The (Si_2O_5)_n^{2n-} anion is obtained when: | vedclass

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(D) In a $SiO_4^{4-}$ tetrahedral unit,there are $4$ oxygen atoms.If $3$ oxygen atoms are shared with other tetrahedra,each shared oxygen contributes

Vedclass · Accepted Answer

three oxygen of a $SiO_4^{4-}$ tetrahedron are shared with another $SiO_4^{4-}$ tetrahedron

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(D) In a $SiO_4^{4-}$ tetrahedral unit,there are $4$ oxygen atoms.If $3$ oxygen atoms are shared with other tetrahedra,each shared oxygen contributes $1/2$ to the unit.The number of oxygen atoms per silicon atom is calculated as: $1$ (unshared oxygen) $+ 3 	imes (1/2)$ (shared oxygens) $= 1 + 1.5 = 2.5$.Thus,the ratio of $Si:O$ is $1:2.5$,which is $2:5$.The formula for the sheet silicate is $(Si_2O_5)_n^{2n-}$,which corresponds to the sharing of $3$ oxygen atoms per tetrahedron.

The $(Si_2O_5)_n^{2n-}$ anion is obtained when:

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