Using hybridization, explain how carbon forms both sigma and and pi bonds in graphite, but only sigma bonds in the diamond form?

The ground state electronic configuration of carbon :
1s² 2s² 2px¹ 2py¹ 2pz⁰
Before formation of covalent bond, one of the 2s electrons is promoted to 2p subshell (say, 2pz) by absorption of energy the surroundings. The electronic configuration now becomes :
1s² 2s¹ 2px¹ 2py¹ 2pz¹.
In diamond, the 2s, 2px, 2py and 2pz orbitals in each carbon atom hybridize to form four sp³ orbitals, which are tetrahedral in shape. Each sp3 orbital contains an unpaired electron and forms a single covalent bond (a σ bond) with a neighboring carbon atom. Therefore, carbon atoms in diamond form σ bonds only.
In graphite, the 2s, 2px and 2py orbitals in each carbon atom hybridize to form three sp² orbitals, which are triangular planar in shape. Each sp orbital contains an unpaired electron and forms a single covalent bond (a σ bond) with its four neighboring carbon atom, and thus layers of atoms are formed. The electrons in the unhybridized 2pz orbitals are π electrons which delocalize over and below the layer of atoms. Therefore, carbon atoms in graphite form σ bonds and π bonds.