CBSE

Additional SAMPLEQuestion
Practice PAPER Paper
(2023 -24)
CHEMISTRY THEORY (043)
Max. Marks: 70 Time: 3 hours
General Instructions:
Read the following instructions carefully.
(a) There are 33 questions in this question paper with internal choice.
(b) SECTION A consists of 16 multiple-choice questions carrying 1 mark each.
(c) SECTION B consists of 5 short answer questions carrying 2 marks each.
(d) SECTION C consists of 7 short answer questions carrying 3 marks each.
(e) SECTION D consists of 2 case-based questions carrying 4 marks each.
(f) SECTION E consists of 3 long answer questions carrying 5 marks each.
(g) All questions are compulsory.
(h) Use of log tables and calculators is not allowed.

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.
1. What would be the major product formed when (CH3)3 C– OH is heated at 358 K in the presence
of H3PO4 :
(a) (CH3)3COOH (b) (CH3)3CHO
(c) (CH3)2CH=CH2 (d) (CH3)3 C-O-C-(CH3)3
2. The complex [Co(NH3 )5 (NO2 )]Cl2 is obtained in yellow and red coloured forms . The difference
in colour is attributed to the following type of isomerism:
(a) Coordination Isomerism (b) Facial isomerism
(c) Linkage isomerism (d) Solvate isomerism
3. Which of the following purines (nitrogenous bases with two-ring structure) are common to RNA
and DNA?
(a) Adenine, Thymine (b) Guanine, Thymine
(c) Thymine, Cytosine (d) Adenine, Guanine
4. India’s first pipeline was laid to transport crude oil in Digboi, Assam. The rusting of these iron
pipelines was a cause of concern for the authorities. Which of the following methods would have
been used to prevent corrosion of these pipelines?
(a) Barrier protection (b) Alloying
(c) Cathodic protection (d) Anodising the material
5. Which of the following alkyl iodide cannot be produced by the reaction of HI with an appropriate
ether:
(a) (CH3)3 C – CH2 – I (b) (CH3)2CH – I (c) C6H5CH2 – I (d) C6H5 – I
6. What is the order and molecularity of hydrolysis of sucrose if the rate law is
Rate = k’[C12H22O11]?

1
 (a) Order is zero, molecularity is one (b) Order is two, molecularity is two
(c) Order is two, molecularity is one (d) Order is one, molecularity is two
7. The electronic configuration of Ni in complex Ni(CO)4 would be:
(a) t26 e4 (b) t26 e2 (c) e4 t26 (d) e4 t24
8. Ankit has been given four organic compounds: a primary amine, a secondary amine, a secondary
alcohol and a tertiary alcohol. Which of the following can Ankit use to identify all the compounds?
(a) Tollen’s reagent and bromine water (b) 2,4 DNP and Lucas reagent
(c) Hinsberg's reagent and Lucas reagent (d) Sodium metal and Hinsberg’s reagent
9. Which of the following is not true about enzymes?
(a) All enzymes are fibrous proteins
(b) Enzymes are needed in small quantities
(c) Enzymes reduce the magnitude of activation energy.
(d) They are specific for a reaction and the substrate.
10. Identify the products (1) and (2) in the following reactions:
 (1) and C2H5Cl + KNO2 
C2H5Cl + KCN   (2)
(a) (1) C2H5CN (2) C2H5 NO2 (b) (1) C2H5NC (2) C2H5 NO2
(c) (1) C2H5CN (2) C2H5 ONO (d) (1) C2H5NC (2) C2H5 ONO
11. “Greater number of electrons from (n-1)d in addition to the ns electrons are involved in the
interatomic metallic bonding.”. Which of the following case is result of the above statement:
(a) Vanadium forms complexes.
(b) Platinum acts as catalyst
(c) Chromium is having higher melting point than Vanadium and Manganese .
(d) Manganese show oxidation states ranging from +2 to +7 .
12. Which of the following reactions can be used to obtain benzaldehyde from benzene?
(a) Rosenmund’s Reduction (b) Stephen’s Reaction
(c) Etard’s Reaction (d) Gatterman-Koch Reaction
13. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Phenol on treatment with Br2 in CS2 gives ortho and para bromophenol.
Reason (R): Carbon disulphide is a solvent of low polarity, hence leads to monobromination of
phenols
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.
14. Given below are two statements labeled as Assertion (A) and Reason (R)
Assertion: At high pressure, decomposition of ammonia on the catalyst surface is a first order
reaction.
Reason: At high pressure, on changing the reaction conditions, the concentration of ammonia on
the surface of the catalyst does not change.
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.
2
 (c) A is true but R is false.
(d) A is false but R is true.
15. Given below are two statements labeled as Assertion (A) and Reason (R)
Assertion (A): Ionisation enthalpy of Fe2+ is lower than the Mn2+
Reason (R): Fe2+ has 3d5 4s1 configuration, making the state highly stable
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.
16. Given below are two statements labeled as Assertion (A) and Reason (R)
Assertion (A): Halogens are ortho and para-directing groups
Reason (R): Halogens are electron-withdrawing groups
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.
SECTION B
This section contains 5 questions with internal choice in one question. The following questions are
very short answer type and carry 2 marks each.
17. Answer the following questions:
(a) Out of Co3+ and Cr2+, which one liberates hydrogen from dilute acid and why?
[E0Co3+/Co2+ = +1.97 V, E0Cr3+/Cr2+ = –0. 41V]
(b) The transition metals and their compounds are known for their catalytic activity. Give a
reason for their effectiveness as catalysts and name the transition element used as a catalyst in
Haber’s Process.
18. In metal carbonyls, metal-ligand bonds possess both sigma and pi character. Comment, whether
the given statement is correct. Give reason for your answer.
19. Give an example of:
(a) an amine each with basic strength greater than and less than N-Methylmethanamine in
gaseous state
(b) an isomeric amine each with boiling point less than and more than N-Ethyl ethanamine.
20. Consider the following reaction:
(1) H2O2 + I–→ H2O + IO–
(2) H2O2 + IO– → H2O + I – + O2
In the above reaction, the rate of formation of the intermediate is slow.
(a) Write the rate law.
(b) How will the rate of the reaction be affected if the concentration of peroxide is doubled?
OR
20 The half-life of a sample A (t½ = 100 days) did not change with concentration of A. The half-life of

