Answer
Imagine a chain of tetrahedra starting with $\mathrm{SiO}_{4}^{4-}$ and increases by $\mathrm{SiO}_{3}^{2-}$, yielding $\mathrm{SiO}_{4}^{4-}, \mathrm{Si}_{2} \mathrm{O}_{7}^{6-}, \mathrm{Si}_{3} \mathrm{O}_{10}^{8-} \mathrm{Si}_{4} \mathrm{O}_{13}^{10-}\mathrm{Si}_{5} \mathrm{O}_{16}^{12-}, \mathrm{Si}_{6} \mathrm{O}_{19}^{14-} .$ If the six-unit silicate chain is bent into a hexagonal ring by eliminating one $\mathrm{O}^{2-}$ between the ends of the chain, the result is the anion, $\mathrm{Si}_{6} \mathrm{O}_{18}^{12-}$. In beryl, $3 \mathrm{Be}^{2+}$ and $2 \mathrm{Al}^{3+}$ provide the positive charge of twelve.
Work Step by Step
See answer.