Question

The following pictures represent various silicate anions.

Their formulae are respectively:

• A. $Si{O}_3^{2-}S{i}_3{O}_7^{2-}$
• B. $Si{O}_4^{4-}S{i}_3{O}_{10}^{8-}$
• C. $Si{O}_4^{2-}S{i}_3{O}_9^{2-}$
• D. $Si{O}_3^{4-}S{i}_3{O}_7^{8-}$

Answer 1

The correct option is B

$Si{O}_4^{4-}S{i}_3{O}_{10}^{8-}$

There are numerous examples of minerals that contain discrete $(Si{O}_4{)}^{4-}$ tetrahedra; that is, they no corners. Example: Willemite - $Z{n}_2\left[Si{O}_4\right]$ - where the central metal atom (of the silicate anion) has a coordination number of 4. The same is true for phenacite - $B{e}_2\left[Si{O}_4\right]$.

To solve this problem, there is a simple workaround. Just figure out how many oxygen atoms are shared between tetrahedra. For the pyroxene (sheet silicate) each tetrahedron shares two of its oxygen atoms. The shared oxygens are similar to the oxygen atom in an ether linkage, it bears no charge and has 2 lone pairs and 2 bond pairs. The remaining oxygen atoms will bear a negative charge each. Don't forget that from the perspective of the structure shown, one oxygen is not directly visible. So the sheet silicate shown has exactly 2 oxygen atoms shared. How many silicon atoms are there? There are 3 of them, and corresponding to these silicon atoms, each one has one hidden (not clear from the diagram) oxygen atom. So overall, there are 10 oxygen atoms - 4 in a straight line, 3 hidden, and 3 total on either side of the straight-line oxygen atoms. Hence, this one has a formula of $(S{i}_3{O}_{10}{)}^{8-}$.