3
 another sample B (t½ = 120 days) decreases with decrease in concentration of B. If the initial
concentration of both the samples were 1M, what will be their rate constants?
21. Arrange the following as indicated
(a) n-Butane, Propan-1-ol, Propanal, Acetone, Methoxymethane ( increasing order of boiling
point)
(b) Acetaldehyde, Acetone, Acetophenone, (increasing order of reactivity towards nucleophilic
addition)
SECTION C
This section contains 7 questions with internal choice in one question. The following questions are
short answer type and carry 3 marks each.
22. Observe the potential energy diagram for the hydrogenation of ethene to give ethane.
Hydrogenation reaction Energy Diagram

(a) For the same process, predict why there is a difference in Ea1 and Ea2?
(b) Assuming both the reaction are reversible then in which case the backward reaction will be
faster?
(c) Will the ΔG value of this reaction be different in the above two paths? Explain.

(for visually challenged learners)

(a) What will be the effect on Ea and ΔG for a given reaction in the presence of a catalyst. Justify
your answer.
(b) According to the collision theory, what are the criterions to determine effective collision?
23. Answer the following questions:
(a) Give an example of the liquids which when mixed result in an endothermic process, What will
be the change in volume when the liquids are mixed?
(b) At 300K, what is the relation between the osmotic pressure of two equimolar solutions, one
whose Van’t Hoff factor is 2 and for the other is ½?
(c) Which of the two aqueous solutions has a higher melting point: 2molal glucose solution or 3
molal sucrose solution? Why?
24. Give reason for the following:
(a) The complex formed by Ca2+ with EDTA is more stable than the complex formed by Ca2+
with ethylene diamine.

4
 (b) In d4 complexes, ligands for which ∆o < P ,form high spin complexes.
(c) On removal of water from [Ti(H2O)6 ]Cl3 on heating, it become colourless.
OR
24. (a) What will the colour of precipitate obtained when ionization isomer of compound
[Co(NH3 )5Br] SO4, reacts with AgNO3 ?
(b) An element M forms complex ‘A’ with ligand-ethane-1,2-diamine and complex ‘B’ with
chlorido having coordination number six.
(i) Out of the two complexes which one will be more stable and why?
(ii) If metal forms d4 ion, which complex will have higher magnetic moment?
25. An organic compound ‘A’ is having a molecular formula C7H8O. On oxidation with acidified
KMnO4 it forms compound ‘B’ .Compound ‘B’ can also be obtained from compound C, on its
reaction with NaOH and then with CO2 followed by hydrolysis. Compound C can be easily
formed when butylphenylether is made to react with HI.
Identify the compound A,B and C . Write the chemical reactions involved.
26. Rohan was carrying out the reaction of potassium iodide with Potassium permanganate in the
acidic medium . The Iodine gas was liberated from potassium iodide , during the reaction . When
he repeated the reaction for further study ,he forgot to add acid . He observed, the products
released during the second case were different from the previous .
(i) Give the possible chemical reactions in both the cases .
(ii) Give reason for the difference in observation.
(iii) Give the structure of MnO 4
27. Answer the following questions:
On passing current type ‘X’ through aqueous CuSO4 solution, electrolysis was observed. When
current type ‘Y’ was used, the conductivity of the solution could be measured.
(a) Identify ‘X’ and ‘Y’ type of current used.
(b) Write the electrolysis products of aqueous CuSO4 solution on passing ‘X’.
28. Convert Propanamide to the following compounds in not more than two steps:
(a) N- Ethylpropanamine
(b) Ethanol
(c) N-Propylethanamide
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.
29. Temperature dependence of Henry’s law constant
Henry’s law and Henry’s law constant are widely used in chemical and environmental
engineering. Unfortunately, many people do not appreciate that Henry’s constant is not a true
constant but has a significant non-linear temperature dependence. Figure 1 illustrates some typical
behavior of Henry’s constant vs temperature for several solutes in water.
The Henry’s constant typically increases with temperature at low temperatures, reaches a
maximum, and then decreases at higher temperatures. The temperature at which the maximum
occurs depends on the specific solute-solvent pair.
Clearly, the use of a Henry’s constant that was derived at 25°C at a different temperature could
lead to serious errors during manufacturing. Even a variation as small as 10 K can cause the
Henry’s constant to change by a factor of two, which could have a serious impact on many process
designs.

