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# The charge $x$ of $Si_6O_{18}^x$ will be

$\begin{array}{1 1}4-\\6-\\8-\\12-\end{array}$

Can you answer this question?

$Si_6O_{18}^x=(SiO_4^{4-})_6-6O^{2-}=Si_6O_{18}^{12-}$
answered Jul 25, 2014