The possible product (s) formed in the following reaction is/are:
`IF_(5)+H_(2)O to`?

Options

A. `HIO_(3)`
B. `HIO`
C. `HIO_(4)`
D. `HF`

Answer

To determine the possible products formed when iodine pentafluoride (IF₅) reacts with water (H₂O), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are iodine pentafluoride (IF₅) and water (H₂O).

### Step 2: Understand the Reaction
Iodine pentafluoride is known to react vigorously with water. This reaction typically leads to the formation of hydrofluoric acid (HF) and iodic acid (HIO₃).

### Step 3: Write the Chemical Equation
The reaction can be represented as follows:
\[ IF_5 + H_2O \rightarrow HF + HIO_3 \]

### Step 4: Balance the Reaction
To ensure that the reaction is balanced, we need to adjust the coefficients:
\[ 5 IF_5 + 3 H_2O \rightarrow 5 HF + 3 HIO_3 \]
This indicates that for every 5 moles of IF₅, 3 moles of water produce 5 moles of HF and 3 moles of HIO₃.

### Step 5: Identify the Products
From the balanced equation, we can see that the products formed are:
- Hydrofluoric acid (HF)
- Iodic acid (HIO₃)

### Conclusion
The possible products formed in the reaction of iodine pentafluoride with water are:
- Hydrofluoric acid (HF)
- Iodic acid (HIO₃)

### Final Answer
The correct options are:
- Option A: HIO₃ (Iodic acid)
- Option D: HF (Hydrofluoric acid)

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