$Si$ has 4 valence electrons.
Work Step by Step
1) The electron configuration of $Si$ is $$[Ne]3s^23p^2$$ $Si$ has 4 valence electrons. The electron configuration of $Cl$ is $$[Ne]3s^23p^5$$ $Cl$ has 7 valence electrons. 2) Here we see that both $Si$ and $Cl$ need more electrons to fill an octet ($Si$ needs 4 more electrons and $Cl$ needs 1 more electron), so we would use covalent bonding in $SiCl_4$ formation, which means each atom must share its own electrons so that all can reach an octet. Since each $Cl$ atom only needs 1 more electron to reach an octet, it will share 1 electron and $Si$ also shares 1 electron so that these 2 electrons would be used by both $Si$ and $Cl$ atoms. Therefore, both $Si$ and $Cl$ atoms would acquire 1 more electron to its atom, and now $Cl$ atom reaches an octet. The same process applies in forming covalent bonding of $Si$ with the other 3 $Cl$ atoms. So eventually, $Si$ atom would give out all of its 4 valence electrons to the sharing. However, since each sharing has 2 electrons which can be used by both atoms, after 4 sharings $Si$ atom now has 8 valence electrons and reaches an octet. The diagram of formation of $SiCl_4$ is shown below. (a) As we see from the electron configuration of $Si$, the valence electrons in the $3s$ and $3p$ subshells. There are 2 electrons in subshell $3s$ and 2 electrons in subshell $3p$. Therefore, initially $Si$ has 4 valence electrons.