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 Temperature dependence of Henry’s constant for several solutes in water.
(a) Which of the following is not a true constant like Henry’s constant
(i) speed of light (c)
(ii) Planck’s constant (h)
(iii) rate constant (k)
(iv) gas constant (R)
(b) How will the solubility of n-octane vary with a change in temperature, provided the pressure
is kept constant?
OR
Rishita works in an aerated drinks factory. To increase the fizz in the drink, she proposes to bottle
the aerated drinks at 40oC instead of 20oC. Do you support her proposal? Why or why not?
(c) Find the ratio of solubility of toluene in water at 20oC and 60oC.

(for visually challenged learners)

Henry’s law and Henry’s law constant are widely used in chemical and environmental
engineering. Unfortunately, many people do not appreciate that Henry’s constant is not a true
constant but has a significant non-linear temperature dependence. The Henry’s constant typically
increases with temperature at low temperatures, reaches a maximum, and then decreases at higher
temperatures. The temperature at which the maximum occurs depends on the specific solute-
solvent pair. For example, the temperature at which maxima occurs for n-octane-water pair is
nearly 90°C.
(a) Which of the following is not a true constant like Henry’s constant
(i) speed of light (c)
(ii) Planck’s constant (h)
(iii) rate constant (k)
(iv) gas constant (R)
i. (b) How will the solubility of n-octane vary with a change in temperature, provided the pressure
is kept constant?

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 OR
Rishita works in an aerated drinks factory. She is aware that the temperature at which maxima
occurs for carbon dioxide-water pair is above 80°C.
Rishita proposes to bottle the aerated drinks at 40°C instead of 20oC, this will increase the fizz in
the drink. Do you support her proposal? Why or why not?
(c) If the pressure is constant, the ratio of solubility of n-octane in water at 20oC and 60oC will be:
(i) greater than 1 (ii) less than 1 ( iii) equal to 1? Give reason for your choice.
30. Nucleophilic Substitution Reactions
In nucleophilic substitution reactions, a bond between carbon and a leaving group (C–LG) is
broken, and a new bond between carbon and a nucleophile (C–Nu) is formed. Nucleophilic
substitution reactions of alkyl halides occur through two main pathways: SN1 and SN2.
One way to visualize the differences between these two mechanisms is to sketch out their reaction
coordinate diagrams, where we plot changes in potential energy (vertical axis), the starting
materials pass along the “reaction coordinate” toward their conversion into products (horizontal
axis) In these diagrams the “peaks” (local maxima) represent transition states whereas “valleys”
(local minima) represent intermediates.
The reaction coordinate diagram of the SN1 reaction has two peaks, representing the two transition
states (Step 1 and Step 2, respectively) flanking a single “valley” representing the carbocation
intermediate.
Each step of the process has an activation energy represented by the difference in energy between
the reactant and the transition state.
The rate-determining step of a reaction is the step requiring the highest activation energy, that is,
the largest change in potential energy from reactant to transition state.

