Question & Answers Chemistry Class 12

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## Chemistry Class 12 Sample Paper 2025-26

Time: 3 Hours | Max Marks: 70

SECTION A — MCQ (1 mark each)

Q1. Which of the following is a colligative property?
a) Viscosity
b) Surface tension
c) Osmotic pressure
d) Optical activity
Ans: c) Osmotic pressure

Q2. Van't Hoff factor for Na2SO4 in dilute solution is:
a) 1
b) 2
c) 3
d) 4
Ans: c) 3

Q3. Standard electrode potential of SHE is:
a) 1 V
b) -1 V
c) 0 V
d) 0.5 V
Ans: c) 0 V

Q4. For a first order reaction, a graph of log[A] vs time gives:
a) Straight line with positive slope
b) Straight line with negative slope
c) Parabola
d) Hyperbola
Ans: b) Straight line with negative slope

Q5. Which of the following shows maximum magnetic moment?
a) Fe2+ (d6)
b) Cu2+ (d9)
c) Mn2+ (d5)
d) Zn2+ (d10)
Ans: c) Mn2+ (d5) — 5 unpaired electrons

Q6. IUPAC name of [Co(NH3)5Cl]Cl2 is:
a) Pentaamminechloridocobalt(III) chloride
b) Chloridopentaamminecobalt(II) chloride
c) Pentaamminecobalt chloride
d) None of these
Ans: a) Pentaamminechloridocobalt(III) chloride

Q7. Which reagent converts primary alcohol to aldehyde without further oxidation?
a) KMnO4
b) K2Cr2O7
c) PCC (Pyridinium chlorochromate)
d) Conc. HNO3
Ans: c) PCC

Q8. Assertion: Aniline is a weaker base than methylamine.
Reason: Lone pair of nitrogen in aniline is delocalized into the benzene ring.
a) Both A and R are true and R explains A
b) Both A and R are true but R does not explain A
c) A is true, R is false
d) A is false, R is true
Ans: a)

SECTION B — Very Short Answer (2 marks each)

Q9. What is elevation in boiling point? Write the formula relating it to molar mass of solute.

Ans: When a non-volatile solute is dissolved in a solvent, the boiling point of the solution is higher than that of pure solvent. This increase is called elevation in boiling point.

Formula: Tb = Kb x m

Molar mass: M2 = (Kb x w2 x 1000) / (Tb x w1)

Q10. Define the following:
i) Specific conductivity
ii) Molar conductivity

Ans:

i) Specific conductivity (kappa) is the conductance of a solution kept between two electrodes of 1 cm2 area and 1 cm apart. Unit: S cm-1.

ii) Molar conductivity (Lambda m) is the conductance of all ions produced by one mole of electrolyte in a solution. Unit: S cm2 mol-1.

Lambda m = (kappa x 1000) / Molarity

Q11. The rate constant of a reaction is 1.386 min-1. Find its half life and time for 75% completion.

Ans:

t1/2 = 0.693 / k = 0.693 / 1.386 = 0.5 min

For 75% completion, [A] remaining = 25% = [A]0/4

t = 2 x t1/2 = 2 x 0.5 = 1 min

Q12. Distinguish between Werner's primary and secondary valency in coordination compounds.

Ans:

Primary valency:

It is the oxidation state of the central metal ion.
It is ionisable and is satisfied by negative ions only.
It is non-directional.

Secondary valency:

It is the coordination number of the central metal ion.
It is non-ionisable and is satisfied by neutral molecules or negative ions.
It is directional and determines geometry of the complex.

Q13. Why is -OH group ortho and para directing in electrophilic aromatic substitution?

Ans: The -OH group donates its lone pair into the benzene ring by resonance. This increases electron density at ortho and para positions. The incoming electrophile attacks these positions preferentially. Hence -OH is an ortho and para director.

Q14. Give two differences between SN1 and SN2 reactions.

Ans:

SN1 reaction:

Proceeds in two steps with formation of a planar carbocation intermediate.
Favoured by tertiary alkyl halides and polar protic solvents.
Results in racemisation at chiral centre.

SN2 reaction:

Proceeds in one concerted step with no intermediate formed.
Favoured by primary alkyl halides and polar aprotic solvents.
Results in inversion of configuration (Walden inversion).

SECTION C — Short Answer (3 marks each)

Q15. Explain the following concentration terms with formulae:
i) Mole fraction
ii) Molarity
iii) Molality

Ans:

i) Mole fraction (x) = moles of component / total moles of all components. It is dimensionless and does not change with temperature.

ii) Molarity (M) = moles of solute / volume of solution in litres. Unit: mol L-1. Changes with temperature.

iii) Molality (m) = moles of solute / mass of solvent in kg. Unit: mol kg-1. Does not change with temperature.

