The silicate anion in the mineral X is a chain of four ${SiO}_{4}^{- 4}$ tetrahedral units that share corners with adjacent tetrahedra. The mineral also contains $C a^{2 +}$ ions, ${Al}^{3 +}$ ions and water molecules in a $1 : 1 : 1$ ratio. Mineral is represented as :

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$C a A l S i_{2} O_{13} \cdot H_{2} O$

$C a_{2} A l S i_{4} O_{13} \cdot 2 H_{2} O$

$C a_{2} A l_{2} S i_{4} O_{13} \cdot 2 H_{2} O$

$C a_{3} A l_{3} S i_{4} O_{13} \cdot 2 H_{2} O$

answer is C.

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Detailed Solution

As it shares corner with next so it forms linear structure with ${({SiO}_{3})}_{n}^{- 2 n}$ Therefore, three unit will have total negative charge of eight and positive charge will be same, so formula is - $C a_{2} A l_{2} S i_{4} O_{13} \cdot 2 H_{2} O$