Class 12 Chemistry Sample Paper 2023

Chemistry Class 12 Sample Paper 2023: CBSE has released Class 12 Chemistry Sample Paper 2023 along with solution and marking scheme. Now the students must be buckling up to begin their revision for Chemistry Exam and solving the Class 12 Chemistry Sample Paper could be the best way to boost up yourself and enhance your confidence for the real examination. Are you searching for the Class 12 Chemistry Sample Paper 2023 to practice for your Chemistry Exam as per the new CBSE Exam Pattern? Then you are in the right place, as here we have provided the CBSE Class 12 Chemistry Sample Paper that is issued by the CBSE on their official website as per the latest exam pattern. 

Chemistry Sample Paper Class 12 2023

Central Board of Secondary Education has also released the blueprint and marking scheme for the Class 12 Chemistry Board Exam 2023. In exam is divided into five sections with 35 questions for 70 marks and the complete exam pattern that has been followed is as follows- 

1. Section A consists of 18 Questions carrying 1 mark each [Multiple Choice Questions]. 

2. Section B consists of 7 questions carrying 2 marks each [Very Short Answer Questions]. 

3. Section C consists of 5 questions carrying 3 marks each [Short Answer Questions].

4. Section D consists of 2 Questions carrying 4 marks each [Case Based Questons].

5. Section E consists of 2 Questions carrying 5 marks each [Long Answer Questions].

Chemistry Class 12 Sample Paper 2023 PDF

The students who are doing their revision for the upcoming CBSE Class 12 Chemistry Exam must download the official sample paper PDF along with its solution from the below link. With this CBSE Class 12 Chemistry Sample Paper, it would be easier for the students to attempt the types of question that is going to be asked by CBSE in the Board Exam 2023. The best way is to solve the sample just like you are sitting in the examination hall and then analyze your attempt with the solution provided by CBSE the link for the same is given below. 

CBSE Class 12 Chemistry Sample Paper 2023- Download PDF

CBSE Class 12 Chemistry Sample Paper Solution

CBSE Class 12 Chemistry Sample Paper and Solution 2023

Chemistry is one of the main subjects of the class 12th boards examination, it consists of a total of 100 Marks which further comprises of 70 Marks for Theory and 30 Marks for the Practical test. CBSE has recently released the official sample paper, Below are the sample question from the official paper. Practice them thoroughly so as to score good marks in chemistry, as chemistry is considered one of the scoring subjects in board exams.

Section A 

The following questions are multiple-choice questions with one correct answer. Each question carries 1 mark. There is no internal choice in this section. 

Question1. The major product of acid-catalyzed dehydration of 1-methyl cyclohexanol is:

a. 1-methylcyclohexane

b. 1-methylcyclohexene

c. 1-cyclohexylmethanol

d. 1-methylcyclohexane 

Solution:  b. 1-methyl cyclohexene. According to Saytzeff's rule i.e, highly substituted alkene is the major product. Here dehydration reaction takes place, an alkene is formed due to the removal of a water molecule. 

Question 2. Which one of the following compounds is more reactive towards SN1 reaction?

a. CH2=CHCH2Br 

b. C6H5CH2Br 

c. C6H5CH (C6H5)Br 

d. C6H5CH(CH3) Br

Solution:  c. C6H5CH (C6H5)Br

C6H5CH (C6H5)+ carbocation formed is more stable.

Question 3. KMnO4 is colored due to:

a. d-d transitions

b. charge transfer from ligand to metal

c. unpaired electrons in d orbital of Mn

d. charge transfer from metal to ligand

Solution: b. charge transfer from ligand to metal The Mn atom in KMnO4 has a +7 oxidation state with electron configuration [Ar]3d 04s0 Since no unpaired electrons are present, d−d transitions are not possible. The molecule should, therefore, be colorless. Its intense purple due to L→M (ligand to metal) charge transfer 2p(L) of O to 3d(M) of Mn.

