An element `A` in a compound `ABD` has oxidation number `A^(n-)`. It is oxidised by `Cr_(2)O_(7)^(2-)` in acid medium. In the experiment `1.68xx10^(-3)` moles of `K_(2)Cr_(2)O_(7)` were used for `3.26xx10^(-3)` moles of `ABD`. The new oxidation number of `A` after oxidation is:

Options

A. 3
B. 3-n
C. n-3
D. `+n`

Answer

To find the new oxidation number of element A after oxidation, we will follow these steps:

### Step 1: Understand the oxidation process
Element A in the compound ABD has an oxidation state of \( A^{(n-)} \). It is oxidized by \( Cr_2O_7^{2-} \) in acidic medium. In this reaction, \( Cr^{+6} \) from \( Cr_2O_7^{2-} \) is reduced to \( Cr^{+3} \).

### Step 2: Write the half-reaction
The half-reaction for the oxidation of A can be written as:
\[ A^{(n-)} \rightarrow A^{(x)} \]
where \( x \) is the new oxidation state of A after the reaction.

### Step 3: Determine the n-factor for A
The n-factor for A can be defined as the change in oxidation state multiplied by the number of moles of A involved in the reaction. Since there is one atom of A, the n-factor can be expressed as:
\[ n_{\text{factor}}(A) = |(n - x)| \]

### Step 4: Determine the n-factor for \( Cr_2O_7^{2-} \)
For the dichromate ion \( Cr_2O_7^{2-} \), the change in oxidation state is from \( +6 \) to \( +3 \). The n-factor for \( Cr_2O_7^{2-} \) is:
\[ n_{\text{factor}}(Cr) = 2 \times (6 - 3) = 6 \]

### Step 5: Use the law of equivalence
According to the law of equivalence:
\[ \text{Equivalence of } A^{(n-)} = \text{Equivalence of } Cr_2O_7^{2-} \]
This can be expressed as:
\[ \text{moles of A} \times n_{\text{factor}}(A) = \text{moles of } Cr_2O_7^{2-} \times n_{\text{factor}}(Cr) \]

### Step 6: Substitute the values
We know:
- Moles of \( K_2Cr_2O_7 \) used = \( 1.68 \times 10^{-3} \)
- Moles of \( ABD \) = \( 3.26 \times 10^{-3} \)

Since \( K_2Cr_2O_7 \) provides the same number of moles of \( Cr_2O_7^{2-} \):
\[ 1.68 \times 10^{-3} \text{ moles of } Cr_2O_7^{2-} \]

Now substituting into the equivalence equation:
\[ 3.26 \times 10^{-3} \times |(n - x)| = 1.68 \times 10^{-3} \times 6 \]

### Step 7: Solve for \( |(n - x)| \)
\[ |(n - x)| = \frac{1.68 \times 10^{-3} \times 6}{3.26 \times 10^{-3}} \]
\[ |(n - x)| = \frac{10.08 \times 10^{-3}}{3.26 \times 10^{-3}} \]
\[ |(n - x)| = 3.09 \]

### Step 8: Interpret the modulus
Since we are dealing with modulus, we have two cases:
1. \( n - x = 3.09 \)
2. \( n - x = -3.09 \)

From the first case:
\[ x = n - 3.09 \]

From the second case:
\[ x = n + 3.09 \]

### Step 9: Determine the new oxidation state
Since we are looking for a practical oxidation state, we will consider the first case where:
\[ x = n - 3 \]

### Conclusion
Thus, the new oxidation number of A after oxidation is:
\[ x = 3 - n \]