UKBSE 12 Chemistry Solved Sample Paper 2026

UK Board Class 12 Chemistry Solved Sample 2026 SET-1

Q. No.	Question Summary	Correct Answer
1 (क)	Chemical name of Vitamin B₁	(iv) Thiamine
1 (ख)	Order of reaction if half-life is independent of initial concentration	(ii) First (प्रथम)
1 (ग)	Coordination number of 'Fe' in $K_4[Fe(CN)_6]$	(iii) 6
1 (घ)	Value of Henry's Constant $K_H$	(i) Increases with increase in temperature
1 (ङ)	Faraday's first law of electrolysis	(iii) $W \propto Q$
1 (च)	Formaldehyde + KOH heating product	(i) Methyl alcohol
1 (छ)	Reagent to convert amide to amine	(i) $Br_2 / KOH$
1 (ज)	Chemical formula of Phosgene	(iv) $COCl_2$
1 (झ)	Assertion: d/f blocks produce colored ions. Reason: Unpaired electrons present.	(i) Both A and R are correct and R is the correct explanation of A
1 (ञ)	Assertion: Maltose/Lactose are reducing sugars. Reason: They reduce Fehling's/Tollen's.	(i) Both A and R are correct and R is the correct explanation of A

2. What is the chemical name of Vitamin D?

The chemical name of Vitamin D is Calciferol (specifically Ergocalciferol for D2 and Cholecalciferol for D3).

3. Explain Van’t Hoff factor.

The Van’t Hoff factor ($i$) is defined as the ratio of the observed (experimental) value of a colligative property to the calculated (theoretical) value. It accounts for the extent of association or dissociation of solute particles in a solution.
$$i = \frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}$$

4. In which process does the rusting of iron, tarnishing of silver, and the formation of a green coating on copper occur?

These processes occur due to Corrosion.

5. Give the IUPAC name of coordination compound K₂[Zn(OH)₄].

Potassium tetrahydroxozincate(II)

6. Calculate the half-life of a first-order reaction from their rate constants.

Formula: $t_{1/2} = \frac{0.693}{k}$
(i) For $k = 200 \text{ s}^{-1}$: $t_{1/2} = \frac{0.693}{200} = 0.003465 \text{ s}$
(ii) For $k = 2 \text{ min}^{-1}$: $t_{1/2} = \frac{0.693}{2} = 0.3465 \text{ min}$

7. Explain: (i) Specific conductivity (ii) Standard electrode Potential

(i) Specific conductivity (κ): It is the conductance of a solution of 1 cm length and 1 cm² area of cross-section (i.e., conductance of 1 cm³ of electrolyte).
(ii) Standard electrode Potential (E°): It is the potential difference developed between the metal electrode and the solution of its ions of unit molarity (1 M) at 298 K.

8. Write chemical equations: (i) Carbylamine reaction (ii) HVZ reaction

(i) $R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$ (Isocyanide formation)
(ii) $CH_3COOH \xrightarrow[H_2O]{(i) X_2/\text{Red P}} X-CH_2COOH$ (where X = Cl, Br) OR (i) Why Phenol is acidic: Phenol is acidic because after losing a proton, it forms a phenoxide ion which is resonance-stabilized.
(ii) Auto-oxidation of diethyl ether: $C_2H_5OC_2H_5 + O_2 \xrightarrow{\text{long exposure to air}} CH_3CH(OOH)OC_2H_5$ (Ethoxyethyl hydroperoxide)

9. Explain the followings:
(i) Most of the transition elements show variable oxidation states.
(ii) Most of the compounds of transition elements are coloured.

(i) Transition elements show variable oxidation states because the energy difference between $(n-1)d$ and $ns$ orbitals is very small, allowing electrons from both levels to participate in bonding.
(ii) They are coloured due to d-d transitions. In the presence of ligands, the d-orbitals split, and electrons absorb visible light to jump from lower to higher d-energy levels.

10. What is meant by unidentate and ambidentate legands ? Give example for each.

Unidentate: Ligands that bind to a metal ion through only one donor atom. Example: $Cl^-$, $H_2O$.
Ambidentate: Ligands that have two different donor atoms and can coordinate through either of them. Example: $NO_2^-$ (can bind via N or O), $SCN^-$.