Question 5. The molar conductivity of CH3COOH at infinite dilution is 390 Scm²/mol. Using the graph and given information, the molar conductivity of CH3COOK will be:

a. 100 Scm²/mol

b. 115 Scm²/mol

c. 150 Scm²/ mol

d. 125 Scm²/mol

Solution:  b. 115 Scm² /mol ΛOCH3COOK = ΛO CH3COOH +ΛOKCl - ΛOHCl = 390 +150-425 = 115 Scm²/mol

Question 6. For the reaction, A +2B - AB2, the order w.r.t. reactant A is 2 and w.r.t. reactant B. What will be the change in the rate of reaction if the concentration of A is doubled and B is halved?

a. increases four times

b. decreases four times

c. increases two times

d. no change

Solution: a. increases 4 times, Rate = [A]² If [A] is doubled then Rate’ = [2A]²= 4 [A]² = 4 Rate

Question 7. Arrange the following in the increasing order of their boiling points: A: Butanamine, B: N, N-Dimethylethanamine, C: N- Etthylethanaminamine

a. C<B<A

b. A<B<C

c. A<C<B

d. B<C<A

Solution: d. B<C<A. In primary amine, intermolecular association due to H-bonding is maximum while in tertiary it is minimum. 

Question 8. The CFSE of [CoCl6]3- is 18000 cm-1 the CFSE for [CoCl4] - will be:

a. 18000 cm-1

b. 8000cm-1

c. 2000 cm-1

d. 16000 cm-1

Solution: b. 8000 cm-1 ∆t =(4/9 ) x 18000cm-1 =8000 cm-1

Question 9. What would be the major product of the following reaction?

C6H5 -CH2-OC6H5 + HBr = A + B

a. A= C6H5CH2OH , B= C6H6

b. A=C6H5CH2OH ,B= C6H5Br

c. A=C6H5CH3 ,B= C6H5Br

d. A=C6H5CH2Br , B= C6H5OH

Solution: d. A.=C6H5CH2Br , B = C6H5OH, C6H5CH2OC6H5 H+ C6H5CH2OC6H5

Question 10. Which of the following statements is not correct for amines?

a. Most alkyl amines are more basic than ammonia solution.

b. pKb value of ethylamine is lower than benzylamine.

c. CH3NH2 in reaction with nitrous acid releases NO2 gas.

d. Hinsberg’s reagent reacts with secondary amines to form sulphonamides.

Solution: c. CH3NH2 in reaction with nitrous acid releases NO2 gas Wrong statement. The evolution of nitrogen gas takes place.

Question 11. Which of the following tests/ reactions is given by aldehydes as well as ketones?

a. Fehling’s test

b. Tollen’s test

c. 2,4 DNP test

d. Cannizzaro reaction

Solution: c. 2,4 DNP test Fehling’s, Tollen’s, and Cannizzaro's reactions are shown by alcohols only. 

Question12. Arrhenius's equation can be represented graphically as follows:

The (i) intercept and (ii) slope of the graph are:

a. (i) ln A (ii) Ea/R

b. (i) A (ii) Ea

c. (i)ln A (ii) - Ea/R

d. (i) A (ii) -Ea 

Solution: 12. c.(i)ln A (ii) - Ea/R

Question13. 13. The number of ions formed on dissolving one molecule of FeSO4.(NH4)2SO4.6H2O in water is:

a. 3

b. 4

c. 5

d. 6

Solution: 13. c. 5 1Fe2+, 2 SO4 2- and 2 NH4 + ions

Question14. The oxidation of toluene to benzaldehyde by chromyl chloride is called

a. Etard reaction

b. Riemer-Tiemann reaction

c. Stephen’s reaction

d. Cannizzaro’s reaction

Solution: A. Etard's reaction

Question15. Given below are two statements labeled Assertion (A) and Reason (R)

Assertion (A): An ether is more volatile than alcohol of comparable molecular mass. 

Reason (R): Ethers are polar in nature. 

Select the most appropriate answer from the options given below:

a. Both A and R are true and R is the correct explanation of A 

b. Both A and R are true but R is not the correct explanation of A. 

c. A is true but R is false. 

d. A is false but R is true.

Solution: b Both A and R are true but R is not the correct explanation of A. A and R are two different statements about ethers The correct reason is that hydrogen bonding does not exist amongst ether molecules

Question16. Given below are two statements labeled Assertion (A) and Reason (R)

Assertion (A): Proteins are found to have two different types of secondary structures viz alpha-helix and beta-pleated sheet structure.

Reason (R): The secondary structure of proteins is stabilized by hydrogen bonding.

Select the most appropriate answer from the options given below:

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true but R is not the correct explanation of A.

c. A is true but R is false.

d. A is false but R is true.

Solution: b Both A and R are true but R is not the correct explanation of A.  

Question17. Given below are two statements labeled Assertion (A) and Reason (R)

Assertion: Magnetic moment values of actinides are lesser than the theoretically predicted values.

Reason: Actinide elements are strongly paramagnetic.

Select the most appropriate answer from the options given below:

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true but R is not the correct explanation of A.

c. A is true but R is false.

d. A is false but R is true.

