions too...
View attachment 38368
PVC has single bonds because the double bonds breaks in chloroethene so always the single sigma bonds are stronger right?Though answer is B
View attachment 38369
I thought answer should be simply be A, and besides even in data booklet atomic radii of Neon is not mentioned so how is the answer D?

I am asking many question to you but seems like you are genius at chemistry...

Thanks a tonne!

Q23 - When bond length increases, bond strength decreases. The change from chloroethene to PVC is from double bond to single bond. Double bonds are short meaning it is a strong bond (if you think about it, they are strong because you have sigma and pi bonding, hence more orbital overlap and therefore a stronger bond). Single bonds are long, meaning the bond is weak. Therefore the change is from strong, short double bonds in chloroethene to long, weak single bonds in PVC.

Q12 -Actually the option A is going in the order of decreasing size. Since for N, O and F, all the electrons are in the same principal quantum shell or energy level, the extra number of electrons you are putting from N, O and F doesn´t affect much the size of the atom. However, from N, O and F you are increasing the number of protons, meaning that the nuclear charge will pull those electrons closer to the nucleus, thus reducing the size of the atom. D is the correct answer because if you see, the number of electrons for each is the same, so it now it only depends on the number of protons. The one with the highest nuclear charge will pull those electrons closer to the nucleus, reducing the size of the particle and the one with the lowest nuclear charge won´t pull the electrons as much, so the electrons will be further away from the nucleus and therefore the particle is bigger. The one with the highest number of protons is Na+ and the one with the least is F-.

Hope it helps.

 

[Muskan Achhpilia]

Q23 - When bond length increases, bond strength decreases. The change from chloroethene to PVC is from double bond to single bond. Double bonds are short meaning it is a strong bond (if you think about it, they are strong because you have sigma and pi bonding, hence more orbital overlap and therefore a stronger bond). Single bonds are long, meaning the bond is weak. Therefore the change is from strong, short double bonds in chloroethene to long, weak single bonds in PVC.

Q12 -Actually the option A is going in the order of decreasing size. Since for N, O and F, all the electrons are in the same principal quantum shell or energy level, the extra number of electrons you are putting from N, O and F doesn´t affect much the size of the atom. However, from N, O and F you are increasing the number of protons, meaning that the nuclear charge will pull those electrons closer to the nucleus, thus reducing the size of the atom. D is the correct answer because if you see, the number of electrons for each is the same, so it now it only depends on the number of protons. The one with the highest nuclear charge will pull those electrons closer to the nucleus, reducing the size of the particle and the one with the lowest nuclear charge won´t pull the electrons as much, so the electrons will be further away from the nucleus and therefore the particle is bigger. The one with the highest number of protons is Na+ and the one with the least is F-.

Hope it helps.

Thank you!!

 

[Muskan Achhpilia]

Can anyone please explain the following mcq questions...


Thanks!

 

[midha.ch]

Question 24
someone explain please
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf

 

[AbbbbY]

Question 24
someone explain please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_11.pdf

C.

Idk how to explain so I'll just tell you my thought process.
Oxidising something to get a ketone+dioic acid. So I need 2 primary alcohols and 1 secondary alcohol. I already have 2 primary alcohols in the reactant, I just need a secondary alcohol across the double bond so I'll use steam + H2SO4. Option C.

ALTERNATE APPROACH:
If you're good and have enough practice, just by seeing the reaction and the fact that oxidation is happening, you'll figure out what X is. You're left with A and C. When you add cold KMnO4 you'll just be adding two alcohols across the double bond. That gives you C.

 

[sitooon]

Question 24
someone explain please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_11.pdf

A is wrong because cold acidified KMnO4 changes alkene to diol .. and thats already done
B is wrong because hot acidified KMnO4 is oxidising agent and will oxidise diol > COOH but thats done in the SECOND step
D is wrong because it also oxidise OH to cooH since its primary alcohol

let some one else explain , he might be better than me
currently i am in A2

 

[midha.ch]

C.

Idk how to explain so I'll just tell you my thought process.
Oxidising something to get a ketone+dioic acid. So I need 2 primary alcohols and 1 secondary alcohol. I already have 2 primary alcohols in the reactant, I just need a secondary alcohol across the double bond so I'll use steam + H2SO4. Option C.

