The difference between total number of lone pairs and total number of `sigma`-bonds in `[B_(3)O_(3)(OH)_(6)]^(3-)` molecular ion is:

Explanation

To find the difference between the total number of lone pairs and the total number of sigma bonds in the molecular ion \([B_3O_3(OH)_6]^{3-}\), we will follow these steps:

### Step 1: Determine the structure of \([B_3O_3(OH)_6]^{3-}\)
- The molecular ion consists of 3 boron (B) atoms and 3 oxygen (O) atoms, along with 6 hydroxyl (OH) groups.
- Each boron atom is coordinated with two hydroxyl groups.

### Step 2: Count the lone pairs
- Each oxygen atom in the hydroxyl groups has 2 lone pairs.
- There are 6 hydroxyl groups, so:
- Total lone pairs from hydroxyl groups = \(6 \text{ (OH groups)} \times 2 \text{ (lone pairs per O)} = 12 \text{ lone pairs}\).
- Additionally, the 3 oxygen atoms in the cyclic structure also have 2 lone pairs each:
- Total lone pairs from the 3 oxygen atoms = \(3 \text{ (O atoms)} \times 2 \text{ (lone pairs per O)} = 6 \text{ lone pairs}\).
- Therefore, the total number of lone pairs in the entire structure is:
\[ 12 \text{ (from OH)} + 6 \text{ (from O)} = 18 \text{ lone pairs} \]

### Step 3: Count the sigma bonds
- Each B-O bond and each O-H bond is a sigma bond.
- There are 3 B-O bonds (one for each boron to the cyclic oxygen) and 6 O-H bonds (one for each hydroxyl group).
- Therefore, the total number of sigma bonds is:
\[ 3 \text{ (B-O bonds)} + 6 \text{ (O-H bonds)} = 9 \text{ sigma bonds} \]

### Step 4: Calculate the difference
- Now, we find the difference between the total number of lone pairs and the total number of sigma bonds:
\[ \text{Difference} = \text{Total lone pairs} - \text{Total sigma bonds} = 18 - 9 = 9 \]

### Conclusion
The difference between the total number of lone pairs and the total number of sigma bonds in \([B_3O_3(OH)_6]^{3-}\) is **9**.