CBSE Class 12 Question Paper 2023 Solution Chemistry

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior Secondary School Examination,2023.
SUBJECT: CHEMISTRY (043) (56/1/1)
General Instructions: -
1 You are aware that evaluation is the most important process in the actual and
correct assessment of the candidates. A small mistake in evaluation may lead
to serious problems which may affect the future of the candidates, education
system and teaching profession. To avoid mistakes, it is requested that before
starting evaluation, you must read and understand the spot evaluation
guidelines carefully.
2 “Evaluation policy is a confidential policy as it is related to the
confidentiality of the examinations conducted, Evaluation done and
several other aspects. Its’ leakage to public in any manner could lead to
derailment of the examination system and affect the life and future of
millions of candidates. Sharing this policy/document to anyone,
publishing in any magazine and printing in News Paper/Website etc may
invite action under various rules of the Board and IPC.”
3 Evaluation is to be done as per instructions provided in the Marking Scheme. It
should not be done according to one’s own interpretation or any other
consideration. Marking Scheme should be strictly adhered to and religiously
followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed
for their correctness otherwise and due marks be awarded to them. In
class-XII, while evaluating two competency-based questions, please try to
understand given answer and even if reply is not from marking scheme
but correct competency is enumerated by the candidate, due marks
should be awarded.
4 The Marking scheme carries only suggested value points for the answers
These are in the nature of Guidelines only and do not constitute the complete
answer. The students can have their own expression and if the expression is
correct, the due marks should be awarded accordingly.
5 The Head-Examiner must go through the first five answer books evaluated by
each evaluator on the first day, to ensure that evaluation has been carried out
as per the instructions given in the Marking Scheme. If there is any variation,
the same should be zero after delibration and discussion. The remaining
answer books meant for evaluation shall be given only after ensuring that there
is no significant variation in the marking of individual evaluators.
6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer
CROSS ‘X” be marked. Evaluators will not put right (✓)while evaluating which
gives an impression that answer is correct and no marks are awarded. This is
most common mistake which evaluators are committing.
7 If a question has parts, please award marks on the right-hand side for each
part. Marks awarded for different parts of the question should then be totaled
up and written in the left-hand margin and encircled. This may be followed
strictly.
8 If a question does not have any parts, marks must be awarded in the left-hand
margin and encircled. This may also be followed strictly.

9 If a student has attempted an extra question, answer of the question deserving
more marks should be retained and the other answer scored out with a note
“Extra Question”.
10 No marks to be deducted for the cumulative effect of an error. It should be
penalized only once.
11 A full scale of marks 70 has to be used. Please do not hesitate to award full
marks if the answer deserves it.

12 Every examiner has to necessarily do evaluation work for full working hours
i.e., 8 hours every day and evaluate 20 answer books per day in main subjects
and 25 answer books per day in other subjects (Details are given in Spot
Guidelines).This is in view of the reduced syllabus and number of questions in
question paper.
13 Ensure that you do not make the following common types of errors committed
by the Examiner in the past:-
● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the
title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right
tick mark is correctly and clearly indicated. It should merely be a line.
Same is with the X for incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no
marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect,
it should be marked as cross (X) and awarded zero (0)Marks.
15 Any un assessed portion, non-carrying over of marks to the title page, or
totaling error detected by the candidate shall damage the prestige of all the
personnel engaged in the evaluation work as also of the Board. Hence, in order
to uphold the prestige of all concerned, it is again reiterated that the
instructions be followed meticulously and judiciously.
16 The Examiners should acquaint themselves with the guidelines given in the
“Guidelines for spot Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks
carried over to the title page, correctly totaled and written in figures and words.
18 The candidates are entitled to obtain photocopy of the Answer Book on request
on payment of the prescribed processing fee. All Examiners/Additional Head
Examiners/Head Examiners are once again reminded that they must ensure
that evaluation is carried out strictly as per value points for each answer as
given in the Marking Scheme.

