The number and type of bonds between two carbon atoms in calcium carbide are

In calcium carbide the bonding is predominantly ionic. The solid may be represented as $$\bigl(Ca^{2+}\bigr)\bigl(C_2^{2-}\bigr)$$, where $$C_2^{2-}$$ is called the acetylide (or carbide) ion.

1. Electron count for $$C_2^{2-}$$:
  • Each carbon atom contributes 6 valence electrons → total 12.
  • The overall charge $$2-$$ supplies 2 more electrons.
Hence the ion possesses $$12 + 2 = 14$$ valence electrons, the same number as in the neutral molecule $$N_2$$.

2. Bond order from the MO picture:
  • $$N_2$$, with 14 valence electrons, has a bond order of 3 (one $$\sigma$$ and two $$\pi$$ bonds).
  • Because $$C_2^{2-}$$ has the same electron configuration, its bond order is also 3.

3. Valence-bond (hybridisation) view:
  • Each carbon undergoes $$sp$$ hybridisation, placing one $$sp$$ hybrid along the internuclear axis.
  • The overlap of these two $$sp$$ hybrids gives one $$\sigma$$ bond.
  • Each carbon retains two unhybridised $$p$$ orbitals ($$p_x$$ and $$p_y$$). Sidewise overlaps of the corresponding orbitals produce two mutually perpendicular $$\pi$$ bonds.

Thus a total of three bonds—one $$\sigma$$ and two $$\pi$$—join the two carbon atoms, i.e. a carbon-carbon triple bond.

Therefore, the number and type of bonds between the two carbon atoms in calcium carbide are: one sigma bond and two pi bonds.

Option B which is: One sigma, two pi