RBSE 12th Chemistry Model Paper Set 4 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 4 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 4 with Answers in English

Time: 2 Hours 45 Min
Maximum Marks: 56

General Instruction to the Examinees :

• Candidate must write first his/her roll No. on the question paper compulsory.
• All the questions are compulsory.
• Write the answer to each question in the given answer book only.
• For questions having more than one part the answers to those parts are to be written together in continuity.

Section – A
Multiple Choice Questions (MCQs):

Question 1.
Choose the correct option:
(i) 10 L dilute solution is prepared from 1L of 0;5 M H₂SO₄. The molarity of solution obtained is: (1)
(a) 1M
(b) 0.05 M
(c) 0.5M
(d) 0.1 M
Answer:
(b) 0.05 M

(ii) General formula of alipha6c aldehyde is: (1)
(a) C_nH_2n+2O
(b) C_nH_2nO
(c) C_nH_2n-1O
(d) C_nH_2n-1O
Answer:
(b) C_nH_2nO

(iii) Which of the solution represents positive deviation? (1)
(a) Ethyl bromide + Methyl bromide
(b) Cyclohexane + Ethanol
(c) Chloroform + Acetone
(d) None of the above
Answer:
(b) Cyclohexane + Ethanol

(iv) Whose aqueous solution is formalin? (1)
(a) Formic acid
(b) Formaldehyde
(c) Fluorine
(d) Furfural
Answer:
(b) Formaldehyde

(v) The rate of reaction on doubling the active mass of A by keeping active mass of B constant in the reaction 2A + B → Product will be: (1)
(a) twice as before
(b) four times as before
(c) half as before
(d) one fourth as before
Answer:
(b) four times as before

(vi) Which metal is extracted from its ore by electrolytic method? (1)
(a) Pb
(b) Cu
(c) Al
(d) Ag
Answer:
(c) Al

(vii) The oxidation state of nickel in |Ni(CO)₄] is: (1)
(a) 4
(b) 0
(c) 2
(d) 3
Answer:
(b) 0

(viii) ‘A’ in following reaction is: (1)

(a) CH₃CN
(b) CH₃NC
(c) C₂H₅CN
(d) CH₃NO₂
Answer:
(a) CH₃CN

(ix) How many isomers will be possible for compound having molecular formula C₇H₉N? (1)
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(b) 5

Question 2.
Fill in the Blanks:
(i) Oxidation state of Cr in K₂Cr₂O₇ is _________. (1)
Answer:
+ 6

(ii) Unit of rate constant for first order reaction is _________. (1)
Answer:
s^-1

(iii) DDT is formed by reaction of chlorobenzene with _________. (1)
Answer:
Chloral

(iv) Power alcohol is _________ and _________. (1)
Answer:
alcohol, petrol.

Question 3.
Very Short Answer Type Questions:
(i) What is the two dimensional coordination number of a molecule in square close-packed layer? (1)
Answer:
Two dimensional coordination number of a molecule in square close- packed layer is equal to 4 as each atom is in contact with four neighbouring atoms.

(ii) Write the structure of the compound whose IUPAC name is l-phenylpropan-2-ol. (1)
Answer:
1-phenylpropan-2-ol

(iii) Define the term’amorphous’. Give few examples of amorphous solids. (1)
Answer:
The solids in which the constitiuent particles are not arranged in any regular manner are called amorphous solids. They have short range order. They are isotropic in nature. Example : glass, plastic, polymers, amorphous silica etc.

(iv) How would you convert ethanol to ethene? (1)
Answer:

(v) Write the coordination isomer of [Cu(NH₃)₄][PtCl₄] (1)
Answer:
[Pt(NH₃)₄] [CuCl₄]

(vi) Actinoid concentration is greater from element to element than lanthanoid contraction. Why? (1)
Answer:
The 5f electrons are more effectively shielded from nuclear charge. In other words, the 5f electrons themselves provide poor shielding from element to element in Actinoid series.

(vii) For a reaction A + B → P, the rate law is given by, r = k[A]^1/2 [B]². What is the order of this reaction? (1)
Answer:
For a reaction, A + B → P
Given, rate of a reaction, r = k[A]^1/2 [B]²
Order wrt A = \(\frac{1}{2}\);
Order wrt B = 2
∴ Overall order of a reaction = 1/2 + 2 = 5/2

(viii) Write the structure of 2-hydroxybenzoic acid. (1)
Answer:
2-hydroxybenzoic acid:

Section – B
Short Answer Type Questions:

