Practice Questions

Q32.Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron? (A) NO (B) N2 (C) O2 (D) C2 (E) B2 Choose the most appropriate answer from the options given below : (1) A, B, C only (2) B, C, E only (3) A, C only (4) D only JEE Main 2022 (25 Jun Shift 1) JEE Main Previous Year Paper

Q37.Product ' A ' of following sequence of reactions is (1) (2) (3) (4)

Q37. L-isomer of a compound ′A′(C4H8O4)gives a positive test with [Ag (NH3)2]+ . Treatment of ' A' with acetic anhydride yields triacetate derivative. Compound ' A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and HNO3 respectively. Compound (A) is : (1) (2) (3) (4)

Q38.Among the following marked proton of which compound shows lowest pKa value? (1) (2) (3) (4)

Q38.In the given reaction sequence, the major product ' C' is : HNO3 Br2 alcoholic −−C8H10 → A → B → C H2 SO4 Δ KOH JEE Main 2022 (24 Jun Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)

Q38.The ( ∂T∂E )P of different types of half cells are as follows : A B C D 1 × 10−4 2 × 10−4 0. 1 × 10−4 0. 2 × 10−4 (Where E is the electromotive force). Which of the above half cells would be preferred to be used as reference electrode (1) A (2) B (3) C (4) D

Q39.Which of the following reagents / reactions will convert ' A' to ' B'? (1) PCC oxidation (2) Ozonolysis (3) BH3, H2O2/ −OH followed by PCC oxidation (4) HBr, hydrolysis followed by oxidation by K2 Cr2 O7 .

Q40.The photochemical smog does not generally contain (1) NO (2) NO2 (3) SO2 (4) HCHO

Q40.Consider the following reaction : (1) Electrolysis 2 HSO−4 (aq) (2) Hydrolysis 2 HSO−4 + 2H+ + A The dihedral angle in product A in its solid phase at 110 K is (1) 104° (2) 111. 5° (3) 90. 2° (4) 111. 0°

Q40.The correct order of reduction potentials of the following pairs is A. Cl2 / Cl- B. I2 / I- C. Ag+ / Ag D. Na+ / Na E. Li+ / Li Choose the correct answer from the options given below. (1) A > B > C > E > D (2) A > C > B > E > D (3) A > B > C > D > E (4) A > C > B > D > E

Q44.Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: Mn = 25; Fe = 26) A. [FeF6]3− B. [Fe (CN)6]3− C. [MnCl6]3− (high spin) D. [Mn (CN)6]3− Choose the correct answer from the options given below (1) A < B < D < C (2) B < D < C < A (3) A < C < D < B (4) B < D < A < C

Q45.Two isomers (A) and (B) with Molar mass 184 g/ mol and elemental compositio C, 52. 2%; H, 49% and Br 42. 9% gave benzoic acid and p− bromobenzoic acid, respectively on oxidation with KMnO4 . Isomer ′A′ is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO3 . Isomer ′A′ and ′B′ are, respectively : (1) (2) (3) (4) H3C −CHBr −C6H5 and and H3C −CHBr −C6H5

Q45.The reagent, from the following, which converts benzoic acid to benzaldehyde in one step is (1) LiAlH4 (2) KMnO4 (3) MnO (4) NaBH4

Q46.Isobutyraldehyde on reaction with formaldehyde and K2 CO3 gives compound ' A '. Compound ' A' reacts with KCN and yields compound ' B ', which on hydrolysis gives a stable compound ' C '. The compound ' C ' is (1) (2) (3) (4)

Q46.Decarboxylation of all six possible forms of diaminobenzoic acids C6H3(NH2)2 COOH yields three products A, B and C . Three acids give a product ' A ', two acids gives a product ' B ' and one acid give a product ' C '. The melting point of product ' C ' is (1) 63 °C (2) 90 °C (3) 104 °C (4) 142 °C

Q47.An organic compound ' A ' contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound ' A ' with standard base indicates that the molecular weight of ' A ' is 131 ± 2 . When a sample of ' A ' is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound ' A ' is (1) (2) (3) (4)

Q47.An organic compound 'A' on reaction with NH3 followed by heating gives compound B. Which on further strong heating gives compound C C8H5NO2. Compound C on sequential reaction with ethanolic KOH, alkyl chloride and hydrolysis with alkali gives a primary amine. The compound A is (1) (2) (3) (4)

Q50.When sugar ' X ' is boiled with dilute H2 SO4 in alcoholic solution, two isomers ' A' and ' B' are formed. ' A' on oxidation with HNO3 yields saccharic acid where as ' B' is laevorotatory. The compound ' X' is (1) Maltose (2) Sucrose (3) Lactose (4) Strach

Q61.Let a circle 𝐶 in complex plane pass through the points 𝑧1 = 3 + 4𝑖, 𝑧2 = 4 + 3𝑖 and 𝑧3 = 5𝑖. If 𝑧≠𝑧1 is a point on 𝐶 such that the line through 𝑧 and 𝑧1 is perpendicular to the line through 𝑧2 and 𝑧3, then arg𝑧 is equal to 2 (1) tan-124 - 𝜋 (2) tan-1 - 𝜋 7 √5 (3) tan-13 - 𝜋 (4) tan-13 - 𝜋 4 JEE Main 2022 (25 Jun Shift 1) JEE Main Previous Year Paper 1 1 1 𝐾

Q62.Let for some real numbers α and β, a = α −iβ . If the system of equations 4ix + (1 + i)y = 0 and ¯8(cos 2π3 + i sin 2π3 )x + ay = 0 has more than one solution then αβ is equal to (1) 2 −√3 (2) 2 + √3 (3) −2 + √3 (4) −2 −√3

Q62.Let 𝑆= 𝑧= 𝑥+ 𝑖𝑦: 𝑧- 1 + 𝑖≥𝑧, 𝑧< 2, 𝑧+ 𝑖= 𝑧- 1. Then the set of all values of 𝑥, for which 𝑤= 2𝑥+ 𝑖𝑦∈𝑆 for some 𝑦∈ℝ, is 1 1 1 (2) - (1) -√2, 4 2√2 √2, (3) -√2, 1 (4) - 1 1 2 √2, 2√2

Q62.The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives atleast 4 and atmost 7 candies, C3 receives atleast 2 and atmost 6 candies, is equal to (1) 205 (2) 615 (3) 510 (4) 430 JEE Main 2022 (28 Jun Shift 2) JEE Main Previous Year Paper

Q62.Let (z) represent the principal argument of the complex number z. The, |z| = 3 and arg(z −1) −arg(z + 1) = π4 intersect: (1) Exactly at one point (2) Exactly at two points (3) Nowhere (4) At infinitely many points.

Q62.Consider two G.Ps. 2, 22, 23, … and 4, 42, 43, … of 60 and n terms respectively. If the geometric mean of all 225 the 60 + n terms is (2) 8 , then ∑nk=1 k(n −k) is equal to: (1) 560 (2) 1540 (3) 1330 (4) 2600 n(S) + ∑θ∈S(sec( π4 + 2θ) cosec ( π4 + 2θ)) is equal

Q63.If the constant term in the expansion of (3x3 −2x2 + x5 ) is 2k. l, where l is an odd integer, then the value of k is equal to (1) 6 (2) 7 (3) 8 (4) 9