Question

The general formula of silicate ion present cyclic silicates is:

• A. $Si{O}_4^{4-}$
• B. $S{i}_2{O}_5^{2-}$
• C. $S{i}_2{O}_7^{6-}$
• D. $(Si{O}_3{)}_n^{2n-}$

Answer 1

The correct option is D $(Si{O}_3{)}_n^{2n-}$

Cyclic silicates contain $(Si{O}_3{)}_n^{2n^{-}}$ or $(Si{O}_3^{2-}{)}_n$ ions which are formed by linking three or more tetrahedral $Si{O}_4^{4-}$ units cyclically. Each unit shares two oxygen atoms with other units.