HBSE Class 12 Sample Paper 2027 Chemistry Marking Scheme

FOR HARYANA CLASS 12 PREPARATION

HBSE Class 12 Sample
Paper 2027 Chemistry
Marking Scheme
EXAM YEAR TYPE SUBJECT

Haryana Class 12 2027 Answer Key Chemistry

Notes · Sample Papers · Previous Year Papers · Mock Tests

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a
CLASS : 12th (Sr. Secondary)
MARKING SCHEME 2026-27
CHEMISTRY ( Subject Code: 856)
Paper Code :- ………………..
Q.No. EXPECTED OUTCOMES /VALUE POINTS Marks
SECTION - A (1 Mark each) m
Q.1
m
(D) Depression in boiling point 1
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Q.2 (C) Peptide bond
m .co 1
s e m
Q.3 (C) Thymine
s e 1
l a
a ag
Q.4 (A) Ethylamine 1
Q.5
g l
(C) Propan-1-ol < Butan-2-ol < Butan-1-ol < Pentan-1-ol 1
Q.6 a
(B) Butan-2-ol 1
Q.7 (D) [Co(NH3)4Cl2]+ (heteroleptic) 1
Q.8 (B) Diamminedichloridoplatinum(II) 1
Q.9 (C) [Ar] 3d5 4s1 1
Q.10 Electrophilic substitution reactions 1
Q.11 Peptide linkage 1
Q.12 Rosenmund reduction 1
Q.13 RMgX + CO2 → RCOOMgX → H3O+ → RCOOH 1
m
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Q.14 s⁻¹ 1
Q.15 1
em
S m⁻¹
Q.16
Q.17
(D) Assertion false, Reason true
(A) Both true, Reason correct explanation l as 1
1
Q.18 ag
(A) Both true, Reason correct explanation
SECTION – B (2 Marks each)
1

Q.19
(i) Pentaamminecarbonatocobalt(III) chloride. (1 Mark) 1

(ii) Potassium tri(oxalato)ferrate(III). (1 Mark) 1

Q. 20 (a) Methyl bromide with sodium in dry ether:
2CH3Br + 2Na → C2H6 + 2NaBr (Wurtz reaction) (1 Mark)
m 1
m Ethyl chloride with aqueous KOH: c. o
c. o (b)
C H Cl + KOH(aq) → C H OH + KCl (Ethanol formed) (1 Mark) se
m
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1
s
2 5 2 5

g l a
la Q. 21 (i) Ca + 2e− → Ca 40 g Ca requires 2F
2+
a
ag 20 g Ca requires 1F 1

(ii) Al3+ + 3e− → Al 27 g Al requires 3F
Answer: (i) 1F (ii) 3F (1 mark each) 1

OR Cell Representation:
Mg(s)∣Mg²⁺ (0.130 M)∣∣Ag⁺ (0.0001 M)∣Ag(s)
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Calculation of Ecell : Q = [Ag⁺]² / [Mg²⁺] 1

= (10⁻⁴)² / 0.130 = 1.3 × 10⁻⁷

Ecell = E°cell − (0.0591 / 2) log Q

= 3.17 − 0.02955 log (1.3 × 10⁻⁷) = 3.17 − 0.210 Ecell = 2.96 V 1
Q. 22 Cannizzaro Reaction:
2HCHO + NaOH → HCOONa + CH3OH (1 Mark) 1

HVZ Reaction:
CH3COOH + Br2/P → CH2BrCOOH + HBr (1 Mark) 1

Q.23 First-order reaction:
k = 2.303/t log(a/(a−x)) ½

k = 2.303/5 × log(0.6/0.2) 1
= 2.303/5 × 0.48
= 0.221 min−1 Answer = 0.221 min−1 ½

OR k = 2.54 × 10⁻³ s⁻¹

For 75% decomposition, 25% remains:[R] 0 /[R] = 4 1

t = 2.303/k log [R]₀/[R] 1
t = 544 s
Q.24 (i) Correct Structure (1 Mark) 1
(ii) Ethanol will be formed , if correct equation (1mark) 1
Q.25. (a) Butanal + H2/Ni → Butan-1-ol (1 Mark) 1

