Calcium carbide (CaC\u2082) is an ionic compound composed of Ca\u00b2\u207a ions and C\u2082\u00b2\u207b ions. The question asks about the type of bond between the two carbon atoms within the C\u2082\u00b2\u207b ion.

Step 1: Determine the structure of the C\u2082\u00b2\u207b ion.
The C\u2082\u00b2\u207b ion is also known as the acetylide ion. It is isoelectronic with the N\u2082 molecule. Each carbon atom has 4 valence electrons. With a -2 charge, there are an additional 2 electrons. So, the total number of valence electrons is $4+4+2=10$.

Step 2: Draw the Lewis structure for C\u2082\u00b2\u207b.
To satisfy the octet rule for both carbon atoms with 10 valence electrons, a triple bond must exist between the two carbon atoms. Each carbon atom will also have one lone pair of electrons.
The Lewis structure is: $:C\equiv C:$^{2-}

Step 3: Identify the types of bonds in a triple bond.
A triple bond consists of one sigma ($\sigma$) bond and two pi ($\pi$) bonds.

Option Analysis:

• A) one sigma and one pi: This describes a double bond, not a triple bond.
• B) one sigma and two pi: This correctly describes a triple bond, which is present in the C\u2082\u00b2\u207b ion.
• C) one sigma and one half pi: This is not a standard type of bond.
• D) one sigma: This describes a single bond.

Correct Answer: (B)