In `K[B_(5)O_(6)(OH)_(4)]`, the number of tetrahedral units is /are,

Answer

To determine the number of tetrahedral units in the compound \( K[B_5O_6(OH)_4] \), we can follow these steps:

### Step 1: Understand the Structure
The compound consists of boron (B), oxygen (O), and hydroxyl (OH) groups. Each boron atom can form four bonds due to its sp³ hybridization, which is characteristic of tetrahedral geometry.

### Step 2: Identify Boron Atoms
In the formula \( K[B_5O_6(OH)_4] \), there are 5 boron atoms (B).

### Step 3: Determine Bonding of Boron
Each boron atom in the structure typically forms three bonds with oxygen atoms and one bond with another atom or hydroxyl group. Since boron can form four bonds, we consider how many of these boron atoms can be classified as tetrahedral units.

### Step 4: Count Tetrahedral Units
In this structure, each boron atom contributes to a tetrahedral unit. However, since there are 5 boron atoms, and each boron atom forms a tetrahedral unit, we can conclude that there are 5 tetrahedral units.

### Final Answer
Thus, the number of tetrahedral units in \( K[B_5O_6(OH)_4] \) is **5**.

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