1.

Identify the cyclic silicate ion given in the figure below:(A) [Si4O25]24-(B) [Si6O18]12-(C) [Si4O12]12-(D) [Si6O24]12-

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.



Discussion

No Comment Found

Related InterviewSolutions