Students must start practicing the questions from RBSE 12th Chemistry Model Papers Board Model Paper 2022 with Answers in English Medium provided here.
RBSE Class 12 Chemistry Board Model Paper 2022 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) On heating zinc oxide turns yellow. The crystalline defect arises in the process is: (1)
(a) Metal excess defect
(b) Impurity defect
(c) Metal deficiency defect
(d) Schottky defect
Answer:
(a) Metal excess defect
(ii) The group of oxide ores is: (1)
(a) Bauxite, Siderite
(b) Bauxite, Cuprite
(c) Haematite, Siderite
(d) Haematite, Copper pyrites
Answer:
(b) Bauxite, Cuprite
(iii) The oxidation state of Ni in [Ni (CO)4] is: (1)
(a) 0
(b) +1
(c) +2
(d) +4
Answer:
(a) 0
(iv) The hybridization state of the carbon attached to the halogen in CH2 = CH – Cl is: (1)
(a) sp
(b) sp2
(c) sp3
(d) dsp2
Answer:
(b) sp2
(v)
The name of above reaction on taking copper metal instead of cuprous salt, will be: (1)
(a) Sandmeyer’ s reaction
(b) Gattermann’s reaction
(c) Friedel reaction
(d) Ullmann reaction
Answer:
(b) Gattermann’s reaction
(vi) Name of major product on mixing phenol in mixture of cone. HNO3 and cone. H2SO4 will be: (1)
(a) 2-nitrophenol
(b) 4-nitrophenol
(c) 2,4-di nitrophenol
(d) 2,4,6-trinitrophenol
Answer:
(d) 2,4,6-trinitrophenol
(vii) Aldohexose mono-saccharide is: (1)
(a) Ramnose
(b) Fructose
(c) Glucose
(d) Sucroce
Answer:
(c) Glucose
(viii) Water-soluble vitamin is: (1)
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D
Answer:
(c) Vitamin C
(ix) The number of peptide bonds in the compound formed by the combination of glycine and alanine will be: (1)
(a) 1
(b)2
(c) 3
(d)4
Answer:
(a) 1
Question 2.
Fill in the Blanks:
(i) __________ furnace is used for the smelting of haematite. (1)
Answer:
Blast
(ii) The chemical formula of copper glance is __________. (1)
Answer:
Cu2S
(iii) The general oxidation state of Lanthanoids is __________. (1)
Answer:
+3
(iv) The coordination number of Co in complex [Co (en)3]2+ is __________. (1)
Answer:
6
Question 3.
Very Short Answer Type Questions:
(i) Write the chemical formula of the solid showing both Frenkel and Schottky defects. (1)
Answer:
AgBr shows both Frenkel and Schottky defects.
(ii) Write Henry’s Law. (1)
Answer:
Henery’s law states that “at a constant temperature the solubility of a gas in a liquid is directly proportional to the pressure of the gas”.
(iii) Write the name of bond responsible for dimerisation of ethanoic acid in benzene solution. (1)
Answer:
Hydrogen bond is responsible for dimerisation of ethanoic acid in benzene solution.
(iv) FeO + SiO2 → [A]
Write the chemical formula of compound [A] in the above reaction. (1)
Answer:
(v) Explain the reason for Lanthanoid contraction. (1)
Answer:
Lanthanoid contract on occurs due to inclusion of fourteen elements of lanthanoid series between lanthanum and hafnium.
There is regular decrease in atomic size from cerium to lutetiam due to poor shielding of 4/-electrons.
(vi) Write the chemical formula of the main product obtained when ethanol is heated with concentrated H2SO4 at 413 K. (1)
Answer:
(vii) Write the chemical equation for the diazotisation of aniline. (1)
Answer:
(viii) Draw the orbital diagram of Trimethylamine. (1)
Answer:
Section – B
Short Answer Type Questions
Question 4.
Atoms of element B form hep lattice and those of the element A occupy 1/3rd of tetrahedral voids. What is the formula of the compound formed by the element A and B? (1½)
Answer:
Let, number of atoms in hep lattice = N
Number of atoms in tetrahedral voids = 2N
As element B is arranged in hep lattice,
Hence, the number of atoms of B = N
As elements A occupies \(\frac{1}{3}\) rd of tetrahedral voids
Number of atoms of A = \(\frac{1}{3}\) × Tetrahedral voids = \(\frac{1}{3}\) × 2N = \(\frac{2N}{3}\)
Ratio of elements = A : B = \(\frac{2N}{3}\) : N = 2 : 3
Hence, formula of compound is = A2B3
Question 5.