Question 4.
Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass. (1½)
Answer:
20% C₂H₆O₂ means 20 g of C₂H₆O₂ is present in 100 g of solution.
Mass of C₂H₆O₂, W_B = 20 g
Mass of water, W_A = 100 – 20 = 80 g
Molar mass of C₂H₆O₂ (M_B)
= 2 × 12 + 6 × 1 + 2 × 16
= 62 g mol

Molar mass of H₂O (M_A)
= 18 g mol^-1

Question 5.
Write the reactions involved in the following:
(a) Hell-Volhard Zelinsky reaction
(b) Decarboxylation reaction (¾ + ¾ = 1½)
Answer:

Question 6.
Define the following terms:
(a) Mole fraction (χ)
(b) Molality of a solution (m) (¾ + ¾ = 1½)
Answer:
(a) Mole fraction (χ):
The mole fraction of a component is the ratio of the number of moles of the component to the total num¬ber of moles of all the components present in the solution.

Molality of solution (m):
Molality is defined as the number of moles of the solute per kilogram of the solvent. It is represented by m.
= \(\frac{\text { Number of moles of the component }}{\text { Total number of moles of all the components }}\)

For a binary solution, mole fraction of component A,
χ_A = \(\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\)
Similarly, for B, χ_B = \(\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\)
and χ_A + χ_B = 1

(b) Molalily of solution (m): Molal it y is defined as the number of moles of the solute per kilogram of the solvent. It is represented by in Molality (in)
= \(=\frac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)

Question 7.
Carried out the following conversions in not more than two steps:
(a) Benzoic acid to benzaidehyde
(b) Ethyl benzene to Benzok acid
(c) Propanone to Propene (1½)
Answer:

Question 8.
Calculate the overall order of reaction which has the rate expression: (¾ + ¾ = 1½)
(a) Rate = k[A]^1/2[B]^3/2
Answer:
Rate =k[A]^x[B]^y
Order of reaction =x
Rate = k[A]^1/2[B]^-1
Order = \(\frac{1}{2}+\frac{3}{2}\) = 2
i.e., this reaction is of second order.

(b) Rate = k[A]^3/2[B]^-1.
Answer:
Rate = k[A]^3/2 [B]^-1
Order = \(\frac{3}{2}\) + (-1)
= \(\frac{1}{2}\)
i.e., this reaction is of half order.

Question 9.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. (1½)
Answer:
t_1/2 = 60 minutes = 60 60 s
For the first order reaction.
K = \(\frac{0.693}{t_{1 / 2}}=\frac{0.693}{60 \times 60}\)
= 1925 × 10_-4 s_-1
So, the rate constant
= 1.925 × 10_-4 s_-1

Question 10.
Write structures of different isomers corresponding to the molecular formula, CsHghl. Write IUPAC names of isomers which will liberate nitrogen gas on treatment with nitrous acid. (1½)
Answer:
Four structural isomers are possible: 1° Amines

Only primary (10) amines react with HNO₂ to liberate nitrogen gas.

Question 11.
Explain the principle of the method of electrolytic refining of metals. Give one example. (1½)
Answer:
Electrolytic refining: In this method, the impure metals is made anode and a strip of pure form of same metal is made cathode. Aqueous solution of salt of same metal is taken as electrolyte. On passing electric current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal while an equivalent amount of metal dissolves from the anode and goes into the electrolyte solution as metal ions.

Electrolytic refining of copper: Here, anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate. Due to . electrolysis, pure copper from anode is transferred to cathode.
At anode, Cu → Cu²⁺ + 2e^–
At cathode, Cu²⁺ + 2e^– → Cu
Impurities from the blister copper deposit as anode mud which contains impurities like antimony, selenium, tellurium, silver, gold and platinum.

Question 12.
What is the role of depressant in forth floatating process? (1½)
Answer:
When more than one sulphides ore to be concentrated them depressents are used. They change the oil to water tendency of ore or gangue for example : NaCN is used as depressant in concentration of PbS and ZnS. NaCN reacts with ZnS and converted into Na₂ [Zn(CN)₄] which is soluble in water. In this way wetting tendency of PbS with oil increases while with water decreases. On the other hand wetting tendency of ZnS with wafer increases.

Question 13.
Give reasons: (¾ + ¾ = 1½)
(a) Aniline is a weaker base than cyclohexylamine.
Answer:
In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen decreases.
But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of π-electrons. Hence aniline is weaker base than cyclohexylamine.

(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Answer:
Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further 2° and 3° alkyl amine.

Question 14.
On what ground you can say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not? (1½)
Answer:
On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d¹), it is regarded as a transition element. On the other hand, zinc atom has completely filled d-orbitals (3d¹⁰) in its ground state as well as in its excited state, hence it is not regarded as a transition element.

