Question

$C-C\left(I\right),C=C\left(II\right)andC\equiv C\left(III\right)$ are three bond types, then their bond enthalpy will be respectively,

• A. I > II > III
• B. I = II = III
• C. I < II < III
• D. Cannot determine

Answer 1

The correct option is C

I < II < III

Generally, if two same covalently bonded are there with single, double and triple bonds, then we can say that for Bond Enthalpy that Triple bond > Double bond > Single bond

Therefore, it take more energy to break a triple bond because there we are breaking three bonds

$\to$ In double bond, we are breaking two double bond.

Bond Enthalpy $\propto$ Bond Order