The state of hybridisation of central atom in dimer of `BH_(3)` and `BeH_(2)` IS `:`

Explanation

To determine the state of hybridization of the central atom in the dimers of BH3 and BeH2, we will analyze each molecule separately.

### Step 1: Analyze BH3
1. **Valence Electrons**: Boron (B) has 3 valence electrons.
2. **Bond Formation**: Boron forms three bonds with hydrogen atoms. Each hydrogen contributes one electron, making a total of 6 electrons involved in bonding.
3. **Hybridization**: To accommodate these three bonds, boron undergoes hybridization. The 2s and two 2p orbitals combine to form three equivalent sp² hybrid orbitals.
4. **Geometry**: The geometry of BH3 is trigonal planar, with bond angles of 120°.

### Step 2: Analyze BeH2
1. **Valence Electrons**: Beryllium (Be) has 2 valence electrons.
2. **Bond Formation**: Beryllium forms two bonds with two hydrogen atoms. Each hydrogen contributes one electron, making a total of 4 electrons involved in bonding.
3. **Hybridization**: To form two bonds, beryllium undergoes hybridization. The 2s and one 2p orbital combine to form two equivalent sp hybrid orbitals.
4. **Geometry**: The geometry of BeH2 is linear, with a bond angle of 180°.

### Conclusion
- The hybridization of boron in BH3 is **sp²**.
- The hybridization of beryllium in BeH2 is **sp**.

### Final Answer
The state of hybridization of the central atom in the dimer of BH3 and BeH2 is **sp² and sp**, respectively.

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