$\text{The oxidation number of iron in the compound formed during brown ring test for NO}_3^-\text{ ion is \_\_\_\_\_\_.}$
Correct Answer: 1
Approach Solution - 1
To solve the problem of determining the oxidation number of iron in the compound formed during the brown ring test for nitrate ion (NO3-), we first need to understand the chemical reaction and its ensuing step. The brown ring test involves the reaction of nitrates with ferrous sulfate (FeSO4) in the presence of concentrated sulfuric acid. This leads to the formation of a brown ring complex where the nitrate ion is reduced, and iron plays a crucial role in this complex formation.
The chemical equation for the formation of the brown ring complex is:
\[ [Fe(H_2O)_5NO]^{2+} \]
In this complex, iron is in a low oxidation state. To determine this oxidation state:
- The complex ion is [Fe(H2O)5NO]2+.
- Let x be the oxidation number of Fe.
- The nitrosyl (NO) group behaves as a neutral ligand maintaining a charge of 0.
- Each water molecule, being neutral, also contributes 0.
- Overall, the complex has a charge of +2. Thus, the equation can be set up as: \( x + 0 \times 5 + 0 = +2 \).
- Solving, we find \( x = +1 \).
Thus, the oxidation number of iron in the brown ring compound is +1.
Checking the range (1,1) confirms that +1 is the correct and only possible value within this range.
Approach Solution -2
In the brown ring test for nitrates, the compound formed is:
\([\text{Fe(H}_2\text{O)}_5(\text{NO})]^{2+}.\)
The oxidation number of Fe in this complex is:
\(+1\)
The Correct Answer is: +1
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