Oswal 36 Sample Question Papers CBSE Class 12 Chemistry Solutions
Section-A
1. (d) A-(iii), B-(iv), C-(ii), D-(i)
The structure of DNA molecule is a double helix structure. In this structure double helix are made up of polynucleotide chains which are held together by H-bonds. In these helixes the adenine (A) base is linked with thymine (T) by two H-bonds and guanine (G) is linked with cytosine (C) by three H-bonds as A ≡ T, and G ≡ C.
2. (c) C6H5CH(C6H5)Br
The structure of DNA molecule is a double helix structure. In this structure double helix are made up of polynucleotide chains which are held together by H-bonds. In these helixes the adenine (A) base is linked with thymine (T) by two H-bonds and guanine (G) is linked with cytosine (C) by three H-bonds as A ≡ T, and G ≡ C.
3. (b) charge transfer from ligand to metal
In KMnO4,
Oxidation state of Manganese = +7
Electronic configuration = [Ar]3d04s0
Due to the absence of unpaired electrons, d-d transition in KMnO4 molecule is not possible. Thus, the molecule should be colourless. Therefore, the intense purple colour of potassium permanganate is due to the charge transfer from ligand to metal, 2p(L) of O to 3d(M) of Mn.
4. (a) 15O
5. (b) 115 S cm2/mol
According to Kohlrausch’s law at infinite dilution, Molar conductivity of weak e lectrolyte CH3COOH will be,
Λ°CH3COOK = Λ°CH3COOH + Λ°KCl – Λ°HCl
= 390 + 150 – 425 = 115 S cm2/mol
6. (a) increases four times
For the reaction, A + 2B → AB2
The order of reaction w.r.t. A is 2 and w.r.t. B is 0.
Hence,
Rate = [A]2
If the concentration of A is doubled, A’ = 2A
Rate’ = [A’]2 = [2A]2 = 4 [A]2
7. (d) B < C < A
Amines have a tendency to form hydrogen bonds with each other. Hence, amines have high boiling point. Since, in primary amine, intermolecular association due to H-bonding is maximum while in tertiary amine, it is minimum therefore, the boiling point will be highest for primary amine and lowest for tertiary amines.
8. (b) 8000 cm–1
According to crystal field theory,
Δt = 4/9 Δ0
Therefore, for the given complex,
Δt = 4/9 × 18000 cm–1
= 8000 cm–1
9. (d) A = C6H5CH2Br, B = C6H5OH
In the case of alkyl aryl ethers, the compound is cleaved at the alkyl-oxygen bond due to the more stable aryl-oxygen bond. The reaction yields phenol and alkyl halide.
Therefore, for the given compound:
C6H5 – CH2 – OC6H5 + HBr → C6H5 – CH2 – Br + C6H5OH
Option (a) is the correct statement because due to the presence of electron donating alkyl groups, most alkyl amines are more basic than ammonia solution.
Option (b) is the correct statement because due to resonance, the lone pair of electrons are not available in benzyl amine. Therefore, pKb value of ethylamine is lower than benzylamine.
Option (c) is the incorrect statement because when primary amines reacts with nitrous acid, the evolution of Nitrogen gas takes place not NO2 gas.
Option (d) is the correct statement because when Hinsberg’s reagent reacts with secondary amines, sulphonamides are formed which is insoluble in an alkali.
11. (a) Yes
EMF of a cell is equal to the maximum potential difference across its electrodes, which occurs when no current is drawn through the cell. It can also be defined as the net voltage between the oxidation and reduction half-reactions.
12. (c) (i) ln A (ii) – Ea/R
Arrhenius equation is, k = Ae–Ea /RT
Taking natural logarithm of both sides of equation
ln k = – Ea/RT + ln A
According to the given equation, The plot of ln k vs 1/T gives a straight line where slope = – Ea/R and
intercept = ln A.
13. (b) Both A and R are true and R is not the correct explanation of A.
