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Sample Papers for Class 12 Chemistry
|CBSE Class 12 Chemistry Sample Paper Set A|
|CBSE Class 12 Chemistry Sample Paper Set B|
|CBSE Class 12 Chemistry Sample Paper Set C|
|CBSE Class 12 Chemistry Term 1 Sample Paper|
|CBSE Class 12 Chemistry Term 1 Sample Paper Set A|
|CBSE Class 12 Chemistry Term 1 Sample Paper Set B|
|CBSE Class 12 Chemistry Term 2 Sample Paper|
|CBSE Class 12 Chemistry Term 2 Sample Paper Set A|
CBSE Class 12 Chemistry Term 1 Sample Paper Set A
1. An organic compound C8H18 on chlorination gives a single monochloride. Write the structure of the hydrocarbon.
Answer. As C8H18 on chlorination gives a single monochloride therefore, all the hydrogen atoms in it are equivalent. The possible structure of the hydrocarbon is
2. Pure silicon is an insulator then, how does it behave as a semiconductor on heating?
Answer. On heating some covalent bonds among silicon atoms break and electrons become free to move under applied field hence, silicon behaves like a semiconductor at high temperature.
3. Explain why the bond angle (C — 0 — C) in ether is slightly greater than the tetrahedral angle (109.5°28’)?
Answer. There are two lone pairs on oxygen which cause lone pair-bond pair repulsions and hence, bond angle (C — 0 — C) should decrease and have value less than 109°28’ but steric hindrance of bulky alkyl groups causes increase in bond angle from 109.5°28’ to 111.7°.
4. Which of the following electrolyte is most effective for the coagulation of Fe(OH)3 sol and why?
Na3PO4, Na2SO4, NaCl
Answer. Fe(OH)3 is a positively charged sol hence, the anion having maximum charge will be more effective in coagulation. Therefore, Na3PO4 having PO3–4 ion (charge = –3) will be most effective.
5. Comment on how sulphuric acid is a dibasic acid.
Answer. H2SO4 forms two series of salts, i.e., both the hydrogen atoms are replaceable.
6. The partial pressure of ethane over a saturated solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what would be the partial pressure of the gas?
Answer. According to Henry’s law, m = KH × p
Case I: 6.56 × 10–3 g = KH × 1 bar
or, KH = 6.56 × 10–3 g bar–1
Case II : 5.00 × 10–2 g = (6.56 × 10–3 g bar–1) × p
7. How will you bring about the following conversions?
(i) Propanone to propene
(ii) Benzoic acid to m-nitrobenzyl alcohol
8. Explain the following observations :
(i) Many of the transition elements are known to form interstitial compounds.
(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29).
Answer. (i) In the crystal lattice, transition elements have interstitial vacant spaces into which small sized non-metal atoms such as H, B, C, or N are trapped. These compounds are known as interstitial compounds. These are non-stoichiometric, neither typically ionic nor covalent, e.g., TiC, Mn4N, Fe3H, etc.
(ii) As we move from left to right along a transition series (from Ti to Cu), the atomic radii decrease due to increase in nuclear charge. Therefore, atomic volume decreases with increase in atomic mass and hence, density increases.
9. For the cell Zn|Zn2+ (aq) || Cu2+(aq)|Cu, derive the relation between E°cell and K at 298 K.
Answer. In this cell, electrons will ow from Zn electrode to Cu electrode.
10. Write the formulae of the following complexes :
(i) Triamminetriaquachromium(III) chloride
(ii) Tris(ethane-1, 2-diammine)cobalt(III) sulphate
(iii) Diamminesilver(I) dicyanoargentate(I)
The spin only magnetic moment of [MnB4r]2– is 5.9 BM. Predict the geometry of the complex ion.
Answer. (i) [Cr(NH3)3(H2O)3]Cl3
Oxidation number of Mn in the complex = +2
Mn(25) : [Ar] 3d54s2
Mn2+ : [Ar] 3d5
11. The edge length of unit cell of a metal having molecular weight 75 g/mol is 5 Å which crystallises in cubic lattice. If the density is 2 g/cc, then find the radius (in pm) of metal atom.
(Given, NA = 6 × 1023).
