Question

$\text{The oxidation number of iron in the compound formed during brown ring test for NO}_3^-\text{ ion is \_\_\_\_\_\_.}$

Answer 1

Correct Answer: 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:

1. The complex ion is [Fe(H₂O)₅NO]²⁺.
2. Let x be the oxidation number of Fe.
3. The nitrosyl (NO) group behaves as a neutral ligand maintaining a charge of 0.
4. Each water molecule, being neutral, also contributes 0.
5. Overall, the complex has a charge of +2. Thus, the equation can be set up as: \( x + 0 \times 5 + 0 = +2 \).
6. 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.

Answer 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