An ether has a higher volatility than an alcohol with the same molecular formula. This is because intermolecular hydrogen bonding occurs in alcohols. It raises the boiling point of alcohol while decreasing its volatility. Others, such as ethers, do not allow for such inter-molecular hydrogen bonding. Thus, both A and R are true but R is not the correct explanation of A.
14. (b) Both A and R are true and R is not the correct explanation of A.
The —NH group of each amino acid residue is hydrogen-bonded to the C = O of an adjacent turn of helix. In the α-helix, the most common ways a polypeptide chain exists and forms all possible hydrogen bonds by twisting into a right-handed screw (helix). Thus, both A and R are true but R is not the correct explanation of A.
15. (b) Both A and R are true and R is not the correct explanation of A.
Due to the presence of large number of unpaired electrons, actinides are paramagnetic in nature. Moreover, Magnetic moment values of actinides are lesser than the theoretically predicted values. This is because 5f electrons of actinides are less effectively shielded which result in quenching of orbital contribution. Thus, both A and R are true but R is not the correct explanation of A.
16. (a) Both A and R are true and R is the correct explanation of A.
The presence of electron donating alkyl group makes the electron density on the alkylamine’s nitrogen (+I effect increases) greater than the nitrogen of ammonium. Tertiary amines have three alkyl groups which cause +I effect due to which basic characters are more in it. Thus, both A and R are true and R is the correct explanation of A.
Section-B
$$\text{17. Given, A → Product}\\[5px]t = [R_0]\\t\space_{80\%}=\frac{[R_0]-80}{100[R_0]}=\frac{20}{100[R_0]}\\[5px]\text{For the first order reaction,}\\[5px]\text{half life t}_\frac{1}{2}=\frac{0.693}{k}\\[5px]\text{k}=\frac{0.693}{k}= 0.01 min.^{–1}\\[5px]\text{Now,}\space k =\frac{2.303}{t}log\frac{[R_0]}{[R]}\\[5px]\text{On putting the values in equation, we get}\\[5px]0.01=\frac{2.303}{t}log\frac{[R_0]}{\frac{20}{100[R_0]}}\\[5px]t=\frac{2.303}{0.01}log\frac{100}{2}\\[5px]\text{t = 230.3 log 5 = 160.9 min.}$$
18. (a) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.
(b) Carbohydrates which reduce Fehling’s solution and Tollens’ reagent are reducing sugars.
When glucose reacts with Fehling’s reagent, it get oxidised to Gluconic acid.
(a) D-glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.
(b) On oxidation with nitric acid, D-glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.
19. (a) When electrophilic substitution reaction takes place, then ortho and para products are formed among them para is considered as the major product and ortho as a minor product because of the steric hindrance. In the ortho product, the groups are present next to each other this causes the hindrance and the group tries to attach at the para position. Hence para product predominates over ortho product.
(b) A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such a mixture is known as racemic mixture. During the SN1 mechanism, intermediate carbocation formed is sp2 hybridized and is planar in nature. This allows the attack of nucleophile from either side of the plane resulting in the racemic mixture.
20. Corrosion is a natural process in which a refined metal is converted to a more chemically stable form, such as oxide, hydroxide, or sulphide.
Corrosion of zinc:
Anode: Zn(s) → Zn2+(aq) + 2e–
Cathode: O2(g) + 2H2O(l) + 4e– → 4OH–(aq.)
Overall: 2Zn (s) + O2(g) + 2H2O(l) → 2Zn2+(aq.) + 4OH–(aq.)
21. (a) A catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst.
The catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier. Therefore, the rate of reaction will increase on addition of a catalyst.
(b) For a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled. At lower temperatures the kinetic energy of molecules decreases thereby the collisions decrease resulting in a lowering of rate of reaction.
2-Methylpropene will be formed.
(c) On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring. This reaction is known as Reimer - Tiemann reaction.
o-Hydroxybenzaldehyde (or salicylaldehyde) will be formed.
