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$\begin{array}{l} {CH}_{4} (g) \to {CH}_{3} (g) + H (g) Q_{1} \\ {CH}_{2} (g) \to CH (g) + H (g) Q_{3} \end{array}$ $\begin{array}{l} {CH}_{3} (g) \to {CH}_{2} (g) + H (g) Q_{2} \\ CH (g) \to C (g) + H (g) Q_{4} \end{array}$
Then the relationship between $Q_{1}, Q_{2}, Q_{3} and Q_{4}$ are

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$bond energies uniformly decreases from Q_{1} to Q_{4}$

$bond energies uniformly increases from Q_{1} to Q_{4}$

$values from Q_{1} to Q_{4} are not uniform, but random$

$bond energies are equal i . e ., Q_{1} = Q_{2} = Q_{3} = Q_{4}$

answer is C.

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Detailed Solution

with breakage of each C-H bond, bond energies will change and they are random order.

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