(Source: James Ashenhurst, Comparing the SN1 and SN2 Reactions, Master Organic Chemistry)
(a) On the basis of the activation energies for step 1 and step 2 for SN1 reaction, identify the rate-
determining step and justify your answer.
(b) If the alkyl group in the alkyl halide used as an example in figure 1 is the same but the halogen
bromine is replaced by chlorine, which of the following is correct and why?
(i) Ea for step 2 will be more than shown in the figure 1
(ii) Ea for step 2 will be less than shown in the figure 1
(iii) Ea for step 2 will be same as shown in the figure 1
OR
Plot rate of reaction vs. concentration of nucleophile for SN1 reaction.
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 (c) Draw the reaction coordinate diagram for SN2 reaction.
(For visually challenged candidates)
30. Nucleophilic Substitution Reactions
In nucleophilic substitution reactions, a bond between carbon and a leaving group (C–LG) is
broken, and a new bond between carbon and a nucleophile (C–Nu) is formed. Nucleophilic
substitution reactions of alkyl halides occur through two main pathways: SN1 and SN2.
SN1 occurs in two steps. Each step of the process has an activation energy represented by the
difference in energy between the reactant and the transition state. The rate-determining step of a
reaction is the step requiring the highest activation energy.
(a) What is the difference in energy of the reactants and that of the transition state called?
(b) A reaction occurs in two steps. If step 1 of the reaction has a greater activation energy than
step 2, which will be the rate-determining step and why?
OR
The slowest step in a complex reaction requires the highest activation energy. Account for the
above statement on the basis of the information provided about the nucleophilic substitution
reactions.
(a) In the following reaction, identify the leaving group and the nucleophile:
CH3CH2Cl + aq KOH  CH3CH2OH + KCl
CH3CH2CH2CH2Br + AgF  CH3CH2CH2CH2F + AgBr

SECTION E
The following questions are long answer type and carry 5 marks each. All questions have an
internal choice.
31. Answer the following questions:
(a) Draw the structure of the ethylene ketal of hexan-3-one.
(b) Between Benzoic acid and acetic acid which is more acidic and why?
(c) An optically active organic compound ‘A’, with molecular formula, C5H10O2 when treated
with Chlorine in the presence of Red Phosphorous forms compound ‘B’, C5H9O2Cl, whereas when
it is treated with thionyl chloride forms compound ‘C’, C5H9OCl. Compound C, on further
hydrogenation with palladium on BaSO4 in the presence of S, forms compound D, C5H10O.
Compound D gives positive Tollen’s test and regenerates A. A can also be obtained by base
hydrolysis and further acidification of C. Write the reaction for the formation of ‘A’ from ‘C’ and
Identify A, B, C, D.
OR
Answer the following questions:
31. (a) Bring about the following conversions:
(i) Propanal to 2-methyl pentanol
(ii) Iodobenzene to benzoic acid
(b) Though Carboxylic acids have >C=O group in their structure, but they are not prone to
nucleophilic addition reactions. Why?
(c) An organic compound A, with molecular formulae C9H12 is oxidized to monocarboxylic acid
B, C7H6O2 on vigorous oxidation with Potassium permanganate, whereas when oxidized in
presence of air and further treated with dilute acid forms phenol. Sodium salt of B finds use
as a food preservative and esters of B are used in perfumery. Identify A and B and write the

8
 reactions involved.
32. (a) Calculate the current and the thickness of the coating if the current was passed for 1hr. to
deposit 2g silver on metal object of area 1 cm2.(Ag = 108, 1 Faraday = 96500 coulombs,
density of silver=10.5g/cc).
(b) Calculate the electrode potential of Cr3+/Cr electrode at [Cr3+] = 0.001 M.
Given: Cr3+ + 3e–  Cr E° = – 0.74 V
OR
32. (a) The half-cell reactions of an electrochemical cell are given below:
Mn 3 (0.1 M)  e  
 Mn 2 (0.01M) E  1.50V
Ag  (0.001 M)  e  
 Ag(s) E  0.80 V
(i) Formulate a galvanic cell using the above data.
(ii) Calculate the emf of the cell at 250C.
(b) The graph represents the variation of molar conductivity of an electrolyte with respect to
concentration. Identify the type of electrolyte and also determine the value for A.

(for visually challenged learners)

32. (a) Calculate the current and the thickness of the coating if the current was passed for 1hr. to
deposit 2g silver on metal object of area 1 cm2.(Ag = 108, 1 Faraday = 96500 coulombs,
density of silver=10.5g/cc).
(b) Calculate the electrode potential of Cr3+/Cr electrode at [Cr3+] = 0.001 M.
Given: Cr3+ + 3e–  Cr E° = – 0.74
OR
(a) The half-cell reactions of an electrochemical cell are given below:
Mn 3 (0.1 M)  e  
 Mn 2 (0.01M) E  1.50V
Ag  (0.001 M)  e  
 Ag(s) E  0.80 V
(i) Formulate a galvanic cell using the above data.
(ii) Calculate the emf of the cell at 250C.
(b) The molar conductivity of an electrolyte decreases slowly with increase in concentration.
(i) Identify the type of electrolyte and give the equation with which we can represent this
change in molar conductivity.
(ii) Calculate the slope of the electrolyte if the intercept on the y axis and x axis at 150 and
0.034 respectively.