Solution: b Both A and R are true but R is not the correct explanation of A. The magnetic moment is less as the 5f electrons of actinides are less effectively shielded which results in quenching of orbital contributions, they are strongly paramagnetic due to the presence of unpaired electrons

Question18. Given below are two statements labeled Assertion (A) and Reason (R)

Assertion (A): Tertiary amines are more basic than corresponding secondary and primary amines in a gaseous state.

Reason (R): Tertiary amines have three alkyl groups that cause the +I effect.

Select the most appropriate answer from the options given below:

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true but R is not the correct explanation of A.

c. A is true but R is false.

d. A is false but R is true. 

Solution: Both A and R are true and R is the correct explanation of A.

Section B

This section contains 7 questions with internal choice in two questions. The following questions are very short answer types and carry 2 marks each.

Question 19. A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given: log 5 =0.6990, log 8 = 0.9030, log 2 = 0.3010)

Solution: 19.Half-life t1/2 = 0.693 /k

k= 0.693/69.3 = 1/100 = 0.01 min-1

For first-order reaction

𝑘 = 2.303/𝑡 𝑙𝑜𝑔 [𝑅𝑜]/ [𝑅]

𝑡 = 2.303 /0.01 𝑙𝑜𝑔 100/20 

𝑡 = 230.3 log 5 (log 5 =0.6990)

 t= 160.9 min

Question20. Account for the following: 

a. There are 5 OH groups in glucose 

b. Glucose is a reducing sugar 

Solution: 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 

OR

Question20. What happens when D – glucose is treated with the following reagents

a. Bromine water

b. HNO3

Question 21. Give a reason for the following: 

a. During the electrophilic substitution reaction of haloarenes, the para-substituted derivative is the major product. 

b. The product formed during SN 1 reaction is a racemic mixture.

Solution:  a. At the ortho position, higher steric hindrance is there, hence para isomer is usually predominate and is obtained in the major amount. 

b.During the SN 1 mechanism, intermediate carbocation formed is sp2 hybridized and planar in nature. This allows the attack of nucleophiles from either side of the plane resulting in a racemic mixture.

OR 

Question 21. Give a reason for the following: 

a. Name the suitable alcohol and reagent, from which 2-Chloro-2-methyl propane can be prepared. 

b. Out of the Chloromethane and Fluoromethane, which one has a higher dipole moment and why? 

Solution:  a. Tert butyl alcohol or 2-methyl propane-2-ol using Lucas reagent, a mixture of conc. HCl and ZnCl2 the reaction will follow the SN 1 pathway. 

b.Chloromethane is having higher dipole moment. Due to the smaller size of fluorine, the dipole moment of flouromethane is comparatively lesser. 

Question22.The formula Co(NH3)5CO3Cl could represent a carbonate or a chloride. Write the structures and names of possible isomers. 

Solution: [Co(NH3)5CO3]Cl and [Co(NH3)5Cl]CO3 

Pentaaminecarbonatocobalt(III)chloride 

Pentaaminechloridocobalt(III)carbonate 

Question 23. Corrosion is an electrochemical phenomenon. The oxygen in moist air reacts as follows:

O2(g) + 2H2O(l) + 4e– → 4OH– (aq).

Write down the possible reactions for corrosion of zinc occurring at the anode, cathode, and overall reaction to form a white layer of zinc hydroxide. 

Solution: Anode: Zn (s)→ Zn2+ (aq) + 2 e- 

Cathode: O2(g) + 2H2O(l) + 4e– → 4OH–(aq). 

Overall: 2 Zn (s) + O2(g) + 2H2O(l) →2 Zn2+ (aq) + 4OH– (aq)

2 Zn (s) + O2(g) + 2H2O(l) →2 Zn(OH)2 (ppt) 

Question24. Explain how and why will the rate of reaction for a given reaction be affected when

a. a catalyst is added 

b. the temperature at which the reaction was taking place is decreased

Solution: The rate of reaction will increase. The catalyst decreases the activation energy of the reaction therefore the reaction becomes faster. (1/2+1/2) b.The rate of reaction will decrease. At lower temperatures, the kinetic energy of molecules decreases thereby the collisions decrease resulting in a lowering of the rate of reaction.

Question 25. Write the reaction and IUPAC name of the product formed when 2-Methylpropanal (isobutyraldehyde) is treated with ethyl magnesium bromide followed by hydrolysis.

Solution:

Section C

This section contains 5 questions with internal choice in two questions. The following questions are short answer types and carry 3 marks each. 

