Question

The charge on the cyclic silicate anion \[{[S{i_3}{O_9}]^{n - }}\] is:
A.-2
B.-3
C.-4
D.-6

Answer 1

Hint: To solve this question, we need to first calculate for the oxidation of silicon and oxygen in the compound. Then we need to equate these values with the total charge on the given anion.

Formula Used:
(Charge on Si) (no. of atoms of Si) + (Charge on O) (no. of atoms of O) = (Charge on the compound)

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
The atomic number of silicon is 14, and hence the electronic configuration of silicon can be given as: \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^2}\] . Hence, we can see that there are only 2 electrons present in both the 3s orbital and the 3p orbital each. This means that to obtain a stable octet electronic configuration, the silicon atom can either donate these 4 electrons or gain 4 new electrons. In this particular compound, Silicon is forming a compound with oxygen. Since oxygen is a more electronegative element, silicon would choose to lose the 4 valence electrons, thus getting a charge of +4. On the other hand, the oxidation state of oxygen is (-2). Hence, the charge on the compound can be calculated as:
(Charge on Si) (no. of atoms of Si) + (Charge on O) (no. of atoms of O) = (Charge on the compound)
 \[\begin{array}{*{20}{l}}
  {\left( { + 4} \right)\left( 3 \right) + \left( { - 2} \right)\left( 9 \right) = - n} \\
  {12-18 = - n} \\
  { - 6 = - n}
\end{array}\]
Hence, \[n = 6\]
Hence, the charge on the silicate anion can be calculated to be (-6)

Hence, Option B is the correct option

Note: Cyclic silicates contain \[{[Si{O_3}]_n}^{2n - }\] ions which are formed by linking three or more tetrahedral \[Si{O_4}^{4 - }\] units cyclically. Each unit shares two oxygen atoms with other units.