Ocr f321 23 may 2013~official discussion thread

I remember doing the 8o2 in the exam though.
Something about that is wrong I can't even remember

At lowest a very high B, at best around 90%

really want an A what do you think it will be??

How much UMS would 54/60 give myself?


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so annoyed with myself, i put tsilver nitrate and yellow precipitate then crossed it out later

same here! The more I read the question, the more I confused myself :L Luckily they were only 1 markers...

At lowest a very high B, at best around 90%

UNOFFICIAL MARKSCHEME F321

1a)i) The mass of an atom of an isotope of an element compared with 1/12 of the mass of an atom of c-12.

a)ii) 39K ~19 protons 20 neutrons 19 electrons
41K+ ~19 protons 22neutrons 18electrons
b) 1s2, 2s2, 2p6, 3s2, 3p2
c)i) 3.01 x10 ^22

c)ii) Alternating K+ and Cl- Ions (But simply + and – is also acceptable.
2)a)i) 2 Main ions are Al(3+) and (SO4)2-
ii) Al203 (s) + 3H2SO4 (aq) ---> Al2(SO4)3 (aq) + 3H2O (l)
iii) x represents the number of water molecules present in the crystalline structure (water of crystallisation)
iv) x=16
b)i) It turns red because HCl is an Acid
ii) ClO-


3)a)i) P has been both reduced and oxidised. Oxidised from 0 to +3 and reduced from 0 to -1.
ii) volume of PH3 = 360cm^3
b) 4PH3 + 802 ---> P4H10 + 6H20
c)i) Because the H+ ions have been replaced with a metal ion, in this case Zn.
ii) Moles of H3PO4 = 0.0015
iii) 22.5cm^3
d)i) NH3 = Hydrogen Bonding, PH3 = Permant Dipole-Dipole attraction
ii) Because hydrogen bonding is a stronger intermolecular force requiring more energy to break
e)i) A shared pair of electrons supplied by one of the bonding atoms only.
ii) Essentially the same as in the diagram on the earlier page on the dative covalent bond is labelled between nitrogen and boron.
iii) BF3 = 120*, H3NBF3 = 109.5*
iv) To put it shortly NH3 has 3 bond pairs and 1 lone pair giving it an angle of 107*, but now in H3NBF3 it has 4 bond pairs and no lone pairs making it assume the tetrahedral shape with an angle of 109.5*.

4a) -increase nuclear charge outweighed by increase atomic radius
-more shells;more electrons;more inner shielding effect by inner electrons;
decrease nuclear attraction. easier to remove outer elect to react with halogens. reactivity increases hence reaction more vigorous down the group
b)i) AgNO3 (Silver Nitrate)
ii) A yellow precipitate would be produced
iii) Ag+ (aq) + I- (aq) ---> AgI (s)
iv) concentrated NH3 (ammonia)

5)a)i) Across the period, atomic radius decreases and nuclear charge increases making the attraction between the nucleus and outermost electron stronger and require more energy to break. (The electron shielding effect also remains the same).
ii) It is unnecessary because first ionisation energy takes place when the atoms are gaseous, so there would be no giant covalent structure anyway.

5b) Fill the boxes in of the table stating the properties of lithium, carbon and fluorine (6)

Li = Bottom box: Delocalised electrons
C = Top box: Giant Covalent, Middle Box: Covalent bonds, Bottom Box: Again covalent bond or a shared pair of electrons
F2 = Middle box: van der Waal’s forces, Bottom Box: None since there are no particles between the molecules.

You're saying 49/60 will be a B. So according to you 81% is a B?

You're saying 49/60 will be a B. So according to you 81% is a B?

Could you please tell me if the paper has been posted? If so what page is it on?

In the past 52 has been the A boundary. Raw marks are converted to UMS marks (/90) depending on how the cohort found the exam.

This basically maintains the proportion of people that get an A each year and stops people being disadvantaged if their exam was particuarly difficult.

Therefore, 49/60 which is indeed over 80% in raw marks MAY not be an A after UMS scaling if in general people found the paper slightly easier than average

I got 43-45 what grade would that be?

I just want B grade to be 44/46 marks, all I want is a B on the paper. High B in unit 2 and get an A overall :/