Identify the cyclic silicate ion given in the | vedclass

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(B) In cyclic silicates,two oxygen atoms of each $SiO_4^{4-}$ tetrahedron are shared to form a ring with the general formula $[SiO_3]_n^{2n-}$.The fig

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$[Si_6O_{18}]^{12-}$

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(B) In cyclic silicates,two oxygen atoms of each $SiO_4^{4-}$ tetrahedron are shared to form a ring with the general formula $[SiO_3]_n^{2n-}$.The figure shows a ring consisting of $6$ silicon atoms.Substituting $n = 6$ into the general formula,we get $[SiO_3]_6^{2(6)-} = [Si_6O_{18}]^{12-}$.Thus,the cyclic silicate ion is $[Si_6O_{18}]^{12-}$.

Identify the cyclic silicate ion given in the figure below:

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