Consider the following reaction: `6NaOH+3Cl _2 →5NaCl+A+3H_ 2O`.
What is the oxidation number of chlorine in A?

Answer

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 compound A
From the balanced reaction, we can deduce that the compound A is sodium chlorate, which has the formula \( \text{NaClO}_3 \).

### Step 2: Write the formula for sodium chlorate
The formula for sodium chlorate is:

\[ \text{NaClO}_3 \]

### Step 3: Assign oxidation states
In the compound \( \text{NaClO}_3 \):
- The oxidation state of sodium (Na) is +1.
- The oxidation state of oxygen (O) is -2.

### Step 4: Set up the equation for the oxidation state of chlorine
Let the oxidation state of chlorine (Cl) be \( x \). The overall charge of the compound is neutral (0), so we can set up the equation:

\[ \text{Charge of Na} + \text{Charge of Cl} + 3 \times \text{Charge of O} = 0 \]

Substituting the known values, we have:

\[ +1 + x + 3(-2) = 0 \]

### Step 5: Simplify the equation
Now, simplify the equation:

\[ +1 + x - 6 = 0 \]

This simplifies to:

\[ x - 5 = 0 \]

### Step 6: Solve for x
Now, solve for \( x \):

\[ x = +5 \]

### Conclusion
The oxidation number of chlorine in compound A (sodium chlorate, \( \text{NaClO}_3 \)) is +5.