$C_{1} - H, C_{2} - H, C_{3} - H$ and $C_{4} - H$ homolytic bond dissociation energy is in order:

C_{2} - H > C_{3} - H > C_{4} - H > C_{1} - H

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C_{1} - H > C_{4} - H > C_{2} - H > C_{3} - H

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C_{2} - H > C_{3} - H > C_{1} - H > C_{4} - H

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C_{1} - H > C_{4} - H > C_{3} - H > C_{2} - H

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Solution

The correct option is C $C_{1} - H > C_{4} - H > C_{3} - H > C_{2} - H$
In $C - H$ bond, the carbon atom is more electronegative than hydrogen atom. When several alkyl groups are attached to carbon atom (such as $C_{2}$ ), due to the positive inductive effects of alkyl group, the electronegativity difference between carbon and hydrogen decreases. The hold of the carbon on the shared electron pair of $C - H$ bond decreases. The $C - H$ bond strength decreases. Hence, the homolytic bond dissociation energy decreases.
The order in which the number of alkyl groups attached to the carbon atom increases is $C_{1} - H < C_{4} - H < C_{3} - H < C_{2} - H$
The order of the bond dissociation energy is $C_{1} - H > C_{4} - H > C_{3} - H > C_{2} - H$ .

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Similar questions

Which carbon atom will show minimum electronegativity -

H $^{1} C$ $\equiv$ $^{2} C$ - C $^{3} H$ - C $^{4} H$ - C $^{5} H$

C_{0} . C_{2} + C_{1} . C_{3} + C_{2} . C_{4} + . . . . . . . . + C_{n - 2} . C_{n}

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . . + C_{n - 2} C_{n} =

(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + . . . + (^{21} C_{10} -^{10} C_{10})

y = C_{1} + C_{2} e^{x} + C_{3} e^{- 2 x + C_{4}}

C_{1}, C_{2}, C_{3}, C_{4}

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