Consider the following reaction
`6NaOH_(("Hot. Conc.")) + 3Cl_2 to 5NaCl + A + 3H_2O`. What is the oxidation number of chlorine in "A"? 

Options

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

Answer

To determine the oxidation number of chlorine in compound "A" from the reaction:

\[ 6 \text{NaOH (Hot, Conc.)} + 3 \text{Cl}_2 \rightarrow 5 \text{NaCl} + A + 3 \text{H}_2\text{O} \]

we can follow these steps:

### Step 1: Identify the products of the reaction
In this reaction, we know that chlorine reacts with hot and concentrated sodium hydroxide (NaOH) to produce sodium chloride (NaCl), water (H2O), and another compound "A". The products indicate that chlorine is being oxidized.

### Step 2: Determine the nature of compound "A"
From the known chemistry of chlorine with hot and concentrated NaOH, we can infer that compound "A" is sodium chlorate (NaClO3). This is a common product formed when chlorine reacts with concentrated alkali.

### Step 3: Write the formula for "A"
The formula for sodium chlorate is NaClO3.

### Step 4: Assign oxidation states
Now, we need to assign oxidation states to the elements in NaClO3:
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

### Step 5: Set up the oxidation state equation
Let the oxidation state of chlorine in NaClO3 be \( x \). The overall charge of the compound is neutral (0). Therefore, we can set up the equation:

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

Substituting the known values:

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

### Step 6: Solve for \( x \)
Now, simplify the equation:

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

\[ x - 5 = 0 \]

\[ x = +5 \]

### Conclusion
The oxidation number of chlorine in compound "A" (NaClO3) is +5.

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