23. Oxidation state of Mn is x + 6(–1) = –3
x = +3
Electronic configuration of Mn (in ground state) = [Ar]3d54s2
Electronic configuration of Mn3+ (in excited state) = [Ar]3d4
Six pairs of electrons, one from each CN molecule, occupy the six hybrid orbitals. Thus, the complex has
octahedral geometry and is paramagnetic because of the absence of unpaired electron. In the formation
of this complex, since the inner d-orbital (3d) is used in hybridisation, the complex, [Mn(CN)6]3– is called
an inner orbital or low spin or spin paired complex.
(a) Type of hybridization – d2sp3
(b) Magnetic moment value –√n(n+2)=√2(2+2)= 2.87 BM
(c) Type of complex – inner orbital
24. (a) Henry’s law states that: ‘‘The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution’’. Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure under water. Increased pressure, increases the solubility of atmospheric gases in blood.
When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
(b) According to henry’s law :
p = KHX
Mole fraction of argon in water x = p/KH
$$\frac{6}{40}×10^3=1.5×10^{-4}$$
25. Step I: Molecular formula of [A] C4H8Cl2, hence it must be a substituted alkane.
Step II: It gives C4H8O [B] on hydrolysis and [B] does not gives a positive Tollen’s test. Hence [B] must be a ketone.
26. ‘X’ must be chloroform (CHCl3) as it produces phosgene, a highly poisonous gas on being exposed to air and light.
$$\underset{[A]}{2CHCl_3+o_2}\space\ce{->[Lihgt] }\underset{\text{Phosgene}}{2COCl_2}+2HCl$$
27. The structures of different isomers corresponding to the molecular formula, C3H9N are given below:
28. The central metal ion in all the three complexes is the same with same oxidation number of (+2); Ni+2. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series is as follows:
H2O < NH3 < NO2–
Thus, the amount of crystal-field splitting observed will be in the following order:
Δa(H2O) Δa<(NH3 )Δa<(NO2)
Hence, the wavelenghts of absorption in the visible region will be in the order:
[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4–
as wavelength l is inversely proportional to the energy of the radiation.
$$E ∝\frac{1}{\lambda}.$$
Section-D
29. (a) In a segment of DNA, number of Adenine = 100.
So the number of thymine = 100.
And number of cytosine bases = 150.
So the number of guanine will be = 150
Therefore, Total nucleotides = 100 + 100 + 150 + 150 = 500
(b) Scientist studied the nucleotide composition of DNA. It was the same so they concluded that the samples belong to same species.
(c) Percentage of Adenine = Thymine = 20% But the percentage of Guanine is not equal to Cytosine. So, double helix is ruled out. The bases pairs are ATGC and not AUGC so it is not RNA. The virus is a single helix DNA virus.
OR
According to Chargaff’s rule, all double helix DNA will have the same amount of Adenine and thymine as well as cytosine will be same amount as guanine. If this is not the case then the helix is single stranded.
30. (a) For a pure liquid, The melting point of ice is the freezing point of water Using the depression in freezing point property in this case.
3rd reading for 0.5 g, there has to be an increase in depression of freezing point and therefore decrease in freezing point. So, also decrease in melting point when amount of salt is increased but the trend is not followed on this case.
(b) By collecting two sets of readings, help to avoid error in data collection and give more objective data.
(c) Expression for depression in freezing point - ΔTf = iKf m $$\text{For glucose molecule}\\[5px]\Delta T_f (glucose) = 1 × K_f ×\frac{0.6 ×1000}{180×10}…(i)\\[5px]\text{For NaCl molecule,}\\[5px]\Delta T_f (NaCl) = 2 × K_f ×\frac{0.6 ×1000}{58.5×10}\\[5px]3.8 = 2 × K_f ×\frac{0.6 ×1000}{58.5×10}…(ii)\\[5px]\text{Divide equation (i) by (ii),}\\[5px]\frac{\Delta T_f (glucose)}{3.8}=\frac{585}{2×180}\\[5px]\Delta T_f (glucose) = 0.62\\[5px]\text{Freezing point or Melting point = – 0.62°C}$$
OR
Since, depression in freezing point is directly proportional to molality (mass of solute when the amount of solvent remains same).
