Question

A silicate mineral has three $Si{O}_4$ tetrahedral that share corners with adjacent tetrahedral. Mineral also contains $C{a}^{2+},C{u}^{+}$ and $H_{2}O$ in $1:1:1$ molar ratio. Thus, total number of atoms of $Cu,Si$ and $Ca$ per unit of mineral is :

Answer 1

When three $Si{O}_4$ tetrahedral share corners with adjacent tetrahedral, $(S{i}_3{O}_9{)}^{6-}$ ion is obtained. The -6 charge is balanced by two $C{u}^{+}$ ions and two $C{a}^{2+}$ ions. Thus, there are 3 Si atoms, 2 Cu atoms and 2 Ca atoms per unit of mineral. Hence, the total number of atoms of Cu, Si and Ca is $3+2+2=7$.