CBSE Class 12 Question Paper 2026 Solution Chemistry

FOR CBSE CLASS 12 EXAM PREPARATION

CBSE Class 12 2026
Question Paper
Solution · Chemistry
EXAM YEAR TYPE SUBJECT

CBSE Class 12 2026 Question Paper Solution Chemistry

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Strictly Confidential
(For Internal and Restricted use only)
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Senior Secondary School Examination, 2026 (XIIth)
SUBJECT NAME: - CHEMISTRY (043) (Q.P. CODE: 56/1/1
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General Instructions: -
1 The CBSE has decided to introduce On Screen Marking (OSM) for the evaluation of
Class XII answer Book with the 2026 Examination.
2 You are aware that evaluation is the most important process in the actual and
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correct assessment of the candidates. A small mistake in evaluation may lead to
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serious problems which may affect the future of the candidates, education system
and teaching profession. To avoid mistakes, it is requested that before starting
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evaluation, you must read and understand the spot evaluation guidelines carefully.
e policy is a confidential policy as it is related to the confidentialitygl as
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“Evaluation
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of the
Its
examinations conducted, evaluation done and several other aspects.
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 Newspaper/Website, etc. may invite action under various rules of
the Board and IPC.”
4 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
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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,

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while evaluating two competency-based questions, please try to understand

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given answer and even if reply is not from marking scheme but correct

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competency is enumerated by the candidate, due marks should be awarded.
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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.
6 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 deliberation and discussion. The remaining answer books meant for
evaluation shall be given only after ensuring that there is no significant variation in
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the marking of individual evaluators.
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7 Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS „X‟

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impression that answer is correct and no marks are awarded. This is most

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common mistake which evaluators are committing.

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If a question has parts, please award marks on the right-hand side for each part in
a the OSM Portal. Marks awarded for different parts of the question will be totaled up
by the OSM System.
9 If a question does not have any parts, marks must be awarded in the left-hand
margin in the OSM Portal. This may also be followed strictly.

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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 __________ (example 0 to 80/70/60/50/40/30 marks as given
in Question Paper) 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 :-
● 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 The Examiners should acquaint themselves with the guidelines given in the
“Guidelines for Spot Evaluation” before starting the actual evaluation.
16 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.
17 If a candidate attempts both alternatives/options in a question where only one
option/ alternative is required to be attempted, the Evaluator shall award
marks in both the options. The system will take the higher of two scores and
disregard the other response.
18 In a question having two options/alternatives, if a candidate has attempted
only one, then the evaluator shall mark “NA” (Not attempted) against the
option that has not been attempted by the candidate.

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CHEMISTRY (Subject Code-043)
(PAPER CODE : 56/1/1) (26-01-43N)

Q.No. EXPECTED OUTCOMES/VALUE POINTS Marks
SECTION - A
1. (B) 1
2. (B) 1
3. (C) 1
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4. (B)
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5. (D) 1
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6. (C) 1
7. (D)
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8. (D) 1
9. (B) 1
10. (C) 1
11. (A) 1
12. (C) 1
13. (A) 1
14. (D) 1
15. (B) 1
16. (C) 1
SECTION - B
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17.  Negative deviation ½

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Solute -solvent interactions are stronger than solute-solute and solvent-
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solvent. / The intermolecular hydrogen bonding between phenolic
proton and lone pair on nitrogen atom of aniline is stronger than the 1
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respective intermolecular hydrogen bonding between similar molecules
 Boiling point of the solution increases. ½
18. Because of, 11
(i) CX acquires partial double bond due to resonance
(ii) sp2 hybridization of C of CX make it more electronegative and hold the
electron pair tightly.
(iii) repulsion between nucleophile and electron rich arenes.
(iv) Instability of phenyl cation
(Any two)
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19(a). (i) Diamminesilver(I) dicyanidoargentate(I) 1
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(ii) Potassium trioxalatoferrate(III) 1

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OR
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s e (i) [Co(NH3)5(SO4 )]Cl with AgNO3 solution gives white ppt of AgCl
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g la whereas [Co(NH3)5Cl]SO4 does not / [Co(NH3)5Cl]SO4 with a
a BaCl2 gives white ppt of BaSO4 whereas [Co (NH3)5(SO4)] Cl does
not.
(ii) Formation of stable complexes by di or polydentate ligand with
½
single metal atom/ ion.
For example: [Co(en)3]3+ (or any other correct example)
½
20. (i) It is an amide linkage formed between two amino acids through –CONH- / 1
an amide linkage formed between -COOH and -NH2 group whereas in
glycosidic linkage two or more monosaccharides are joined together by
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an oxygen atom.

