What is atomicity of gaseous compound formed in the Mond's process when Ni reacts with CO?

Answer

To determine the atomicity of the gaseous compound formed in the Mond's process when nickel (Ni) reacts with carbon monoxide (CO), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Mond's Process**: The Mond process is a method used for refining nickel. In this process, nickel reacts with carbon monoxide to form a gaseous compound.

2. **Identify the Reaction**: The chemical reaction that occurs in the Mond process can be represented as:
\[ \text{Ni} + 4 \text{CO} \rightarrow \text{Ni(CO)}_4 \]
Here, nickel (Ni) reacts with four moles of carbon monoxide (CO) to form nickel carbonyl, which is represented as Ni(CO)₄.

3. **Determine the Formula of the Compound**: The compound formed is nickel carbonyl, which has the formula Ni(CO)₄.

4. **Calculate the Atomicity**: Atomicity refers to the number of atoms present in a molecule of a compound. In the case of Ni(CO)₄:
- There is 1 atom of nickel (Ni).
- There are 4 carbon monoxide (CO) units, and each CO unit consists of 1 carbon atom and 1 oxygen atom. Therefore, 4 CO contributes:
- 4 carbon atoms
- 4 oxygen atoms

Thus, the total number of atoms in Ni(CO)₄ is:
\[ 1 \text{ (Ni)} + 4 \text{ (C)} + 4 \text{ (O)} = 1 + 4 + 4 = 9 \text{ atoms} \]

5. **Conclusion**: The atomicity of the gaseous compound Ni(CO)₄ is 9.

### Final Answer:
The atomicity of the gaseous compound formed in the Mond's process when nickel reacts with carbon monoxide is **9**.

---