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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 : 33

Total: 57

For the following reaction CH4(g)+Cl2(g)

‌ light ‌

────▶

CH3Cl(g)+HCl(g) the correct statement is

Initiation step is exothermic with ∆H∘=−58kcalmol−1
Propagation step involving ‌●CH3 formation is exothermic with ∆Ho=−2kcalmol−1.
Propagation step involving CH3Cl formation is endothermic with ∆H∘=+27kcalmol−1.
The reaction is exothermic with ∆Ho=−25kcalmol−1.

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