Compare paramagnetism and ferromagnetism. (1½)
Answer:
Paramagnetism | Ferromagnetism |
1. The substances, which are attracted by the external magnetic field are known as paramagnetic substances and this property is known as paramagnetism. | 1. The substances which have permanent magnetism even in the absence of magnetic field are called ferromagnetic substances and this property is known as ferromagnetism. |
2. Paramagnetism exhibited by those substances which have impaired electrons. | 2. Ferromagnetism exhibited by those substances which have domains. |
3. Example: O2, Cu2+, Fe3+ | 3. Example: Fe, Co, Ni |
Question 6.
Draw a diagrammatic representation of the process of reverse osmosis. (1½)
Answer:
Question 7.
Explain the ideal solution on the basis of following points – (¾ + ¾ = 1½)
(a) Change in enthalpy
Answer:
Change in enthalpy: An ideal solution is formed by mixing two components with similar properties. Hence, no heat should be absorbed or evolved when the two components are mixed so that enthalpy change of an ideal solution should be zero.
∆H(mixing) = 0
(b) Change in Volume
Answer:
Change in volume: In ideal solution intermolecular interactions between the components (A – B or solvent-solute) are of the same magnitude as the intermolecular intractions found in the pure components.
So, the volume of solution should be equal to the sum of volume of two components, means volume change on mixing should be zero.
∆V(mixing) = 0
Question 8.
Explain the reason for the value of order of reaction is 1, while the value of molecularity is 2 in above reaction. (1½)
Answer:
The given reaction
is a pseudo first order reaction.
Hence, molecularity of reaction is 2 as two molecules of ester and water taking part in reaction, but here water present in excess means it does not affect rate of reaction.
So that order of reaction is first.
Question 9.
Show thatin a first order reaction, time required for completion of 99.9% is 10 times of half-life (t½) of the reaction. (1½)
Answer:
For the first order reaction,
K = \(\frac{2.303}{t}\)log\(\frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
Time required for 99.9% completion of reaction is
We know that for the first order reaction, half life period is
t½ = \(\frac{0.693}{K}\)
Comparing the equations (1) and (2)
\(\frac{t}{t_{1 / 2}}=\frac{6.909}{\mathrm{~K}} \times \frac{\mathrm{K}}{0.693}\)
\(\frac{t}{t_{1 / 2}}=\frac{6.909}{0.693}\) = 10 = 10 × t½
Thus, it is proved that time required for completion of 99.9% is 10 times of half life of the reaction.
Question 10.
Explain the following properties exhibiting by transition elements: (¾ + ¾ = 1½)
(a) Variable oxidation state
Answer:
Variable oxidation state: Transition metals show a large number of oxidation states. The variable oxidation state of a transition metal is due to the participation of (n – 1)d and ns electrons in bonding because the energy difference of ns and (n – 1)d subshell are very less.
(b) Catalytic property
Answer:
Catalytic properties: Many transition elements and their compounds act as catalyst. They act as catalyst because of the following reasons:
(i) They have variable oxidation states.
(ii) They easily absorb and re-emit the wide range of energies to provide the necessary activation energy.
Question 11.
Calculate the “Spin only” magnetic moment for bi-positive ion of an element M having atomic number 26. (1½)
Answer:
26M 1s2 2s2 2p6 3s2 3p6 3d6 4s2
26M2+ 3s2 3p6 3d6 4s0
Number of impaired electrons (n) = 4
Formula of ‘spin only’ magnetic moment
μ = \(\sqrt{n(n+2)}\) B.M.
μ = \(\sqrt{4(4+2)}\)
μ = \(\sqrt{24}\) = 4.89BM.
Question 12.
Draw the orbital diagram of carbonyl group showing the n bond. (1½)
Answer:
Question 13.
Write the chemical formula of [A], [B] and [C] in above reaction sequence. (1½)
Answer:
Question 14.
Arrange HCOOH, CH3COOH, ClCH2COOH in ascending order of their acidic strengths. (1½)
Answer:
Ascending order of acidic strengths:
HCOOH, CH3COOH, ClCH2COOH
Question 15.
Nitro group present at ortho and positions increases the acidic strength of phenol. Explain. (1½)
Answer:
Nitro group present on ortho and para positions increase the acidic strength of phenol because nitro group (-NO2) is an electron with drawing group, thus it decrease electron density of ring. When electron density of ring decrease oxygen atom of phenol withdraw more elctrons from hydrogen and hydrogen released in the form of H+ ion.
This electron withdrawing nature of nitro group acts only when it is present at ortho and para positions, because negative charge of resonating electron comes only at ortho and para positions, not at meta positions.
