Consider the following reaction `6NaOH+3Cl_(2) to 5NaCl+A+3H_(2)O`. What is the oxidation number of chlorine in ''A'' ?

Options

A. `+5`
B. `+4`
C. `+3`
D. `+1`

Explanation

To determine the oxidation number of chlorine in the compound "A" formed in the reaction \(6 \text{NaOH} + 3 \text{Cl}_2 \rightarrow 5 \text{NaCl} + A + 3 \text{H}_2\text{O}\), we will follow these steps:

### Step 1: Identify the Product "A"
In the given reaction, we need to balance the number of atoms on both sides. We have:
- Sodium (Na): 6 on the left (from 6 NaOH) and 5 on the right (from 5 NaCl). Therefore, "A" must contain 1 Na.
- Chlorine (Cl): 6 on the left (from 3 Cl₂) and 5 on the right (from 5 NaCl). Therefore, "A" must contain 1 Cl.
- Oxygen (O): 6 on the left (from 6 NaOH) and 3 on the right (from 3 H₂O). Therefore, "A" must contain 3 O.
- Hydrogen (H): 6 on the left (from 6 NaOH) and 3 on the right (from 3 H₂O). Therefore, "A" must contain 0 H.

Thus, "A" can be identified as \( \text{NaClO}_3 \) (sodium chlorate).

### Step 2: Assign Oxidation Numbers
Now that we have identified "A" as \( \text{NaClO}_3 \), we can assign oxidation numbers:
- Sodium (Na) typically has an oxidation number of +1.
- Oxygen (O) typically has an oxidation number of -2.

### Step 3: Set Up the Equation
Let the oxidation number of chlorine (Cl) in \( \text{NaClO}_3 \) be \( x \). The compound is neutral, so the sum of oxidation numbers must equal zero:
\[ \text{Oxidation number of Na} + \text{Oxidation number of Cl} + 3 \times \text{Oxidation number of O} = 0 \]
Substituting the known values:
\[ +1 + x + 3(-2) = 0 \]

### Step 4: Solve for \( x \)
Now, we can simplify the equation:
\[ +1 + x - 6 = 0 \]
\[ x - 5 = 0 \]
\[ x = +5 \]

### Conclusion
The oxidation number of chlorine in "A" (which is \( \text{NaClO}_3 \)) is +5.