Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{SiCl}_{4}\) from \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms.

To find the Lewis symbols, let's start by looking at the positions of both silicon (Si) and chlorine (Cl) in the periodic table. Silicon, with the atomic number 14, lies in Group 14 (or IV) and therefore has four valence electrons. Meanwhile, chlorine, with the atomic number 17, is in Group 17 (or VII) and has seven valence electrons. The Lewis symbols for the atoms are: Si: \(\cdot \overset{\displaystyle..}{\underset{\displaystyle..}{\mathrm{Si}}} \cdot \) Cl: \(\overset{\displaystyle..}{\underset{\displaystyle..}{\mathrm{Cl}}} \longrightarrow \)

In order to form a stable compound, silicon (Si) needs to achieve a full octet (eight electrons in its outermost shell), while each chlorine (Cl) atom also needs to reach a full octet. Silicon can do this by forming a single covalent bond with each of the four chlorine atoms, allowing both elements to share electrons and achieve a full outer shell. The Lewis structure for silicon tetrachloride will have Si at the center, bound to four Cl atoms: \(\mathrm{SiCl}_{4}\): \(\overset{\displaystyle..}{\underset{\displaystyle..}{\mathrm{Cl}}}\) - \(\cdot \overset{\displaystyle..}{\underset{\displaystyle..}{\mathrm{Si}}} \cdot \) - \(\overset{\displaystyle..}{\underset{\displaystyle..}{\mathrm{Cl}}}\) | \(\overset{\displaystyle..}{\underset{\displaystyle..}{\mathrm{Cl}}}\) The structure above shows the silicon atom at the center, sharing one pair of electrons with each chlorine atom. This results in a complete octet for both Si and each Cl atom. Consequently, in \(\mathrm{SiCl}_{4}\), each chlorine atom forms a single covalent bond with the central silicon atom, satisfying the octet rule for all atoms involved and illustrating the formation of the compound using Lewis symbols and Lewis structures.