Section – C
Long Answer Type Questions:
Question 16.
Define order of reaction. Derive integrated rate equation for zero order reaction. (1 + 2 = 3)
OR
Define rate of reaction. Explain temperature dependence of rate of reactionon the basis of activation energy (Ea). (1 + 2 = 3)
Answer:
Order of reaction:
The sum of powers of the concentration of reactants in the rate law expression is called the order of that chemical reaction.
For a general chemical reaction
aA + bB → cC + dD
Rate law expression is
Rate = k[A]x [B]y
Order of reaction = x + y
Order of a reaction can be 0, 1, 2, 3 and even a fraction.
Rate equation for zern order reaction:
Consider a general reaction:
A → Product
at t = o Concentration [A]0
at t = 0 Concentration = [A]
For zero order reaction
rate = k[A]° …(1)
rate = \(\frac{-d[\mathrm{~A}]}{d t}\) …..(2)
Since, both the equations (1) and (2) gives the rate of reaction, therefore
k[A]° = \(\frac{-d[\mathrm{~A}]}{d t}\)
k = \(\frac{-d[\mathrm{~A}]}{d t}\)
-kAt =d[A]
d[A] = -kAt …..(3)
Integrating both the sides, ∫ d[A] = ∫-kdt.
[A] = -kt + c …(4)
Here, C is the integration constant. If t = 0 and [A] = [A]0, then substituting these values in equatoin (4)
[A]0 = -K × 0 + C
[A]0 = C
Substituting the value of C in equation (4)
[A] = -kt + [A]0
kt = [A]0 – [A]
k = \(\frac{[A]_{0}-[A]}{t}\) ……..(5)
Equation (5) is the integrated equation for zero order reaction.
OR
Rate of reaction:
When a chemical reaction proceed, concentration of reactants decreases and that of products increases.
Thus, change in the molar concentration of any one of the reactants or products per unit time is called rate of reaction.
Units of rates of reaction are mol L-1s-1.
Rate of reaction = \(=\frac{\text { Change in concentration of reactant or product }}{\text { Time taken in change }}\)
Rate of reaction = ±\(\frac{d x}{d t}\)
Effect of temperature:
On increasing the temperature of a reaction, the kinetic energy of reactant molecule increases. Hence, rate of reaction also increases.
Collision between the reactant molecules which posses energy less than the threshold energy do not form products. This means the reaction involves some energy barrier which must be crossed before the reactants are converted into products. The energy barrier is called activation energy barrier.
Thus, some extra energy inform of temperature has to be supplied to the reactants to cross this energy barrier.
So, when we increase temperature of reactants it helps them to attain threshold energy requirement or activation energy requirement full filled by temperature.
It has been found that reaction rate almost becomes double for every 10°C rise in temperature.
Question 17.
(i) Write the nature of carbon-halogen (C-X) bond.
(ii) Compound [X] is obtained on benzene react with Cl2 in presence of FeCl3. Compound [Y] is obtained on mixing compound [X] with sodium in presence of dry ether. Write the names of compounds [X] and [Y] and write equations of chemical reactions involved. (1 + ½ + ½ + 1 = 3)
OR
(i) Write the full name of DDT.
(ii) Compound [X] is obtained on ethanol react with PCl5. Compound [Y] is obtained on
heating compound [X] with alcoholic KOH. Write the names of compounds [X] and [Y] and write the equations of chemical reactions involved. (1 + ½ + ½ + 1 = 3)
Answer:
(i) Nature of Carbon-Halogen bond is polar. This is because of electronegativity difference between carbon and halogen atom.
OR
(i) DDT—p, p’-Dichlorodiphenyl trichioroethane
Question 18.
(i) Give resonating strudures of anilinium cation.
(ii) Arrange
order of their basic strengths.
(iii) Write the reason for solubility of Butan-1-oI in water is more as compare to Butan-1-Amine. (1 + 1 + 1 = 3)
OR
(i) Write the structural formula and chemical name of Hinesburg’s reagent.
(ii) Write the balanced chemical equation of carbyl amine reaction exhibiting by ethanamine.
(iii) Write the reason for basic strength of CH3NH2 is more as compare to C6H5NH2 (1 + 1 +1 =3)
Answer:
(ii) Ascending order of basic strength CH3
(iii) Butan-l-ol is more water soluble than butan-l-amine due to strong hydrogen bond. Butan-l-ol has oxygen atom while butan-l-amine has nitrogen atom which is less electronegative than oxygen atom.
Less partial charge on nitrogen atom leads to weak hydrogen bond formation with water.