XII_39_043_56/1/1_Chemistry # Page-2

MARKING SCHEME
Senior Secondary School Examination, 2023
CHEMISTRY (Subject Code–043)
[ Paper Code: 56/1/1]

Q. No. EXPECTED ANSWER / VALUE POINTS Marks

SECTION-A

1. (a) 1
(c)
2. 1
3. (d) 1
4. (c) 1

5. (c) 1

6. (c) / Full mark to be awarded for any option 1

7. (b) 1

8. (c) 1

9. (b) 1

10. (a) 1

11. (b) 1

12. (c) 1

13. (c) / Award full mark if attempted (Printing error) 1

14. (c) 1

15. (b) 1

16. (c) 1

17. (a) 1

18. (a) 1

SECTION- B
19. Henry’s law states that the partial vapour pressure of a gas is directly proportional to
the mole fraction of the gas in the solution / p = KH x where p = partial pressure 1
of gas, x = mole fraction in solution, and kH is Henry’s constant.
Application:
To increase the solubility of CO2 in soft drinks and soda water, the bottle is
sealed under high pressure / to minimize the painful effects accompanying the
XII_39_043_56/1/1_Chemistry # Page-3

decompression of deep-sea divers (bends). Oxygen diluted with less soluble
helium gas is used for breathing / At higher altitudes, low blood oxygen causes
climbers to become weak and unable to think clearly (anoxia). (Any one) 1
(or any other correct application)

20. ‘B’ is a strong electrolyte. 1

1

OR
A = r2 = 3·14  (0·5)2 = 0·785 cm2 , ℓ = 50 cm
ℓ ½
k=𝑅×𝐴
50 1
= 0·785 𝑋 ( 5.55 𝑋 103)

= 11·47 X 10-3 S cm-1 ½
(or by any other correct method)

21. – –
(a) 2 Mn𝑶−𝟒 + 5 N O 2 + 6 H+ ⎯⎯→ 2 Mn2+ + 5 N O 3 + 3 H2O 1
2–
(b) Cr2 O7 + 14 H+ + 6 e– ⎯⎯→ 2 Cr3+ + 7 H2O 1
22. CH3 – CH – CH3
(a) A =
|
/ 2-Chloropropane

Cl
½x4

(b) A = CH3 – CH = CH2 / Propene
B = CH3 – CH – CH3 / 2-Bromopropane
|
Br
23. (a)
(i) Because phenoxide ion is more stable due to resonance than alkoxide ion.
(or any other correct explanation) 1
(ii) Because branching decreases the surface area / the van der Waals force decreases
with a decrease in surface area. 1

OR

XII_39_043_56/1/1_Chemistry # Page-4

(b)
(i)

½+½

(For any two correct steps)
(ii)

1
/

24. (a) Aliphatic and aromatic primary amines on heating with chloroform and ethanolic
potassium hydroxide form isocyanides or carbylamines which are foul-smelling 1
substances. /

Isocyanide with an offensive smell is formed.
(Explanation or reaction)
(b) Phthalimide on treatment with ethanolic potassium hydroxide forms potassium
salt of phthalimide which on heating with alkyl halide followed by alkaline
hydrolysis produces the corresponding primary amine /

1

(Explanation or reaction)

XII_39_043_56/1/1_Chemistry # Page-5

(a)
25.

1

(b) Peptide linkage 1

SECTION-C

26. (a)
Ideal Solution Non-ideal solution

The solution obeys Raoult’s law at The solution does not obey Raoult’s law.
all concentrations. 1
Vmixing = 0 and Hmixing = 0 Vmixing ≠ 0 and Hmixing ≠ 0.
(Any one)
(or any other correct difference)
WB
o –P
PA A MB
(b) =
P o WB WA
A MB
+ MA
½

30 30
PA 60 PA 60 1
1– = 846 or 1 – =
23 ·8 23 ·8 846 +30
18 18 60
46.5 47
PA =  23·8 = 23·5 mm Hg or PA =  23·8 = 23·5 mm Hg
47 47.5 ½
(Full marks may be awarded if the student substitutes MB for molar mass as the
molar mass of urea is not given in the question).