(b) CH3CH=CH2 + H2O/H+ → CH3CH(OH)CH3
Product: Propan-2-ol (1 Mark) 1

OR (a)Boiling point of alcohols increases with increase in molecular mass and
decreases with branching. 1
Methanol<Ethanol<Propan-1-ol<Butan-2-ol<Butan-1-ol<Pentan-1-ol
(b)Boiling point increases with stronger intermolecular forces:n- 1
Butane<Ethoxyethane<Pentanal<Pentan-1-ol

Q.26 1
(i) Phenol → 2,4,6-Tribromophenol Reagent: Bromine water. (1 Mark) 1
(ii) Propene → Propan-1-ol
Reagent: BH3·THF followed by H2O2/OH− (Hydroboration-oxidation). (1 Mark) 1
(iii) Butan-2-one → Butan-2-ol Reagent: NaBH4 or LiAlH4. (1 Mark)

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OR (i) Acid-catalysed hydration of alkene: a
2
Step 1: Protonation of alkene to form carbocation.

Step 2: Attack of water molecule.

Step 3: Deprotonation to form alcohol.

m
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Thus, alkene gives the corresponding alcohol.
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(ii) RCH₂OH (Carboxylic acid is reduced to primary alcohol.).
m
1
s e m
Q.27
e
Moles of ethylene glycol = 45/62 = 0.726 mol
s l a
g l a
Mass of water = 0.600 kg
Molality = 0.726/0.600 = 1.21 m Formula (1 Mark) ag1
a
ΔTf = Kf × m
= 1.86 × 1.21 = 2.25°C Calculation (1 Mark) 1
Freezing point = 0 − 2.25
= −2.25°C Answer (1 Mark) 1
2− 2+ 8
Q.28 [Ni(CN)4] : Ni = 3d
CN− is strong field ligand.
Pairing occurs and dsp2 hybridisation formed.
Square planar and diamagnetic. (1.5 Marks) 1½
m
[NiCl4]2− : Cl− is weak field ligand.
m .co
No pairing occurs. sp3 hybridisation formed.
s e
la
Tetrahedral and paramagnetic. (1.5 Marks) 1½

Q.29 (a) Hydrolysis products: ag
Sucrose → Glucose + Fructose (1 Mark) 1
Lactose → Glucose + Galactose (1 Mark) 1

(b) Vitamin C is water-soluble and excess amount is excreted through urine, 1
therefore it cannot be stored in the body. (1 Mark)

Q.30 k = 1.15 × 10−3 s−1 Formula (1 Mark) 1
t = 2.303/k log(a/a−x) m
c om
t = 2.303/(1.15×10−3) × log(5/3) Calculation (1 Mark)
m.co 1

m . = 2002.6 × 0.222 = 444.6 s
s e
s e
Answer ≈ 445 s Answer (1 Mark)
g l a 1

g la OR Given: t = 40 min a
a 30% decomposed ⇒ 70% remains

For a first-order reaction, t = (2.303/k) log (a/(a − x)) 2

40 = (2.303/k) log (100/70) 40 = (2.303/k) × 0.155
k = 8.92 × 10⁻³ min⁻¹

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Now, t₁/₂ = 0.693/k
1
t₁/₂ = 0.693/(8.92 × 10⁻³) = 77.7 min Answer: t₁/₂ ≈ 78 min.

Q.31 (a) Complete the following equations:

(i) CH₃CONH₂ ──(i) LiAlH₄ / (ii) H₂O──→ CH₃CH₂NH₂ 1

(ii) p-Nitrotoluene ──Fe + HCl──→ p-Toluidine (p-CH₃C₆H₄NH₂) 1

(b) Primary amines have higher boiling points than tertiary amines because 1
primary amines form stronger intermolecular hydrogen bonds.

(c) Classification of amines:
• C₂H₅NHCH₃ – Secondary amine 1
• CH₃NH₂ – Primary amine
• (CH₃)₂NC₂H₅ – Tertiary amine

OR

(c) Butane-1-ol is more soluble in water than Butane-1-amine because the –OH 1
group forms stronger hydrogen bonds with water than the –NH₂ group.
Q.32 (a) Calculation of mass of CaCl₂

Formula: π = (i nRT) / V
1
n = (πV) / (iRT) n = (0.70 × 2.46) / (2.59 × 0.082 × 300)

n = 0.0271 mol

Mass of CaCl₂: m = n × M

m = 0.0271 × 111 m = 3.01 g 1

Answer: Mass of CaCl₂ = 3.01 g

(b) Endosmosis 1

OR

(b) Osmotic pressure can be measured accurately for very dilute solutions at
room temperature. 1

