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JEE Advanced 2021 Paper 2

Question Stem for Question Nos.32 and 33

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 s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :

Cl−Cl(g)⟶Cl∙(g)+Cl∙(g)‌∆H∘=58kcalmol−1
H3C−Cl(g)⟶H3C∙(g)+Cl∙(g)‌∆H∘=85kcalmol−1
H−Cl(g)⟶H∙(g)+Cl*(g)‌∆H∘=103kcalmol−1

Question : 32

Total: 57

Correct match of the C−H bonds (shown in bold) in Column J with their BDE in Column K is(S) H−C≡CH

Column J Molecule	Column K BDE (kcal mol ‌−1 )
(P) H−CH(CH3)2	(i) 132
(Q) H−CH2Ph	(ii) 110
(R) H−CH=CH2	(iii) 95
(S) H−C≡CH	(iv) 88

P−iii,Q−iv,R−ii,S−i
P−i,Q−i,R−iii,S−iv
P−iii,Q−ii,R−i,S−iv
P−ii,Q−i,R−iv,S−iii

Validate

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