11. Write Chemical reactions for the preparation of following.
(i) D.D.T. (ii) Diphenyl.

(i) CCl₃CHO + 2 C₆H₅Cl $\xrightarrow{\text{conc. } H_2SO_4}$ D.D.T + $H_2O$
(ii) 2 C₆H₅Cl + 2Na $\xrightarrow{\text{dry ether}}$ C₆H₅-C₆H₅ + 2NaCl (Fittig Reaction)

12. What happens when (Write Chemical equation only and name the product).
D - Glucose reacts with the following.
(i) HNO₃
(ii) Fehling Solution

Sucrose: $C_{12}H_{22}O_{11} + H_2O \to C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)}$
Lactose: $C_{12}H_{22}O_{11} + H_2O \to C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Galactose)}$

13. An aromatic compound 'A' on treatment with aqueous ammonia and heating form compound 'B' which on heating with Br₂ and KOH forms a compound 'C' of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds (A), (B) and (C).

Compound C is $C_6H_7N$ (Aniline). This is formed via Hoffmann Bromamide degradation from B.
A: Benzoic acid ($C_6H_5COOH$)
B: Benzamide ($C_6H_5CONH_2$)
C: Aniline ($C_6H_5NH_2$)

14. When 10 gm of a substance was dissolved in 100 gm solvent, observed elevation is boiling point was 1^oC. Calculate the molecular weight of the substance. [K_b for solvent is 2.53 k kg mol^-1].

$\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$
$1 = \frac{2.53 \times 10 \times 1000}{M \times 100}$
$M = 2.53 \times 10 \times 10 = 253 \text{ g/mol}$

15. Λ^o_m of HCl, NaCl and CH₃COONa are 426.1, 126.5 and 91.0 mho cm² mol^-1 respectively. Calculate Λ^o_m for CH₃COOH.

According to Kohlrausch's Law:
$\Lambda^o_m(CH_3COOH) = \Lambda^o_m(CH_3COONa) + \Lambda^o_m(HCl) - \Lambda^o_m(NaCl)$
$= 91.0 + 426.1 - 126.5 = 390.6 \text{ mho cm}^2 \text{ mol}^{-1}$

16. Complete the following chemical reactions:

(i) Dehydration/Amination of Ethanol:

C₂H₅OH + NH₃ → (Al₂O₃, 450°C) → C₂H₅NH₂ + H₂O
(Product: Ethylamine)

(ii) Carbylamine Reaction:

RNH₂ + CHCl₃ + 3KOH(alc) → R-NC + 3KCl + 3H₂O
(Product: Alkyl Isocyanide)

(iii) Diazotization Reaction:

C₆H₅NH₂ + HCl → [C₆H₅NH₃]⁺Cl^- --- HNO₂ + 273K → C₆H₅N₂Cl + 2H₂O
(Product: Benzene Diazonium Chloride)

17. (i) Write Cell reactions of the given Cell: Zn(S) | Zn²⁺(aq) || Ag⁺(aq) | Ag (s)

At Anode (Oxidation): Zn(s) → Zn²⁺(aq) + 2e^-
At Cathode (Reduction): 2Ag⁺(aq) + 2e^- → 2Ag(s)
Net Cell Reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
(ii) Describe the construction of a dry cell and write the cell reactions. 2

Construction: A dry cell consists of a Zinc container which acts as the Anode. The Cathode is a Carbon (graphite) rod surrounded by powdered MnO₂ and carbon. The space between the electrodes is filled by a moist paste of NH₄Cl and ZnCl₂ which acts as the electrolyte.

Cell Reactions:

Anode: Zn(s) → Zn²⁺ + 2e^-
Cathode: MnO₂ + NH₄⁺ + e^- → MnO(OH) + NH₃

18. Define the following terms.

(i) Mole fraction (x): It is the ratio of the number of moles of a particular component to the total number of moles of all the components present in the solution.

(ii) Molality (m): It is defined as the number of moles of the solute dissolved per kilogram (1000g) of the solvent.

(iii) Mass Percentage: It is the mass of a component per 100 grams of the total mass of the solution.

OR: Calculate molality of 2.5 gm ethanoic acid (CH₃COOH) in 75 gm of benzene.

1. Calculate Molar Mass of CH₃COOH:

(2 × 12) + (4 × 1) + (2 × 16) = 24 + 4 + 32 = 60 g/mol

2. Calculate Number of Moles:

Moles = Mass / Molar Mass = 2.5 / 60 = 0.0417 mol

3. Calculate Molality (m):

Mass of solvent (Benzene) = 75 g = 0.075 kg

Molality = Moles of solute / Mass of solvent in kg

m = 0.0417 / 0.075 = 0.556 mol/kg (or 0.556 m)

19. Briefly explain the Valence Bond theory. Explain the geometrical and magnetic behavior of complex ion [Fe(CN)6]^4- on the basis of this theory.

