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Day 252: JEE Mains Chemistry

5 questions. Try each one first, then open its answer.

  1. Question 1

    Amines. Asked once: 2021 Paper 14 (same type asked 3 times) See all 3 of this type

    28
    Assertion (A): Primary aromatic amine cannot be prepared by Gabriel phthalimide reaction. Reason (R): Primary aromatic amine cannot be produced by nucleophilic substitution reaction

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  2. Question 2

    Chemical Bonding and Molecular Structure. Asked once: 13 Apr 2023 (same type asked 3 times) See all 3 of this type

    133
    Given below are two statements: Statement I: SO2SO_{2} and H2OH_{2}O both possess VV-shaped structure. Statement II: The bond angle of SO2SO_{2} is less than that of H2OH_{2}O. In the light of the above statements, choose the most appropriate answer from the options given below:

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  3. Question 3

    Chemical Bonding and Molecular Structure. Asked once: 21 Jan 2026 S2 (same type asked 3 times) See all 3 of this type

    39
    The correct increasing order of C−H(A)C - H(A), C−O(B),C=O(C)C - O(B),C = O(C) and C≡N(D)C \equiv N(D) bonds in terms of covalent bond length is:

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  4. Question 4

    Coordination Compounds. Asked once: 25 Jan 2023 (same type asked 3 times) See all 3 of this type

    141
    Total number of moles of AgClAgCl precipitated on addition of excess of AgNO3AgNO_{3} to one mole each of the following complexes [Co(NH3)4Cl2]Cl,[Ni(H2O)6]Cl2,[Pt(NH3)2Cl2]\left\lbrack Co\left( NH_{3} \right)_{4}Cl_{2} \right\rbrack Cl,\left\lbrack Ni\left( H_{2}O \right)_{6} \right\rbrack Cl_{2},\left\lbrack Pt\left( NH_{3} \right)_{2}Cl_{2} \right\rbrack and [Pd(NH3)4]Cl2\left\lbrack Pd\left( NH_{3} \right)_{4} \right\rbrack Cl_{2} is

  5. Question 5

    Equilibrium. Asked once: 25 Jun 2022 (same type asked 3 times) See all 3 of this type

    56
    The standard free energy change (ΔG∘)\left( \Delta G^{\circ} \right) for 50%50\% dissociation of N2O4N_{2}O_{4} into NO2NO_{2} at 27∘C27^{\circ}C and 1 atm pressure is −xJmol−1- xJmol^{- 1}. The value of xx is (Nearest Integer) [Given: R=8.31 J K−1 mol−1,log⁡1.33=0.1239R = 8.31\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1},\log1.33 = 0.1239 ln⁡10=2.3\ln10 = 2.3 ]