 Solution

The extra electron associated with trivalent lanthanum weakens the C–C bond by occupying states with C–C antibonding character. Calcium gives the ten-valence-electron acetylide description $\mathrm{C}{\mathrm{a}}^{2+}[{\mathrm{C}}_2{]}^{2-}$, with formal bond order three, consistent with the short $119.2$ pm distance.

A fully ionic count for $\mathrm{La}{\mathrm{C}}_2$ gives $\mathrm{L}{\mathrm{a}}^{3+}[{\mathrm{C}}_2{]}^{3-}$. The eleventh valence electron then enters a ${\pi }_g^{\ast }$ orbital, giving the simple estimate $3-1/2=2.5$. This explains the longer $130.3$ pm bond. In the actual solid that electron is not necessarily localized on an isolated ${\mathrm{C}}_2^{3-}$ radical: metal–carbon mixing and band delocalization make “triple bond plus partial antibonding occupation” more appropriate than an exact isolated-ion bond order. Bond length alone does not determine a unique numerical bond order.