0.3 g depression is 1.9°C
0.6 g depression is 3.8°C
1.2 g depression will be 3.8 × 2 = 7.6°C
Section-E
31. (a) In a mercury cell,
Anode: Zinc – mercury amalgam
Cathode: A paste of HgO and carbon
Electrolyte: A paste of KOH and ZnO
Anode: Zn(Hg) + 2OH– → ZnO(s) + H2O + 2e–
Cathode: HgO + H2O + 2e– → Hg(l) + 2OH–
The overall reaction is represented by Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
The cell potential is approximately 1.35 V and remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time.
(b) Electrolysis of aqueous KCl,
KCl(aq) → K+(aq) + Cl–(aq)
Cathode: H2O(l) + e– → ½ H2(g) + OH–(aq)
Anode: Cl–(aq) → ½Cl2 (aq) + e–
Net reaction: KCl(aq) + H2O(l) → K+(aq) + OH–(aq) + ½ H2(g) + ½ Cl2(g)
Hence, Hydrogen gas will be released at cathode, chlorine gas at anode and KOH will remain in the
solution as a byproduct.
(c) Electrode reaction: H+ + e– → ½ H2
Given,
E (H+/H2 ) = – 0.59 V and we know that, E°(H+/H2 ) = 0 V
Applying Nernst equation
$$E_{(H^+/H_2 )}= E°_{(H^+ /H_2)}^-–\frac{0.059}{n}log\frac{[H_2]^{\frac{1}{2}}}{[H^+]}\\[5px]\text{Here, n = 1 and [H2] =1 bar}\\[5px]\text{ Putting all the given values,}\\[5px]– 0.59 = 0 –\frac{0.059} {n}log\frac{[H_2]^{\frac{1}{2}}}{[H^+]}\\[5px]– 0.59 = – 0.059 (–log [H^+])\\[5px]Since,\space pH = – log [H^+]\\[5px]Hence, 1 0 = – log [H^+] = pH\\[5px]pH = 10$$
OR
(a) Conductors are those substances that conducts electricity. These substances have high value of conductivity. Out of the two given substances, A has a higher values of conductivity is 5.9 × 103 S/m). As metals are good conductors of electricity and copper is a metal. Hence, substance A should be a metal i.e., copper.
(b) In molten MgCl2, production of metal magnesium takes place as follows:
Mg2+ + 2e– → Mg
1 mole of magnesium ions gains two moles of electrons or 2F to form 1 mole of Mg
Therefore, 24 g Mg will requires 2 F electricity
And 4.8 g Mg requires 2 ×(4.8/2.4)= 0.4 F
= 0.4 × 96500 = 38600C
For CaCl2,
Ca2+ + 2e– → Ca
2 F electricity is required to produce 1 mole = 40 g Ca
0.4 F electricity will produce 8 g Ca
(c) Sn(s) + 2Cu2+(aq) → Sn2+(aq) + 2Cu+(s)
Here n = 2,
Sn2+(aq) + 2e– → Sn(s);E ° (Sn+2 /Sn)= – 0.14V
Cu2+(aq) + e– → Cu+(aq); E °(Cu+2 /Cu)= – 0.15V
Therefore,
E° cell = E° cathode – E° anode
= 0.15 – (–0.14)
= 0.29V
ΔG° = – nF °Ecell = – 2 × 96500 × 0.29 = 55970 J/mol
32. The hydrocarbon A with molecular formula similar to alkene or cycloalkane. But since, A undergoes ozonolysis reaction, it must be an alkene. Compound B and C must be methylated aldehyde or ketones as both of them shown iodoform reaction.
