Question

The number and type of bonds between two carbon atoms in $Ca{C_2}$ are
(A) One sigma ($\sigma $) and one pi ($\pi $) bond
(B) One sigma ($\sigma $) and two pi ($\pi $) bonds
(C) One sigma ($\sigma $) and one and a half pi ($\pi $) bonds
(D) One sigma ($\sigma $) bond.

Answer 1

Hint:When we talk about covalent bonding, we consider a single covalent bond as a sigma ($\sigma $) bond. When we talk about a double covalent bond, we consider one sigma ($\sigma $) and one pi ($\pi $) bond. Lastly, for triple covalent bonds, we consider it as one sigma and two pi ($\pi $) bonds.

Complete answer:
To find out how many sigma and pi bonds are there, let’s consider the structure of calcium carbide $Ca{C_2}$.

Figure: structure of calcium carbide.
The ionic structure of calcium carbide can be shown as below:

As we can see that there is a triple bond between the two carbon atoms and a single bond between each carbon and calcium atom. As we have already discussed in the hint that a triple bond is considered as one sigma and two pi bonds thus we can say that there are a total three sigma and two pi-bonds.

Additional information:The difference in sigma and pi bonds lie in the overlapping of the orbitals. Sigma bonds are formed when there is end to end overlapping of the orbitals. In case of pi bonds, the lobes of atomic orbitals overlap one another, which is known as lateral overlapping. Since sigma bonds form because of end to end overlapping, they are stronger as compared to the pi bonds, because of more extent of overlapping.

Therefore, the correct answer is (B).

Note: Always remember the valency of each element in the compounds and the number of sigma and pi-bonds between the atoms. Also calcium carbide is ionic in nature where calcium is present as a cation and two carbon atoms called ethynediene $(C_2^{2 - })$ acts as an anion.