OR
(i) Benzene sulphonyl chloride (C6H5SO2Cl) is termed as Hinesberg’s reagent.
(iii) Aniline (C6H5NH2) is less basic than methylamine because the lone pair of electrons on nitrogen atom gets delocalised over the benzene ring are unavailable for protonation due to resonance in aniline which is absent in case of alkylamine (CH3NH2).
Section – D
Essay Type Questions :
Question 19.
(i) Define Homoleptic and Hetroleptic complexes.
(ii) Solution of [Ti(H2O)6]3+ is coloured. Givereasons.
(iii) Draw the crystal field splitting of degenerate d-orbitals of free metal ion in octahedral crystal field. (1 + 1+ 1+1 = 4)
OR
(i) Write the type of isomerism exhibited by [Co (NH3)5 (SO4)] Br and [Co (NH3)5Br] SO4 and also define it.
(ii) [Co (NH3)6]3+ is diamagnetic while [CoF6]3- is paramagnetic. Give reasons.
(iii) Draw the crystal field splitting of degenerate d-orbitals of free metal ion in tetrahedral crystal field. (1+ 1 + 1 + 1 = 4)
Answer:
(i) Homoleptic Complex: The complexes in which central metal atom or ion is attached to only one kind of ligands are called homoleptic complexes.
Example: [Ni (NH3)6]3+
(ii) Heteroleptic Complexes: The complexes in which central metal atom or ion is attached to more than one kind of ligands are called heteroleptic complexes.
Example: [NiCl2(H2O)4]
(ii) [Ti(H2O)6]3+ Complex solution is coloured due to presence of unpaired electron in 3d orbital of Ti+3 ion.
[Ti(H2O)6]3+
Oxidation State ⇒ x + 6(0) = + 3
x = + 3
(iii) Crystal field splitting of degenerate d-orbitals:
OR
(i) Complex [Co(NH3)5 (SO4)] Br and [Co (NH3)5 Br] SO4 show ionisation isomerism.
Ionisation isomerism is a type of structural isomerism in that molecular formula of two complexex is same but they form different kind of ions in their solutions.
This is being shown here :
[Co(NH3)5(SO4)]Br → [Co(NH3)5 SO4]+ Br–
[Co(NH3)5Br]SO4 → [Co(NH3)5Br]+2 + SO4-2
Here, oxidation state of cobalt in both of complex is same.
27CO 3d7 4s0
27CO3+ 3d6 4s0
Above electronic configuration shows that cobalt has unpaired electron in its d-orbitals, but in [CO(NH3)6]3+ complex ligands are amine molecules that are strong ligands. In presence of strong ligand unpaired electron of d-orbitals move inward to provide inner orbitals for hybridization. So that no unpaired electron left over. In absence of unpaired electron complex becomes diamagnetic.
In another complex [CoF6]3-, fluorine is a weak ligand, so outer orbitals are used for bond formation. Means impaired electrons remains in inner orbitals. Due to presence of these unpaired electrons complex become paramagnetic.
Question 20.
(i) Write the name and chemical formula of aldehyde obtained from vanilla beans.
(ii) Write the chemical equations for following conversions:
(a) Benzaldehyde from benzene
(b) Acetaldoxime from acetaldehyde
(iii) Give reasons for acidity of oc-hydrogen atom of aldehyde. (½ + ½ + 1 + 1 + 1 = 4)
OR
(i) Write the name and chemical formula of aldehyde obtained from cinnamon.
(ii) Write the chemical equations for following conversions—
(a) Benzaldehyde from benzoyal chloride
(b) Acetone hydrozone from acetone
(iii) Give reasons for boiling point of carboxylic acids is more as compare to aldehydes of comparable molecular mass. (½ + ½ + 1 + 1 + 1 = 4)
Answer:
(i) Aldehyde found vanilla beans is C8H8O3 and name of it is vanillin.
This reaction is known as Gattermann-Koch formylation.
(iii) The acidic character of a-hydrogen can be explained as follows:
Due to electron withdrawing inductive effect of the carbonyl group, it withdraws electrons from the adjacent C-C bond.This makes a-carbon atom electron deficient. When a-carbon atom withdraws electrons from the Cα – H Bond. As a result the electron desity in Cα-H Bond decrease and H atom from a-carbon released in form of H+ ion.
OR
(i) Aldehyde obtained from cinnamon is cinnamaldehyde and chemical formula of it is
C6H5CH = CHCHO
(iii) Carboxylic acids have higher boiling points than those of aldehydes of comparable molecular masses. This occurs due to extensive association of carboxylic acid molecules through intermolecular hydrogen bonding.
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