Question 26. Write the equations for the following reaction:

a. Salicylic acid is treated with acetic anhydride in the presence of conc. H2SO4

b. Tert butyl chloride is treated with sodium ethoxide.

c. Phenol is treated with chloroform in the presence of NaOH

Solution:(i) Aspirin is formed

(iii) o-hydroxybezaldehyde will be formed

Question 27. Using the Valence bond theory, explain the following in relation to the paramagnetic complex [Mn(CN)6]3-

a. type of hybridization

b. magnetic moment value

c. type of complex – inner, outer orbital complex

Solution: [Mn(CN)6] 3- Mn = [Ar] 3d54s2 Mn3+ = [Ar] 3d4 Mn (ground state)

Type of hybridization – d2sp3

Magnetic moment value – √n(n+2) = √(2(2+2)) = 2.87 BM (n= no. of unpaired electrons) 

Type of complex – inner orbital

Question 28. Answer the following questions:

a. State Henry’s law and explain why are the tanks used by scuba divers filled with air diluted with helium (11.7% helium, 56.2% nitrogen, and 32.1% oxygen).

b. Assume that argon exerts a partial pressure of 6 bar. Calculate the solubility of argon gas in water. (Given Henry’s law constant for argon dissolved in water, KH = 40kbar)

Solution:a. Henry’s law: the partial pressure of the gas in the vapor phase (p) is proportional to the mole fraction of the gas (x) in the solution.  The pressure underwater is high, so the solubility of gases in the blood increases. When the diver comes to the surface the pressure decreases so does the solubility causing bubbles of nitrogen in the blood, to avoid this situation and maintain the same partial pressure of nitrogen underwater too, the dilution is done. 

b. p = KH x mole fraction of argon in water x = p/k = 6/ 40 x103 = 1.5 x10-4 

Question 29. Give reasons for any 3 of the following observations:

a. Aniline is acetylated before the nitration reaction.

b. pKb of aniline is lower than the m-nitroaniline.

c. Primary amine on treatment with benzene sulphonyl chloride forms a product that is soluble in NaOH however secondary amine gives a product that is insoluble in NaOH.

d. Aniline does not react with methyl chloride in the presence of an anhydrous AlCl3 catalyst.

Solution:(any 3)

a. Aniline is acetylated, before the nitration reaction in order to avoid the formation of tarry oxidation products and protect the amino group, so that p -nitro derivative can be obtained as the major product. 

b.pKb of aniline is lower than the m-nitro aniline.The basic strength of aniline ismore than m-nitroaniline. pkb value is inversely proportional to basic strength. The presence of an Electron withdrawing group decreases basic strength. 

c. Due to the presence of acidic hydrogen in the N-alkylbenzenesulphonamide formed by the treatment of primary amines. (1)

d. Aniline does not react with methyl chloride in the presence of an AlCl3 catalyst, because aniline is a base and AlCl3 is Lewis acid which leads to the formation of salt. 

Question 30. a. Identify the major product formed when 2-cyclohexylchloroethaneundergoes a dehydrohalogenation reaction. Name the reagent which is used to carry out the reaction.

b. Why are haloalkanes more reactive towards nucleophilic substitution reactions than haloarenes and vinylic halides?

Solution: a. The major product formed when 2-cyclohexylchloroethane undergoes dehydrohalogenation reaction is 1- cyclohexylethene. The reagent which is used to carry out the reaction is ethanolic KOH. (1+1) 

b. Haloalkanes are more reactive than haloarenes and vinylic halides because of the presence of a partial double bond character C-X bond in haloarenes and vinylic halides. Hence they do not undergo nucleophilic reactions easily. 

OR

a. Name the possible alkenes which will yield 1-chloro-1-methylcyclohexane in their reaction with HCl. Write the reactions involved.

b. Allyl chloride is hydrolyzed more readily than n-propyl chloride. Why?

Solution: b. Allyl chloride shows high reactivity as the carbocation formed in the first step is stabilized by resonance while no such stabilization of carbocation exists in the case of n-propyl chloride.

Section D

The following questions are case-based questions. Each question has an internal choice and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow. 

Question 31. Strengthening the Foundation: 

Chargaff Formulates His "Rules" Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, DNA was first identified in the late 1860s by Swiss chemist Friedrich Miescher. Then, in the decades following Miescher's discovery, other scientists--notably, Phoebus Levene and Erwin Chargaff--carried out a series of research efforts that revealed additional details about the DNA molecule, including its primary chemical components and the ways in which they joined with one another. Without the scientific foundation provided by these pioneers, Watson and Crick may never have reached their groundbreaking conclusion of 1953: that the DNA molecule exists in the form of a three-dimensional double helix. 