Valence Bond Theory (VBT)

Proposed by Linus Pauling, VBT explains the formation, structure, and magnetic properties of coordination compounds based on the following postulates:

• The central metal ion provides a number of empty orbitals (s, p, and d) equal to its coordination number.
• These empty orbitals hybridize to form a set of equivalent orbitals of definite geometry (Octahedral, Tetrahedral, etc.).
• Ligands donate a pair of electrons into these vacant hybridized orbitals to form coordinate covalent bonds.
• If inner d-orbitals $(n-1)d$ are used, it is an inner orbital complex. If outer d-orbitals $(n)d$ are used, it is an outer orbital complex.

Application to [Fe(CN)₆]^4-

1. Oxidation State and Configuration:

Oxidation state of Iron (Fe) is +2.

Configuration of Fe²⁺: $[Ar] 3d^6 4s^0$

2. Role of Ligand

The Cyanide ion (CN^-) is a strong field ligand. According to VBT, strong field ligands force the unpaired electrons in the 3d orbitals to pair up against Hund's Rule to make inner d-orbitals available for bonding.

3. Hybridization

To accommodate 6 pairs of electrons from 6 CN^- ligands, the Fe²⁺ ion undergoes d²sp³ hybridization using two 3d, one 4s, and three 4p orbitals.

State	3d Orbitals	4s	4p
Fe2+ (free ion)	↑↓ ↑ ↑ ↑ ↑
Fe2+ (paired by CN-)	↑↓ ↑↓ ↑↓ [__] [__]	[__]	[__] [__] [__]
d2sp3 Hybrid Orbitals	Six empty hybrid orbitals ready for CN- lone pairs

Final Characteristics of [Fe(CN)₆]^4-

Geometry: Octahedral (due to d²sp³ hybridization).

Magnetic Behavior: Diamagnetic. All electrons are paired up; hence the magnetic moment is zero.

Type: Inner orbital complex (also called a low-spin complex).

OR 19. Explain the following with example.

(i) Coordination Entity: A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. It is usually enclosed in square brackets. It can be neutral, positively charged, or negatively charged.

Example: In the compound $K_4[Fe(CN)_6]$, the part $[Fe(CN)_6]^{4-}$ is the coordination entity.

(ii) Ligand: The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These act as Lewis bases by donating a pair of electrons to the central metal (Lewis acid) to form a coordinate bond.

Example: In $[Cu(NH_3)_4]^{2+}$, the four $NH_3$ (ammonia) molecules are the ligands.

(iii) Coordination Number (C.N.): The coordination number of a metal ion in a complex is defined as the total number of ligand donor atoms to which the metal is directly bonded. It represents the number of sigma bonds formed between ligands and the central metal.

Example: In $[PtCl_6]^{2-}$, there are six $Cl^-$ ions attached to Pt, so the Coordination Number is 6. In $[Co(en)_3]^{3+}$, since 'en' (ethylenediamine) is didentate, the Coordination Number is $3 \times 2 = 6$.

20. Write short notes on following:

(i) Aldol Condensation: Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali (like NaOH) to form β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols).

2CH₃CHO → (dil. NaOH) → CH₃CH(OH)CH₂CHO

(ii) Ammonolysis: The process of cleavage of the C-X bond in alkyl halides by an ammonia molecule is called ammonolysis. It produces a mixture of primary, secondary, and tertiary amines, and finally quaternary ammonium salts.

R-X + NH₃ → R-NH₂ + HX → (excess R-X) → R₂NH → R₃N → [R₄N]⁺X^-

(iii) Diazotisation: The conversion of primary aromatic amines into diazonium salts using nitrous acid (NaNO₂ + HCl) at low temperatures (273-278 K) is known as diazotisation.

C₆H₅NH₂ + NaNO₂ + 2HCl → (273-278K) → C₆H₅N₂⁺Cl^- + NaCl + 2H₂O

21. Give Chemical equations for following name reactions:

(i) Wolff Kishner Reduction: Carbonyl groups are reduced to -CH₂ groups by treatment with hydrazine followed by heating with KOH in ethylene glycol.

C=O + NH₂NH₂ → (-H₂O) → C=NNH₂ → (KOH/Ethylene glycol, Δ) → CH₂ + N₂

(ii) Williamson Ether Synthesis: An alkyl halide reacts with sodium alkoxide to form an ether (S_N2 mechanism).