For a metal with Z = 2 means that it has body centred cubic (bcc) structure. For bcc structure,
12. Write the chemical equations for synthesis of following polymers :
(i) Teffon (ii) PVC (iii) Glyptal
Answer. (i) Tetrafluoroethylene is the monomer that undergoes addition polymerisation to form telflon or polytetrafluoroethylene.
(ii) Vinyl chloride is the monomer that undergoes addition polymerisation of form polyvinyl chloride (PVC).
(iii) Phthalic acid and ethylene glycol are the monomers that undergo condensation polymerisation to form glyptal.
13. (i) What is meant by van’t Hoff factor?
(ii) The osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm at 27°C. Calculate the van’t Hoff factor.
(R = 0.082 L atm mol–1 K–1).
What conclusion do you draw about the molecular state of the solute in the solution?
Answer. (i) van’t Hoff factor (i) is the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property. When i > 1, there is dissociation of solute in the solution.
When i < 1, there is association of solute in the solution.
When i = 1, there is no association or dissociation of solute in the solution.
(ii) 𝛑 = iCRT
or, 0.70 = i × 0.0103 × 0.082 × (27 + 273)
14. How can reducing and non-reducing sugars be distinguished? Mention the structural feature characterising reducing sugars.
Reducing sugars are those carbohydrates which can reduce reagents like Tollens’ reagent (ammoniacal AgNO3), Benedict’s solution and Fehling’s solution whereas, non-reducing sugars cannot reduce these reagents.
All monosaccharides whether aldoses or ketoses are reducing sugars.
The characteristic structural feature of reducing sugars is that they must have aldehydic group
(—CHO) or ketonic group
in the hemiacetal or hemiketal form.
15. Calculate the standard cell potential of the given galvanic cell in which the following reaction takes place :
2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
Given : Eocr3+ Cr = / 0.74 V,Eo. cd2+ /cd = – 40 V
Also, calculate ΔrG° and equilibrium constant of the reaction.
Calculate the standard cell potential, ΔrG° and
equilibrium constant for the reaction :
E°cell = E°cathode – E°anode
= –0.40 V – (–0.74 V) = + 0.34 V
ΔrG° = –nFE°cell
= –6 × 96500 C mol–1 × 0.34 V
= –196860 C V mol–1 = –196860 J mol–1
rG° = –2.303 RT log Kc
–196860 = –2.303 × 8.314 ×298 log Kc
or log Kc = 34.5014
Kc = Antilog 34.5014 = 3.172 × 1034
E°cell = + 0.80 V – 0.77 V = + 0.03 V
ΔrG° = –nFE°cell
= –(1) × (96500 C mol–1) × (0.03 V)
= –2895 C V mol–1 = –2895 J mol–1
ΔrG° = –2.303 RT log Kc
–2895 = –2.303 × 8.314 × 298 × log Kc
or log Kc = 0.5074
or Kc = Antilog (0.5074) = 3.22
16. Give explanation for the following statements :
(i) Chloroacetic acid has lower pKa value than acetic acid.
(ii) Carboxylic acids have higher boiling points than alcohols of comparable molecular masses.
(iii) Sodium bisulphite is used for the purification of aldehydes and ketones.
Answer. (i) ‘Cl’ in chloroacetic acid shows –I effect which creates less electron density on oxygen of carboxylic acid. Thus, release of proton becomes easier. In case of acetic acid, +I efect of –CH3 group makes the release of proton difficult. Hence, chloroacetic acid is stronger acid than acetic acid. Thus, chloroacetic acid has lower pKa value than that of acetic acid.
(ii) Carboxylic acids and alcohols both undergo hydrogen bonding among their molecules. Hydrogen bonding among acid molecules is far stronger than among alcohol molecules. Hence, carboxylic acids have higher boiling points than alcohols.
Strong hydrogen bonding among carboxylic acid molecules (Dimer formation).
(iii) Sodium bisulphite reacts with aldehydes and ketones to form insoluble, crystalline addition products. ese products can be easily separated in pure state. These products can be decomposed by dilute mineral acid or alkali to give back original aldehyde or ketone.
Hence, it can be used for purification of aldehydes and ketones.
