`FeC_(2)O_(4)overset(Delta)to` products
Number of diamagnetic products=x
Number of unpaired electrons in paramagnetic products=y
report your answer as (x+y)

Explanation

To solve the problem, we need to analyze the thermal decomposition of iron(II) oxalate, FeC₂O₄, and identify the products formed, as well as their magnetic properties.

### Step-by-Step Solution:

1. **Identify the Reaction**:
When FeC₂O₄ is heated, it decomposes into:
\[ \text{FeC}_2\text{O}_4 \xrightarrow{\Delta} \text{FeO} + \text{CO} + \text{CO}_2 \]
The products are iron(II) oxide (FeO), carbon monoxide (CO), and carbon dioxide (CO₂).

2. **Determine Diamagnetic Products**:
- **CO (Carbon Monoxide)**: CO is a stable molecule with no unpaired electrons, making it diamagnetic.
- **CO₂ (Carbon Dioxide)**: CO₂ is also a stable molecule with no unpaired electrons, hence it is diamagnetic.
- **FeO (Iron(II) Oxide)**: We need to check if FeO is diamagnetic or paramagnetic.

Since CO and CO₂ are diamagnetic, we have:
\[ x = 2 \quad (\text{number of diamagnetic products}) \]

3. **Determine the Magnetic Properties of FeO**:
- The oxidation state of Fe in FeO is +2. The electronic configuration of Fe²⁺ is:
\[ \text{Fe}^{2+}: [\text{Ar}] 3d^6 \]
- In the 3d subshell, the distribution of electrons is as follows:
- 5 electrons will fill the first five orbitals (3d), and the sixth electron will occupy one of the orbitals, resulting in:
\[ \uparrow \downarrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow \]
- This means there are 4 unpaired electrons in FeO.

Therefore, FeO is paramagnetic, and we have:
\[ y = 4 \quad (\text{number of unpaired electrons in paramagnetic product}) \]

4. **Final Calculation**:
Now, we need to report the final answer as \( x + y \):
\[ x + y = 2 + 4 = 6 \]

### Final Answer:
The answer is \( 6 \).