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Question

In the Mond process for refining nickel, the volatile intermediate formed is:

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Mond Process for Nickel Refining
	**Related Topics**: Metal carbonyl formation, purification of metals
	**Prerequisites**: Basic understanding of chemical reactions and coordination complexes.
## SOLUTION PATHWAY
**Given Information**:
	The question asks about the volatile intermediate formed in the Mond process for nickel refining.
	The options are different nickel carbonyl compounds.
**Method & Steps**:
	1. Understand the Mond Process: The Mond process is a technique used to purify nickel. It involves reacting impure nickel with carbon monoxide ($CO$) at moderate temperatures and pressures to form a volatile nickel carbonyl compound. This volatile compound is then decomposed at higher temperatures to yield pure nickel.
	2. Identify the volatile intermediate: The volatile intermediate formed is a nickel carbonyl compound. The most common and stable nickel carbonyl is tetracarbonylnickel(0), $Ni(CO)_4$.
	3. Evaluate the options: Option A, $Ni(CO)_4$, is the correct volatile intermediate formed in the Mond process. Options B, C, and D are not the primary volatile compounds formed in this process.
	Result: $Ni(CO)_4$
	Verification: The formation and decomposition of $Ni(CO)_4$ are the key steps in the Mond process.
## OPTION ANALYSIS
[A] ✓: $Ni(CO)_4$ is the volatile intermediate formed in the Mond process. It is a tetrahedral complex where nickel is in the zero oxidation state.
[B] :x: $Ni(CO)_6$ is not a stable compound. Nickel's coordination number is typically 4.
[C] :x: $Ni_2(CO)_4$ is not the primary volatile species in the Mond process. Although dimers can exist, the monomer, $Ni(CO)_4$, is the dominant species.
[D] :x: $Ni(CO)_2$ is not a stable compound under the conditions of the Mond process.
## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing the stoichiometry of the nickel carbonyl compound:  Remembering that nickel commonly forms tetrahedral complexes helps avoid this mistake.
		2.  Choosing incorrect oxidation state of Ni:  Understanding the role of CO as a ligand helps avoid selecting an incorrect oxidation state for nickel.
	**Quick Tips**:
		1. Remember the Mond process involves the formation of a volatile nickel carbonyl compound.
		2. Nickel commonly forms tetrahedral complexes.

A.RAMA NAGESWARARAO · Answer

The correct Option is "$Ni(CO)_4$"

A.RAMA NAGESWARARAO · Answer

$Ni(CO)_6$

A.RAMA NAGESWARARAO · Answer

$Ni_2(CO)_4$

A.RAMA NAGESWARARAO · Answer

$Ni(CO)_2$

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