The heat of hydrogenation of ethene is $x_{1}$ and that of benzene is $x_{2}$
Hence resonance energy of benzene is

x_{1} - x_{2}

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x_{1} + x_{2}

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3 x_{1} - x_{2}

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x_{1} - 3 x_{2}

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Solution

The correct option is C $3 x_{1} - x_{2}$
$C H_{2} == C H_{2} + H_{2} ⟶ C H_{3} - - C H_{3} Δ H = x_{1}$
$R E = 3 x_{1} - x_{2}$

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