RBSE 12th Chemistry Model Paper Set 1 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 1 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 1 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) The rate constant of reaction depends on: (1)
(a) temperature
(b) mass
(c) weight
(d) catalyst
Answer:
(a) temperature

(ii) The formula of Haematite is: (1)
(a) Fe₃O₄
(b) Fe₂O₃
(c) FeS₄
(d) FeO
Answer:
(b) Fe₂O₃

(iii)

is an example of: (1)
(a) replacement
(b) addition
(c) rearrangement
(d) elimination
Answer:
(a) replacement

(iv) A transition element containing only one electron in: (1)
(a) ₂₁Sc
(b) ₂₅Mn
(c) ₂Fe
(d) ₂₉Cu
Answer:
(d) ₂₉Cu

(v) Which of the following is a tridentate ligand? (1)
(a) NO₂^–
(b) oxalate ion
(c) glycinate ion
(d) dien
Answer:
(d) dien

(vi) An example of ambidentate ligand is: (1)
(a) Ammine
(b) Aquo
(c) Oxalato
(d) Thiocyanato
Answer:
(d) Thiocyanato

(vii) Chloroethane (C₂H₅Cl) reacts with A to form diethylether. The compound A may be: (1)
(a) NaOH
(b) H₂SO₄
(c) C₂H₅ONa
(d) Na₂S₂O₃
Answer:
(c) C₂H₅ONa

(viii) At room temperature formaldehyde is: (1)
(a) gas
(b) liquid
(c) solid
(d) none of these
Answer:
(a) gas

(ix) The best conditions to get maximum yield of C₂H₅Cl is: (1)

Answer:

Question 2.
Fill in the Blanks:
(i) Amalgum is the example of ____________ in ____________. (1)
Answer:
Liqluid, solid

(ii) The process of extraction of metals from their ores is called ____________. (1)
Answer:
metallurgy

(iii) Ethylene diamine is a ____________ ligand. (1)
Answer:
Bidentate

(iv)

Answer:
CH₃-CH₂-CH₂-Br

Question 3.
Very Short Answer Type Questions:
(i) What type of solids are electrical conductors, malleable and ductile? (1)
Answer:
Metallic solids.

(ii) Express the rate of the following in terms of ammonia. (1)
Answer:
N₂(g) + 3H₂(g) → 2NH₃(g)
Rate = \(\frac{-d\left[\mathrm{~N}_{2}\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_{2}\right]}{d t}\) = + \(\frac{1}{2} \frac{d\left[\mathbf{N H}_{3}\right]}{d t}\)

(iii) Define order of a reaction.
Answer:
Order of a reaction: The sum of the powers of the concentration i.e. sto-ichiometric coefficients of reactants of a chemical reaction in rate law ex-pression is called the order of that chemical reaction. For example, for the reaction given below:
a A + bB → Cc
Rate = k[A]^x[B]^y
Order of reaction = x + y
Where x and y are actual stoichiometric coefficients for the reaction.

(iv) On what principle is the method of zone-refining of metals based? (1)
Answer:
Zone-refining: It is based on the fact that the impurities are more soluble in the molten state than in the solid state of the metal.

(v) Name the depressant which is used to separate ZnS and PbS ores in froth floatation process. (1)
Answer:
Sodium cyanide is the depressant that is used to separate ZnS and PbS ores in the froth flotation process.

(vi) Sc(21), is a transition element but Ca(20) is not. Why? (1)
Answer:
Sc(21) with electronic configuration [Ar]3d¹4s² have incompletely filled d-orbitals whereas Ca(20).does not. Thus, Sc(21) is a transition element.

(vii) Given the IUPAC name of H₂N – CH₂ – CH₂ – CH = CH₂. (1)
Answer:
IUPAC name: But-3-ehe-l -amine

(viii) Arrange the following compounds in an increasing order of their solubility in water: (1)
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer:
C₆H₅NH₂ < (C₂H₅)₂ NH < C₂H₅NH₂

Section – B
Short Answer Type Questions

Question 4.
Use the data to answer the following and also justify giving reason: (¾ + ¾ = 1½)

(a) Which is a stronger reducing agent in aqueous medium, Cr²⁺ or Fe²⁺ and why?
Answer:
Reactivity series is made on the basis of standard reduction potential (E°) and E° for Cr is more negative than that of Fe. Thus, Fe get reduced and Cr get oxidised, in other words Cr is a stronger reducing agent

(b) Which is the most stable ion in + 2 oxidation and why?
Answer:
Among the given ions, the ion with more negative value of E° (red.) will loose the electron more easily, thus is more stable in (+) 2 oxidation state. Hence Mn in (+) 2 oxidation state is the most stable species.

