My textbook says that $\ce{C_2}^-$ has a lower ionization energy than both $\ce{C_2}$ and $\ce{C_2}^+$. I calculated that the bond orders of $\ce{C_2}^+$, $\ce{C_2}$, and $\ce{C_2}^-$ are 1.5, 2, and 2.5, respectively.

Why doesn't $\ce{C_2}^-$ have a higher ionization than the other two molecules? Shouldn't the molecule with the larger bond order have the larger ionization energy, because it would be more difficult to remove an electron from the stable bond?


1 Answer 1


Bond order has to do nothing with ionization energy, it is the bond dissociation energy which has to be dealt with bond order. For minimum ionization energy, the electron has to be lost from the highest energetic orbital (outer).

In $\ce{C2^-}$, the electron is lost from high energetic $\sigma(\ce{2p_x})$ (if you take the bonding along x-axis) orbital, whereas, in the other two molecules electrons are lost from low energetic $\pi(\ce{2p_y})$ or $\pi (\ce{2p_z})$ orbitals. So $E_{\infty} - E_\text{initial}$ is higher for the latter case, which makes $\ce {C2^-}$ having lower ionization energy.


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