Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

  • There are 35 questions in this question paper with internal choices.
  • Section A consists of 18 multiple-choice questions carrying 1 mark each.
  • Section B consists of 7 very short answer questions carrying 2 marks each.
  • Section C consists of 5 short answer questions carrying 3 marks each.
  • Section D consists of 2 case-based questions carrying 4 marks each.
  • Section E consists of 3 long answer questions carrying 5 marks each.
  • All questions are compulsory.
  • Use of log tables and calculators are not allowed.

SECTION – A (18 Marks)
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mart There is no internal choice in this section.)

Question 1.
Which of the following is not the factor for electrode potential of an electrode? [1]
(a) Nature of metal and ions
(b) Temperature
(c) Concentration of ions
(d) Volume
Answer:
(d) Volume

Explanation:
Factors affecting the electrode potential of an electron assembly are the nature of metal and ions, temperature and on the concentration of ions in the solution.

 

Question 2.
Reactions involving the cleavage of C—OH bond follows the order: [1]
(a) 3° alcohol> 2° alcohol> 1° alcohol
(b) 2° alcohol>l° alcohol>3° alcohol
(c) 1° alcohol>2° alcohol>3° alcohol
(d) 3° alcohol<2° alcohol>1 alcohol
Answer:
(a) 30 alcohol > 20 alcohol > 10 alcohol

Explanation:
This order of reactivity can be understood by the +I effect shown by the alkyl groups attached to the carbon carrying -OH group. Higher the number of alkyl groups attached to the carbon carrying -OH group, more is the polarity of C-O bond and easier is its cleavage, leads to the greater reactivity of the alcohol.

Question 3.
Mention the type of solution which follows Raoult’s Law. [1]
(a) Ideal solutions
(b) Non-ideal solutions
(c) Azeotropes
(d) Binary solutions
Answer:
(a) Ideal solutions

Explanation:
An ideal solution is the one in which the molecules of different species are easily distinguishable and their molecules exert force on each other and Raoult’s law states that “for a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction”.

Related Theory:
Henry’s law states that “at a constant temperature the solubility of gas in a liquid is directly proportional to the pressure of gases”.

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Question 4.
Which amine does not react with Hinsberg’s Reagent? [1]
(a) Primary amines
(b) Secondary amines
(c) Tertiary amines
(d) Quaternary amines
Answer:
(c) Tertiary amines.

Explanation:
A chemical test that is most commonly used for the identification of primary, secondary and tertiary amines is called Hinsberg’s test. An amine in the presence of an aqueous alkali interacts with a Hinsberg’s reagent i.e; benzene sulphonyl chloride.

Related Theory:
Now a days, instead of benzene sulphonyl chloride, p- toluene sulphonyl chloride is commonly used in the Hinsberg’s test because of cost and availability issues related to benzene sulphonyl chloride

Question 5.
The standard reduction potential values of the elements P,Q, and R are shown in graph: [1]
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 1
The order of reducing power is:
(a) P > Q > R
(b) Q > P > R
(c) R > P > Q
(d) P < Q > R
Answer:
(a) P > Q > R

Explanation:
A reduction reaction that involves the addition of electrons is conceivable, according to a positive reduction potential value. R will acquire an electron and function as an oxidising agent since it has a positive electrode potential. Therefore, out of the three, it has the least decreasing capabilities.

The possibility of oxidation involving the loss of electrons is suggested by a negative reduction potential. P has a stronger reducing agent than Q because it has a larger negative reduction potential than R and Q. As a result, it will lose an electron.

Question 6.
Identify the liquid pairs forming ideal solutions. [1]

I II
(a) Chlorobenzene Bromobenzene
(b) Sulphuric acid Water
(c) Water Methanol
(d) Carbon Tetrachloride Toluene

Answer:
(a) I. Chlorobenzene; II. bromobenzene

Explanation:
Chlorobenzene and bromobenzene is the example of the ideal solution , rest are the examples of non-ideal solutions.