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(ii)Those which cannot be synthesised in the body and must be obtained 1
through diet known as essential amino acids. The amino acids, which can
be synthesised in the body, are known as nonessential amino acids.

21. Rate  k [A]2 [B]
1 rd
If the volume is decreased to of original concentration increases 3 times
3
 Rate  k [3A]2 [3B]  27 k [A]2 [B]
Hence rate will be increased by 27 times. 1
Order remains same. 1
SECTION – C
22.

1

1

1
23(a). (i) Because at equilibrium , Ecell  0 1
(ii) Metal ‘A’ 1
(iii) 2PbSO4(s) 2H2O(l)  PbO2(s)  Pb(s)  2H2SO4(aq) 1
OR
23(b).  Mercury cell – Primary cell 1
 It provides constant potential difference throughout its life. 1
 Zn(Hg)  HgO(s)  ZnO(s)  Hg (l) 1
24.
½

½

1

1
25. (a) 13

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(b)

(c)

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26. (a) Due to extensive association of carboxylic acid molecules through 1
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intermolecular hydrogen bond./ Dimer formation takes place
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(b) l Due to the strong electron withdrawing effect of the carbonyl group /
g resonance stabilisation of their conjugate base. a
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(c) Protonation of nucleophile takes place in strongly acidic medium. 1
27.

1
(a)

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(b)
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1
(c)

28. ½½
X
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(a) Y
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(b)

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(c) Y, Because reaction is SN2/due to rear side attack of nucleophile. ½½
SECTION – D
29. (a) (i) Due to combined effect of the inductive effect, solvation effect and 1
steric hinderance of the alkyl group.

(ii) A= CH3NH2 B= CH3NHCOC6H5 ½+½

CH3CH2NHCH3 / N – Methylethanamine 1
(b)
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OR
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(b) 1

(c) The nitration reaction can be controlled and the p-nitro derivative can be
obtained as the major product / To prevent the formation of anilinium ion 1
which is meta directing.
30. (a)  dx2  y2 and dz2 1
 In Octahedral Crystal field, ligands approach metal atom /ion along the
axis due to which dx2  y2 and dz2 orbitals experience more repulsion. 1

½
(b) (i) [CoF6 ]3 Co3+ = 3d 6
(ii) [Co(NH3 )6 ]3+ Co3+ = 3d 6 ½
(c)
In [NiCl4]2, Cl- is a weak field ligand and does not pair up the unpaired ½
electrons in 3d-orbital and hence paramagnetic.
In [Ni(CO)4], CO being a strong field ligand pair up the unpaired electrons ½
and hence diamagnetic.
OR
(c)
Hybridisation- d2sp3
½½
Magnetic behaviour: Paramagnetic
SECTION – E
31(a). (i) (I) 1

(II) 1

(III) 1

(ii) (I)
1
/

(II)
1

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OR
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31(b).
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(i)

1
(ii) Di-tert-butyl ketone < Acetone < Acetaldehyde
1
(iii) Add NaHCO3 to both, benzoic acid will give a brisk effervescence while

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ethyl benzoate will not.

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1
(i)DIBALH (ii) H3O / (i) SnCl2HCl (ii)H3O
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(iv)
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(i) l(I) Due to the irregular variation of ionisation enthalpies ( iH  iH ) a 1
agand sublimation enthalpies.
32(a). 1 2

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(II) Due to low hyd H and high a H of Cu.
(III) Because Mn2 is highly stable due to half-filled 3d5 configuration. 1
  2
(ii) 2MnO 4  10I  16H  2Mn  8H2O  5I2 1
 
2MnO 4  I  H2O  2MnO2  IO3  2OH 1

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32(b). (i) (I) Cerium ½½
(II) Europium / Ytterbium
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(Or any other correct example)
(ii) Because of their ability toashow multiple oxidation states / complex 1
formation / Due to large surface area.
(iii) Involvement of greater number of electrons from (n-1)d and ns electrons in
interatomic metallic bonding. 1
(iv)
/ 1
Reduction reaction 1

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Redox reaction
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½
½
1

OR
33(b). ½
1

½

1

1

1
-oOo-

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