A cyclic slicate having structural formula `[Si_(6)O_(18)]^(n-)`. What is vlaue of n ?

Answer

To find the value of \( n \) in the cyclic silicate with the structural formula \([Si_6O_{18}]^{n-}\), we can follow these steps:

### Step 1: Understand the General Formula for Cyclic Silicates
The general formula for cyclic silicates is given by:
\[ Si_nO_{3n} \]
This means that for every \( n \) silicon atoms, there are \( 3n \) oxygen atoms.

### Step 2: Compare with the Given Formula
In our case, we have:
\[ Si_6O_{18} \]
We can see that this matches the general formula if we set \( n = 6 \):
\[ Si_6O_{3 \times 6} = Si_6O_{18} \]

### Step 3: Determine the Charge
Cyclic silicates are typically negatively charged due to the presence of oxygen. Each silicon atom typically has a valency of +4, and each oxygen atom has a valency of -2.

### Step 4: Calculate the Total Charge
Let's calculate the total charge contributed by silicon and oxygen:
- Total charge from silicon:
\[ 6 \text{ Si} \times (+4) = +24 \]
- Total charge from oxygen:
\[ 18 \text{ O} \times (-2) = -36 \]

### Step 5: Set Up the Equation for Charge Balance
The overall charge of the silicate can be expressed as:
\[ \text{Total Charge} = \text{Charge from Si} + \text{Charge from O} + n \]
Substituting the values we calculated:
\[ n - 12 = 24 - 36 \]
\[ n - 12 = -12 \]
Thus, we can solve for \( n \):
\[ n = 12 \]

### Conclusion
The value of \( n \) in the cyclic silicate \([Si_6O_{18}]^{n-}\) is:
\[ \boxed{12} \]