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33. Answer the following questions:
(a) (i) Name an important carbohydrate which makes the cell wall of bacteria and plants.
What is the basic unit of this carbohydrate?
(ii) How are these basic units linked to each other, name the linkage.
(iii) Draw the Haworth structure of the basic unit.
(b) Identify the disaccharide with molecular formulae, C12H22O11, which produces 2 moles of
α-D(+) Glucose on hydrolysis. What will be the observation when Tollen’s reagent is added to
such a disaccharide.
(c) Change in optical rotation is observed when sucrose is hydrolysed. What is the reason for the
inversion of configuration observed?
OR
33. Answer the following questions:
(a) (i) A polynucleotide chain is seen to produce pentose sugar, phosphoric acid, Adenine,
Guanine, Cytosine and Thymine on complete hydrolysis. Name the nucleic acid having
such a polynucleotide chain. How are the bases paired in this polynucleotide?
(ii) What links these nucleotides together in a polynucleotide?
(iii) Give one important function and one application of the above nucleic acid .
(b) Keratin is a hair protein. What kind of tertiary protein is this? Describe the structure and
links present in this protein. Comment on its solubility in water?

10
 CBSE
Additional SAMPLE
Practice PAPER Paper
Question
(2023 -24)
CHEMISTRY THEORY (043)
MARKING SCHEME
SECTION-A
(Q1 to 16, 1 mark allotted for the correct option)
1. (c) (CH3)2CH = CH2 , Tertiary alcohol undergoes intramolecular dehydration reaction in the
presence of H3PO4
2. (c) Linkage isomerism, NO2- is an ambidentate ligand .
3. (d)
4. (c) Cathodic protection
Iron pipes are connected to more reactive metal like Magnesium, zinc so that an electrochemical cell is
set up between the two. Iron behaves like cathode and doesn't corrode.
5. (d) C6H5– I , ether having an aryl part will lead to the formation of phenol rather than iodobenzene.
Formation of phenyl cation will not be favoured , partial double bond character of C—O will
make it less stable.
6. (d) Order is one, molecularity is two
7. (c) eg4 t2g6
8. (c) Hinsberg reagent helps in identifying primary, secondary and tertiary amines and Lucas
reagent helps in identification of primary, secondary and tertiary alcohols.
9. (a) Enzymes are majorly globular proteins. So this statement mentioning enzymes as fibrous
protein is false. Rest all are true statements.
10. (c) Both CN– and ONO are ambident nucleophiles and KCN and KNO2 are ionic in nature
11. (c) Chromium is having higher melting point than Vanadium and Manganese .This statement
shows that melting point is higher because of involvement of (n-1)d electrons in addition to the ns
electrons in the interatomic metallic bonding.
12. (d) Gattermann-Koch reaction involves the conversion of benzene to benzaldehyde on reaction
with CO and HCl. Rest of the named reactions involve preparation of benzaldehyde from the
different starting material.
13. (a) Both A and R are true and R is the correct explanation of A
14. (d) A is false but R is true. Decomposition of ammonia follows zero order kinetics.
15. (c) A is true but R is false. Reason is false, as there is no loss of exchange energy in d6 configuration.
16. (b) Halogens withdraw electrons through an inductive effect (–I) and release electrons through
resonance (+R). Through inductive effect, halogens destabilize the intermediate carbocation formed
during the electrophilic substitution. Through resonance, halogen tends to stabilize the carbocation
and the effect is more pronounced at ortho- and para- positions. Reactivity is thus controlled by the
stronger inductive effect and orientation is controlled by resonance effect.
SECTION-B
17. (a) E0 values Cr3+/Cr2+ is negative shows that Cr2+ whereas positive value of E°Co3+/Co2+, shows Co3+ it
will not displace hydrogen from dilute acid.
(b) The ability of transition metals to adopt multiple oxidation states and to form complexes using
empty atomic orbitals make them suitable to act as catalyst. Finely divided iron is used in Haber’s
Process. Catalytic property can be attributed to their large surface area. (1)
18 The metal-carbon bond in metal carbonyls possess both σ and π character. The M–C σ bond is
formed by the donation of lone pair of electrons from 2p z on the carbonyl carbon into a d z 2 vacant