17. Explain what happens when
(i) ethanol vapours are passed over alumina at 600 K.
(ii) excess of ethanol is heated with conc. H2SO4 at 413 K.
(iii) phenol is treated with acetyl chloride.
Answer. (i) Dehydration will take place and ethene will be formed.
18. Explain the following phenomenon giving reasons :
(i) Rate of physical adsorption decreases with rise of temperature.
(ii) Brownian movement
(iii) Colloidal particles scatter light.
Answer. (i) Physical adsorption is an exothermic process and the equilibrium is represented as Gas (Adsorbate) + Solid (Adsorbent) ⇔ Gas adsorbed on Solid + Heat As the temperature is increased, equilibrium shifts in the backward direction, i.e., adsorption decreases (Le Chatelier’s principle).
(ii) Brownian movement is due to bombardment of colloidal particles by the molecules of the dispersion medium with unequal forces from different directions. As a result, there is a resultant force acting on them causing the particles to move in zig-zag directions.
(iii) Scattering of light by colloidal particles or Tyndall effect is observed when the diameter of the dispersed particles is not much smaller as compared to the wavelength of the light. So, colloidal particles can scatter light as they have desirable particle size of 10 Å – 10,000 Å.
19. Give reasons for the following statements :
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH3—X.
Answer. (i) The rate of SN2 reaction depends on the tendency of leaving group to leave and the tendency of attacking nucleophile to attack. Iodide is a better leaving group because of its larger size than bromide therefore, ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
(ii) (±) 2-butanol is a racemic mixture. It is a mixture which contains two enantiomers in equal proportions and thus, have zero optical rotation. Therefore, it is optically inactive.
(iii) Due to delocalisation of lone pair of electrons of the X atom over the benzene ring, C—X bond in halobenzene acquires some double bond character while in CH3—X, C—X bond is a pure single bond. Therefore, C—X bond in halobenzene is shorter than in CH3—X.
20. Consider the following chromatogram of column chromatography :
Answer the following questions :
(i) Which substance can act as stationary phase?
(ii) Which of the three components A, B and C is eluted first of all?
(iii) What is the principle of this technique?
Answer. (i) Silica gel and alumina can act as stationary phase.
(ii) Component C is weakly adsorbed component thus, travels faster than A and B components with mobile phase and hence, eluted first.
(iii) Different components of the mixture are adsorbed at different rates on the surface of adsorbent. Thus, in mobile phase each component move on stationary phase to different levels and get separated.
21. Write short notes on :
(i) Ionisation isomerism
(ii) Linkage isomerism
(iii) Coordination isomerism.
Answer. (i) Ionisation isomerism : When two complexes have same molecular formula but give different ions in solution, they are said to be ionisation isomers and the phenomenon is known as ionisation isomerism. This isomerism arises due to the exchange of ions between coordination sphere and ionisation sphere. For example,
Pentaamminebromidocobalt (III) Sulphate
Pentaamminesulphatocobalt (III) bromide
(ii) Linkage isomerism : When any mono dentate ligand in a complex has more than one donor atoms, it may be bonded to metal ion through either of the atoms and gives different isomers. For example, ligand SCN– may be bonded to the metal either through sulphur or through nitrogen to give two different isomers.
[Co(NH3)5SCN]Cl2 and [Co(NH3)5NCS]Cl2
cobalt(III) chloride cobalt(III) chloride
(iii) Coordination isomerism : This type of isomerism arises when both positive and negative ions of a compound are complex ions. There may be an exchange of ligand molecules between the two coordination spheres of two ions to give two coordination isomers.
[Cu(NH3)4 [PtCl4] and [Pt(NH3)4][CuCl4
Tetraamminecopper (II) Tetraammineplatinum(II)
tetrachloridoplatinate (II) tetrachloridocuprate(II)
22. Account for the following statements :
(i) Cerium (atomic number = 58) forms tetra positive ion, Ce4+ in aqueous solution.
(ii) The second and third members in each group of transition element have similar atomic radii.
