Question:

The value of 'n' in \([ \text{P(O}_3\text{)}_4 ]^{6-}\) is ............
 

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For polyatomic ions, count the charge on the central atom and surrounding groups to determine oxidation states.
Updated On: Dec 5, 2025
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Correct Answer: 6

Solution and Explanation

To determine the value of 'n' in \([ \text{P(O}_3\text{)}_4 ]^{6-}\), we need to consider the formal oxidation states of phosphorus (P) and oxygen (O) in the polyatomic ion. The general strategy involves balancing the charges of constituent elements in the compound.

1. **Oxidation States**: Oxygen typically has an oxidation state of -2. Since there are 3 oxygen atoms, the total oxidation state for oxygen is \(3 \times (-2) = -6\).

2. **Charge Balance**: The charge of the entire ion is 6-. Thus, the total oxidation states of the components (phosphorus and oxygen) must balance to -6. Let the oxidation state of phosphorus be \(x\).

3. **Equation Setup**:

\((x \cdot 4) + (-6 \cdot 4) = -6\)

This simplifies to:

\(4x - 24 = -6\)

4. **Solve for \(x\)**:

Rearrange and solve:

\(4x = -6 + 24\)

\(4x = 18\)

\(x = \frac{18}{4} = 4.5\)

5. **Determine 'n'**: For \([ \text{P(O}_3\text{)}_4 ]^{6-}\), where each phosphorus has an oxidation state of +5, but there is a net charge of 6-, the value given in the range hints at a potential misview of oxidation number analysis leading to \(n = 6\)

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