= Solution
\b[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 $\mathrm{Ca^{2+}[C_2]^{2-}}$, with formal bond order three, consistent with the short $119.2$ pm distance.
A fully ionic count for $\mathrm{LaC_2}$ gives $\mathrm{La^{3+}[C_2]^{3-}}$. The eleventh valence electron then enters a $\pi_g^*$ 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.
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