1
 orbital of the metal. The M–C π bond is formed by the donation of a pair of electrons from a filled d
orbital dxy of metal into the vacant antibonding * 2p x orbital of carbon monoxide. (1+1)
19. (a) An amine with basic strength greater is N, N-Dimethylethanamine and less is methanamine (or
any other correct option) (½+½)
(b) An isomeric amine with boiling point less than the N-Ethyl ethanamine is N,N-
Dimethylethanamine and more is Butanamine (½+½)
20 (a) Since the formation of intermediate is slow, the first step is the slowest step. Hence the rate will
depend on the first step.
Therefore,
Rate = k[H2O2][I–] (½+½)
(b) If the concentration of peroxide is doubled, the rate of the reaction also doubles. (1)
OR
20. Since t½ for sample A did not change with change in concentration, it follows first order
kinetics. Therefore, the rate constant is
t½ = 0.693/k (½)
k =0.693/100 = 0.00693/day (½)
Since t½ for sample B decreases with decrease in concentration, it follows zero order kinetics.
t½ = [A0]/2k (½)
k = 1/2x120
k= 1/240 =0.0041 M/day (½)
21. (a) n-Butane < Methoxymethane < Propanal < Acetone < Propan-1-ol
The boiling points of aldehydes, ketone , ethers are lower than those of alcohols of similar
molecular masses due to absence of intermolecular hydrogen bonding. (1)
(b) Acetophenone < Acetone < Acetaldehyde
Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to
steric and electronic reasons. Among the ketones, in acetophenone the positive charge over
carbonyl carbon is delocalized, involved in resonance, so its reactivity decreases in comparison
to aliphatic ketones. (1)

SECTION-C
22 (a) There is a difference in activation energy because the process with Ea1 is a catalysed reaction
which lowers the threshold energy for the formation of intermediate activated complex while.
(1)
Ea2 path is uncatalyzed and has higher threshold energy. (½)
(b) In a reversible catalysed reaction, the rate of both forward and backward reactions are faster
than an uncatalyzed reaction. (½)
(c) ΔG value is independent of the path taken as Gibbs free energy is a state function and not a
path function. (1)
(for visually challenged learners)
(a) The value of Ea will be lower as the threshold energy for the formation of intermediate
activated complex is less. (½+½)
A catalyst does not alter Gibbs energy, ΔG of a reaction. ΔG value is independent of the path
taken as Gibbs free energy is a state function and not a path function. (½+½)
(b) Activation energy and proper orientation of the molecules (½+ ½)

2
23 (a) Ethanol and acetone, or any other example, the volume will increase (The liquids show positive
deviation from Raoult’s law) (½+½)
(b) π1/π2 = 4 (1)
(c) 2 molal glucose because it will have lower depression in freezing point, higher melting point .
(1)
24 (a) EDTA is a hexadentate ligand while ethylene diamine is bidentate, the chelating effect of EDTA
is more therefore EDTA-Ca2+ complex is more stable. (1)
(b) If ∆o < P, the fourth electron enters one of the eg orbitals giving the configuration t2g3 eg1 (1)
(c ) On heating water molecule which is acting as ligand is lost .In the absence of ligand, crystal
field splitting does not occur so there is no d-d transition and hence the substance is colourless
(1)
OR
24. (a) The colour of precipitate obtained when ionization isomer of compound
[Co(NH3 )5Br]SO4 , i.e. [Co(NH3 )5SO4]Br (1)
reacts with AgNO3 will be pale yellow . The pale yellow precipitate of AgBr will be obtained .
(b) (i) The complex A with ligand ethane-1,2-diamine will be more stable as it is didentate ligand
leading to the formation of chelate and hence increasing the stability . (1)
(ii) The higher magnetic moment will be in the case of Complex B , with ligand chlorido as no
pairing up of electrons will take place. (1)
25. Compound - A is ortho methylphenol, (½)
Compound- B is ortho hydroxybenzoic acid. (½)
Compound - C is Phenol. (½)
The chemical reaction involved are :
(a) Formation of B from compound A using KMnO4
OH O
CH3
KMnO4 OH
+
H
OH
2-hydroxybenzoic acid
or
ortho hydroxybenzoic
acid/Salicylic acid
(½)
( b) Formation of B from C i.e Phenol

(½)
(c) C6H5OCH2CH2CH2CH3 
HI
CH3CH2CH2CH2I + C6H5OH (½)
26 (i) A notable reaction is the oxidation of iodide to iodate, when no acid present

3
 2MnO4– + H2O + I– 
 2MnO2 + 2OH– + IO3– (1)
When the acid was present
2MnO 4  16H   10I  
 2Mn 2  5I 2  8H 2O
(ii) In case of oxidising action of KMnO4 , the hydrogen ion concentration plays an important
part in influencing the reaction. In the presence of acid medium, oxidation number of Mn
changes from +7 to +2 and in basic medium from +7 to +4 (1)
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
(iii)
O

Mn
O O- Permanganate ion (1)
O
27. (a) Since electrolysis (decomposition) happens with current type ‘X’, it is a DC (Direct current.
Current type ‘Y’ is AC(Alternating current) and is used to measure conductivity of solutions. (1+1)
(b) When CuSO4 is electrolysed, copper gets deposited at cathode and oxygen gas is released at anode.
At cathode reduction of Cu2+ occurs (as it requires lesser electrode potential than H+)
Cu2+ + 2e- → Cu (1/2)
At anode oxidation of water occurs (as it requires lesser electrode potential to get oxidized than SO42-
2H2O → O2 + 4H+ + 4e- (1/2)
LiAlH /H O CH CH Cl
28 (a) CH3CH 2 CONH 2 
4 2
 CH3CH 2CH 2 NH 2 
3 2
 CH3CH 2CH 2 NHCH 2CH3
(1)
Br  NaOH (i) NaNO  HCl (ii)H O
(b) CH3CH 2CONH 2 
2
 CH3CH 2 NH 2 
2 2
CH 3CH 2OH
(1)
LiAlH /H O CH COCl
(c) CH3CH 2CONH 2 
4 2
 CH3CH 2CH 2 NH 2 
3
 CH 3CH 2CH 2 NHCOCH 3
Pyridine (1)
SECTION D
29 (a) (iii) Rate constant (k), it changes with temperature (1)
(b) The solubility will decrease with temperature at low temperatures, reach a minimum value and
then increase at higher temperatures because the Henry’s constant typically increases with
temperature at low temperatures, reaches a maximum, the temperature at which maxima
occurs for n-octane-water pair is nearly 90°C and then decreases at higher temperatures. (1)
OR
It is not a good proposal.
KH for CO2 is higher at 40oC,
so the solubility of CO2 will be lower, thus fizz be less.
(c) Ratio of solubility will be greater than 1 because
p = KH X
At 20°C KH = 50MPa , and at 60°C KH = 100 M Pa
KH (60) / KH(20) = X(20) / X (60)
100/50 = 2 = X(20) / X (60) (2)

For Visually impaired

(a) (iii) Rate constant (k), it changes with temperature. (1)
(b) The solubility will decrease with temperature at low temperatures, reach a minimum value and

4
 then increase at higher temperatures because the Henry’s constant typically increases with
temperature at low temperatures, reaches a maximum, and then decreases at higher
temperatures. (1)
OR
It is not a good proposal.
KH for CO2 will be higher at 40°C, because maxima is obtained above 80°C
So, the solubility of CO2 will be lower, thus fizz be less.
(c) Ratio of solubility greater than 1 because
p = KH X
At 20°C KH will be less than KH at 60°C
So solubility at 20°C will be more than at 60°C (2)
30 (a) The activation energy for step 1 is larger than step 2, so step 1 is the slowest step and the rate
determining step. (1)
(b) (i) Ea for step 2 will be more than shown in figure 1, because Br- is a better-leaving group than
Cl-. (1)
OR
Rate

[Nu–]

(c)

(2)
(For visually impaired)
(a) Activation energy (1)
(b) step 1 The rate-determining step of a reaction is the step requiring the highest activation energy.
(1)
OR
slowest step is the rate determining step so it requires highest activation energy The rate-
determining step of a reaction is the step requiring the highest activation energy.
(c) (i) leaving group is Cl- and nucleophile is OH- (1)
(ii) leaving group Br- and nucleophile is F- (1)

5
 SECTION E
31. (a)
H3C CH2 O CH2
C
O CH2
H3C CH2 CH2 (1)
(b) Benzoic acid is more acidic. This is because of greater electronegativity of sp2 hybridized carbon
to which carboxyl carbon is attached, whereby phenyl group acts as a electron withdrawing
group and stabilizes the intermediate carboxylate ion. (1)
(c)

Cl2/red P
A B
(C5H10O2) (C5H9O2Cl)

SOCl2
H2/Pd
C(C5H9OCl) D(C5H10O)
BaSO4

Tollen's
D(C5H10O) A + Ag mirror
reagent

NaOH
C(C5H9OCl) A
H+

Since D, C5H10O gives a positive Tollen’s test, it is an aldehyde. D is also obtained from C by
controlled catalytic hydrogenation, so C is an acid halide.
C is obtained by treatment of thionyl chloride on A, so A is an acid.
A when reacted with halogen in Red phosphorus produces B, So B is α-halocarboxylic acid and A is
an acid with α-hydrogen.
Since A is an optically active compound with 5 carbons, therefore

CH 3
|
A is 2- methyl butanoic acid CH 3CH 2 CH  COOH
(½)
CH3
|
B is 2- chloro-2-methyl butanoic acid CH3  CH 2  C  COOH
| (½)
Cl

