Ionistable H atom in `H_(3)PO_(3)` is x and in `H_(3)PO_(2)` is y. Then ratio of x:y is

Answer

To solve the problem of finding the ratio of ionizable hydrogen atoms in \( H_3PO_3 \) and \( H_3PO_2 \), we will follow these steps:

### Step 1: Identify the Structure of \( H_3PO_3 \)

1. **Draw the Lewis structure of \( H_3PO_3 \)**:
- Phosphorus (P) is the central atom.
- It is bonded to three hydrogen (H) atoms and one oxygen (O) atom.
- The oxygen atom is further bonded to another oxygen atom, which has a double bond with phosphorus.

2. **Count the Ionizable Hydrogen Atoms**:
- In \( H_3PO_3 \), the hydrogen atoms that are attached to the oxygen atoms can ionize.
- There are two hydrogen atoms attached to the oxygen atoms.

Thus, for \( H_3PO_3 \), the number of ionizable hydrogen atoms \( x = 2 \).

### Step 2: Identify the Structure of \( H_3PO_2 \)

1. **Draw the Lewis structure of \( H_3PO_2 \)**:
- Again, phosphorus (P) is the central atom.
- It is bonded to two hydrogen (H) atoms and two oxygen (O) atoms.
- One of the oxygen atoms is bonded to phosphorus with a double bond, while the other oxygen atom is bonded to one hydrogen atom.

2. **Count the Ionizable Hydrogen Atoms**:
- In \( H_3PO_2 \), only one hydrogen atom is attached to an oxygen atom and can ionize.

Thus, for \( H_3PO_2 \), the number of ionizable hydrogen atoms \( y = 1 \).

### Step 3: Calculate the Ratio of \( x \) to \( y \)

Now that we have determined the values of \( x \) and \( y \):
- \( x = 2 \)
- \( y = 1 \)

The ratio of \( x \) to \( y \) is:

\[ \text{Ratio} = \frac{x}{y} = \frac{2}{1} \]

### Final Answer

The ratio of ionizable hydrogen atoms in \( H_3PO_3 \) to \( H_3PO_2 \) is \( 2:1 \).

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