ALTERNATE APPROACH:
If you're good and have enough practice, just by seeing the reaction and the fact that oxidation is happening, you'll figure out what X is. You're left with A and C. When you add cold KMnO4 you'll just be adding two alcohols across the double bond. That gives you C.

A is wrong because cold acidified KMnO4 changes alkene to diol .. and thats already done
B is wrong because hot acidified KMnO4 is oxidising agent and will oxidise diol > COOH but thats done in the SECOND step
D is wrong because it also oxidise OH to cooH since its primary alcohol

let some one else explain , he might be better than me
currently i am in A2

Thanks

 

[Namehere]

C.

Idk how to explain so I'll just tell you my thought process.
Oxidising something to get a ketone+dioic acid. So I need 2 primary alcohols and 1 secondary alcohol. I already have 2 primary alcohols in the reactant, I just need a secondary alcohol across the double bond so I'll use steam + H2SO4. Option C.

ALTERNATE APPROACH:
If you're good and have enough practice, just by seeing the reaction and the fact that oxidation is happening, you'll figure out what X is. You're left with A and C. When you add cold KMnO4 you'll just be adding two alcohols across the double bond. That gives you C.

Was wondering...

Say for option A that intermediate is correct, would you get the the desired product? Yes, right?

 

[AbbbbY]

Was wondering...

Say for option A that intermediate is correct, would you get the the desired product? Yes, right?

Yeah.

A and C are the same thing.

 

[Muskan Achhpilia]

Can anyone please explain the following mcq questions...

Thanks!

 

[Namehere]

Can anyone please explain the following mcq questions...

Thanks!

Q33 - electrons on sulfur = 16 , electrons on hydrogen = 1 , negative ion (gained one electron) = 1. Thus, 16 + 1 + 1= 18 electrons.
Sulfur is similar to oxygen in that its valence electrons are on the s and p orbitals. Sulfur has 4 electrons on the p orbital and 2 electrons on the s orbital. So as it is now, it has 2 lone pairs (1 on the s and 1 on the p) and 2 single electrons. If you now make this an ion you gain one electron, so an electron comes in and pairs up with a single electron, so now you have 3 lone pairs (i.e. 3 pairs of electrons which are not involved in bonding). If you see, to form the SH- ion, the H comes in with its electron and forms a bond with the remaining electron. So the SH- ion contains 3 lone pairs and one covalent bond.

Q16 - If you see, from option C, phosphorous (Y), has the lowest melting point. This is because silicon is a giant covalent structure and so has the highest melting point of the three. Sulfur is a S8 structure, wheras phosphorous is P4, so out of the those two, phosphours must have the lowest m.p. because it forms less van der Waals forces of attraction because there are less points of contact for the forces to act on (4 molecules on phosphorous, 8 on sulfur).

Phosphours also has the highest first ionisation energy and this is due to the electronic configuration and nuclear charge. For electron configurations, Si is similar to C in that it promotes an electron from the s orbital to the p orbital so that it has 4 single electrons in each orbital. Sulfur will have the lowest first ionisation energy since in its p orbital there are 2 electrons paired together, which gives rise to electron spin pair repulsion and so those electrons are easier to remove. For silicon and phosphorus the electron configuration is pretty much stable, so the only factor here is the nuclear charge. Phosphours has higher nuclear charge than Silicon and so it is harder to remove an electron because they are more attracted towards the nucleus compared to Silicon.

Hope it helps.

 

[Muskan Achhpilia]

Q33 - electrons on sulfur = 16 , electrons on hydrogen = 1 , negative ion (gained one electron) = 1. Thus, 16 + 1 + 1= 18 electrons.
Sulfur is similar to oxygen in that its valence electrons are on the s and p orbitals. Sulfur has 4 electrons on the p orbital and 2 electrons on the s orbital. So as it is now, it has 2 lone pairs (1 on the s and 1 on the p) and 2 single electrons. If you now make this an ion you gain one electron, so an electron comes in and pairs up with a single electron, so now you have 3 lone pairs (i.e. 3 pairs of electrons which are not involved in bonding). If you see, to form the SH- ion, the H comes in with its electron and forms a bond with the remaining electron. So the SH- ion contains 3 lone pairs and one covalent bond.

