Structure of `Na_(2)B_(4)O_(7).10H_(2)O` contains ---

Options

A. Two triangular and two tetrahedral units
B. Three triangular and one tetrahedral units
C. All tetrahedral units
D. All triangular units.

Answer

To determine the structure of `Na₂B₄O₇·10H₂O`, we will analyze the composition and connectivity of the atoms in the compound step by step.

### Step-by-Step Solution:

1. **Identify the Components**:
- The compound consists of sodium (Na), boron (B), oxygen (O), and water molecules (H₂O).
- The formula indicates that there are 2 sodium ions, 4 boron atoms, 7 oxygen atoms, and 10 water molecules.

2. **Understanding Boron Coordination**:
- Boron typically forms three bonds due to its electron deficiency. In this structure, boron can exhibit different coordination geometries.
- Boron can exist in two forms: trigonal planar (sp² hybridization) and tetrahedral (sp³ hybridization).

3. **Constructing the Borate Structure**:
- The basic unit of borate is the boron-oxygen bond. In `Na₂B₄O₇`, the boron atoms can be connected through oxygen atoms.
- The structure can be visualized as a network of boron and oxygen atoms where some boron atoms are bonded to three oxygen atoms (trigonal) and others to four oxygen atoms (tetrahedral).

4. **Counting the Units**:
- In the structure, we can identify:
- Two boron atoms that are in a trigonal planar arrangement (sp²).
- Two boron atoms that are in a tetrahedral arrangement (sp³).
- Therefore, we have a total of two triangular (trigonal) units and two tetrahedral units.

5. **Final Structure Analysis**:
- The complete structure of `Na₂B₄O₇·10H₂O` shows that it contains:
- 2 triangular units (sp²)
- 2 tetrahedral units (sp³)

### Conclusion:
The structure of `Na₂B₄O₇·10H₂O` contains **two triangular and two tetrahedral units**.