27. (a) CH3I / Iodomethane / Methyl iodide
(b)

1x3

/ Picric acid / 2,4,6-Trinitrophenol / 2,4,6-Trinitrobenzenol
(c) CH3CH2CH =CH2 / But-1-ene
28. (a)

(b)

1x3

XII_39_043_56/1/1_Chemistry # Page-6

(c)

(d)

(or by any other correct method)
(a)
29.
(i) Because it is an electron-withdrawing group / deactivating group / -R effect ,
electrophilic substitution takes place at the m-position.
1x3
(ii) Because aldehydes & ketones form addition compound with NaHSO3 which on
hydrolysis forms pure aldehydes & ketones.
(iii) Due to resonance, carboxylic carbon becomes less electrophilic.
OR
(b)

1x3
(or explanation with correct structures of A, B, and C)
30. (a) (i) Glucose and Galactose (ii) Glucose and Glucose 1+1
(b) Starch is a polymer of -glucose while cellulose is a polymer of -glucose
(or any other correct structural difference) 1
SECTION-D
31. (i) Change in the concentration of a reactant or product per unit time. 1
(ii) Concentration of reactants, Surface area, catalyst and temperature (any two). 1
(iii) (1) rate is independent of the concentration of reactant(s) /rate remains constant /
rate = k
(2) mol L–1 s–1 1+1
OR
(iii) (1) 3/2 / 1.5 1
(2) A reaction that appears to be of higher order but follows first-order kinetics.
Example: Hydrolysis of an ester (or any other correct example) ½,½
32. (i) [Pt(NH3)2Cl2] 1
(ii) 6 1
(iii) (1) Fe4[Fe(CN)6]3
(2) Pentamminechloridocobalt(III) chloride. 1,1
OR
2
(iii) dsp , diamagnetic 1,1
SECTION-E
33. (a)
(i) Limiting molar conductivity of an electrolyte can be represented as the sum of the 1
individual contributions of the anion and cation of the electrolyte.
m (CH3COOH) = CH3COO– + H+ 1

XII_39_043_56/1/1_Chemistry # Page-7

(ii) rG = – nF E cell
Maximum work = – rG = nF E cell ½
–1
= 2  96500 C mol  (0·80 + 0·25) V ½
–1
= 2  96500  1·05 J mol
–1 –1
= 202,650 J mol or 202·65 kJ mol 1
nE cell
log Kc = ½
0·059
2  1·05
= = 35 ·6 ½
0·059
OR
(b) (i) It states that the mass of a substance deposited /liberated at the electrodes is 1
directly proportional to the charge/quantity of electricity passed through the
electrolyte. 1
2F charge is required.
0·0591 [Mg 2 + ]
(ii) Ecell = E cell – log
2 [Cu 2 + ] 1
0·0591 0·1
= 2·71 V – log
2 0·01
0·0591 1
= 2·71 V – log 10
2
= 2·71 V – 0·0295
= 2·68 V. (Deduct ½ mark for no or incorrect unit) 1
34. (i) Due to the participation of all 3d and 4s electrons in bond formation /due to the
1
presence of maximum number of unpaired electrons.
(ii) Due to variable oxidation state / due to the ability to adopt multiple oxidation
states / due to the large surface area / due to complex formation. 1
(iii) Cr2+ changes from d4 to stable half-filled t2g3 configuration while Mn3+
1
changes to stable half-filled d5 configuration.
(iv) Due to the absence of unpaired electrons and weak interatomic interactions. 1
(v) Cu+ ion (aq.) undergoes disproportionation to Cu2+ (aq.) and Cu /
2 Cu+ (aq.) ⎯⎯→ Cu2+ (aq.) + Cu (s) 1
35. (a) (i)
(1)
dil. NaOH 
CH3CHO CH3CH—CH2—CHO CH3—CH —
—CH—CHO 1
| —H2O
OH
(2)

1

XII_39_043_56/1/1_Chemistry # Page-8

(ii)

1x3

OR
(b) (i) (1) Add Iodine (I2), NaOH, and heat both the test tubes containing the given
organic compounds. Butanone gives yellow precipitate (CHI3) while butanal will not 1
give the positive iodoform test.
(2) Add NaHCO3 in both the test tube containing the given organic compounds.
Ethanoic acid will give brisk effervescence of CO2 and ethanal will not. 1
(or any other suitable chemical test)
(ii)

1

(iii) A = CH3COCl, B = CH3CHO, C = (CH3)2CH(OH), D = CH3CH2OH ½x4
***

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