Now, since compound B shows silver mirror test, it must be methylated aldehyde and C must be methylated ketone
OR
(a) (i) In aldehydes the intermolecular forces of attraction are dipole-dipole interactions, which are far weaker than the hydrogen bondings present in the alcohols. Hence, aldehydes are more volatile than alcohols.
(ii) The carbonyl group of ketones and aldehydes is more electrophilic than in carboxylic acids. This is because the lone pairs on oxygen atom attached to hydrogen atom in the −COOH are involved in resonance thereby making the carbon atom less electrophilic.
The carboxylic group in carboxylic acids is a resonance hybrid of the given structures:
Hence, the carbonyl carbon of the resonance hybrid, of structure (I), (II) and (III) is less electrophilic than the carbonyl carbon of aldehydes and ketones. Therefore, it does not give some characteristic reactions 2-4 DNP, oxime test etc., like other carbonyl compounds.
(iii) On treating benzaldehyde with HCN, nucleophilic addition take place and two isomers of hydroxyphenyl- acetonitrile are formed.
The carbon in this intermediate is a chiral carbon and is sp3 hybridised.
These two isomers are mirror images of each other called enantiomers (compounds with nonsuperimposable mirror images). They have same chemical and physical properties so they cannot be separated even by fractional distillation.
(b) (i) Steric hindrance at the carbonyl carbon and presence of more electron-donating groups i.e., (+ I) effect, lowers down the reactivity towards a nucleophilic reaction.
The +I effect of the alkyl group increases in the order:
Ethanal < Propanal < Propanone < Butanone
This is because the electron density at the carbonyl carbon increases with the increase in the + I effect. Thus the affinity of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is :
Butanone < Propanone < Propanal < Ethanal
(ii) Presence of electron with drawing group (–I) effect, increases the reactivity for nucleophilic reactions.
The + I effect is more in ketone than in aldehyde. Therefore acetophenone is the least reactive. Among aldehydes, the + I effect is the highest in p-tolualdehyde because of the presence of the electron-donating – CH3 group and the lowest in p-nitrobenzaldehyde because of the presence of the electron-withdrawing – NO2 group. Hence, the increasing order of the reactivities of the given compounds is:
Acetophenone < p-tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde.
33. (a) The iodide ion being a bulky ion, and strong reducing agent reduces copper iodide to cupric iodide.
2CuI2 → 2CuI + I2
Or the cupric ions oxidised the iodide ions into the molecular iodine and itself get reduced to cuprous ions. Hence, all copper halides known except that copper iodide.
(b) Vanadium having atomic number 23 and electronic configuration - [Ar]4s23d3.
Electronic configuration of V+2 – [Ar]3d3.
It has half filled t2g configuration due to which it quite stable and hence V+3 readily reduces to V+2.
(c) It reacts with the titration indicator potassium permanganate (KMnO4) and oxidises it resulting into the release of chlorine gas. At the end of the process, the final value of the indicator used can differ greatly from the total value of the reaction. This will cause the entire experiment to be disrupted. In essence, in volumetric analysis, HCl cannot be used to acidify KMnO4.
(d) Across the period, among d-block elements, the atomic size first decreases with increase in atomic number, then remains constant and then slightly increases.
At the end of each period among d-block elements, the atomic size slightly increases. It is because of the presence fully filled d-orbital, when pairing takes place,the e– – e– repulsion between the electrons increases.
(e) The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. Increasing the pH (in basic solution)of dichromate ions a colour change from orange to yellow is observed as dichromate ions change to chromate ions.
(f) General electronic configuration of transition elements: (n – 1) d1–10 ns1–2
(g) Copper.
CBSE 36 Sample Question Papers Science Stream (PCM)
All Subjects Combined for Class 12 Exam 2024
CBSE 36 Sample Question Papers Science Stream (PCB)
All Subjects Combined for Class 12 Exam 2024
The dot mark ◉ field are mandatory, So please fill them in carefully
To download the Sample Paper (PDF File), Please fill & submit the form below.
