Assertion. Beryllium hydride is a covalent hydride.
Reason. The electronegativity difference between Be and H is very hig.

Explanation

### Step-by-Step Solution

1. **Understanding the Assertion**: The assertion states that beryllium hydride (BeH₂) is a covalent hydride. We need to determine if this statement is true or false.

2. **Analyzing Beryllium Hydride**: Beryllium hydride consists of beryllium (Be) and hydrogen (H). To classify it as a covalent or ionic compound, we must look at the properties of the elements involved.

3. **Electronegativity Values**: The electronegativity of beryllium (Be) is approximately 1.57, while that of hydrogen (H) is about 2.20.

4. **Calculating Electronegativity Difference**: The difference in electronegativity between beryllium and hydrogen is calculated as follows:
\[ \text{Electronegativity difference} = |1.57 - 2.20| = 0.63 \]

5. **Interpreting the Electronegativity Difference**: A lower electronegativity difference (typically less than 1.7) suggests that the bond is more covalent in nature. Since 0.63 is less than 1.7, this indicates that beryllium hydride is indeed a covalent hydride.

6. **Understanding the Reason**: The reason provided states that the electronegativity difference between beryllium and hydrogen is very high. However, we calculated that the difference is 0.63, which is not considered high.

7. **Conclusion**: The assertion that beryllium hydride is a covalent hydride is correct. However, the reason given is incorrect because the electronegativity difference is not very high.

8. **Final Answer**: Therefore, the correct conclusion is that the assertion is true, but the reason is false.