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Bond energy of N–N is x kJ mol^–1. Then bond energy of N $\equiv$ N is

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> 3x kJmol^–1

x kJmol^–1

< 3x kJmol^–1

3x kJmol^–1

answer is B.

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Detailed Solution

In N≡N , one sigma and two pi bonds are present.

sigma bond energy is higher than pi-bond energy.

Hence Then bond energy of N≡N is <3x kjmol^-1

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