R-X + R'-ONa → R-O-R' + NaX

(iii) Kolbe Schmitt Reaction: Phenoxide ion reacts with CO₂ followed by acidification to produce Salicylic acid.

C₆H₅ONa + CO₂ → (400K, 4-7 atm) → Sodium Salicylate → (H⁺) → C₆H₄(OH)COOH

22. Write Chemical reactions for the preparation of chloroform in the laboratory.

Chloroform (CHCl₃) is prepared by heating ethanol or acetone with a paste of bleaching powder [CaOCl₂].

Step 1: Hydrolysis of Bleaching Powder: CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

Step 2 (Using Acetone):

CH₃COCH₃ + 3Cl₂ → CCl₃COCH₃ (Trichloroacetone) + 3HCl
2CCl₃COCH₃ + Ca(OH)₂ → 2CHCl₃ (Chloroform) + (CH₃COO)₂Ca (Calcium acetate)

23. Write Short notes on following:

(i) Zwitter ion: In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton, giving rise to a dipolar ion known as a Zwitter ion. It is neutral but carries both positive and negative charges.

H₂N-CH(R)-COOH ↔ ⁺H₃N-CH(R)-COO^-

(ii) Nucleotide: A nucleotide is the basic building block of nucleic acids (DNA/RNA). It consists of three components: a nitrogenous base, a pentose sugar, and a phosphoric acid group.

(iii) Glycosidic bond: It is the oxide linkage formed between two monosaccharide units by the loss of a water molecule. This bond joins sugar molecules to each other in disaccharides and polysaccharides.

24. (i) Derive the integrated rate equation for the first order reaction.

Consider a first order reaction: R → P

The rate of reaction is given by: $$Rate = -\frac{d[R]}{dt} = k[R]^1$$

Rearranging the equation: $$\frac{d[R]}{[R]} = -k dt$$

Integrating both sides: $$\int \frac{d[R]}{[R]} = -\int k dt$$ $$\ln[R] = -kt + I \quad \text{--- (Eq. 1)}$$

At $t = 0$, $[R] = [R]_0$ (where $[R]_0$ is initial concentration). Substituting these values in Eq. 1: $$\ln[R]_0 = -k(0) + I \implies I = \ln[R]_0$$

Substituting the value of $I$ back into Eq. 1: $$\ln[R] = -kt + \ln[R]_0$$ $$kt = \ln[R]_0 - \ln[R] = \ln \frac{[R]_0}{[R]}$$

Converting natural log ($\ln$) to common log ($\log_{10}$): $$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$$

(ii) Calculate half life of the reaction (K = 5.5 x 10^-14 s^-1).

For a first order reaction, the half-life ($t_{1/2}$) is given by:

$$t_{1/2} = \frac{0.693}{k}$$

Substituting the value of $k$:

$$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} = 1.26 \times 10^{13} \text{ s}$$

OR (i) Show that $t_{99.9\%}$ is 10 times $t_{1/2}$.

Step 1: For $t_{99.9\%}$

Initial concentration $[R]_0 = 100$, Final concentration $[R] = 100 - 99.9 = 0.1$

$$t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{2.303}{k} \times 3 \quad \text{--- (Eq. A)}$$

Step 2: For $t_{1/2}$

$$t_{1/2} = \frac{0.693}{k} = \frac{2.303 \log 2}{k} = \frac{2.303 \times 0.3010}{k} \quad \text{--- (Eq. B)}$$

Step 3: Comparison

Dividing Eq. A by Eq. B:

$$\frac{t_{99.9\%}}{t_{1/2}} = \frac{3}{0.3010} \approx 9.96 \approx 10$$

Therefore, $t_{99.9\%} \approx 10 \times t_{1/2}$.

OR (ii) Kinetics study of decomposition of H₂O₂.

The decomposition of Hydrogen Peroxide ($2H_2O_2 \rightarrow 2H_2O + O_2$) is a first-order reaction. It is studied by titration:

• A known volume of the reaction mixture is withdrawn at different time intervals ($t$).
• The reaction is "quenched" (stopped) by adding it to ice-cold water or dilute $H_2SO_4$.
• The amount of unreacted $H_2O_2$ is titrated against a standard solution of Potassium Permanganate (KMnO₄).
• The volume of $KMnO_4$ used ($V_t$) is proportional to the concentration of $H_2O_2$ remaining at that time.