Q16. Derive the integrated rate law for a first order reaction. Write its characteristics.

Ans:

For: A → Products

Rate = -d[A]/dt = k[A]

On integrating:

ln[A] = ln[A]0 - kt

or k = (2.303/t) log ([A]0/[A])

Characteristics:

Unit of k is s-1 or min-1.
Half life t1/2 = 0.693/k, independent of initial concentration.
Plot of log[A] vs time is a straight line with slope = -k/2.303.

Q17. Explain the following properties of transition elements:
i) Variable oxidation states
ii) Formation of coloured compounds
iii) Catalytic properties

Ans:

i) Transition metals have (n-1)d and ns electrons with similar energies. Both sets of electrons can be used in bond formation, giving multiple oxidation states. Example: Mn shows +2 to +7.

ii) Transition metal ions have partially filled d orbitals. d-d transitions occur when electrons absorb visible light and jump to higher d levels. The complementary colour is observed. Example: Cu2+ appears blue.

iii) Transition metals can provide a surface for reactants to adsorb and react. They also offer variable oxidation states allowing them to form intermediate compounds. Example: Fe in Haber process, V2O5 in Contact process.

Q18. What are enantiomers? Explain with example of lactic acid. State the optical activity of racemic mixture.

Ans:

Enantiomers are non-superimposable mirror images of each other. They have the same physical and chemical properties but differ in the direction of rotation of plane polarised light.

Example: Lactic acid (2-hydroxypropanoic acid) has one chiral carbon (C2). Its two enantiomers are:
d-lactic acid — rotates light to the right (dextrorotatory)
l-lactic acid — rotates light to the left (laevorotatory)

A racemic mixture contains equal amounts of both enantiomers. It is optically inactive because the rotations cancel each other.

Q19. Explain the following reactions of aldehydes:
i) Cannizzaro reaction
ii) Aldol condensation
iii) Tollens test

Ans:

i) Cannizzaro reaction: Aldehydes without alpha hydrogen undergo self oxidation-reduction with concentrated NaOH. One molecule is oxidised to acid salt and another is reduced to alcohol.
2HCHO + NaOH → HCOONa + CH3OH

ii) Aldol condensation: Aldehydes with alpha hydrogen react with dilute NaOH to form beta-hydroxy aldehyde (aldol), which on heating gives alpha-beta unsaturated aldehyde.
CH3CHO + CH3CHO → CH3CH(OH)CH2CHO

iii) Tollens test: Aldehyde reduces Tollens reagent (ammoniacal AgNO3) to give silver mirror on the inner wall of the test tube. Ketones do not give this test.
RCHO + 2[Ag(NH3)2]+ → RCOO- + 2Ag + 4NH3 + H3O+

Q20. Write a note on the following biomolecules:
i) Structure of glucose
ii) Difference between DNA and RNA

Ans:

i) Structure of glucose:
Molecular formula: C6H12O6
Open chain: It is an aldohexose with -CHO group at C1 and -OH groups at C2, C3, C4, C5 and -CH2OH at C6.
Cyclic (Haworth): Glucose exists predominantly as cyclic pyranose ring formed by reaction of -OH at C5 with -CHO at C1.
It shows mutarotation due to alpha and beta anomers.

ii) DNA vs RNA:

DNA:

Sugar present is deoxyribose.
Bases are Adenine, Guanine, Cytosine, Thymine.
Double stranded helix structure.
Found mainly in nucleus; carries genetic information.

RNA:

Sugar present is ribose.
Bases are Adenine, Guanine, Cytosine, Uracil (no Thymine).
Usually single stranded.
Found in nucleus and cytoplasm; involved in protein synthesis.

SECTION D — Case Based Questions (4 marks each)

Q21. Read the passage and answer:

Electrochemical cells convert chemical energy into electrical energy. In a Daniel cell, zinc acts as anode and copper as cathode. The cell reaction is spontaneous when standard cell potential is positive. The relationship between Gibbs energy and EMF is: delta G = -nFE. Nernst equation relates EMF to concentration of ions.

i) Write cell representation of Daniel cell.
ii) Write overall cell reaction.
iii) Calculate EMF if E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V.
iv) Write Nernst equation for the Daniel cell.

