Plus Two Chemistry Previous Year Questions And Answers PDF Free Download, Higher Secondary Plus Two Chemistry Question Bank (Solved) by Anil Kumar K L.
Plus Two Chemistry Previous Year Questions And Answers PDF
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Chemistry Previous Year Question Paper Class 12 with Solutions – 2021
Section – A
Question 1.
Write balanced chemical equations for the following processes:
(a) Cl2 is passed through slaked lime.
(b) SO2 gas is passed through an aqueous solution of Fe (III) salt. [2]
OR
(a) Write two poisonous gases prepared from chlorine gas.
(b) Why does the Cu2+ solution give a blue colour on reaction with ammonia?
Answer:
(a) Cl2 is passed through slaked lime to give bleaching powder [Ca(OCl)2]
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O
(b) When SO2 gas is passed through a Fe(III) aqueous solution, Fe(III) is reduced to Fe(II) ion:
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+
OR
(a) Two poisonous gases prepared from chlorine – Phosgene (COCl2) and tear gas (CCl3NO2).
(b) Nitrogen in ammonia has a lone pair of electrons, which makes it a Lewis base. It donates the electron pair and forms linkage with metal ions-
Question 2.
Give reasons:
(a) Cooking is faster in a pressure cooker than in cooking the pan.
(b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water. [2]
Answer:
(a) Boiling points increase in increasing the pressure in case of liquids. Water used for cooking attains a higher temperature than the usual boiling temperature inside the pressure cooker due to the existing high pressure inside the pressure cooker vessel. This leads to a faster flow of water inside the vegetables or grains etc. resulting in faster cooking of food in a pressure cooker than in the cooking pan.
(b) Red blood cells shrink when placed in saline water because of exosmosis, i.e., water comes out from the cell to surrounding (more concentrated) to equate the concentration. Whereas, when placed in distilled water concentration within the cell becomes more than the surrounding, hence water comes inside and endosmosis takes place to equate the concentrations.
Question 3.
Define the order of the reaction. Predict the order of reaction in the given graphs:
where [R]0 is the initial concentration of reactant and t1/2 is a half-life. [2]
Answer:
It is defined as the sum of powers to which the concentration terms are raised in the rate law equation.
(a) In this graph, as t1/2 is independent of initial reactant concentration, it is a first-order reaction.
(b) In this graph, as tin is directly proportional to the initial concentration of reactant hence, it is a zero-order reaction.
Question 4.
When FeCr2O4 is fused with Na2CO3 in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured compound (C). An acidified solution of compound (C) oxidises Na2O3 to (D). Identify (A), (B), (C) and (D). [2]
Answer:
Question 5.
Write IUPAC name of the complex [Co(en)2(NO2)Cl]+. What type of structural isomerism is shown by this complex? [2]
OR
Using IUPAC norms, write the formulae for the following complexes:
(a) Hexaaquachromium (III) chloride
(b) Sodium trioxalatoferrate (III)
Answer:
IUPAC name of [Co(en)2(NO2)Cl]+ is Chlorobis(ethane-1, 2-diamine)nitrocobalt(III).
This compound shows geometrical isomerism.
OR
(a) Hexaaquachromium(III) chloride- [Cr(H2O)6]Cl3
(b) Sodium trioxalatoferrate(III)- Na3[Fe(C2O4)3]
Question 6.
(a) Although both [NiCl4]2- and [Ni(CO)4] have sp3 hybridisation yet [NiCl4]2- is paramagnetic and [Ni(CO)4] is diamagnetic. Give reason. (Atomic no. of Ni = 28).
(b) Write the electronic configuration of d5 on the basis of crystal field theory when
(i) ∆0 < P and (ii) ∆0 > P [2]
Answer:
(a) [NiCl4]2- is a high spin complex and there are two impaired electrons with 3d8 electronic configuration of a central metal atom, hence it is paramagnetic. Whereas in [Ni(CO)4] Ni is in zero oxidation state and contains no unpaired electrons, hence it is diamagnetic in nature.
(b) (i) Electronic configuration of d5 when ∆o < P is given as t2g3 eg2
(ii) Electronic configuration of d5 when ∆o > P is given as t2g5 eg0
Question 7.
Write structures of main compounds A and B in each of the following reactions: [2]
Answer:
Question 8.
Arrange the following in decreasing order of basic character: [1]
C6H5NH2, (CH3)3N, C2H5NH2
Answer:
Decreasing order of basic character:
CH3CH2NH2 > (CH3)3N > C2H5NH2
Question 9.
What type of colloid is formed when a solid is dispersed in a liquid ? Give an example. [1]
Answer:
Sols are formed when a solid is dispersed in a liquid.