25. An Organic compound [A] whose molecular formula is C₃H₆O, gives Iodoform reaction and forms compound [B]. Compound [B], when heated with silver powder converts into compound [C] reacts with dil. Sulphuric acid and Mercuric Sulphate to obtain compound [D], which gives Aldol condensation reaction. Write down the name of all compounds from [A] to [D] and write chemical equation for each step.

Based on the reactions described, here is the identification of the compounds:

[A] Acetone (Propanone): $CH_3COCH_3$ (Gives iodoform test)
[B] Iodoform: $CHI_3$ (Yellow precipitate formed from A)
[C] Acetylene (Ethyne): $HC \equiv CH$ (Formed by heating B with Ag powder)
[D] Acetaldehyde (Ethanal): $CH_3CHO$ (Formed by hydration of C; gives Aldol reaction)

Chemical Equations:

1. Iodoform Reaction (A to B):

CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ (B) + CH₃COONa + 3NaI + 3H₂O

2. Reaction with Silver Powder (B to C):

2CHI₃ + 6Ag → (Δ) → HC≡CH (C) + 6AgI

3. Hydration (C to D):

HC≡CH + H₂O → (dil. H₂SO₄ / HgSO₄) → CH₃CHO (D)

4. Aldol Condensation (of D):

2CH₃CHO → (dil. NaOH) → CH₃CH(OH)CH₂CHO (3-Hydroxybutanal)

25. OR: Name the products and write chemical equations.

(i) Acetone + Bleaching Powder:
Product: Chloroform ($CHCl_3$)

2CH₃COCH₃ + 3Ca(OCl)Cl → 2CHCl₃ + (CH₃COO)₂Ca + 2Ca(OH)₂ + CaCl₂

(ii) Formaldehyde + Ammonia:
Product: Urotropine (Hexamethylenetetramine)

6HCHO + 4NH₃ → (CH₂)₆N₄ + 6H₂O

(iii) Sodium Acetate + Sodalime:
Product: Methane ($CH_4$)

CH₃COONa + NaOH → (CaO, Δ) → CH₄ + Na₂CO₃

(iv) Benzoyl Chloride + Hydrogen (Pd/BaSO₄):
Product: Benzaldehyde ($C_6H_5CHO$) - Rosenmund Reduction

C₆H₅COCl + H₂ → (Pd/BaSO₄) → C₆H₅CHO + HCl

26. Read the following passage carefully and answer the questions given below.

The f block elements are those in which the differentiating electron enters the (n-2) f orbital. There are two series of f-block elements corresponding to fitting of 4f and 5f orbitals. The series of 4f orbital is called lanthanoids. Lanthanoids show different oxidation states depending upon stability of f^o, f⁷, and f¹⁴ configurations though the most common oxidation state is +3. There is a regular decrease in the size of Lanthanoids ions with increase in atomic number which is known as lanthanoid contraction.

(i) The atomic number of three lanthanoid elements X, Y and Z are 65, 68 and 70 respectively, their Ln³⁺ electronic configuration is.
(a) f⁸ f¹¹ f¹³
(b) f¹¹ f⁸f¹³
(c) f^of²f¹¹
(d) f³f⁷f⁹

(ii) Lanthanoid Contraction is observed in
(a) Gd
(b) At
(c) Xe
(d) Te

(iii) Name a number of the lanthanoids series which is well known to exhibit +4 oxidation state. (a) Cerium, Z = 58
(b) Europium, Z = 63
(c) Lanthanum, Z = 57
(d) Gadolinium Z = 64

(iv) Which of the following is not the configuration of Lanthanoid ?
(a) [Xe] 4f¹⁰ 6s²
(b) 2 [Xe] 4f¹ 5d¹ 6s²
(c) 2 [Xe] 4f¹⁴ 5d¹⁰ 6s²
(d) [Xe] 4f⁷ 5d¹ 6s²

(i) (a) f⁸ f¹¹ f¹³
(ii) (a) Gd (Part of the series)
(iii) (a) Cerium (Ce⁴⁺ is stable due to $f^0$)
(iv) (c) [Xe] 4f¹⁴ 5d¹⁰ 6s² (This is Mercury/Transition, not Lanthanoid)

Must read

UKBSE Chemistry Sample Paper 2026 set-1

UKBSE Chemistry Solved Sample Paper 2026 set-1

UKBSE Chemistry Sample Paper 2026 set-2

UKBSE Chemistry Question Paper 2025

UK Board Class 12 Chemistry PYQs with Answer