(iii) Write the structures of
(a) MnO–4 (b) CrO5
Answer. (i) The electronic configuration of Ce
(Z = 58) is 58Ce = [Xe] 4f15d16s2
Cerium can lose four electrons (4f15d16s2)
in aqueous solution to acquire stable configuration of rare gas xenon. Due to small size and high charge, Ce4+ ion has high hydration energy
(ii) the second and third members in each group of transition elements have very similar atomic radii due to lanthanoid contraction which arises due to poor shielding effect of f-electrons.
23. A local resident, Shyam observed that some people wash clothes in the Ganga river everyday and some people wash their cars, scooters, rickshaw etc. too. At the same time, local residents consumed the same water. Shyam held a resident’s meeting to educate people to stop polluting the river water. During the meeting a local resident, Vidhu resisted this by arguing that running water of river which is in huge amount does not get polluted by washing clothes and vehicles.
(i) How does river water get polluted by washing clothes and vehicles using detergents?
(ii) What values are displayed by Shyam?
(iii) How many categories of detergents are there?
Answer. (i) Detergents get accumulated in the river water causing foaming in river and thus, water gets polluted and becomes unfit for drinking.
(ii) Leadership, awareness and concern for people and environment, are the values displayed by Shyam.
(iii) Detergents are mainly classified into three categories :
(a) Anionic detergents, which are sodium salts of sulphonated long chain alcohols or hydrocarbons.
(b) Cationic detergents, which are quarternary ammonium salts of acetates, chlorides or bromides as anions.
(c) Non-ionic detergents which do not contain any ion in their constitution.
24. (i) The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is a first order reaction.
(a) After 50 seconds at 400 K, the pressure of Cl2O7 falls from 0.062 to 0.044 atm.
Calculate the rate constant.
(b) Calculate the pressure of Cl2O7 after 100 sec of decomposition at this temperature.
(ii) One-fourth of a first order reaction is completed in 32 minutes. What is the half life period of this reaction?
Explain the following statements :
(i) Average rate of reaction does not give the true picture of the reaction rate.
(ii) A lump of coal burns at a moderate rate in air while coal dust burns explosively.
(iii) It takes more time to boil an egg or cook rice at higher altitudes.
(iv) Hydrogen and oxygen do not react at room temperature.
(v) How is rate constant related to concentration of the reactants?
Answer. (i) a ∝ P0 and (a – x) ∝ Pt
(i) Average rate of reaction is obtained by dividing the change in concentration of a particular species by the time interval. As the concentration changes are not uniform, the average rate does not give a true picture of the reaction rate.
(ii) Coal dust has greater surface area than a lump of coal. Thus, coal dust has greater ease of coming in contact with air and therefore burns explosively.
(iii) Water boils at temperature lower than 100°C due to low atmospheric pressure at higher altitudes. Thus, when eggs or rice are boiled
in water at higher altitudes, the water boils earlier i.e., at lower temperature. Hence, to cook the food completely the food should be heated for more time.
(iv) At room temperature, there is no reaction between hydrogen and oxygen as the activation energy of the reaction is very high. They require very high temperature to react which can be produced generally by electric spark or explosion.
(v) Consider the reaction
A + B → C
Then, according to rate law,
Rate = k [A]x[B]y
where k = rate constant [A] and [B] are concentrations of A and B and x and y are the order of the reaction w.r.t. A and B respectively.
Now, as the concentration of A and B changes, the rate of reaction changes such that rate constant remains unchanged. Thus, rate constant of a reaction does not depend on the concentration of the reactants.
25. (i) Arrange the following in decreasing order of property indicated :
(a) H2O, H2S, H2Se, H2Te (Boiling point)
(b) NH3, PH3, AsH3, SbH3 (Base strength)
(ii) Account for the following observations :
(a) The +5 oxidation state of phosphorus is stable but not for bismuth.
(b) Nitrogen forms a number of oxides while fewer number of oxides are possible for other elements of the group.
(c) Hydrogen fluoride has higher boiling point than hydrogen chloride.
(i) Account for the following statements :
(a) Decomposition of O3 molecule is a spontaneous process.
(b) SF6 is inert towards hydrolysis.
(c) H2S is less acidic than H2Te.
(d) SO2 is an air pollutant.
(ii) What happens when white phosphorus is heated with conc. NaOH solution in an inert gas atmosphere?