Example – Paints.
Question 10.
Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why?
Answer:
Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In Chlorobenzene the carbon bearing the halogen is a part of an aromatic ring and is electron-rich due to the electron density in the ring.
Question 11.
What is the basic structural difference between starch and cellulose? [1]
OR
Write the products obtained after hydrolysis of DNA.
Answer:
Starch consists of two components- amylose and amylopectin. Amylose is a long linear chain of α-D -(+)-glucose units joined by C1-C4 glycosidic linkage (α-link). Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C1-C4 glycosidic linkage and the branching occurs by C1-C6 glycosidic linkage. On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1-C4 glycosidic linkage (β-link).
OR
Hydrolysis of DNA yields a pentose sugar (β-D-2deoxyribose), phosphoric acid and nitrogen-containing heterocyclic compounds called bases (Adenine, Guanine, Cytosine and Thymine).
Question 12.
(a) The conductivity of 0.001 mol L-1 acetic acid is 4.95 × 10-5 S cm-1. Calculate the dissociation constant if ∧0m for acetic acid is 390.5 S cm2 mol-1.
(b) Write Nest equation for the reaction at 25°C:
2Al(s) + 3Cu2+ (aq) → 2 Al3+ (aq) + 3Cu (s) (c)
What are secondary batteries? Give an example. [5]
OR
(a) Represent the cell in which the following reaction takes place:
2Al (s) + 3 Ni2+ (0.1M) → 2Al3+ (0.01M) + 3 Ni (s)
Calculate its emf if E0cell = 1.41 V.
(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity (∧m0) for weak electrolyte?
Answer:
(a) Conductivity ∧m of a solution is given by the following equation:
∧m = kc,
where k is dissociation constant and c is the concentration of the solution.
Here, given.
Conductivity, k = 4.95 × 10-5 S cm-1
Limiting molar conductivity,
∧0m = 390.5 S cm2 mol-1
Concentration,
c = 0.001 mol L-1 = 1 × 10-3 mol L-1
Substituting the given values in above equation
Molar conductivity,
(b) Nemst equation for the given reaction can be written as
(c) A secondary battery can be recharged after use, bypassing current through it in the opposite direction so that it can be used again.
Example: The most important secondary cell is lead storage cell. It consists of a lead anode and a grid of lead packed with lead dioxide as a cathode. A 38% solution of sulphuric acid is used as an electrolyte.
OR
(a) The cell can be represented as
(b) For strong electrolytes, the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With a decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ∧m0.
In the case of weak electrolytes as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in solution and thus, there is an increase in molar conductivity, also there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at infinite dilution (zero concentration).
So, limiting molar conductivity for weak electrolytes are obtained by using Kohlrausch law, from the limiting molar conductivities of individual ions (λ0).
Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.
∧m0 = λ0+ + λ0
Question 13.
(a) Give the equation of the following reactions:
(i) Phenol is treated with cone. HNO3.
(ii) Propene is treated with B2H6 followed by H2O2/OH-.
(iii) Sodium t-butoxide is treated with CH3Cl.
(b) How will you distinguish between butane-l-ol and butane-2-ol?
(c) Arrange the following in increasing order of acidity: Phenol, ethanol, water [5]
OR
(a) How can you obtain Phenol from (i) Cumene, (ii) Benzene sulphonic acid, (iii) Benzene diazonium chloride?
(b) Write the structure of the major product obtained from the denitration of 3-methyl phenol.
(c) Write the reaction involved in Kolbe’s reaction.
Answer:
(a) (i) Phenol is treated with conc. HNO3 to obtain 2,4,6-trinitrophenol picric acid.
(ii) Propene undergoes hydroboration-oxidation when treated with B2H6 followed by hydrogen peroxide in basic medium to give propan-1-ol.
(iii) Methyl tert-butyl ether is produced when sodium tert-butoxide is treated with methyl chloride.
(b) Butan-l-ol and Butan-2-ol can be distinguished using Lucas reagent (ZnCl2+HCl), where butan-2-ol would react with Lucas reagent in around 5 minutes to give a white precipitate of 2-chlorobutane, whereas butan-l-ol won’t give any reaction at room temperature.
(c) Increasing order of acidity can be given as Ethanol < water < phenol
OR
(a) (i) Phenol from cumene
(ii) Phenol from benzene sulphonic acid
(iii) Phenol from benzene diazonium chloride
(b) The combined influence of -OH and -CH3 groups determine the position of the entering groups, also the sterically hindered positions are not substituted.
(c) In Kolbe’s reaction phenol is reacted with CO2 in the presence of sodium hydroxide, followed by acidification, to give a carboxylic acid group on 2-position of phenol-
