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Identify the cyclic silicate ion given in the figure below:(A) [Si4O25]24-(B) [Si6O18]12-(C) [Si4O12]12-(D) [Si6O24]12- |
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Answer» As we know, cyclic silicates have general formula = \([siO_3]_n^{-2n}\) therefore, option (B) [Si6O18]-12 can be written as \([SiO_3]_6^{-12}\) or \([SiO_3]_6^{-{2\times6}}\) (n = 6) Hence, [Si6O18]-12 is a cyclic silicate. |
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