Question 7.
Equivalent conductivity is _________ proportional to the specific conductivity. [1]
(a) Equal
(b) Inversely
(c) Directly
(d) Cannot be predicted
Answer:
(c) Directly

Explanation:
The equivalent conductivity (Aeq) is related to the specific conductivity (k) by the following equation.
Λeq = \(\frac{\kappa}{C_{e q}}\)

Question 8.
The combustion of hydrogen takes place in a hydrogen-oxygen fuel cell to: [1]
(a) generate heat.
(b) create potential difference between the two electrodes.
(c) produce high purity water.
(d) remove adsorbed oxygen from electrode surfaces.
Answer:
(b) create potential difference between the two electrodes.

Explanation:
The oxygen is reduced at the cathode and the hydrogen is oxidised at the anode in a hydrogen-oxygen fuel cell. These two processes combine to form the hydrogen combustion reaction, which produces a lot of heat.

Question 9.
In an octahedral structure, the pair of d- orbitals involved in d2sp3 hybridisation are:
(a) dxy, dyz
(b) dz2, dx2-y2
(c) dz2, dxy
(d) dx2-y2, dxy
Answer:
(b) dz2, dx2-y2

Explanation:
We know that the bonds are in the shape of an octahedron with the bonds parallel to the x, y and z axes. d2sp3 or sp3d2 are hybridisations for the octahedral geometry and thus, 2d orbitals are involved in the formation of the hybrid orbitals i.e., dx2-y2 and dz2. The other 3d orbitals (i.e., dxy, dxz and dyz are not parallel to the axis but they are formed in between the axes).
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 2

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Question 10.
Vinylic halide are compounds in which halogen atom is bonded to sp2 hybridized carbon atom of:
(a) an alkane
(b) an alkene
(c) an alkyne
(d) an aromatic ring
Answer:
(a) an alkane

Explanation:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 3

Question 11.
Which of the following compounds has tetrahedral geometry? 1]
(a) [Ni(CN)4]2-
(b) [Pd(CN)4]2-
(c) [PdCl4]2-
(d) [NiCl4]2-
Answer:
(d) [NiCl4]2-

Explanation:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 4

Question 12.
In a coordination complex’s the central metal atom acts as
(a) Bronsted-Lowry acid
(b) Lewis base
(c) Lewis acid
(d) Bronsted-Lowry base
Answer:
(c) Lewis acid

Explanation:
A Lewis acid is a species that has the ability to receive an electron pair. The central metal atom is a Lewis acid because it accepts electrons due to the presence of empty d-orbitals.

Question 13.
Glucose reacts with HI to form:
(a) n-Hexane
(b) Gluconic acid
(c) Fructose
(d) Saccharic acid
Answer:
(a) n-Hexane

Explanation:
Glucose on prolonged heating with the HI gives the n- hexane and n- hexane has 6 carbon atoms connected with each other in the straight chain thus this concludes that all the 6 carbon atoms in the glucose molecule are also connected linearly in the form of a straight chain.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 5

Question 14.
Glucose on treatment with NH2OH undergoes:
(a) condensation
(b) reduction
(c) hydrolysis
(d) oxidation
Answer:
(a) condensation

Explanation:
Glucose on reaction with NH2OH will give the glucoxime undergoing the condensation reaction along with the removal of the water molecule. As shown by the reaction given below:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 6

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(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

Question 15.
Assertion: The electrical resistance of any object increases with increase in its length.
Reason: Electrical resistance of any object increases with increase in its area of cross-section. [1]
Answer:
(c) (A) is true but (R) is false

Explanation: R

Related Theory
There are two types of ions formed during a chemical reaction the carbocation and the carbanion. A carbocation is a molecule in which a carbon atom has a positive charge and three bonds whereas carbanion a carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge usually with three substituents for a total of eight valence electrons.

Question 16.
Assertion: Oxidation of glucose by Br2– water gives saccharic acid.
Reason: Br2-water oxidises —CHO group in the structure of glucose. [1]
Answer:
(d) (A) is false but (R) is true

Explanation:
Oxidation of glucose by Br2- water gives Gluconic acid. Br2-water oxidises only —CHO group in the structure of glucose. Thus Assertion is wrong statement but reason is correct statement.

