Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2021-Paper-2 SECTION 3 • This section contains TWO (02) p…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2021-Paper-2 Single Correct MCQ
Published on: August 12, 2026

SECTION 3

• This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions.

• Each question has FOUR options

A
, P - iii, Q - iv, R- ii, S -i Initiation step is exothermic with H° = -58 kcal mol -1 .
B
, P - i, Q - ii, R- iii, S - iv Propagation step involving *CH 3 formation is exothermic with H° = -2 kcal mol -1 .
C
and P - iii, Q - ii, R-i, S - iv Propagation step involving CH 3 C1 formation is endothermic with H° = +27 kcal mol -1 .
D
. ONLY ONE of these four options is the correct answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme: Full Mark +3 If ONLY the correct option is chosen; Zero Marks 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks -1 In all other cases. Paragraph The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by ^-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below: (i) Correct match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is Column J Column K Molecule BDE (kcl mol -1 ) (P) H-CH(CH 3 ) 2 (i) 132 (Q) H-CH 2 Ph (ii)no (R) H-CH = CH 2 (iii) 95 (S) H-C CH (iv)88 P - ii, Q - i, R- iv, S - iii (ii) For the following reaction CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) The correct statement is The reaction is exothermic with H° = -25 kcal mol -1 .

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CHECK THE SOLUTION.

(i) (ii)

(i) As S character increases bond dissociation energy increases

H-C = CH H-CH = CH 2

sp sp 2

H-CH(CH 3 ) 2 H-CH 2 -Ph

sp 3 sp 3

And is more stable then

(ii) CH 4 + C1 2 CH 3 C1 + HC1 this reaction is obtained from given reaction.

CH 3 -H ...(1)

C1-C1 ...(2)

CH 3 -Cl ...(3)

H-C1 ...(4)

(1) + (2)-(3)-(4)

Hence H-105+ 58-85-103--25 KCal/mole

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