(c) Reverse Osmosis (RO) 1

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Q,33 (i) Propanone → Propene a
CH3COCH3 → CH3CHOHCH3 → CH3CH=CH2 (2 Marks) 2

(ii) Benzoic acid → Benzaldehyde
Use SOCl2 then H2/Pd-BaSO4 (Rosenmund reduction) (1 Mark) 1

(iii) Benzene → m-Nitroacetophenone
Acetylation followed by nitration. (1 Mark) 1
m
Reactivity toward HCN: m .co
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(CH3)3C-CO-C(CH3)3 < CH3COCH3 < CH3CHO (1 Mark) 1
s e m
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OR (a) Chemical Tests : (i) Acetophenone and Benzophenone 1
g
a Test:
Iodoform
• Acetophenone → Yellow precipitate of CHI₃ (positive test)
• Benzophenone → No precipitate

(ii) Propanal and Propanone 1

Tollens’ Test:
• Propanal → Silver mirror formed
• Propanone → No reaction m
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(iii) Pentan-2-one and Pentan-3-one
s em 1

Iodoform Test:
g la
a of CHI₃
• Pentan-2-one → Yellow precipitate
• Pentan-3-one → No precipitate

(b) (i) Stronger Acid :- CH₂FCOOH > CH₃COOH 1

Reason: F atom shows strong –I effect, which stabilizes the carboxylate ion and
increases acidity.

om
(ii) Increasing order of boiling points: 1
m . c
c. o CH₃CH₂CH₃ < CH₃CHO < CH₃COCH₃ < CH₃CH₂OH s e m
s emQ.34 Given: g l a
g la κ = 1.06×10 S cm −2 −1
a
a Molar conductivity:
Λm = κ ×1000/M
Λm calculation (2 Marks) 2

= 1.06×10−2 ×1000/0.1 = 106 S cm2 mol−1
Λ°m = 50.1 + 76.5 = 126.6 cm2 mol−1 Λ°m calculation (2 Mark) 2
Degree of dissociation:
α = 106/126.6 = 0.837 Degree of dissociation (1 Marks) 1

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OR (a) Given:

κ₁ = 1.29 × 10⁻² S cm⁻¹ (for 0.1 M KCl) , R₁ = 100 Ω, R₂ = 300 Ω
1
Cell constant = κ₁ × R₁ = (1.29 × 10⁻²) × 100 = 1.29 cm⁻¹

Conductivity of 0.01 M KCl: κ₂ = Cell constant / R₂
1
= 1.29 / 300 = 4.3 × 10⁻³ S cm⁻¹

Molar conductivity: Λₘ = (κ × 1000) / C 1

= (4.3 × 10⁻³ × 1000) / 0.01 = 430 S cm² mol⁻¹

(b) (i) Any two advantages of H₂–O₂ fuel cell: 1

• High efficiency.

• Environment-friendly; water is the only product.

(ii) Cell potential of mercury cell remains constant because the overall reaction 1
does not involve any change in concentration of reactants or products.

Q.35 Preparation of KMnO4:
1. Pyrolusite (MnO2) fused with KOH in air to form K2MnO4.
2. Oxidation of manganate to permanganate. 2
3. Crystallisation of KMnO4.
Oxalate ion reaction:
2MnO4− + 5C2O4 2− +16H+ → 2Mn2+ +10CO2 + 8H2O 1½
Iodide ion reaction:
2MnO4− +10I− +16H+ → 2Mn2+ +5I2 +8H2O 1½

OR (i) (a) Transition metals form complex compounds because they have vacant d- 1
orbitals and small size with high charge density.

(b) Zn²⁺ salts are colourless because Zn²⁺ has completely filled 3d¹⁰ orbitals, 1
while Ni²⁺ salts are coloured due to partially filled 3d orbitals causing d-d
transitions.

(c) Cr²⁺ is a strong reducing agent because it readily loses an electron to form the 1
more stable Cr³⁺ ion.

(ii) The f-block elements (Lanthanoids and Actinoids) are called inner transition
elements. 1

General electronic configuration: (n − 2) f1-14 (n − 1) d0-2 ns² 1

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