Problem 3 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following rection \(4 \mathrm{P}+3 \mathrm{KOH}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{KH}_{2} \mathrm{PO}_{2}+\mathrm{PH}_{3}\) (a) \(\mathrm{P}\) is only oxidized (b) \(\mathrm{P}\) is only reduced (c) \(\mathrm{P}\) is both oxidized as well as reduced (d) none of these

Step-by-Step Solution

Answer

Option (c): \(\mathrm{P}\) is both oxidized as well as reduced.

Step 1: Identify Oxidation States

First, determine the initial and final oxidation states of the element \( \mathrm{P} \) (phosphorus) in the chemical equation. Initially, phosphorus in \( \mathrm{P} \) is in the oxidation state of zero. In \( \mathrm{PH}_3 \), it has an oxidation state of \(-3\), and in \( \mathrm{KH}_2\mathrm{PO}_2 \), it is in the oxidation state of \(+1\).

Step 2: Analyze the Change in Oxidation States

Phosphorus's oxidation state changes from 0 to \(+1\) in \( \mathrm{KH}_2\mathrm{PO}_2 \). This indicates an oxidation process. Meanwhile, the oxidation state changes from 0 to \(-3\) in \( \mathrm{PH}_3 \), indicating a reduction process. Since both oxidation and reduction occur, phosphorus undergoes both processes.

Step 3: Conclusion

Since \( \mathrm{P} \) is both oxidized (0 to \(+1\)) and reduced (0 to \(-3\)), option (c) 'P is both oxidized as well as reduced' is correct. This reaction is an example of disproportionation where an element simultaneously undergoes oxidation and reduction.

Key Concepts

Oxidation States

Understanding oxidation states is crucial for analyzing chemical reactions. An oxidation state, sometimes called an oxidation number, is a representation of the degree of oxidation of an atom. It helps you understand how electrons are distributed around an atom in a chemical compound. In a neutral element, such as phosphorus (P) in its elemental form, the oxidation state is zero.
When phosphorus forms compounds, its oxidation state can change. For example, in the compound \(\text{PH}_3\), phosphorus has an oxidation state of \(-3\). Here, phosphorus gains electrons, resulting in a negative oxidation state. Conversely, in \(\text{KH}_2\text{PO}_2\), the oxidation state of phosphorus is \(+1\), indicating the loss of electrons, which is a sign of participating in oxidation. Understanding these states allows you to deduce how electrons shift in a reaction, giving insight into whether the atom has been oxidized or reduced.

Oxidation Process

The oxidation process involves an increase in the oxidation state, signifying the loss of electrons. Remember the simple mnemonic: "OIL" - Oxidation Involves Loss (of electrons). In an oxidation reaction, an element becomes more positive or less negative, indicating it has lost control over its electrons. In our exercise, the element phosphorus (P) undergoes oxidation. Initially, in its elemental state, P has an oxidation state of zero. As it transitions into \(\text{KH}_2\text{PO}_2\), its oxidation state increases to \(+1\). This increase tells us that phosphorus has lost electrons, characteristic of oxidation. Identifying these changes is critical to understanding how substances interact chemically, especially in reactions involving electron transfer.

Reduction Process

Conversely, the reduction process involves a decrease in oxidation state. This signifies the gain of electrons, and you can keep this simple with the mnemonic "RIG" - Reduction Involves Gain (of electrons). During a reduction reaction, an element becomes more negative or less positive, indicating an increase in electron ownership.In the given reaction, phosphorus also experiences reduction. From its initial neutral state with an oxidation state of zero, phosphorus becomes \(-3\) in \(\text{PH}_3\). This decrease in oxidation state from zero to \(-3\) demonstrates a gain of electrons. Hence, phosphorus is reduced. This dual nature of undergoing both oxidation and reduction in a single reaction is what defines disproportionation, highlighting the unique behavior of molecules like phosphorus in specific chemical reactions.