`Til_(4)`heating gives

Options

A. `TiI_(2) + I_(2)`
B. `Ti + 2I_(2)`
C. `TiI_(3) + 1/2 I_(2)`
D. None of these

Answer

To solve the question regarding the product formed when heating Titanium(IV) iodide (TiI4), we can follow these steps:

### Step 1: Identify the oxidation states
- In TiI4, Titanium (Ti) has an oxidation state of +4, and each Iodine (I) has an oxidation state of -1.
- Therefore, the overall oxidation states can be summarized as:
- Ti: +4
- I: -1

### Step 2: Determine the reaction upon heating
- When TiI4 is heated, it undergoes a decomposition reaction. The titanium in the +4 oxidation state is reduced, while the iodine in the -1 oxidation state is oxidized.
- The expected products of this reaction are elemental titanium (Ti) and diatomic iodine (I2).

### Step 3: Write the balanced equation
- The balanced equation for the decomposition of TiI4 upon heating is:
\[ \text{TiI}_4 \rightarrow \text{Ti} + 2 \text{I}_2 \]
- Here, one mole of TiI4 decomposes to produce one mole of titanium and two moles of iodine.

### Step 4: Identify the changes in oxidation states
- In this reaction:
- Titanium is reduced from +4 to 0 (elemental state).
- Iodine is oxidized from -1 to 0 (elemental state).

### Conclusion
- The products formed when TiI4 is heated are titanium (Ti) and iodine (I2).

### Final Answer
The answer is: **Ti + 2 I2**

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