Question 5.
What are interstitial compounds? Why are such compounds well known for transition metals? (1½)
Answer:
Interstitial compounds:
The compounds in which small atoms like H, C, N, He etc. occupy interstitial sites in the crystal lattice are called interstitial compounds.
These compounds are well known for transition metals because small atoms can easily occupy the positions in the voids present in the crystal lattices of transition metals.

Question 6.
(a) Write down the IUPAC name of the following complex: [Cr (en)₃]Cl₃
(b) Write the formula for the following complex: Potassiumtrioxalato chromate (III) (¾ + ¾ = 1½)
Answer:
(a) The IUPAC name of the complex, [Cr (en)₃]Cl₃ is tris (ethane-1, 2 diamine) chromium (III) chloride.
(b) The formula for the complex, potassiumtrioxalatochromate (III) is K₃[Cr(C₂O₄)₃].

Question 7.
Name the following coordination compounds and draw their structures. (¾ + ¾ = 1½)
(a) [CoCl₂(en)₂]Cl
(b) [Pt (NH₃)₂ Cl(NO₂)]
(Atomic no. of Co = 27, Pt = 78)
Answer:
(a) [CoCl₂(en)₂]Cl
IUPAC name Dichlorido fo’s-(ethane-1,2-diamine) cobalt (HI) chloride.

Structures There are two possible structures of [CoCl₂(en)₂] Cl, one is cis and other is trans that can be represented as:

(b)[Pt(NH₃)₂Cl(NO)₂]
IUPAC name
Diamminechloridonitrito-N- platinate (II).

Structures
There are two structures of [Pt(NH₃)₂ Cl (NO₂)], one is as form and other is trans-form that can be represented as:

Question 8.
Write the equations involved in the following reactions : (¾ + ¾ = 1½)
(a) Wolff-Kishner reduction
Answer:
Wolff-Kishner reduction:

(b) Etard reaction
Answer:
Etard reaction:

Question 9.
Write structures of compounds A, B and C in each of the following reactions: (1½)

Answer:

A = Ethanal
B = 3-Hydroxybutanal (Aldol)
C = But-2-enal

Question 10.
Give the structure and IUPAC name of the product formed when propanone is reacted with methylmagnesium bromide followed by hydrolysis. (1½)
Answer:

Question 11.
Carboxylic acids do not give characteristic reactions of carbonyl group. Explain why? (1½)
Answer:
The carboxylic carbon is less elec-trophilic than carbonyl carbon be-cause of the possible resonance structure.

Question 12.
Give the chemical tests to distinguish between the following pairs of compounds: (¾ + ¾ = 1½)
(a) Ethylamine and Aniline
Answer:
(a) Ethylamine and aniline:
By Azo dye test: It involves the reac-tion of any aromatic primary amine with HNO₂ (NaNO₂ + dil. HCI) at 273-278 K followed by treatment with an alkaline solution of 2-naph- thol when a brilliant yellow, orange or red coloured dye is obtained.

(b) Aniline and Benzylamine
Answer:
Distinction between Aniline and Benzylamine: By Nitrous acid test: Benzylamine reacts with HNO₂ to form a diazonium salt which being unstable even at low temperature, decom poses with evolution of N₂ gas.

Aniline on the other hand, reacts with HNO₂ to form benzene diazo¬nium chloride which is stable at 273¬278 K and hence does not decompose to-evolve N₂ gas.

Question 13.
Identify A and B in each of the following processes : (¾ + ¾ = 1½)

Answer:

Question 14.
What are the hydrolysis products of: (¾ + ¾ = 1½)
(a) sucrose
(b) lactose?
Answer:

Question 15.
What is the basic structural difference between starch and cellulose? (1½)
Answer:
Starch is not a single compound. It consists of amylose (15-20%) and amylopectin (80-85%). In contrast cellulose is a single compound. Amylose is a linear polymer of α-D- glucose while cellulose is a Onden- sation polymer of β-D-glucose. In amylose, C -1 of one glucose unit is joined to C – 4 of the other unit through glycosidic linkage. In cellu-lose, C -1 of one glucose unit is joined to C – 4 of the other emit through β- glycosidic linkage. Amylopectin has highly branched structure.

