The reaction of NaCl and `K_(2)Cr_(2)O_(7)` mixture with conc. `H_(2)SO_(4)` result in the formation of:

Options

A. `CrO_(2)Cl_(2)`
B. `CrOCl_(2)`
C. `CrO_(2)Cl`
D. `(CrOCl_(2))_(2)`

Explanation

To solve the question regarding the reaction of NaCl and K2Cr2O7 with concentrated H2SO4, we can break it down into a series of steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Sodium Chloride (NaCl)
- Potassium Dichromate (K2Cr2O7)
- Concentrated Sulfuric Acid (H2SO4)

### Step 2: Reaction of K2Cr2O7 with H2SO4
When potassium dichromate reacts with concentrated sulfuric acid, it forms potassium hydrogen sulfate (KHSO4) and chromium trioxide (CrO3) along with water (H2O). The unbalanced reaction can be written as:
\[ K_2Cr_2O_7 + H_2SO_4 \rightarrow KHSO_4 + CrO_3 + H_2O \]

### Step 3: Balancing the Reaction
To balance the reaction:
- There are 2 potassium (K) on the left, so we need 2 KHSO4 on the right.
- There are 2 chromium (Cr) on the left, so we need 2 CrO3 on the right.
- The number of hydrogen atoms needs to be balanced as well.

The balanced equation becomes:
\[ K_2Cr_2O_7 + 4 H_2SO_4 \rightarrow 2 KHSO_4 + 2 CrO_3 + 2 H_2O \]

### Step 4: Reaction of NaCl with H2SO4
Next, sodium chloride reacts with concentrated sulfuric acid to form sodium hydrogen sulfate (NaHSO4) and hydrochloric acid (HCl):
\[ NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \]

### Step 5: Combining the Products
Now we have two products from the reactions:
1. From the first reaction: CrO3
2. From the second reaction: HCl

These two products will react together:
\[ CrO_3 + 6 HCl \rightarrow CrO_2Cl_2 + 3 H_2O \]

### Step 6: Final Product
The final product of the reaction is chromyl chloride (CrO2Cl2), which is a red vapor.

### Conclusion
Thus, the reaction of NaCl and K2Cr2O7 mixture with concentrated H2SO4 results in the formation of **chromyl chloride (CrO2Cl2)**.

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