reactant molecules during collision","rate constant at two different temperatures","rate constant at standard temperature","probability of collision"],"answer":2,"why":"The Arrhenius equation relates two rate constants at two temperatures to Eₐ.","simPath":"Class-12/Chemistry/Chapter-3/example-3.10-theory-arrhenius-equation-and-activation-energy","simTitle":"Activation Energy"},{"qNum":60,"topic":"Chemical Kinetics","question":"Which plot of ln k vs 1/T is consistent with the Arrhenius equation?","options":["a straight line with positive slope","a straight line with negative slope","an increasing curve","a line rising from the origin"],"answer":2,"why":"ln k = ln A − (Eₐ/R)(1/T), a straight line with negative slope.","figureNote":"Four ln k vs 1/T plots.","simPath":"Class-12/Chemistry/Chapter-3/problem-3.22-arrhenius-plot-for-n2o5-decomposition","simTitle":"Arrhenius Plot"},{"qNum":61,"topic":"p-Block Elements","question":"Statement I: The boiling points of hydrides of Group 16 follow the order H₂O > H₂Te > H₂Se > H₂S. Statement II: On the basis of molecular mass, H₂O is expected to have a lower boiling point than the other members, but due to extensive H-bonding it has a higher boiling point. Choose the correct answer.","options":["Statement I is true but Statement II is false","Statement I is false but Statement II is true","Both Statement I and Statement II are true","Both Statement I and Statement II are false"],"answer":3,"why":"The BP order and the hydrogen-bonding explanation for water are both correct.","simPath":"Class-11/Chemistry/Chapter-4/prompt-43-hydrogen-bond","simTitle":"Hydrogen Bonding"},{"qNum":62,"topic":"Biomolecules","question":"The reagents with which glucose does NOT react to give the corresponding tests/products are: A. Tollen’s reagent, B. Schiff’s reagent, C. HCN, D. NH₂OH, E. NaHSO₃.","options":["B and E","E and D","B and C","A and D"],"answer":1,"why":"Glucose does not respond to Schiff’s reagent (B) or form a bisulfite adduct with NaHSO₃ (E), indicating the aldehyde exists largely in cyclic form.","simPath":"Class-12/Chemistry/Chapter-10/problem-10.09-glucose-reactions-hi-br2-hno3","simTitle":"Glucose Reactions"},{"qNum":63,"topic":"Periodic Properties","question":"Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N.","options":["Li < Be < C < B < N","Li < Be < N < B < C","Li < Be < B < C < N","Li < B < Be < C < N"],"answer":4,"why":"Boron’s 2p electron is easier to remove than beryllium’s filled 2s, giving Li < B < Be < C < N.","simPath":"Class-11/Chemistry/Chapter-3/prompt-18-second-period-ionization-order","simTitle":"Ionization Enthalpy"},{"qNum":64,"topic":"Periodic Properties","question":"Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si.","options":["O < F < N < C < Si","F < O < N < C < Si","Si < C < N < O < F","Si < C < O < N < F"],"answer":3,"why":"Electronegativity increases across a period and up a group: Si < C < N < O < F.","simPath":"Class-11/Chemistry/Chapter-3/prompt-24-ege-vs-electronegativity","simTitle":"Electronegativity"},{"qNum":65,"topic":"d- and f-Block","question":"The E° value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of:","options":["d⁴ to d⁵ configuration","d³ to d⁵ configuration","d⁵ to d⁴ configuration","d⁵ to d² configuration"],"answer":1,"why":"Reduction of Mn³⁺ (d⁴) to Mn²⁺ (d⁵) gives the extra-stable half-filled d⁵, making E° strongly positive.","simPath":"Class-12/Chemistry/Chapter-4/example-4.07-mn3plus-mn2plus-reduction-potential","simTitle":"Reduction Potentials"},{"qNum":66,"topic":"Chemical Bonding","question":"Match List I (Compound) with List II (Shape/geometry): A. NH₃, B. BrF₅, C. XeF₄, D. SF₆ with I. Trigonal Pyramidal, II. Square Planar, III. Octahedral, IV. Square Pyramidal.","options":["A-III, B-IV, C-I, D-II","A-II, B-III, C-IV, D-I","A-I, B-IV, C-II, D-III","A-II, B-IV, C-III, D-I"],"answer":3,"why":"NH₃ trigonal pyramidal (

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