Which of the following is correctly matched :-
`{:(,"Compound"," Oxidation number"),((a),underlineAlF_(3),("P")+2.5),((b),underline(P)_(4)O_(10),(Q)+7),((c),HunderlineClO_(4),(R)+5),((d),underline(S)_(4)O_(6)^(2-),(S)+3):}`

Options

A. `{:("(a)",(b),(c),(d)),(P,Q,R,S):}`
B. `{:("(a)",(b),(c),(d)),(S,R,Q,P):}`
C. `{:("(a)",(b),(c),(d)),(S,R,P,Q):}`
D. `{:("(a)",(b),(c),(d)),(P,Q,R,S):}`

Answer

To solve the problem of matching the compounds with their respective oxidation numbers, we will calculate the oxidation state of each underlined element step by step.

### Step 1: Calculate the oxidation state of Aluminium in AlF₃
1. Let the oxidation state of Aluminium (Al) be \( x \).
2. Fluorine (F) has an oxidation state of -1. Since there are 3 fluorine atoms, the total contribution from fluorine is \( 3 \times (-1) = -3 \).
3. The compound is neutral, so the sum of oxidation states must equal 0:
\[ x + (-3) = 0 \]
4. Solving for \( x \):
\[ x = +3 \]
Thus, the oxidation state of Aluminium in AlF₃ is +3.

### Step 2: Calculate the oxidation state of Phosphorus in P₄O₁₀
1. Let the oxidation state of Phosphorus (P) be \( x \).
2. Oxygen (O) generally has an oxidation state of -2. For 10 oxygen atoms, the total contribution is \( 10 \times (-2) = -20 \).
3. The compound is neutral, so:
\[ 4x + (-20) = 0 \]
4. Solving for \( x \):
\[ 4x = 20 \implies x = +5 \]
Thus, the oxidation state of Phosphorus in P₄O₁₀ is +5.

### Step 3: Calculate the oxidation state of Chlorine in HClO₄
1. Let the oxidation state of Chlorine (Cl) be \( x \).
2. Hydrogen (H) has an oxidation state of +1, and for 4 oxygen atoms, the total contribution is \( 4 \times (-2) = -8 \).
3. The compound is neutral, so:
\[ x + 1 + (-8) = 0 \]
4. Solving for \( x \):
\[ x - 7 = 0 \implies x = +7 \]
Thus, the oxidation state of Chlorine in HClO₄ is +7.

### Step 4: Calculate the oxidation state of Sulfur in S₄O₆²⁻
1. Let the oxidation state of Sulfur (S) be \( x \).
2. For 6 oxygen atoms, the total contribution is \( 6 \times (-2) = -12 \).
3. The overall charge of the ion is -2, so:
\[ 4x + (-12) = -2 \]
4. Solving for \( x \):
\[ 4x - 12 = -2 \implies 4x = 10 \implies x = +2.5 \]
Thus, the oxidation state of Sulfur in S₄O₆²⁻ is +2.5.

### Summary of Results
- **AlF₃**: Al = +3
- **P₄O₁₀**: P = +5
- **HClO₄**: Cl = +7
- **S₄O₆²⁻**: S = +2.5

### Matching
Now we can match the oxidation states with the options given:
- (a) AlF₃: +3 (P)
- (b) P₄O₁₀: +5 (Q)
- (c) HClO₄: +7 (R)
- (d) S₄O₆²⁻: +2.5 (S)

The correct matches are:
- a → P
- b → Q
- c → R
- d → S

Thus, the correct answer is **option (b)**.