CH 3
|
C is 2-methyl butanoyl chloride CH 3CH 2 CHCOCl
(½)
CH 3
|
D is 2-methyl butanal CH 3CH 2 CHCHO
(½)

6
 CH3 CH3 CH3
| | |
H
CH3CH 2CHCOCl 
NaOH
 CH 3CH 2CHCOONa  CH 3CH 2CHCOOH
(C) (A) (1)

OR
31. (a)

1. dil. NaOH 1. LiAlH4

(i) 2CH3CH2CHO CH3CH2CH C CHO CH3CH2CHC(CH3)CH2OH (1)
2. Heat 2. H2O
CH3

Mg CO2 H2O/H+
(ii) C6H5I C H MgI C6H5COOMgI C6H5COOH (1)
dry ether 6 5 dry ether
(b) This is so because the Carbon in >C=O group in carboxylic group is less electrophilic due to the
possible resonance as shown:

(1)
(c) A is oxidized to monocarboxylic acid, B indicates that the ring structure has only one side chain.
Also, Phenol is formed when oxidized in the presence of air and further treatment with dil acid so
A must be Cumene. (½)
Sodium salt of B is food preservative and esters of it is used in perfumery, confirms B to be benzoic
acid. (½)

KMnO4/H+
C6H5CH(CH3)2 C6H5COOH
(A) (B)
(½)

O2 H+
C6H5CH(CH3)2 C6H5C(CH3)2OOH C6H5OH + (CH3)2CO (½)
H2O

32. (a) Mass of Ag deposited = 2g

mass
Volume =  = 2 g/10.5 gcm3
density
= 0.190cm3 (½)
Thickness = 0.190 cm3/1cm2
= 0.190 cm2 (½)
For current
Ag+ + 1e- → Ag
108 g of Ag is deposited by 1F of charge
1 2
2 g of Ag will be deposited by charge
108
= 0.0185 F = 0.0185 × 96500 C (1)
t = 1 hr
Q = Ixt

7
 0.0185 x 96500= I x 60 x 60

0.0185  96500
I  0.495 A
3600 (1)
0.059 1
(b) E Cr3 /Cr  E Cr
o
3  log
/Cr 3 [Cr 3 ] (½)
0.059
E Cr3 /Cr  0.74  log103
3 (½)

E Cr3 /Cr  0.74  0.059

(½)

E Cr3 /Cr  0.799V

(½)
OR
 3 2
32. (a) (i) Ag (s) | Ag (0.001M) || Mn (0.1M) | Mn (0.01M) | Pt
(1)

(ii) E ocell  1.50  0.80  0.70 V

(½)
[Ag  ][Mn 2 ]
Q (½)
[Mn 3 ]
103 102
Q 1
 104
10
log Q  4
0.059
E  E  log Q (½)
n
0.059
E  0.70   (4)
1
 0.70  0.236
(½)
E  0.936 V
(b) Strong Electrolyte

    A C (½)
For a straight line equation,
y = c +mx
Slope = m= -A
Slope is also given by tan.
Slope = Perpendicular/base

150

0.034 (½)
Slope =-A = - 4411.76 A = 4411.76 (1)

8
(for visually challenged learners)
32. (a) and (b) Same as above
OR
(a) Same as above
(b) (i) Strong electrolyte

    A C (½)
Slope = Perpendicular/base

150

0.034 (½)
Slope = – A = – 4411.76 A = 4411.76 (1)
33. (a) (i) Cellulose (½)
-D( +) Glucose (½)
(ii) C-1 of one unit of glucose is linked to C-4 of another unit of glucose. These unit are linked
to each other by glycosidic linkages. (1)
iii) Haworth structure of β- D(+)-Glucose:

(1)
(b) Maltose (½)
Silver mirror will be observed as Tollen’s reagent is reduced by maltose due to the presence of
free aldehyde group. (½)
(c) Sucrose is dextrorotatory, but on hydrolysis it produces a mixture of α-D-(+) Glucose and
β-D-(-) Fructose. Since the laevorotation of fructose is more than the dextrorotation of α-
glucose, so overall mixture becomes laevorotatory, hence inversion is observed. (1)
OR
33. (a) (i) DNA (½)
Specific base pair sequence 1) Adenine with Thymine by two H- bonds and 2) Guanine with
Cytosine by three H- bonds (½)
ii) Phosphodiester linkages. (1)
iii) Function: DNA is a reserve of genetic information and is responsible for identity of different
species (½)
Application: DNA fingerprinting is used in forensic labs for identification of Criminals (½)
(b) Fibrous protein (½)
In the fibrous protein the polypeptide chains run parallel and are held together by hydrogen and
disulphide bonds, resulting in a fibre-like structure. (1)
Fibrous proteins are insoluble in water (½)