Practice Questions

Q55.Lanthanoid ions with 4f 7 configuration are : (A) Eu2+ (B) Gd3+ (C) Eu3+ (D) Tb3+ (E) Sm2+ Choose the correct answer from the options given below : (1) (A) and (D) only (2) (B) and (C) only (3) (A) and (B) only (4) (B) and (E) only

Q55.Consider the reaction X2Y( g) = X2( g) + 12 Y2( g) The equation representing correct relationship between the degree of dissociation (x) of X2Y(g) with its equilibrium constant Kp is ______ . Assume x to be very very small. (1) (2) x = 3√2Kpp x = 3√2Kp2p (3) (4) x = 3√Kpp x = 3√Kp2p 2025 (23 Jan Shift 2) JEE Main Previous Year Paper

Q55.Match List - I with List - II. Choose the correct answer from the options given below : (1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II) (2) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Q55.Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct? (1) acetophenone < benzaldehyde < p-tolualdehyde (2) benzaldehyde < acetophenone < p- < p- nitrobenzaldehyde nitrobenzaldehyde < p-tolualdehyde (3) p- nitrobenzaldehyde < benzaldehyde < p- (4) acetophenone < p-tolualdehyde < benzaldehyde tolualdehyde < acetophenone < p- nitrobenzaldehyde

Q55.Given below are two statements : Statement (I) : The first ionization energy of Pb is greater than that of Sn . Statement (II) : The first ionization energy of Ge is greater than that of Si . In the light of the above statements, choose the correct answer from the options given below : (1) Statement I is false but Statement II is true (2) Statement I is true but Statement II is false (3) Both Statement I and Statement II are true (4) Both Statement I and Statement II are false

Q55.The correct order of stability of following carbocations is : (1) C > B > A > D (2) A > B > C > D (3) B > C > A > D (4) C > A > B > D

Q55.Match the LIST-I with LIST-II Choose the correct answer from the options given below: (1) A-IV, B-I, C-III, D-II (2) A-IV, B-II, C-I, D-III (3) A-II, B-IV, C-III, D-I (4) A-III, B-II, C-I, D-IV

Q55.Which one of the following, with HBr will give a phenol? 2025 (29 Jan Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)

Q56.Identify correct statements : (A) Primary amines do not give diazonium salts when treated with NaNO2 in acidic condition. (B) Aliphatic and aromatic primary amines on heating with CHCl3 and ethanolic KOH form 2025 (28 Jan Shift 2) JEE Main Previous Year Paper carbylamines. (C) Secondary and tertiary amines also give carbylamine test. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride very easily. Choose the correct answer from the options given below : (1) (A) and (B) only (2) (D) and (E) only (3) (B) and (D) only (4) (B) and (C) only

Q56.Given below are two statements : Consider the following reaction (1) Both Statement I and Statement II are false (2) Statement I is true but Statement II is false (3) Statement I is false but Statement II is true (4) Both Statement I and Statement II are true

Q56. The steam volatile compounds among the following are : (A) (B) (C) (D) Choose the correct answer from the options given below : (1) (B) and (D) Only (2) (A) and (C) Only (3) (A), (B) and (C) Only (4) (A) and (B) Only

Q56.The products A and B in the following reactions, respectively are Ag−NO2 AgCN −A ⟷CH3 −CH2 −CH2 −Br → B (1) CH3 −CH2 −CH2 −NO2, CH3 −CH2 −CH2 −CN (2) CH3 −CH2 −CH2 −ONO, CH3 −CH2 −CH2 −NC (3) CH3 −CH2 −CH2 −ONO, CH3 −CH2 −CH2 −CN (4) CH3 −CH2 −CH2 −NO2, CH3 −CH2 −CH2 −NC

Q56. The maximum number of RBr producing 2-methylbutane by above sequence of reactions is ________ - (Consider the structural isomers only) (1) 5 (2) 4 (3) 3 (4) 1

Q56.Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option. (1) Both ΔH and ΔS are (-ve) (2) ΔH is (-ve) but ΔS is (+ve) (3) ΔH is (+ve) but ΔS is (−ve) (4) Both ΔH and ΔS are (+ve)

Q56.Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of H2 gas. Statement II : Four g of propyne reacts with NaNH2 to liberate NH3 gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below: (1) Statement I is incorrect but Statement II is (2) Both Statement I and Statement II are correct correct (3) Statement I is correct but Statement II is (4) Both Statement I and Statement II are incorrect incorrect 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

Q56.Given below are two statements : Statement (I) : NaCl is added to the ice at 0∘C, present in the ice cream box to prevent the melting of ice cream. Statement (II) : On addition of NaCl to ice at 0∘C, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true

Q56.Identify correct statement/s : (A) −OCH3 and −NHCOCH3 are activating group. (B) -CN and -OH are meta directing group. (C) -CN and −SO3H are meta directing group. (D) Activating groups act as ortho - and para directing groups. (E) Halides are activating groups. Choose the correct answer from the options given below : (1) (A) only (2) (A), (B) and (E) only (3) (A) and (C) only (4) (A), (C) and (D) only

Q56.What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol? (Given molar mass in gmol−1 of C, H, O, Br are 12, 1, 16, 80 respectively ) (1) 20 .44 g (2) 4.0 g (3) 6.0 g (4) 10.22

Q57.For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth ? Where N - Number of Bacteria at any time, N0 - Initial number of Bacteria. (1) (2) (3) (4)

Q57.In a multielectron atom, which of the following orbitals described by three quantum numbers will have same energy in absence of electric and magnetic fields? A. n = 1, l = 0, m1 = 0 B. n = 2, l = 0, m1 = 0 C. n = 2, l = 1, m1 = 1 D. n = 3, l = 2, m1 = 1 E. n = 3, l = 2, m1 = 0 Choose the correct answer from the options given below: (1) B and C Only (2) A and B Only (3) C and D Only (4) D and E Only

Q57.Given below are two statements : Statement (I) : For a given shell, the total number of allowed orbitals is given by n2 . Statement (II) : For any subshell, the spatial orientation of the orbitals is given by −l to +l values including zero. In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are false (2) Statement I is true but Statement II is false (3) Both Statement I and Statement II are true (4) Statement I is false but Statement II is true

Q57.Given below are two statements : Statement ( I): The radii of isoelectronic species increases in the order. Mg2+ < Na+ < F−< O2− Statement (II): The magnitude of electron gain enthalpy of halogen decreases in the order. Cl > F > Br > I In the light of the above statements, choose the most appropriate answer from the options given below : (1) Statement I is incorrect but Statement II is (2) Statement I is correct but Statement II is correct incorrect (3) Both Statement I and Statement II are incorrect (4) Both Statement I and Statement II are correct 2025 (29 Jan Shift 1) JEE Main Previous Year Paper

Q57.Match List - I with List - II : Choose the correct answer from the options given below : (1) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (2) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (4) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Q57.The species which does not undergo disproportionation reaction is : (1) ClO−3 (2) ClO− (3) ClO−2 (4) ClO−4

Q57.Preparation of potassium permanganate from MnO2 involves two step process in which the 1st step is a reaction with KOH and KNO3 to produce (1) K3MnO4 (2) K4 [Mn(OH)6] (3) KMnO4 (4) K2MnO4