The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatohromate (III) is Y. Then the value of X + Y is _________ .

Answer

To solve the problem, we need to determine the number of geometrical isomers for two complexes: triamminetrinitrocobalt(III) and trioxalatohromate(III).

### Step-by-Step Solution:

1. **Identify the Complexes**:
- The first complex is triamminetrinitrocobalt(III), which can be represented as [Co(NH₃)₃(NO₃)₃].
- The second complex is trioxalatohromate(III), represented as [Cr(C₂O₄)₃].

2. **Determine the Number of Geometrical Isomers for [Co(NH₃)₃(NO₃)₃]**:
- In this complex, cobalt is surrounded by three ammonia (NH₃) ligands and three nitrate (NO₃) ligands.
- The ligands can arrange themselves in two ways:
- **Cis**: where similar ligands (NH₃) are adjacent to each other.
- **Trans**: where similar ligands are opposite each other.
- For this complex, we can have two geometrical isomers:
- **Cis isomer**: All three NH₃ ligands are on one side, and all three NO₃ ligands are on the opposite side.
- **Trans isomer**: The NH₃ ligands are arranged such that one NH₃ is opposite to another NH₃, and similarly for the NO₃ ligands.
- Therefore, the number of geometrical isomers (X) for [Co(NH₃)₃(NO₃)₃] is **2**.

3. **Determine the Number of Geometrical Isomers for [Cr(C₂O₄)₃]**:
- In this complex, chromium is surrounded by three bidentate oxalate (C₂O₄) ligands.
- Since oxalate is a bidentate ligand, it binds through two coordination sites, creating a situation where the arrangement does not allow for cis or trans isomers.
- Bidentate ligands do not produce geometrical isomers because they occupy two positions at once, leading to a symmetrical arrangement.
- Therefore, the number of geometrical isomers (Y) for [Cr(C₂O₄)₃] is **0**.

4. **Calculate X + Y**:
- Now, we can sum the values of X and Y:
- X = 2 (from [Co(NH₃)₃(NO₃)₃])
- Y = 0 (from [Cr(C₂O₄)₃])
- Therefore, X + Y = 2 + 0 = **2**.

### Final Answer:
The value of X + Y is **2**.