sp2 hybridized carbon versus sp3 hybridized silicon
Carbon — and all the other elements of the second period, for that matter — is exceptionally good at forming double bonds. A double bond is formed if there is a &#x3c3; bond and a &#x3c0; bond. Forming a &#x3c3; bond is not hard: You have two orbitals coming from two different atoms pointing at each other. Forming a &#x3c0; bond is much harder, since the orbitals are pointing orthogonally away from the bond-axis. Thus, the overlap of &#x3c0; bonds is often a lot weaker than for &#x3c3; bonds.

This gets more important if you consider third-period elements rather than second-period ones. They have a total of eight more core electrons making them much larger in size and the orbitals more diffuse. A &#x3c3; bond is still formed easily, there will be the point when two diffuse orbitals are close enough to interact in a &#x3c3; fashion. To arrive at favourable interactions for a &#x3c0; bond would require the atoms to come much closer than what would be comfortable; unfavourable interactions between the core electrons would already be a thing.

In fact, I was taught in my Concepts of Inorganic Chemistry class, that disilene $\ce{Si2H4}$, the heavier homologue of ethene, does not form a classical &#x3c0; bond. Instead, it can be understood as a par of two singlett silene fragments $\ce{SiH2}$ which approach each other so that the empty p-orbital of one silene can interact with the lone pair in an $\mathrm{sp^2}$ orbital of the other silene fragment. Obviously, the symmetry is lower and the bond is much weaker.

So the reason behind it is that $\ce{Si=Si}$ double bonds are hard to create and this gets even harder for heavier homologues while $\ce{C=C}$ double bonds are reasonably stable and easy to generate.