Q16 - If you see, from option C, phosphorous (Y), has the lowest melting point. This is because silicon is a giant covalent structure and so has the highest melting point of the three. Sulfur is a S8 structure, wheras phosphorous is P4, so out of the those two, phosphours must have the lowest m.p. because it forms less van der Waals forces of attraction because there are less points of contact for the forces to act on (4 molecules on phosphorous, 8 on sulfur).

Phosphours also has the highest first ionisation energy and this is due to the electronic configuration and nuclear charge. For electron configurations, Si is similar to C in that it promotes an electron from the s orbital to the p orbital so that it has 4 single electrons in each orbital. Sulfur will have the lowest first ionisation energy since in its p orbital there are 2 electrons paired together, which gives rise to electron spin pair repulsion and so those electrons are easier to remove. For silicon and phosphorus the electron configuration is pretty much stable, so the only factor here is the nuclear charge. Phosphours has higher nuclear charge than Silicon and so it is harder to remove an electron because they are more attracted towards the nucleus compared to Silicon.

Hope it helps.

Oh, Thank you!

 

[Muskan Achhpilia]

Hey can anyone help me in the following questions




Thanks!

 

[Namehere]

Hey can anyone help me in the following questions
View attachment 38407

View attachment 38408

Thanks!

Q5 - A is correct. Iodine is I2, a molecule, hence it has covalents bonds between the atoms. It also has van der Waals forces of attraction between I2 molecules.

Q33 - Anything from Period 1 to Period 2 cannot have an expansion of the octet. Nitrogen is in Period 2 and so cannot expand its octect, whereas Phosphorus can. This is because Phosphorous d orbitals can also be used in bonding since there isn´t a big enough of an energy gap, whereas Nitrogen´s d orbital are far away and so a lot of energy is needed for the electrons to occupy them, which doesn´t favour Nitrogen.

Q37 - Ammonium ion is NH4+. If you count the electrons properly (4 for the hydrogens, 1 less because it has formed a positive ion and 7 for nitrogen) you should get 10 electrons. Because NH4+ has bonds to 4 things and has no lone pairs, the bonds are spaced out tetrahedrally (since its the most energetically stable geometry for 4 bonded things and no lone pairs). Tetrahedral arrangements have bond angles of 109.5 degrees.

Hope it helps.

 

[A star]

http://maxpapers.com/syllabus-materials/chemistry-9701-a-level/attachment/9701_w13_qp_5/
9701_w13_qp_5 | Max Papers

is ka 9701/53/O/N/13
ka Q3 ka graph check karo kaesae banae ga?

 

[A star]

asd syed1995 daredevil and all othersssss

 

[Muskan Achhpilia]

Q5 - A is correct. Iodine is I2, a molecule, hence it has covalents bonds between the atoms. It also has van der Waals forces of attraction between I2 molecules.

Q33 - Anything from Period 1 to Period 2 cannot have an expansion of the octet. Nitrogen is in Period 2 and so cannot expand its octect, whereas Phosphorus can. This is because Phosphorous d orbitals can also be used in bonding since there isn´t a big enough of an energy gap, whereas Nitrogen´s d orbital are far away and so a lot of energy is needed for the electrons to occupy them, which doesn´t favour Nitrogen.

Q37 - Ammonium ion is NH4+. If you count the electrons properly (4 for the hydrogens, 1 less because it has formed a positive ion and 7 for nitrogen) you should get 10 electrons. Because NH4+ has bonds to 4 things and has no lone pairs, the bonds are spaced out tetrahedrally (since its the most energetically stable geometry for 4 bonded things and no lone pairs). Tetrahedral arrangements have bond angles of 109.5 degrees.

Hope it helps.

Thanks

 

[Muskan Achhpilia]

I also have these questions as doubts-



Thank you!!!

 

[sidbloom1995]

Thanks. I think i'll be able to do 29. And for 24, how do we know it's a Nucleophilic Reaction?

nucleophile is a group with lone pairs of electrons and OH- is a nucleophile!
we also know that during this reaction the OH will displace the Br from bromoehtane so it will most likely be by the SN (nucleophilic substitution), and as the name suggest, in this one nucleophile displaces the other so we know that both are nucleophile!

 

[ahmed abdulla]

Went through my chem paper 4 and had this doubts .. Any help ?
3 e (iv )
5 b (iv )
7 c (iv )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf