The Student Room Group

Aqa chemistry unit 1 unofficial mark scheme

1i. Silicon ( 1mark), many/ strong (1 mark) covalent bonds which required a large amount of energy to break so a high melting point ( 1mark)

ii) Argon ( 1mark), smaller atomic radius/more protons (1 mark) and similar shielding (1 mark)

iii) Chlorine (1 mark)

iv) ClF3 had a t-shape (3 bond pairs and 2 lone pairs). Trigonal planar should also be accepted. CCl2 had a bent shape (2 bond pairs and 1 lone pairs). (2 marks)

v) CCl2 had a bent shape. (1 mark)

vi) 0.5 Cl2 + 1.5 F2 goes to ClF3 (1 mark)

2i) 5s2 4d10 5p4 (Any order) (1 mark)

ii) Calculation and you should get 127.8 (1d.p.) (3 marks)

iii) Presence of other isotopes ( 1 mark)

iv) Te+ (g) + e- goes to Te (g) (1 mark)

v) The 128 Te + had the largest relative abundance ( 1mark) and current is proportional to abundance as more of the isotope = more electrons transferred from the detector to the ion = greater current produced ( 1mark)

vi) The 128 Te isotope ( 1mark) had two electrons knocked out (ionised twice) so it has a m/z ratio of 64. (1 mark) (Have to say 128 isotope)

vii) Same atomic radius ( 1mark) same number of protons and electrons so same attraction between the protons and the electrons so same atomic radius ( 1mark)

3i) Macromolecular ( 1mark)

ii) No delocalised electrons to flow through the macromolecular crystal and conduct electricity ( 1 mark)

iii) Hydrogen bonding ( 1mark)

iv) Bond angle in H2O2 is 104.5 ( 1mark)

v) Show hydrogen bonding between h2o2 and h2o. ALL partial charges ( 1mark). ALL lone pairs so 2 on each oxygen atom ( 1 mark) and attraction between delta + on hydrogen atom and lone pair on the Oxygen atom in the water molecule. H-O-H should be linear (1 mark).

vi) Oxygen is more electronegative than Sulfur (1 mark) due to less shielding etc. This means H2O2 has hydrogen bonding while H2S2 only has pd-pd interactions. This means that the melting point of H2O2 is greater than H2S2 ( 1 mark)

Now I have forgotten the order of the questions lol so sorry.


Saturated - Contains only C-C single bonds (No double C=C bonds) (1 mark)

Hydrocarbons - Contains carbon and hydrogen atoms only (1 mark)

The word ONLY must be used in both of these definitions to gain both marks (or words to that effect)



Fuels containing Sulfur are removed so that SO2 is not produced ( 1mark) from the combustion of these fuels. SO2 contributes to acid rain ( 1 mark).

Poly (propene) can be made from propene. ( 1mark)

Functional group isomer of propene was cyclopropane ( 1mark).

2,4-dichloro-2,4-dimethylhexane (1 mark)

The water of crystallization has 6 water molecules so x is 6 ( 3marks)

Evaporate the niso4.6h2o to constant mass ( 2marks)

SiO2 + 6HF -----> H2SiF6 + 2H2O ( 1mark)

C16H34 ----------> C8H18 + C2H4 + 2 C3 H6 ( 1mark)

C16H34 + 16.5O2-> 16CO + 17H2O ( 1 mark)

Draw 2,3,3-trimethylypentane (1mark)
3 HNO2----------> HNO3+ 2 NO + H20 (1mark) (Multiples accepted)

Ionic bonding ( 1 mark) with strong electrostatic forces of attraction ( 1mark) between the oppositely charged ions (Na+ and (N3)-)) ( 1mark)

Arrow from the N-N^- bond donated to the negative Nitrogen Ion ( 1mark)

N2O ( 1mark) (Other compounds will be accepted but this was the most obvious one)

MgN6 ( 1 mark)
(edited 8 years ago)

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Reply 1
Not finished yet btw but I will post calculations here.
6 marker:
So there was 550g of nan3 produced i think. 95% yield and you have to work out the mass of nan3.

so yield of nan3 was 550g and thats 95 %. That means that 578.9 are produced of nan3

so mr of nan3 is 23+14*3 = 65.

moles of nan3 = 578.9/65 = 8.907 moles.
then from the equation 2nanh2 : 1 nan3
so moles of nanh2 is 8.907 times 2= 17.8 moles.

mr of nanh2 = 23+14+2 =39.
Mass = moles * mr = 17.8 times 39 = 694.7g
695g (3s.f.).


Some people misread the question in the 6 marker and thought that 550g was 100% yield so 522.5 is 95 % yield. THIS IS WRONG AS 550g was 95% yield. People who did this will get an answer of around 627g. Again, THIS IS WRONG. However, you will only lose 1-2 marks out of the 6 marks so don't worry too much :smile:. THE BOLD PART IS WRONG BUT I AM JUST TELLING PEOPLE WHERE THEY WENT WRONG.



5 marker:
so 7.5 times 10^-2 m^3, 150kPa, 35 degrees.
pV= nRT n= pV/RT ( 1 mark)
150kPa =150,000 Pa AND 35 degrees = 308K (1 Mark)
n = (150,000 times 0.075)/ (308 times 8.31)
n= 4.395 moles for whatver the substance was ( 1mark)

2 nan3: 3 whatever it was
moles of nan3 = 4.395 divided by 3 then times by 2 = 2.93 moles (1 mark)
Mr of nan3 = 65. Mass = moles times mr = 2.93 times 65 which equals 190.5g ( 1 mark)

4 marker:

so there was 150g of nan3. Mr= 65. Moles = 150/65 = 2.31 moles ( 2 marks)
2NaN3: 2Nitrous acid so moles of nitrous acid = moles of nan3 = 2.31 ( 1mark)
n=cv
2.31 = 0.5c
Therefore, c= 4.62 mol/dm^3 (1 mark)
(edited 8 years ago)
Original post by PS4
1i. Silicon ( 1mark), many/ strong (1 mark) covalent bonds which required a large amount of energy to break so a high melting point ( 1mark)

ii) Argon ( 1mark), smaller atomic radius/more protons (1 mark) and similar shielding (1 mark)

iii) Chlorine (1 mark)

iv) ClF3 had a t-shape (3 bond pairs and 2 lone pairs). Trigonal planar should also be accepted. CCl2 had a bent shape (2 bond pairs and 1 lone pairs). (2 marks)

v) CCl2 had a bent shape. (1 mark)

vi) 0.5 Cl2 + 1.5 F2 goes to ClF3 (1 mark)

2i) 5s2 4d10 5p4 (Any order) (1 mark)

ii) Calculation and you should get 127.8 (1d.p.) (3 marks)

iii) Presence of other isotopes ( 1 mark)

iv) Te+ (g) + e- goes to Te (g) (1 mark)

v) The 128 Te + ( 1mark) had two electrons knocked out (ionised twice) so it has a m/z ratio of 64. (1 mark) (Have to say 128 isotope)

vi) Same atomic radius ( 1mark) same number of protons and electrons so same attraction between the protons and the electrons so same atomic radius ( 1mark)

3i) Macromolecular ( 1mark)

ii) No delocalised electrons to flow through the macromolecular crystal and conduct electricity ( 1 mark)

iii) Hydrogen bonding ( 1mark)

iv) Show hydrogen bonding between h2o2 and h2o. ALL partial charges ( 1mark). ALL lone pairs so 2 on each oxygen atom ( 1 mark) and attraction between delta + on hydrogen atom and lone pair on the Oxygen atom in the water molecule. H-O-H should be linear (1 mark).

v) Oxygen is more electronegative than Sulfur (1 mark) due to less shielding etc. This means H2O2 has hydrogen bonding while H2S2 only has pd-pd interactions. This means that the melting point of H2O2 is greater than H2S2 ( 1 mark)

Now I have forgotten the order of the questions lol so sorry.






Fuels containing Sulfur are removed so that SO2 is not produced ( 1mark) from the combustion of these fuels. SO2 contributes to acid rain ( 1 mark).

Poly (propene) can be made from propene. ( 1mark)

Functional group isomer of propene was cyclopropane ( 1mark).

2,4-dichloro-2,4-dimethylhexane (1 mark)





















Balance the equation ( 1mark)

Arrow from the N-N^- bond donated to the negative Nitrogen Ion ( 1mark)

N2O ( 1mark) (Other compounds will be accepted but this was the most obvious one)

MgN6 ( 1 mark)


Couldn't you also accept argon for most electronegative, as it has an electronegativity of 3.2 Pauling compared to chlorine's 3.16 (according to Google!).
There was a question about which ion would cause the largest current and I think it was the 128 isotope because it had the largest abundance and current is proportional to abundance?
Reply 4
Question on how the student could have made sure all water evaporated?
Reply 5
cheers mate
I did better than I thought I did :h:
Reply 6
Original post by Cadherin
Couldn't you also accept argon for most electronegative, as it has an electronegativity of 3.2 Pauling compared to chlorine's 3.16 (according to Google!).


Argon doesnt form covalent bonds?
Reply 7
Original post by Cadherin
Couldn't you also accept argon for most electronegative, as it has an electronegativity of 3.2 Pauling compared to chlorine's 3.16 (according to Google!).


I spoke to all my friends and they said it was Chlorine but I am not sure.
Reply 8
Original post by thehollowcrown
There was a question about which ion would cause the largest current and I think it was the 128 isotope because it had the largest abundance and current is proportional to abundance?

I think thats right :smile:
Reply 9
Think the dative covalent arrow should be from the ion to the triple bonded nitrogens
Original post by mickel_w
I think thats right :smile:


thank you :smile:
Original post by Hill1998
Think the dative covalent arrow should be from the ion to the triple bonded nitrogens


Nope, arrow goes right to anion
For the question about the most electronegative the answer was probably Chlorine. The definition of electronegativity is the power of an atom to attract electrons toward itself in a COVALENT BOND. Argon has a full outer shell, therefore it can't form a covalent bond with another element.
(edited 8 years ago)
Reply 13
Original post by Hill1998
Think the dative covalent arrow should be from the ion to the triple bonded nitrogens


No. It's the other way round. The negative Nitrogen Ion has 6 electrons in its outer shell so it is electron deficient. The middle Nitrogen has a lone pair and is donated to the the Nitrogen Ion
SiO2 + 6HF -----> H2SiF6 + 2H2O
Original post by PS4
Not finished yet btw but I will post calculations here.
6 marker:
So there was 550g of nan3 produced i think. 95% yield and you have to work out the mass of nan3.

so yield of nan3 was 550g and thats 95 %. That means that 578.9 are produced of nan3

so mr of nan3 is 23+14*3 = 65.

moles of nan3 = 578.9/65 = 8.907 moles.
then from the equation 2nanh2 : 1 nan3
so moles of nanh2 is 8.907 times 2= 17.8 moles.

mr of nanh2 = 23+14+2 =39.
Mass = moles * mr = 17.8 times 39 = 694.7g
695g (3s.f.).




5 marker:
so 7.5 times 10^-2 m^3, 150kPa, 35 degrees.
pV= nRT n= pV/RT ( 1 mark)
150kPa =150,000 Pa AND 35 degrees = 308K (1 Mark)
n = (150,000 times 0.075)/ (308 times 8.31)
n= 4.395 moles for whatver the substance was ( 1mark)

2 nan3: 3 whatever it was
moles of nan3 = 4.395 divided by 3 then times by 2 = 2.93 moles (1 mark)
Mr of nan3 = 65. Mass = moles times mr = 2.93 times 65 which equals 190.5g ( 1 mark)

4 marker:

so there was 150g of nan3. Mr= 65. Moles = 150/65 = 2.31 moles ( 2 marks)
2NaN3: 2Nitrous acid so moles of nitrous acid = moles of nan3 = 2.31 ( 1mark)
n=cv
2.31 = 0.5c
Therefore, c= 4.62 mol/dm^3 (1 mark)

Good job remembering all those questions ahha!
Original post by cc262626
SiO2 + 6HF -----> H2SiF6 + 2H2O


Also this
Reply 17
Original post by mickel_w
Also this


Thank you :smile:

I also think there was C16H34 to C8H18 + C2H4 + 2C3H6? Not sure though :smile:
v) CCl2 had a bent shape. (1 mark)

V shape also ok?

Thanks for the mark scheme!
Reply 19
Original post by FlyingPigFiler
v) CCl2 had a bent shape. (1 mark)

V shape also ok?

Thanks for the mark scheme!


No problem :smile:

I believe that is fine but I am not sure so I won't put it in just in case it is wrong and I am misleading people. I think it is right however

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