Question 15.
What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in aqueous solution. (1½)
Answer:
The disproportionation reactions are those in which the same substance gets oxidised as well as reduced. When a particular oxidation state becomes less stable relative to other oxidation states (one lower and one higher), it undergoes disproportionation, e.g. Mn (VI) becomes unstable relative to Mn (VII) and Mn (IV) in acidic solution.
M⁺⁶nO₄^2- + 4H⁺ → 2M⁺⁷nO^–₄ + M⁺⁴nO₂ + 2H₂O

Section – C
Long Answer Type Questions:

Question 16.
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it. (3)
Or
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain ’ with the help of a suitable example. (3)
Answer:
When a cation of higher valence is added to anionic solid then impurity defect is produced. In impurity defect, two or more cations of lower valency are replaced by a cation of higher valency in order to maintain the electrical neutrality. Hence, some vacancies are created.For example : In ionic crystal of NaCl, impurity of Sr²⁺ is added as SrCl₂. Then two Na⁺ ions left its lattice site leaving behind holes. One of these hole will be occupied by one Sr²⁺ ion to maintain electrical neutrality and other hole will remain vacant.

Question 17.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether

(a) What would be the major product of this reaction?
(b) Write a suitable reaction for the preparation of f-butyl ethyl ether? (3)
Or
Give the structures and IUPAC names of products expected from the following reactions:
(a) Catalytic reduction of butanal
(b) Hydration of propene in presence of concentrated sulphuric acid.
(c) Reaction of propanone with methylmagnesium bromide followed by hydrolyse. (3)
Answer:
Since the alkyl halide is 3°, (f-butyl chloride) and C₂H₅ONa is strong base, therefore, elimination occurs to form 2-methylprop-l-ene.

To prepare 1-butyl ethyl ether, the alkyl halide should be 1°, i.e., chloroethane and the nucleophile should be sodium f-butoxide.

Question 18.
How do you explain the absence of aldehyde group in the pentaacetate of D – glucose? (3)
Or
The melting points rnd solubility in water of a-amino acids are higher than those of the corresponding halo; ids. Explain. (3)
Answer:
When glucose (a and p) is treated with acetic an- hydride, it forms a pentaacetyl derivatives which does not contain a free hydroxyl (-OH) group at C -1, it cannot be hydrolysed in aqueous solution to give the open chain aldehyde group and hence glucose pentaacetate does not react with NH₂OH to form glucose oxime.

Section – D
Essay Type Questions

Question 19.
Explain why:
(a) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(b) Alkyl halides, though polar are immiscible with water?
(c) Grignard reagents should be prepared under anhydrous conditions? (4)
Or
Predict al? the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.
(a) 1-Bromo-l-methylcyclo- hexane,
(b) 2-Chloro-2-methyl butane,
(c) 3-Bromo-2,2,3-trimethyl pentane. (4)
Answer:
(a)

The polarity of C —Cl bond in chlorobenzene is less than that of same bond in cyclohexyl chloride because of carbon atom involved in chlorobenzene is more electro-negative due to greater s-character as compared to the carbon atom in cyclohexyl chloride with lesser s- character. Therefore, the dipole moment of chlorobenzene is less than cyclohexyl chloride.

(b) Alkyl halides or haloalkanes are polar molecules, therefore their molecules are held together with dipole-dipole interactions. In H20 the molecules are held together by intermolecular H-bonding. Since the new forces of attraction between H20 and haloalkane molecules are weaker than the forces of attraction already existing between haloalkane haloalkane molecules and water- water molecules, therefore haloalkanes or alkyl halides are immiscible with water.

(c) Grignard reagents (R – Mg – X) should be prepared under anhydrous conditions because these are readily decomposed with water to give alkanes.

That is why ether used as solvent in the preparation of Grignard reagents in completely anhydrous in nature.

Question 20.
Write the IUPAC name and draw the structure of each of the following complex entities: (4)

(b)[Cr(CO)₆]
(c)[PtCl₃(C₂H₄)]
(Atomic no. of Cr = 25, Co = 27, Pt = 78)
Or
Name the following complexes and draw the structure of one possible isomer of each: (4)
(a) [Cr(C₂O₄)₃]^3-
(b) [Pt (NH₃)₂ Cl₂]
(c) [Co (en)₂ Cl₂]⁺
(where en = ethane-1,2-diamine or ethylene diamine).
Answer:

IUPAC name Trioxalatocobaltate (III) ion

Structure:

(b)[Cr(CO)₆]
IUPAC name Hexacarbonyl-chromium(0)

Structure:

(c) [PtCl₃(C₂H₄)]
IUPAC name : Trichloridoe- theneplatinum (IV)

Structure: