Comprehension

The amount of energy required to break a bond is the same as the amount of energy released when the same bond is formed. In a gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by the 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:
homolytic cleavage

Question: 1

Correct match of the C–H bonds (shown in bold) in Column J with their BDE in Column K is
 

Column J 
Molecule

Column K 
BDE (kcal mol–1 )

PH–CH(CH3)2\text{H--CH(CH}_3\text{)}_2i132
QH–CH2Ph\text{H--CH}_2\text{Ph}ii110
RH–CH=CH2\text{H--CH}=CH_2iii95
SH–C≡CH\text{H--C≡CH}iv88

Updated On: May 28, 2024
  • P – iii, Q – iv, R – ii, S – i

  • P – i, Q – ii, R – iii, S – iv

  • P – iii, Q – ii, R – i, S – iv

  • P – ii, Q – i, R – iv, S – iii

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The Correct Option is A

Solution and Explanation

Order of stability
Order of stability of free radical
Q>P>R>SQ > P > R > S
Stability of free radical α1Bond energy\text{Stability of free radical } \alpha \frac{1}{\text{Bond energy}}

∴ Order of bond energy :
S>R>P>QS > R > P > Q

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Question: 2

For the following reaction,
CH4(g)+Cl2(g)lightCH3Cl (g)+HCl(g)\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \overset{\text{light}}{\longrightarrow} \text{CH}_3\text{Cl (g)} + \text{HCl(g)}
the correct statement is

Updated On: May 24, 2024
  • Initiation step is exothermic with Δ\DeltaH° = –58 kcal mol–1

  • Propagation step involving ·CH3 formation is exothermic with Δ\DeltaH° = –2 kcal mol–1

  • Propagation step involving CH3Cl formation is endothermic with Δ\DeltaH° = +27 kcal mol–1

  • The reaction is exothermic with Δ\DeltaH° = –25 kcal mol–1

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The Correct Option is D

Solution and Explanation

Endothermic
Step (1) → Endothermic (bond breaking)

Step (2) → ∆H = 105 – 103 = 2 kcal/mol (Endothermic)

Step (3) → ∆H = 58 – 85 = –27 kcal/mol (Exothermic)

For complete reaction

CH4(g)+Cl2(g)lightCH3Cl (g)+HCl(g)\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \overset{\text{light}}{\longrightarrow} \text{CH}_3\text{Cl (g)} + \text{HCl(g)}

∆H = 58 + 105 – 85 – 103

= –25 kcal/mol

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