Explanation

(c)
In benzene all the six carbon atoms are $sp^{2}$ hybridised. Out of these three $sp^{2}$ hybrid orbitals of each C-atom, two orbitals overlap with $sp^{2}$ hybrid orbitals of adjacent C-atoms to form six $C - C$ single bonds. The remaining $sp^{2}$ orbital of each C-atom overlaps with $s$-orbitals of each H-atom to form six $C - H$ single sigma bonds. Each C-atom is now left with one unhybridised $p$-orbital perpendicular to the plane of the ring.