what is the main reason for the difference in hyberdization of alkane,alkene and alkyne?

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When one s orbital and three p orbitals (s, p, p, p) are hybridized (mixed to create new orbitals with both s and p character), the results can be: (A)  4 sp³ orbitals (when all of the s and p orbitals are hybridized), (B) 3 sp² orbitals (when one s and two p's are hybridized and the last p remains unhibridized) or (C) 2 sp orbitals (when one s and one p are hybridized and the last two p's remain unhibridized). Think of hybridization as blending juice. If you put on one bottle of s juice and three bottles of p juice in a blender, then pour that back into the original bottles, then you have 4 bottles worth of sp³ juice (4 sp³ orbitals). Now, if you pour one bottle of s juice and only 2 bottles of p juice (not putting the last p bottle in the blender at all), then you will end up with 3 bottles of sp² juice and one unmixed bottle of p juice (3 sp² orbitals and one unhybridized p orbital).  The superscripts indicate which bottles have been mixed. Example: sp³ = s¹p³ = mixture of one s and three p orbitals  Addition of the superscripts indicates how many hybrid oribitals have been made. Example: 3 + 1 = 4 sp³ orbitals    Alkanes (C single bonded to C, 1 sigma bond and 0 pi bonds) have 4 sp³ orbitals and no unhybridized p orbitals.    Alkenes (C double bonded to C, 1 sigma bond and 1 pi bond) have 3 sp² orbitals and one unhybridized p orbital.    Alkynes (C triple bonded to C, 1 sigma bond and 2 pi bonds) have 2 sp orbitals and two unhybridized p orbitals.    The unhybridized p orbitals of sp² and sp hybridized atoms explain why those atoms can form double and triple bonds, respectively. Side to side overlap of unhybridized p orbitals between atoms creates a pi bond. This also explains why sp³  hybridized orbitals cannot form pi bonds and are therefore alkanes.  No unhybridized p orbitals means no chance for a pi bond.  edit