Chargaff, an Austrian biochemist, as his first step in this DNA research, set out to see whether there were any differences in DNA among different species. After developing a new paper chromatography method for separating and identifying small amounts of organic material, 

Chargaff reached two major conclusions: (i) the nucleotide composition of DNA varies among species. (ii) Almost all DNA, no matter what organism or tissue type it comes from maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is similar to the amount of thymine (T), and the amount of guanine (G) approximates the amount of cytosine (C). 

In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. This conclusion is now known as "Chargaff's rule." Chargaff’s rule is not obeyed in some viruses. These either have single-stranded DNA or RNA as their genetic material. 

Answer the following questions: 

a. A segment of DNA has 100 adenine and 150 cytosine bases. What is the total number of nucleotides present in this segment of DNA? 

b. A sample of hair and blood was found at two sites. Scientists claim that the samples belong to the same species. How did the scientists arrive at this conclusion? 

c. The sample of a virus was tested and it was found to contain 20% adenine, 20% thymine, 20 % guanine, and the rest cytosine. Is the genetic material of this virus (a) DNA- double helix (b) DNA-single helix (c) RNA? What do you infer from this data? 

Solution: a. A = 100 so T = 100 C=150 so G = 150 Total nucleotides = 100+100+150+150 =500 

 b. They studied the nucleotide composition of DNA. It was the same so they concluded that the samples belong to the same species.  

c. A = T = 20% But G is not equal to C so double helix is ruled out. The base pairs are ATGC and not AUGC so it is not RNA The virus is a single helix DNA virus.

 OR

Question 31. How can Chargaff’s rule be used to infer that the genetic material of an organism is double-helix or single-helix? 

Solution:  According to Charagraff's rule, all double helix DNA will have the same amount of A and T as well as C will be the same amount as G. If this is not the case then the helix is single-stranded. 

Question 32. Henna is investigating the melting point of different salt solutions. She makes a salt solution using 10 mL of water with a known mass of NaCl salt. She puts the salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts. She repeats each experiment.

S.NoMass of the salt used in gMelting point in 0C
  Readings Set 1Readings Set 2
1.0.3-1.9-1.9
2.0.4-2.5-2.6
3.0.5-3.0-5.5
4.0.6-3.8-3.8
5.0.8-5.1-5.0
6.1.0-6.4-6.3

Assuming the melting point of pure water as 0oC, answer the following questions:

a. One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.

b. Why did Henna collect two sets of results?

c. In place of NaCl, if Henna had used glucose, what would have been the melting point of the solution with 0.6 g glucose in it? 

Solution: The melting point of ice is the freezing point of water. We can use the depression in freezing point property in this case

a. 3rd reading for 0.5 g there has to be an increase in the depression of freezing point and therefore decrease in freezing point so also decrease in melting point when the amount of salt is increased but the trend is not followed in this case. 

b.two sets of reading help to avoid errors in data collection and give more objective data. 

c. ΔTf (glucose) = 1 x Kf x 0.6 x 1000 /180 x 10 

 ΔTf (NaCl) = 2 x Kf x 0.6 x 1000 /58.5 x 10 

3.8 = 2 x Kf x 0.6 x 1000 /58.5 x 10

 Divide equation 1 by 2

 ΔTf (glucose)/ 3.8 = 58.5 / 2x 180 

 ΔTf (glucose = 0.62 Freezing point or Melting point = - 0.62 °C

OR

Question 32. What is the predicted melting point if 1.2 g of salt is added to 10 mL of water? Justify your answer. 

Solution: Depression in freezing point is directly proportional to molality (mass of solute when the amount of solvent remains the 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 x2 = 7.6° C

Section E

The following questions are long answer types and carry 5 marks each. Two questions have an internal choice.

Question33. a. Why does the cell voltage of a mercury cell remain constant during its 10 lifetimes?

b. Write the reaction occurring at the anode and cathode and the products of electrolysis of aq KCl.

c. What is the pH of the HCl solution when the hydrogen gas electrode shows a potential of -0.59 V at standard temperature and pressure?

