To solve the question, we need to evaluate each statement regarding borazine and diborane (B2H6) to determine which are true.
### Step-by-Step Solution:
1. **Evaluate Statement I: Borazine is aromatic.**
- Borazine (B3N3H6) has a cyclic structure with alternating boron and nitrogen atoms.
- For a compound to be aromatic, it must be planar, have a cyclic structure, and follow Huckel's rule (4n + 2 π electrons).
- In borazine, there are 6 π electrons contributed by the nitrogen atoms, satisfying Huckel's rule (n=1).
- Therefore, **Statement I is true.**
2. **Evaluate Statement II: There are four isotopic disubstituted borazine molecules B3N3H4X2.**
- Borazine has four reactive hydrogen atoms. When substituting two of these with isotopes (X), there are multiple combinations possible.
- The number of isomers for di-substituted borazine can be calculated as combinations of the four hydrogen atoms taken two at a time, which gives us four distinct isomers.
- Therefore, **Statement II is true.**
3. **Evaluate Statement III: Borazine is more reactive towards addition reactions than benzene.**
- Benzene is a stable aromatic compound, while borazine has polar bonds due to the difference in electronegativity between boron and nitrogen.
- The presence of polar bonds in borazine makes it more reactive towards addition reactions compared to the non-polar benzene.
- Therefore, **Statement III is true.**
4. **Evaluate Statement IV: Banana bonds in B2H6 are longer but stronger than normal B-H bonds.**
- In diborane (B2H6), the so-called "banana bonds" (three-center two-electron bonds) are indeed longer than typical B-H bonds.
- However, these banana bonds are weaker than the normal B-H bonds due to the nature of the bonding.
- Therefore, **Statement IV is false.**
### Conclusion:
Based on the evaluations:
- Statements I, II, and III are true.
- Statement IV is false.
Thus, the correct answer is that **Statements I, II, and III are true.**
