Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The number and type of bonds between two carbon atoms in calcium carbide are

Have you ever looked at a simple chemical formula and wondered what secrets it hides? Calcium carbide, written simply as $Ca{C}_2$, is one of those molecules that often tricks students. At first glance, you might be tempted to draw a linear covalent molecule, but the reality is far more interesting. Let's dive into the true nature of the bonds hidden within this compound.

The first trap to avoid is assuming that $Ca{C}_2$ is a purely covalent molecule. Calcium is an alkaline earth metal, and carbon is a non-metal. The significant difference in their electronegativities means that calcium will readily donate its two valence electrons.

Therefore, calcium carbide is an ionic compound. It exists as a crystal lattice composed of calcium cations ($C{a}^{2+}$) and carbide anions ($C_2^{2-}$). To understand the bonding between the two carbon atoms, we must isolate the carbide ion and analyze it independently.

Let's focus our attention entirely on the carbide ion, $C_2^{2-}$. To determine the type of bonds holding these two carbon atoms together, we need to count the total number of valence electrons available for bonding.

Each carbon atom brings 4 valence electrons to the table. The $-2$ charge indicates the presence of 2 additional electrons. Total valence electrons = $4+4+2=10$.

Interestingly, this 10-valence-electron system makes the carbide ion isoelectronic with the nitrogen molecule ($N_2$) and the carbon monoxide molecule ($CO$). To satisfy the octet rule for both carbon atoms using only 10 electrons, they must share 6 electrons between them. This sharing of 6 electrons results in a triple bond, leaving one lone pair of electrons on each carbon atom.

Now that we have established the presence of a triple bond ($[:C\equiv C:{]}^{2-}$), we can easily break down its components.

According to the principles of valence bond theory, a single bond is always a sigma ($\sigma$) bond, formed by the head-on overlap of orbitals along the internuclear axis. Any additional bonds formed between the same two atoms must be pi ($\pi$) bonds, resulting from the lateral (sideways) overlap of unhybridized p-orbitals.

Therefore, a triple bond universally consists of one sigma ($\sigma$) bond and two pi ($\pi$) bonds. This elegant combination of ionic and covalent bonding principles leads us directly to our final answer.