Answer. (i) (a) H2O > H2Te > H2Se > H2S Because the boiling points of hydride increase with the increase in molecular mass of the elements. H2O has exceptionally high boiling point because of intermolecular hydrogen bonding.
(b) NH3 > PH3 > AsH3 > SbH3 With the increase in size of the metal, the electron density of the lone pair gets diffused over a large region and ability to donate lone pair gets reduced hence, basic character decreases.
(ii) (a) In bismuth, due to inert pair effect only 6p3 electrons are used, thus Bi3+ is formed and Bi5+ does not exist (except in BiF5). However, in phosphorus, 3s2 and 3p3 electrons can take part in bonding thus, P5+ can exist.
(b) Nitrogen has a strong tendency to form pp-pp multiple bonds between N and O atoms and thus, it forms number of oxides which have no P, As, Sb or Bi analogues as these elements of group 15 do not have tendency to form pπ-pπ multiple bonds.
(c) In hydrogen fluoride, strong intermolecular hydrogen bonding is present, due to small size and high electronegativity of fluorine and hence, HF is a liquid having higher boiling point than HCl which is a gas due to the absence of hydrogen bonding. OR
(i) (a) For spontaneity of a reaction, ΔG must be negative. Decomposition of O3 is an exothermic process (ΔH = –ve) and occurs with increase in entropy (ΔS = +ve). These two effects reinforce each other which results in large negative Gibbs energy change according to the relation, ΔG = ΔH – TΔS Thus, its decomposition is a spontaneous reaction.
(b) In SF6, S is surrounded by 6 F– octahedrally. Therefore, attack of water molecule on S is sterically hindered. Hence, it is an inert substance.
(c) In a group, on moving down the group there is decrease in E–H bond dissociation enthalpy which increases acidic character. Thus, H2S is less acidic than H2Te.
(d) SO2 is water soluble therefore, it dissolves in rain water causing acid rain. Moreover, when released in air, it leads to several diseases like eye irritation, redness in eyes, asthma, bronchitis, etc. Thus, it is considered as an air pollutant.
(ii) When white phosphorus is heated with conc. NaOH solution in an inert gas atmosphere, phosphine gas is produced.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
26. (i) An organic compound (A) with molecular formula C9H13N dissolves in dil. HCl and releases N2 with nitrous acid giving an optically active alcohol (B). The alcohol (B) on oxidation gives dicarboxylic acid which on heating forms an anhydride. Identify the compounds
(A) and (B).
(ii) Give plausible explanations for the following :
(a) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.
(b) Tertiary amines do not undergo acylation reaction.
(i) An aromatic compound, C7H8 (A) on nitration gives (B) as a major product which on reduction with Sn/HCl gives (C). The compound (C) on treatment with NaNO2/HCl at 273 K following by subsequent treatment with CuCl/HCl gives (D). When (D) is oxidised, it forms ortho-substituted monocarboxylic acid with the formula C7H5O2Cl. Predict the structures of (A), (B), (C) and (D).
(ii) Outline the steps for the conversion of methylcyanide to methylamine
Answer. (i) (a) Since (A) releases N2 with nitrous acid, it must be a primary amine
(b) The alcohol (B) is optically active, therefore, it should contain a chiral carbon atom. Keeping in view, the molecular formula of (A) and (B) as well as the fact that (B) on oxidation gives a dicarboxylic acid, i.e., has two side chains on the benzene ring, the compound (B) can be
(c) Since the dicarboxylic acid on heating forms an anhydride, the two side chains as well as the two –COOH groups are at ortho-position to each other, i.e.,
(ii) (a) Gabriel phthalimide reaction involves the nucleophilic attack of the phthalimide anion on organic halogen compound.
Due to resonance in aryl halides, there is some double bond character in Ar – X because of which it is difficult to break. Therefore, aryl halides do not undergo nucleophilic substitution reactions easily and aromatic primary amines cannot be prepared by Gabriel synthesis.
(b) Tertiary amines do not undergo acylation reaction because they do not have hydrogen attached to nitrogen.
(i) The final product is an ortho-substituted monocarboxylic acid. Therefore, it is monocarboxylic aromatic acid with chloro group at the ortho-position. Therefore, it is
Thus, the sequence of the reactions can be visualised as follows