The oxidation numbers of phosphorus in `Ba(H_2PO_2)_2` and xenon in `Na_4XeO_6` are respectively

Options

A. `+3 and +4`
B. `+2 and +6`
C. `+1 and +8`
D. `-1 and -6`

Answer

To find the oxidation numbers of phosphorus in \( Ba(H_2PO_2)_2 \) and xenon in \( Na_4XeO_6 \), we will follow these steps:

### Step 1: Determine the oxidation number of phosphorus in \( Ba(H_2PO_2)_2 \)

1. **Identify the components of the compound**:
- The compound can be dissociated into \( Ba^{2+} \) and \( H_2PO_2^{-} \).

2. **Assign oxidation states**:
- Barium (Ba) has an oxidation state of +2.
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

3. **Set up the equation**:
- The overall charge of \( H_2PO_2^{-} \) is -1.
- Let the oxidation state of phosphorus (P) be \( x \).
- The equation based on the oxidation states is:
\[ 2(+1) + x + 2(-2) = -1 \]
- Simplifying this gives:
\[ 2 + x - 4 = -1 \]
\[ x - 2 = -1 \]
\[ x = +1 \]

### Step 2: Determine the oxidation number of xenon in \( Na_4XeO_6 \)

1. **Identify the components of the compound**:
- The compound can be analyzed as \( 4Na^{+} \) and \( XeO_6^{2-} \).

2. **Assign oxidation states**:
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of xenon (Xe) be \( y \).

3. **Set up the equation**:
- The overall charge of \( XeO_6^{2-} \) is -2.
- The equation based on the oxidation states is:
\[ y + 6(-2) = -2 \]
- Simplifying this gives:
\[ y - 12 = -2 \]
\[ y = +10 \]

### Final Answer:
- The oxidation number of phosphorus in \( Ba(H_2PO_2)_2 \) is **+1**.
- The oxidation number of xenon in \( Na_4XeO_6 \) is **+10**.