In cyclic silicate ion `Si_(6)O_(18)^(12-)` the number of oxygen atoms shared.

Answer

To determine the number of oxygen atoms shared in the cyclic silicate ion \( \text{Si}_6\text{O}_{18}^{12-} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure of Cyclic Silicates**:
- Cyclic silicates consist of silicate units where silicon atoms are surrounded by oxygen atoms. Each silicon atom is typically bonded to four oxygen atoms in a tetrahedral arrangement.

2. **Identify the Number of Silicate Units**:
- The given formula is \( \text{Si}_6\text{O}_{18}^{12-} \). This indicates that there are 6 silicon atoms in the structure.

3. **Determine the Number of Oxygen Atoms in Each Silicate Unit**:
- Each silicate unit (SiO4) shares oxygen atoms with adjacent silicate units. In a cyclic structure, each silicate unit shares 2 of its oxygen atoms with neighboring units.

4. **Calculate the Total Number of Shared Oxygen Atoms**:
- Since there are 6 silicate units, and each unit shares 2 oxygen atoms, the total number of shared oxygen atoms can be calculated as:
\[ \text{Total shared oxygen atoms} = \text{Number of silicate units} \times \text{Oxygen atoms shared per unit} \]
\[ \text{Total shared oxygen atoms} = 6 \times 2 = 12 \]

5. **Adjust for Double Counting**:
- However, since each shared oxygen atom is counted twice (once for each silicate unit it connects), we need to divide the total by 2 to find the actual number of unique shared oxygen atoms:
\[ \text{Unique shared oxygen atoms} = \frac{12}{2} = 6 \]

### Final Answer:
The number of oxygen atoms shared in the cyclic silicate ion \( \text{Si}_6\text{O}_{18}^{12-} \) is **6**.