The oxidation states of P in the following compounds `H_3PO_3,PCl_3,Ca_3P_2,Na_3PO_4,POF_3` respectively by

Options

A. `+3,+3,-3,+5,+5`
B. `+5,+5,-3,+3,+3`
C. `+4,+3,-3,+4,+5`
D. `+3,+3,-2,+4,+5`

Answer

To determine the oxidation states of phosphorus (P) in the compounds \( H_3PO_3, PCl_3, Ca_3P_2, Na_3PO_4, \) and \( POF_3 \), we will analyze each compound step by step.

### Step 1: Determine the oxidation state of P in \( H_3PO_3 \)

1. Let the oxidation state of phosphorus be \( x \).
2. The oxidation state of hydrogen (H) is \( +1 \) (3 H atoms contribute \( 3 \times +1 = +3 \)).
3. The oxidation state of oxygen (O) is \( -2 \) (3 O atoms contribute \( 3 \times -2 = -6 \)).
4. The overall charge of the molecule is 0.

Setting up the equation:
\[ 3 + x - 6 = 0 \]
\[ x - 3 = 0 \implies x = +3 \]

### Step 2: Determine the oxidation state of P in \( PCl_3 \)

1. Let the oxidation state of phosphorus be \( x \).
2. The oxidation state of chlorine (Cl) is \( -1 \) (3 Cl atoms contribute \( 3 \times -1 = -3 \)).
3. The overall charge of the molecule is 0.

Setting up the equation:
\[ x - 3 = 0 \]
\[ x = +3 \]

### Step 3: Determine the oxidation state of P in \( Ca_3P_2 \)

1. Let the oxidation state of phosphorus be \( x \).
2. The oxidation state of calcium (Ca) is \( +2 \) (3 Ca atoms contribute \( 3 \times +2 = +6 \)).
3. The overall charge of the compound is 0.

Setting up the equation:
\[ 6 + 2x = 0 \]
\[ 2x = -6 \implies x = -3 \]

### Step 4: Determine the oxidation state of P in \( Na_3PO_4 \)

1. Let the oxidation state of phosphorus be \( x \).
2. The oxidation state of sodium (Na) is \( +1 \) (3 Na atoms contribute \( 3 \times +1 = +3 \)).
3. The oxidation state of oxygen is \( -2 \) (4 O atoms contribute \( 4 \times -2 = -8 \)).
4. The overall charge of the molecule is 0.

Setting up the equation:
\[ 3 + x - 8 = 0 \]
\[ x - 5 = 0 \implies x = +5 \]

### Step 5: Determine the oxidation state of P in \( POF_3 \)

1. Let the oxidation state of phosphorus be \( x \).
2. The oxidation state of oxygen is \( -2 \) (1 O atom contributes \( -2 \)).
3. The oxidation state of fluorine (F) is \( -1 \) (3 F atoms contribute \( 3 \times -1 = -3 \)).
4. The overall charge of the molecule is 0.

Setting up the equation:
\[ x - 2 - 3 = 0 \]
\[ x - 5 = 0 \implies x = +5 \]

### Summary of Oxidation States
- \( H_3PO_3 \): +3
- \( PCl_3 \): +3
- \( Ca_3P_2 \): -3
- \( Na_3PO_4 \): +5
- \( POF_3 \): +5

### Final Answer
The oxidation states of P in the compounds are:
- \( H_3PO_3 \): +3
- \( PCl_3 \): +3
- \( Ca_3P_2 \): -3
- \( Na_3PO_4 \): +5
- \( POF_3 \): +5