Ans:

i) Zn | Zn2+(1M) || Cu2+(1M) | Cu

ii) Zn + Cu2+ → Zn2+ + Cu

iii) E(cell) = E(cathode) - E(anode) = 0.34 - (-0.76) = +1.10 V

iv) E(cell) = E(cell) - (0.0591/2) log ([Zn2+]/[Cu2+])

Q22. Read the passage and answer:

Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups. They are classified as primary, secondary, or tertiary depending on the number of hydrogen atoms replaced. Amines are basic in nature. Aryl amines are weaker bases than alkyl amines. Diazonium salts are important intermediates in organic synthesis derived from primary aromatic amines.

i) Arrange in increasing order of basicity: NH3, CH3NH2, (CH3)2NH, C6H5NH2
ii) Why is aniline less basic than methylamine?
iii) How is benzene diazonium chloride prepared from aniline?
iv) Name one reaction where diazonium salt acts as electrophile.

Ans:

i) C6H5NH2 < NH3 < CH3NH2 < (CH3)2NH

ii) In aniline, the lone pair of nitrogen is delocalized into the benzene ring by resonance. This reduces its availability for protonation. In methylamine, the methyl group donates electrons to nitrogen, increasing its electron density and basicity.

iii) Diazotisation:
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl- + NaCl + 2H2O
(Reaction done at 0-5 degrees Celsius)

iv) Coupling reaction: Diazonium salt acts as electrophile and reacts with activated aromatic compounds like aniline or phenol to form azo dyes.
C6H5N2+Cl- + C6H5OH → C6H5-N=N-C6H4-OH + HCl

SECTION E — Long Answer (5 marks each)

Q23 a) Explain Valence Bond Theory for [Co(NH3)6]3+. Why is it diamagnetic? What is its geometry?

Ans:

Co is in +3 oxidation state: Co3+ has d6 configuration.

Electronic configuration of Co3+: [Ar] 3d6

NH3 is a strong field ligand. It causes pairing of all d electrons in inner d orbitals.

The 6 lone pairs of NH3 are donated to d2sp3 hybrid orbitals of Co3+ (using two 3d + one 4s + three 4p orbitals).

All 6 electrons are paired in 3d level → 0 unpaired electrons → diamagnetic.

Hybridisation: d2sp3
Geometry: Octahedral

Q23 b) Explain the following:
i) Spectrochemical series
ii) Crystal Field Splitting in octahedral complex
iii) Why is [Fe(CN)6]4- low spin but [FeF6]3- high spin?

Ans:

i) Spectrochemical series is the arrangement of ligands in increasing order of their crystal field splitting power:
I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO

ii) In an octahedral complex, the five d orbitals split into two sets:
t2g (dxy, dyz, dxz) — lower energy, 3 orbitals
eg (dx2-y2, dz2) — higher energy, 2 orbitals
Energy difference = delta o (crystal field splitting energy)

iii) CN- is a strong field ligand, large delta o, electrons pair in t2g → [Fe(CN)6]4- is low spin, diamagnetic.
F- is a weak field ligand, small delta o, electrons remain unpaired → [FeF6]3- is high spin, paramagnetic.

Q24 a) Explain mechanism of addition of HBr to propene in presence and absence of peroxide. State Markovnikov's rule and anti-Markovnikov's addition.

Ans:

Without peroxide (ionic mechanism — Markovnikov's rule):

Markovnikov's rule: The negative part of the adding molecule (Br-) goes to the carbon carrying lesser number of hydrogen atoms.

Step 1: H+ adds to CH2=CHCH3 → secondary carbocation CH3CH+CH3 (more stable)
Step 2: Br- attacks the carbocation → CH3CHBrCH3 (2-bromopropane)

With peroxide (free radical mechanism — Anti-Markovnikov):

Peroxide generates free radicals. Br radical adds to CH2 end forming more stable secondary radical at C2.
Result: CH3CH2CH2Br (1-bromopropane)

This is anti-Markovnikov addition, also called peroxide effect or Kharasch effect.

Q24 b) Explain the following reactions giving equations:
i) Hell-Volhard-Zelinsky reaction
ii) Decarboxylation of carboxylic acid
iii) Reaction of acetic acid with PCl5

Ans:

i) Hell-Volhard-Zelinsky (HVZ) reaction:
Carboxylic acids react with Cl2 or Br2 in presence of red phosphorus to give alpha-halo acid.
CH3COOH + Br2 / P → BrCH2COOH + HBr

ii) Decarboxylation:
Carboxylic acids lose CO2 when their sodium salts are heated with soda lime (NaOH + CaO).
CH3COONa + NaOH → CH4 + Na2CO3

iii) Reaction with PCl5:
Acetic acid reacts with PCl5 to form acetyl chloride (acid chloride).
CH3COOH + PCl5 → CH3COCl + POCl3 + HCl

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