Concept: The structure of calcium carbide (${\mathrm{CaC}}_2$) involves an ionic bond between the calcium ion (${\mathrm{Ca}}^{2+}$) and the dicarbide ion (${\mathrm{C}}_2^{2-}$). The dicarbide ion is an anion formed by two carbon atoms. To determine the number and type of bonds between the two carbon atoms, we need to analyze the electronic configuration and bonding within the ${\mathrm{C}}_2^{2-}$ ion.

Why (B) is correct:

1. Formation of ${\mathrm{C}}_2^{2-}$ ion: The calcium atom loses two electrons to form ${\mathrm{Ca}}^{2+}$. These two electrons are gained by two carbon atoms to form the dicarbide ion, ${\mathrm{C}}_2^{2-}$.

2. Electronic configuration of Carbon: A neutral carbon atom has 4 valence electrons. In the ${\mathrm{C}}_2^{2-}$ ion, each carbon atom effectively has 5 valence electrons (4 from itself + 1 from the gained electrons, shared between the two carbons). The total number of valence electrons in ${\mathrm{C}}_2^{2-}$ is $(2\times 4)+2=10$.

3. Bonding in ${\mathrm{C}}_2^{2-}$: To achieve a stable octet, the two carbon atoms form a triple bond between them. A triple bond consists of one sigma ($\sigma$) bond and two pi ($\pi$) bonds. The Lewis structure for ${\mathrm{C}}_2^{2-}$ is $[\mathrm{C}\equiv \mathrm{C}{]}^{2-}$.

4. Conclusion: Therefore, between the two carbon atoms in ${\mathrm{CaC}}_2$, there is one sigma bond and two pi bonds.

Option Analysis:

• A) One sigma and one pi bond: This describes a double bond, which is incorrect for ${\mathrm{C}}_2^{2-}$.
• B) One sigma and two pi bonds: This correctly describes a triple bond, which is present in the ${\mathrm{C}}_2^{2-}$ ion.
• C) One sigma and one and a half pi bond: This would imply a bond order of 2.5, which is not the case here.
• D) One sigma bond: This describes a single bond, which is incorrect for ${\mathrm{C}}_2^{2-}$.

Correct Answer: (B)