Question 17.
Assertion: Assertion: Order of the following elementary reaction,
2NO(g) + 2H2(g) → 2H2O(g) + N2(g) is 4.
Reason: Order of the elementary reaction with respect to given reactant is the power of the reactant’s concentration in the rate equation. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Order of the following elementary reaction NO (g) + H(g) → HO(g) + N(g) is 4. As we know that order of the elementary reaction with respect to given reactant is the power of the reactants concentration in the rate equation. Thus, both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Question 18.
Assertion: The solubility of aldehydes and ketones in water decreases with the increase in the size of the alkyl group.
Reason: Alkyl groups are electron releasing groups. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
The solubility decreases due to the increase in bulkiness, due to this, steric hindrance around the carbonyl group increases, so, the water molecules cannot interact with the carbonyl group easily and have less tendency of a formation of hydrogen bond.

SECTION – B (14 Marks)
(The following questions are very short answer type with internal choice in two questions and carry 2 marks each.)

Question 19.
For the standard cell
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s) [E°Ni2+/Ni = – 0.25 and E°Ag2+/Ag = 0.080 V]
(A) Identify the cathode and the anode as the current is drown from the cell.
(B) Write the reaction taking place at the electrodes.
Answer:
Since E°Ni2+/Ni is lower than E°Ag2+/Ag, nickle electrode has higher tendency to undergo

(A) Ag electrode have higher tendency to underg° reduction s° must act as cathode, oxidition and must act as anode. Hence
Anode = Ni | Ni2+
Cathode : Ag+ | Ag
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 7

Question 20.
An organic compound with the molecular formula C8H16O2 was hydrolysed with dilute H2SO4 to give a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid produced B. C on dehydration gives but-l-ene. Write the equations for the reactions involved. [2]
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 8

Question 21.
State Henry’s law of solubility of gases in liquids.
OR
(A) 1 gram of a non-electrolyte solute is dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg/ mol-1. Find the molar mass of solute.
(B) State the two important properties of ideal solutions. [2]
Answer:
The mole fraction of a gas in the solution is directly proportional to the partial pressure of the gas over the solution.
i.e; P ∝ X
P = KH. X
Where P = partial pressure of the gas
X = mole fraction of the gas
KH = Henry’s law constant

Explanation:
The solubility of gases in liquids increase with the increase in the pressure of the gas whereas there is no practical change in the solubility of liquids and solids. Henry’s law explains this concept of the solubility of gases in liquids and this law is applied in the process of production of carbonated beverages.
OR
(A) Given that, WB = 1 g, WA = 50 g, Kf = 5.12 k Kg mol-1, ΔTf = 0.40 k, MB = ?
MB = \(\frac{K_f \times W_B \times 1000}{\Delta T_f \times W_A}=\frac{5.12 \times 1 \times 1000}{0.40 \times 50}\) = 256 gm-1
Molar mass of solute (MB) = 256 g mol-1.

(B) The two important properties of ideal solutions are:
(1) Ideal solutions follow Raoult’s law over entire range of concentrations.
(2) The enthalpy and volume of mixing of the pure components to form a solution are zero- ΔmixH = 0; ΔmixH = 0.

Question 22.
Define the following:
(A) Molar conductivity
(B) Specific conductance [2]
Answer:
(A) Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is denoted by λ (lambda).
(B) Specific conductivity is the measure of the ability of that material to conduct electricity. It is represented by the symbol “K” known as kappa.

Related Theory:
Molar conductivity and specific conductivity are interrelated as:
Specific conductance is given by K and molar
conductance is given by l
The relation between the two terms is:

Λc = K × 1000/M
Also Λc = {(1/R) × (l/a)} × 1000/M Where
Λc = Molar conductance
K = Specific conductance
R = Resistance
M = Molarity of the solution l = length
a = area of cross section

Question 23.
How do you convert the following:
(A) Prop-l-ene to 1-fluoropropane
(B) Chlorobenzene to 2-chlorotoluene
OR
Write the main products when:
(A) n-butyl chloride is treated with alcoholic KOH.
(B) 2,4, 6-trinitrochlorobenzene is subjected to hydrolysis. [2]
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 9
OR
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 10
(B) 2, 4, 6-trinitrochlorobenzene under mild hydrolysis conditions (H2O/323 K) gives 2, 4, 6-tri-nitrophenol or picric acid.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 11

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Question 24.
(A) Give a reason why Mn3+ is oxidizing in nature.
(B) Name a transition element which unlike others does not exhibit variable oxidation state. [2]
Answer:
(A) Mn3+ is oxidising in nature as it gets reduced to Mn2+ state in the process and attains a d5 configuration of half-filled d-orbital which has extra stability.

(B) Scandium (atomic no. 21) does not exhibit variable oxidation state.

Question 25.
What happens when:
(A) aromatic amines react with nitrous acid at low temperatures.
(B) ethanamine reacts with acetyl chloride in presence of a base. [2]
Answer:
(A) Aromatic amines react with nitrous acid at low temperatures to form diazonium salts.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 12
(B) Ethanamine undergoes acylation:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 13

SECTION – C (15 Marks)
(The following questions are short answer type with internal choice in two questions and carry 3 marks each.)

Question 26.
Account for the following:
(A) For a first order reaction, show that time required for completion of 99.9% of reaction is 3 times the time required for completion of the 90% of the reaction.
(B) For a chemical reaction R → P, the variation in the concentration (R) vs. time (t) plot is given as:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 14
(i) What is the order of the reaction?
(ii) What is the slope of the graph of the variation in the concentration (R) vs. time (t) plot?
OR
(A) What is the effect of adding catalyst on the free energy of a reaction?
(B) What is the order of reaction whose rate constant has the same units as the rate of reaction?
(C) Why is it difficult to find more number of reactions with higher order? [3]
Answer:
(A) For a first order reaction,
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 15

(B) (i) It is a zero-order reaction
(ii) Slope = -k.
OR
(A) There is no effect of adding catalyst on the free energy (ΔG) of a reaction. A catalyst lowers the activation energy of a reaction.
(B) Zero order reaction.
(C) A reaction takes place due to collision of molecules. The chances for a large number of molecules or ions colliding simultaneously to give a productive reaction are less. Hence, the reactions of higher order are less in number.

Question 27.
(A) What is the order of the reaction for the rate constant k = 3.5 × 10-5 L mol-1 s-1.
(B) Units of rate and rate constant of a reaction are the same for which order of the reaction?
(C) State the factors which affect the rate of the reaction. [3]
Answer:
(A) The order of the reaction is given as:
3.5 × 10-5 L mol-1 s-1
Unit of rate = mol L-1 s-1
Unit of zero order = mol L-1 s-1
Thus, it is the zero order reaction

(B) For zero order reaction the rate law will be:
R = k[A]
Unit of rate = mol L-1 t-1
Unit of zero order = mol L-1 t-1

(C) Factors affecting the rate of the reaction are:

  • Concentration of reactants
  • Temperature
  • Presence of a catalyst
  • Surface area

Question 28.
Account for the following:
(A) Aniline does not undergo Friedal – Craft’s reaction.
(B) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(C) Gabriel phthalimide synthesis is preferred for synthesising primary amines only. [3]
Answer:
(A) Aniline does not undergo Friedal -Crafts reaction due to salt formation with anhydrous AlCl3 (Lewis acid), aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. Thus, it does not undergo Friedal – Craft’s reaction.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 16

(B) Diazonium salts of aromatic amines are more stable than those of aliphatic amines due to the resonance, there is dispersal of the positive charge on the benzene ring, which in turn accounts for the stability of the diazonium ion and this is not found in the aliphatic diazonium salts.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 17

(C) Gabriel phthalimide synthesis is preferred for synthesising primary amines only because this synthesis results in the formation of the primary (10) amines only, secondary (20) and tertiary amines (30)
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 18

Question 29.
Write the major product of the following reactions:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 19
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 20

Question 30.
0.85% aqueous solution of NaNO3 is apparently 90% dissociated at 27°C. Calculate its osmotic pressure. (R = 0.082 L atm k-1 mol-1). [3]
Answer:
NaNO3 dissociates in solutions as
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 21
At equilibrium (1 – α) moles a moles a moles
Total number of moles in solution = 1 – α + α + α
= 1 + α
van’t Hoff factor,
i = \(\frac{\text { No. of moles in solution }}{\text { No. of moles added }}\)
= \(\frac{1+\alpha}{1}\) = 1 + α

Since the given salt dissociates 90% in the solution,
α = \(\frac{90}{100}\) = 0.9

Hence, for the given salt, i = 1 + α = 1 + 0.9 = 1.9
For the given solution, mass of solute = 0.85 g
∴ Moles of solute added = \(\frac{0.85}{85}\) = 0.01
(normal molecular mass of NaNO3
= 23 + 14 + 48 = 85)
Volume of solution = 100 mL = 0.1 L

According to the modified equation of osmotic pressure,
πV = i n RT
π = \(\frac{\text { in } R T}{V}\)
= \(\frac{1.9 \times 0.01 \times 0.0821 \times 300}{0.1}\)
= 4.68 atm
Hence, the osmotic pressure of the given solution is 4.68 atm.