Solution: The cell potential remains constant during its life as the overall reaction does not involve any ion in a solution whose concentration can change during its lifetime.

b. KCl (aq) à K+ (aq) + Cl- (aq) cathode: H2O(l) + e- à ½ H2 (g) + OH- (aq) (1/2) anode: Cl- (aq) à ½ Cl2 (aq) + e- (1/2) net reaction: KCl (aq) + H2O (l) à K+ (aq) +OH- (aq) + ½ H2 (g) + ½ Cl2 (g) 

c. Given, the potential of hydrogen gas electrode = −0.59 V 

Electrode reaction: H+ + e– → 0.5 H2 

Applying Nernst equation, E (H+ /H2) = Eo (H+ /H2) – 0.059 log [H2] 1/2 n [H+ ] (1) Eo (H+ /H2) = 0 V E (H+ /H2) = -0.59 V n = 1 [H2] =1 bar −0.59 = 0 - 0.059 ( - log [H+ ] ) (1/2) −0.59 = −0.059pH 

∴ pH = 10 

OR

a. Molar conductivity of substance “A” is 5.9×103 S/m and “B” is 1 x 10-16 S/m. Which of the two is most likely to be copper metal and why?

b. What is the quantity of electricity in Coulombs required to produce 4.8 g of Mg from molten MgCl2? How much Ca be produced if the same amount of electricity was passed through molten CaCl2? (Atomic mass of

Mg = 24 u, atomic mass of Ca = 40 u).

c. What is the standard free energy change for the following reaction at room temperature? Is the reaction spontaneous?

Sn(s) + 2Cu2+ (aq) à Sn2+ (aq) + 2Cu+(s)

Solution: a. “A” is copper, metals are conductors and thus have a high value of conductivity. 

b. Mg2+ + 2e-à Mg 1 mole of magnesium ions gains two moles of electrons or 2F to form 1 mole of Mg

24 g Mg requires 2 F electricity

4.8 g Mg requires 2 x4.8/24 = 0.4 F = 0.4 x96500 = 38600C 

Ca2+ + 2e-→ Ca

2 F electricity is required to produce 1 mole =40 g Ca

0.4 F electricity will produce 8 g Ca (1)

c. F = 96500C, n=2,

Sn2+ (aq) + 2e– → Sn(s) –0.14V

Cu2+(aq) + e- → Cu+ (aq) 0.15 V

E°cell = E°cathode – E° anode

= 0.15 – (-0.14) = 0.29V (1)

ΔG° = -nFE°cell

= -2 x96500x 0.29 = 55970 J/mol

Question34. A hydrocarbon (A) with molecular formula C5H10 on ozonolysis gives two products (B) and ( C). Both (B) and (C) give a yellow precipitate when heated with iodine in presence of NaOH while only (B) gives a silver mirror on reaction with Tollen’s reagent.

a. Identify (A), (B), and (C).

b. Write the reaction of B with Tollen’s reagent

c. Write the equation for the iodoform test for C

d. Write down the equation for the aldol condensation reaction of B and C.

OR

An organic compound (A) with molecular formula C2Cl3O2H is obtained when (B) reacts with Red P and Cl2. The organic compound (B) can be obtained from the reaction of methyl magnesium chloride with dry ice followed by acid hydrolysis.

a. Identify A and B

b. Write down the reaction for the formation of A from B. What is this reaction called?

c. Give any one method by which organic compound B can be prepared from its corresponding acid chloride.

d. Which will be the more acidic compound (A) or (B)? Why?

e. Write down the reaction to prepare methane from the compound (B).

Question 35. Answer the following:

a. Why are all copper halides known except that copper iodide?

b. Why is the Eo (V3+/V2+) value for vanadium comparatively low?

c. Why HCl should not be used for potassium permanganate titrations? 

d. Explain the observation, at the end of each period, there is a slight increase in the atomic radius of d block elements.

e. What is the effect of pH on dichromate ion solution?

Solution: .a. Cu2+ oxidizes iodide ion to iodine. 

b. The low value for V is related to the stability of V2+ (half-filled t2g level) 

c. Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidized to chlorine.

d. The d orbital is full with ten electrons and shields the electrons present in the higher s-orbital to a greater extent resulting in an increase in size.

e. The chromates and dichromates are interconvertible in an aqueous solution depending upon the pH of the solution. Increasing the pH (in basic solution)of dichromate ions a color change from orange to yellow is observed as dichromate ions change to chromate ions

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Chemistry Sample Paper Class 12 2023: FAQs

Ans. Yes, CBSE has released Chemistry Class 12 Sample Paper 2023 along with its solution.

Ans. In the above article, we have provided Class 12 Chemistry Sample Paper PDF for 2022-23 Exam for your reference.

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