the overall charge present on the cyclic silicate anion `[Si_(6)O_(18)]^(n-)` is:

Answer

To find the overall charge present on the cyclic silicate anion \([Si_6O_{18}]^{n-}\), we can follow these steps:

### Step 1: Identify the oxidation state of Silicon (Si)
Silicon belongs to Group 14 of the periodic table, which means it has 4 valence electrons. In silicates, silicon typically exhibits a +4 oxidation state.

### Step 2: Calculate the total positive charge from Silicon
Since there are 6 silicon atoms in the anion, the total positive charge contributed by silicon is:
\[ \text{Total positive charge from Si} = 6 \times (+4) = +24 \]

### Step 3: Identify the oxidation state of Oxygen (O)
Oxygen typically has a -2 oxidation state.

### Step 4: Calculate the total negative charge from Oxygen
There are 18 oxygen atoms in the anion, so the total negative charge contributed by oxygen is:
\[ \text{Total negative charge from O} = 18 \times (-2) = -36 \]

### Step 5: Set up the charge balance equation
The overall charge of the anion can be expressed as the sum of the total positive and negative charges:
\[ \text{Overall charge} = \text{Total positive charge} + \text{Total negative charge} \]
Substituting the values we calculated:
\[ \text{Overall charge} = +24 + (-36) = +24 - 36 = -12 \]

### Step 6: Relate the overall charge to \(n\)
Since the overall charge is given as \([Si_6O_{18}]^{n-}\), we can equate this to -n:
\[ -n = -12 \]
Thus, we find:
\[ n = 12 \]

### Conclusion
The overall charge present on the cyclic silicate anion \([Si_6O_{18}]^{n-}\) is \(-12\), which means it can be represented as \([Si_6O_{18}]^{-12}\).

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