Answer
In C-C single bonds, each C atom is bonded with four atoms. All bonds are sigma ($\sigma$) bonds. Every C atom is $sp^{3}$ hybridized i.e. bond angle is $109.5^{\circ}$. Compounds having C-C single bond are tetrahedral in shape. C-C single bond length is maximum than C=C or $C\equiv C$ bond length.
In C=C double bonds, each C atom is bonded with three atoms. Double bond means one bond is $\sigma$ and another bond is $\pi.$ Each C atom here has $3 \sigma$ and$ 1 \pi $ bond. Every C atom is $sp^{2}$ hybridized i.e. bond angle is $120^{\circ}$. Compounds having C=C double bond are planar in shape. Here, bond length is lower than single bond length and higher than the triple bond length.
In $C\equiv C$ triple bonds, each C atom is bonded with two atoms only. Triple bond means one bond is $\sigma$ and other two bonds are $\pi$ bonds. Each C atom has $2\sigma$ and $2\pi$ bonds. Every C atom is $sp$ hybridized i.e. bond angle is $180^{\circ}$. Compounds having $C\equiv C$ triple bond are linear in shape. Bond length is minimum.
Work Step by Step
To find out the hybridization of C atom we have to count the $\sigma$ bonds associated with the C atom. If the total no. of sigma bond = 4 then it is $sp^{3}$ hybridized ; no. of sigma bond = 3 it is $sp^{2}$ hybridized; no. of sigma bond = 2, it is $sp$ hybridized.
From the hybridization of the atom, the shape of the molecule, bond angle between atoms, bond lengths etc. are known.
