Paper 1 Inorganic and Physical Chemistry

AS CHEMISTRY 7404/1

Paper 1 Inorganic and Physical Chemistry

Mark scheme June 2019

Version: 1.0 Final

*196A74041/MS*

MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students' responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students' scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students' reactions to a particular paper. Assumptions about future mark schemes on the basis of one year's document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from .uk

Copyright ? 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third-party even for internal use within the centre. 2

MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

Level of response marking instructions

Level of response mark schemes are broken down into levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are marks in each level. Before you apply the mark scheme to a student's answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme.

Step 1 Determine a level

Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student's answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content.

Step 2 Determine a mark

Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student's answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner's mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the Indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks.

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MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

Question

Marking guidance

Fluoride ion has (two) fewer protons/lower nuclear charge

Weaker attraction between nucleus and (outer) electrons 01.1

(Electrostatic) forces of attraction between oppositely charged ions/Na+ and F?

01.2

Lots of energy needed to overcome/break forces

Type of Bond: Coordinate bond / dative (covalent) bond

01.3

Explanation: A (lone) pair of electrons is donated from F

Shape +

01.4

Name of shape

Octahedral

Bent / V-shaped / angular

Additional Comments/Guidelines

Mark

Do not allow fluorine, but allow fluorine ion

1

Any reference to different numbers of electrons in

the ions loses M1

1

Allow answers in terms of sodium ion but must be explicit. Ignore references to atomic radius

Mention of IMF, covalent, macromolecular, metallic,

electronegativity of ions loses both marks

1

Allow strong ionic bonding

1

Allow strong forces/bonds of attraction (need to be

broken)

If just covalent, then do not award M1 but mark on

1

Allow both electrons (in the shared pair) come from

1

F

Lone pairs on H2F+ are essential (can be shown in

1

lobes)

Ignore missing charges

1

1

1 Mark independently

4

H = H(Bonds broken) - H(Bonds Formed)

-179 = 2(412) + 837 + 2(562) ? [348 + 4(412) + 2(C--F)]

-179 = 2785 ? (1996 +2(C--F)) 01.5

2(C--F) = 968

C--F = 484

MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

Allow M1 if 2785 and 1996 seen (or allow M1 if 1961 and 1172 seen)

1

1 M3 consequential on any M2 if it is clear that M2 is for 2(C-F)

1 -484 scores 2

5

Question

Marking guidance

(Sample is) dissolved (in a volatile solvent)

(Injected through) needle/nozzle/capillary at high voltage/positively charged

02.1

Each molecule/particle gains a proton/H+

02.2 02.3

C3H6O2N+ / C3H5O2NH+ Ge(g) + e? Ge+(g) + 2 e? OR Ge(g) Ge+(g) + e?

MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

Additional Comments/Guidelines

Mark

Allow named solvent (eg water/methanol)

1

Ignore pressure

1

1 Allow M3 from a suitable equation (ignore state symbols) Do not allow atoms gain a proton for M3

Ignore references to electron gun ionisation

Mark each point independently

Must be charged

1

State symbols essential

1

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MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

M1 v = length/t = 0.96 / 4.654 x 10 ?6 v = 206274 m s?1

Notes:

1

M1 = working (or answer)

m = 2KE/v2

M2 mass of one ion = 1.146 x 10?25 kg

M2 = answer conseq on M1

1

02.4

M3 = M2 x 6.022 x 1023

1

M3 mass of 1 mole ions = 1.146 x 10?25 x 6.022 x 1023 = (0.06901 kg)

1

M4

= 69(.01) g

M4 = M3 x 1000

1 M3/M4 could be in either order

M5 mass number = 69

M5 must have whole number for mass no

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Question

Marking guidance

03.1 03.2

03.3

(1s2) 2s2 2p6 3s2 3p6 3d5 4s1 Or (1s2) 2s2 2p6 3s2 3p6 4s13d5

2567

% of 4th isotope = 3.6

M2: (52 x 82.8) + (53 x 10.9) + (54 x 2.7) + (3.6) = 52.09

100

M3: = 49.97 OR 179.9 = 3.6 and = 50 (evidence of working)

03.4 03.5 03.6

03.7

+6 / VI / six / 6+ 2 I? I2 + 2 eCr2O72? + 14 H+ + 6 e? 2 Cr3+ + 7 H2O Cr2O72? + 14 H+ + 6 I? 2 Cr3+ + 7 H2O + 3 I2

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MARK SCHEME ? AS CHEMISTRY ? 7404/1 ? JUNE 2019

Additional Comments/Guidelines

Ignore commas Do not penalise capitals and subscripts

Mark 1

Allow mass number and atomic number on RHS of Fe

1

Allow alternative methods

1

M2

(52 x 82.8) + (53 x 10.9) + (54 x 2.7) + (50 x 3.6) =5209

M3

1

Ar = 5209/100 = 52.09

Or

1

M2

(52 x 82.8) + (53 x 10.9) + (54 x 2.7) + (50) = 52.09

100

M3 awarded for 50 = 179.9 and then = 3.6 (evidence of working)

1

Allow multiples / ignore ss

1

Allow multiples / ignore ss

1

Allow multiples / ignore ss

Allow Cr2O72? + 14 H+ + 8 I? 2 Cr2+ + 7 H2O + 4 I2 as ecf to

1

03.6

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