Section – C
Long Answer Type Questions:

Question 16.
Write the products of the following reactions:

OR
Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions: (1½ + 1½ = 3)
(a) The four isomeric bromobutanes
(b) C₆H₅CH₂Br, CH₅CH(CH₅)Br, CH₅CH(CH₃)Br, CH₅C(CH₃) (CH₅)Br
Answer:

Question 17.
Identify the reaction order from each of the following rate constant: (1½ + 1½ = 3)
(a) k = 2.3 × 10^-5 L mol^-1 s^-1
(b) k = 3 × 10^-4 4 s^-1
OR
Write two differences between ‘order of reaction’ and molecularity of reaction. (3)
Answer:
(a) k = 2.3 × 10^-5 L mol^-1 s^-1
We know that unit of k
= (mol L^-1)^1-ns^-1
where n is the order of reaction. Thus,
L mol^-1 s^-1 = (mol L^-1)^1-n s^-1
or L mol^-1 = (mol L^-11)^1-n
or (mol L^-1)^-1 = (mol L^-1)^1-n

Comparing powers on both sides
-1 = 1 – n
n = 1 + 1 = 2
i.e., the reaction is of second order,

(b) k = 3 × 10^-4 4 s^-1
s^-1 = (mol L^-1)^1-n s^-1
or (mol L^-1)⁰ s^-1 = (mol L^-1)^1-ns^-1
or (mol L^-1)⁰ = (mol L^-1)^1-n

Comparing powers on both the sides
0 = 1 – n
n = 1
i.e., the reaction is of first order.

Question 18.
Write structure of products of the following reactions: (1 + 1 + 1 = 3)

OR
Ortho and para-nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions. (3)
Answer:

(b) NaBH₄ is a weak reducing agent. It reduces the aldehyde or ketones but not the esters.

Section – D
Essay Type Questions:

Question 19.
How will you distinguish between the following pairs of terms: (4)
(a) Hexagonal close-packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
OR
Explain the following terms with suitable examples: (4)
(a) Schottky defect
(b) Frenkel defect
(c) Interstitials
(d) F-centres.
Answer:
(a) Differences between hexagonal close packing and cubic close packing

Hexagonal close packing	Cubic close packing
1. In this type of close packing tetrahedral voids of the second layer is covered by the spheres of third layer.	1. In this type of dose packing octahedral voids of the second layer is covered by the spheres of the third layer.
2. In this case, the spheres of the third 1aye are exactly aligned with those of the first layer.	2. In this case, the spheres of the third layer are not aligned with those of the first layer.
3. It has ABAD — type pattern.	3. It has ABCABC — type pattern.
4. It is found in Mg and Zn	4. It is found in Cu, Ag, Au etc

(b) Difference between crystal lattice and unit cell

Crystal lattice	Unit cell
The three dimensional arrangement of constituent particles of a substance like atoms, ions or molecules is called crystal lattice. It is formed by the regular repetition of unit cell.	It is the smallest three dimensional portion of a crystal lattice which when repeated in three dimensions give the crystal, is called unit cell. It is formed by the regular repetition of lattice points in three dimensions.

(c) Differences between Tetrahedral void and Octahedral void

Tetrahedral void	Octahedral void
1. This type of void is formed by four spheres.	1. This type of void is formed by six spheres.
2. \(\frac{\text { Radius of Tetrahedral void }(\mathrm{r})}{\text { Radius of sphere }(\mathrm{R})}\) = 0.225	2. \(\frac{\text { Radius of Octahedral void(r) }}{\text { Radius of sphere }(\mathrm{R})}\) = 0.414
3. (shape of tetrahedral void)	3. (shape of octahedral void)

Question 20.
(a) Calculate the freezing point of solution when 1.9 g of MgCl₂ (M = 95 g mol^-1) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionisatin.
(K_f for water = 1.86 K kg mol^-1)
(b) (i) Out of 1M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of the solution? (2 + 2 = 4)
OR
(a) When 2.56 g of sulphur was dissolved in 100 g of CS₂ the freezing point gets lowered by 0.383 K. Calculate the formula of sulphur (S_x).
(K_f for CS₂ = 3.83 K kg mol^-1, Atomic mass of sulphur = 32 g mol^-1)
(b) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place ‘ blood cells in a solution containing:
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
Answer:
(a) MgCl₂ → Mg²⁺ + 2Cl^–
1 mol of MgCl₂ gives 3 moles of particles.
∴ i = 3
∆T_f = iK_fm

Given, W_A = Weight of H₂O (solvent)
= 50 g

W_B = Weight of MgCl₂ (solute)
= 1.9g
T°_f = 273.15 K
K_f = 1.86 K kg mol_-1
MB = Molar mass of solute
= 95gmol_-1
∆T_f = \(\frac{i \mathrm{~K}_{f} \times 1000 \times \mathrm{W}_{b}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{a}}\)
= \(\frac{3 \times 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1000 \times 1.9 \mathrm{~g}}{95 \mathrm{~g} \mathrm{~mol}^{-1} \times 50 \mathrm{~g}}\)
∆T_f = 2.232 K

Also ∆T_f = T°_f – T_f
T_f = T°_f – ∆T_f
= 273.15 – 2.232
= 270.918 K

(b) (i) 2M glucose has higher boiling point because more the concentration, more is the elevation in boiling point.
(ii) When the external pressure ap plied becomes more than the osmotic pressure of solution, reverse osmosis takes place.