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.
The interaction of the -OH group with the carbonyl group results in the intramolecular hemiacetal or hemiketal production of pentose and hexose. The real structure consists of a ring with five or six members and an oxygen atom. All pentoses and hexoses are present in pyranose form in the free state (resembling pyran). However, some of them have five-membered cyclic structures known as furanose in the combined state (resembling furan).
The cyclic structure of glucose is represented by Haworth structure:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 22
(A) Give an example of aldohexose. [1]
(B) Write the reaction of glucose with NH2OH. [1]
(C) (i) Give an example of globular protein,
(ii) Give an example of fibrous protein.
OR
Answer the following questions:
(A) Name the carbohydrate molecule which is an aldohexose and also known as dextrose.
(B) What is secondary structure of protein molecules, name its two types. [2]
Answer:
(A) Glucose

Explanation:
Glucose is aldohexose since it contains 6 carbon atoms.

(B) Glucose on reaction with NH2OH forms the glucose oxime and undergoes the condensation reaction as shown by the reaction given below
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 23

(C) (i) Haemoglobin
Explanation:
Haemoglobin is the complex globular protein made up of globules structure of 4 sub units.

(ii) Fibroin
Explanation:
Fibroin has the sheet like structure made up of long chain of polypeptide chains thus it is a fibrous protein.
OR
(A) Glucose is also known as dextrose and is an aldohexose.
(B) The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures namely, a-helix and b-pleated sheet structure.

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Question 32.
Discrete coordination compounds and coordination polymers are top research fields in inorganic chemistry, crystal engineering, solid- state chemistry, and materials science, due to their potential applications in adsorption, separation, catalysis, electrical, magnetic, and optical applications. Beginning with the basic concepts of coordination compounds and coordination polymers. The important factors (such as temperature, pH, template, additive, solvent, and counter-ion) that have influences on the self-assembly and crystallization, as well as solvothermal/hydrothermal in situ metal/ligand reactions for coordination compounds and coordination polymers, as well as post synthetic modifications for porous coordination polymers, through selected examples in the literature.
(A) Explain why toxic metal ions are removed by the chelating ligands. [1]
(B) Why [Pt(NH3)3Cl] will not show geometrical isomerism. [1]
(C) Name the following:
(i) The coordination complex used in gold plating.
(ii) The coordination complex used in the treatment of cancer.
OR
State any two limitations of valence bond theory. [2]
Answer:
(A) When a solution of chelating ligand is added to a solution containing toxic metals ligands, it chelates the metal ions by the formation of a stable complex.
(B) Since there is no chirality in the molecule and due to the restricted rotation around the central atom it will not show the geometrical isomerism and no cis and trans form will be there for the molecule.
(C) (i) K[Au(CN)2]
(ii) Cisplatin- [PtCl2(NH3)2]
OR
(i) It is unable to interpret the absorption spectra of complexes.
(ii) It is unable to account for the detailed magnetic properties of complexes.

Question 33.
(A) Account for the following:
(i) Write a chemical test to distinguish between formaldehyde and acetic acid.
(ii) Aldehydes are more reactive than ketones in nucleophilic reactions.
(B) With the help of suitable reactions explain what happens when:
(i) Ketones react with semi carbazide.
(ii) Acetone reacts with cone. H2SO4.
(C) Explain what is shown in the graph along with the reason for the same.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 24
Answer:
(A) (I) Formic acid is a reducing agent It gives silver mirror test with toLlen’s reagent
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 25
Acetic acid do not give this test.

(ii) Inductive effect and the steric effect makes the aldehydes and ketones more reactive than ketones in the nucleophilic reactions.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 26
Presence of more alkyl groups (-R) in ketones hinders the attack of nucleophiles on the carbonyl group and thus the ketones are less reactive than aldehydes.

