The number and type of bonds in C$$_2^{2-}$$ ion in CaC$$_2$$ are:

The question is about finding the number and type of bonds in the C22- ion present in calcium carbide (CaC2). Calcium carbide is an ionic compound made up of Ca2+ ions and C22- ions. We will focus on the C22- ion.

First, we determine the total number of valence electrons in C22-. Each carbon atom has 4 valence electrons, so two carbon atoms contribute 8 electrons. The ion has a charge of -2, meaning it has gained 2 extra electrons. Therefore, the total valence electrons are 8 + 2 = 10 electrons.

Next, we draw the Lewis structure for C22-. To satisfy the octet rule for both carbon atoms, we need each carbon to be surrounded by 8 electrons. A triple bond between the two carbon atoms uses 6 electrons (3 pairs). The remaining 4 electrons are placed as lone pairs, with two electrons (one lone pair) on each carbon atom. So, the Lewis structure is [:C≡C:]2-. In this structure, each carbon atom has a triple bond (equivalent to 6 electrons) and one lone pair (2 electrons), completing the octet.

A triple bond consists of one sigma (σ) bond and two pi (π) bonds. The sigma bond is formed by the head-on overlap of orbitals along the internuclear axis, and the two pi bonds are formed by the sidewise overlap of orbitals perpendicular to the axis. Therefore, in the C22- ion, there is one σ-bond and two π-bonds.

We can confirm this using molecular orbital theory. For the C22- ion with 10 electrons, the molecular orbital configuration is: (σ1s2)(σ*1s2)(σ2s2)(σ*2s2)(π2px2)(π2py2)(σ2pz2). The core orbitals (σ1s and σ*1s) are ignored as they are filled. The bonding orbitals are σ2s, π2px, π2py, and σ2pz, while σ*2s is antibonding.

Calculating the bond order: Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2. Electrons in bonding orbitals: σ2s2 (2 electrons), π2px2 and π2py2 (4 electrons), and σ2pz2 (2 electrons) give a total of 8 bonding electrons. Electrons in antibonding orbitals: σ*2s2 (2 electrons). So, bond order = (8 - 2) / 2 = 6 / 2 = 3, indicating a triple bond.

This triple bond comprises one σ-bond (from the σ2pz orbital) and two π-bonds (from the π2px and π2py orbitals). The σ2s and σ*2s orbitals cancel each other's bond order contribution, leaving only the σ2pz and the two π bonds.

Hence, the C22- ion has one σ-bond and two π-bonds.

Now, comparing with the options:
A. Two σ-bonds and two π-bonds
B. One σ-bond and two π-bonds
C. One σ-bond and one π-bond
D. Two σ-bonds and one π-bond

So, the correct answer is Option B.