(B) (i) Ketones react with semi carbazone to give semi carbazone which is the white solid.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 27

(C) Acetone reacts with conc. H2SO4. to give 1, 3, 5-trimethyl benzene.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 28
(iii) The graph shows that the boiling points of the alcohols is higher than that of the corresponding alkanes because of the intermolecular hydrogen bonding in the alcohol molecules, which makes the bond more strong and thus the boiling point increase.

Question 34.
Complete the following reactions- (A)
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 29
(C) Explain why (CH3)2C = CH2 is insoluble in water whereas (CH3)2C = O is highly soluble in water.
(D) Arrange the following in increasing order of their acidity:
FCH2COOH, BrCH2COOH, CH3CH2COOH, HCOOH
(E) What is Tollen’s reagent, whot is it used for?
OR
(A) What is Clemmenson reduction?
(B) What is the general name of carboxylic acid found in:
(i) Lemon (ii) Yoghurt
(C) Give the IUPAC names of following:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 30
(D) Complete the following reaction:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 31
Answer:
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 32
(C) The presence of oxygen with its non¬bonding electron pairs makes acetone hydrogen-bond acceptors, hence a better bonding with water molecules resulting in increased water solubility relative to hydrocarbons, like 2-methyl propene in this case.

(D) The compounds in their increasing order of acidity would be:
CH3CH2COOH < HCOOH < BrCH2COOH < FCH2COOH

(E) TolLen’s reagent is ammoniacal, solution of silver (I) nitrate. When reacted with aldehydes it oxidises aldehdes to corresponding acids and in turn silver ion reduces to silver metal. giving a shing film of silver metal. on the test tube wall, which confirms the presence of aldehgde group in analyte organic compound.

OR

(A) The Ctemmensen reduction is an organic reaction used to reduce an aldehyde or ketone to an alkane using amalgamated zinc and hjdrochloric acid.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 33
(B) The general. name of carboxylic acid found in:
(i) Lemon is Citric acid
(ii) Yoghurt is Lactic acid
(C) 2-Bromo-4,4-dimethyl cdl.ohexa none
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 34

CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions

Question 35.
Assign reasons for each of the following:
(A) Transition metals generally form coloured compounds.
(B) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements.
(C) Cr2+ is reducing in nature while with the same d-orbital configuration (d4) Mn3+ is an oxidising agent.
(D) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series.
OR
Explain the following observations:
(A) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(B) Transition elements and their compounds are generally found to be good catalysts in chemical reactions.
(C) Transition metals and their compounds are generally found to be good catalysts.
(D) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series.
(E) Why the ionization enthalpy values slowly increases for the first transition series as shown by the graph? Explain this trend.
CBSE Sample Papers for Class 12 Chemistry Set 8 with Solutions 35
Answer:
(A) Transition metals generally form coloured compounds because of the presence of unpaired d e[ectrons, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours.

(B) Manganese exhibits highest oxidation of +7 among 3d series of transition elements because all the oxidation states are exhibited from +2 to +7 by Mn and no other element of this series show this highest state of oxidation.

(C) Cr2 has the configuration 3d4 which easily changes to d3 due to stable half-filled t2g orbitaIs. Therefore, Cr2 is reducing agent. While Mn2 has stable half-filled d5 configuration. Hence, Mn3 easily changes to Mn2 and acts as oxidising agent

(D) Due to presence of more unpaired electrons and use of all 4s and 3d electrons in the middle of series.
OR
(A) From titanium to copper the atomic size of elements decreases and mass increases as result of which density increases.

(B) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitaIs and variable oxidation states.

(C) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.

(D) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series as these have their electrons of outer most shell at greater distance from the nucleus, as compared to atoms of 3d transition metals.

(E) In the first transition series, the first and second ionization enthalpy values increases slowly from left to right with certain irregularities. As the nuclear charge increases, the attraction between the nucleus and outer electron increases due to which the ionization enthalpy increases. However the electron is added to (n-1) d subshell, the screening effect increases. Thus the increase in the nuclear charge is opposed by the increase in the screening effèct. Hence, the ionization enthalpy values slowly increases.