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

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

Step 1: Assign Oxidation States

Before we determine the oxidation states, remember that in any compound like \( \mathrm{KOH} \) or \( \mathrm{H_2O} \), the sum of the oxidation states equals the net charge of the molecule, which is zero for neutral molecules. In \( \mathrm{KOH} \), \( \mathrm{K} \) is typically +1, \( \mathrm{O} \) is -2, and \( \mathrm{H} \) is +1. In \( \mathrm{H_2O} \), \( \mathrm{H} \) is +1, and \( \mathrm{O} \) is -2. For the products, in \( \mathrm{PH_3} \), \( \mathrm{H} \) is +1, making \( \mathrm{P} \) -3 to balance, and in \( \mathrm{KH_2PO_2} \), \( \mathrm{P} \) will be determined next.

Step 2: Oxidation State of Phosphorus in Reactants

In the reactants, phosphorus (\( \mathrm{P} \)) in its elemental form has an oxidation state of 0 because it is in its elemental state.

Step 3: Calculate Oxidation State of Phosphorus in Products

In \( \mathrm{KH_2PO_2} \), assign known oxidation states: \( \mathrm{K} \) is +1, and \( \mathrm{H} \) is +1 each.Calculate \( \mathrm{P} \) by setting the sum of oxidation states to 0 (since \( \mathrm{KH_2PO_2} \) is a neutral molecule):\[ +1 + 2(+1) + x -4 = 0 \]Solving gives \( x = +1 \), so \( \mathrm{P} \) in \( \mathrm{KH_2PO_2} \) is +1. In \( \mathrm{PH_3} \), we established \( \mathrm{P} \) is -3.

Step 4: Analyze Changes in Oxidation States

Compare the oxidation states of \( \mathrm{P} \) in reactants and products:- From \( 0 \) to \( +1 \): \( \mathrm{P} \) is oxidized in forming \( \mathrm{KH_2PO_2} \).- From \( 0 \) to \( -3 \): \( \mathrm{P} \) is reduced in forming \( \mathrm{PH_3} \).Thus, \( \mathrm{P} \) undergoes both oxidation and reduction.

Key Concepts

Oxidation State

Understanding oxidation states is like finding a molecule's accounting balance. It shows how electrons are distributed among atoms. An oxidation state indicates how many electrons an atom gains or loses during a reaction compared to its elemental form. For example, in a molecule, you could think of oxidation states as charges that atoms earn or pay based on electron transactions. In elemental form, like phosphorus ( "P" ) in our reaction, the oxidation state is always zero. Once phosphorus combines with other elements, its oxidation state changes to reflect its new electron environment. Some rules for assigning oxidation states include:

• Elements in their standard state have an oxidation state of 0.
• For simple ions, the oxidation state equals the ion charge: e.g., Na⁺ is +1.
• Oxygen typically has an oxidation state of -2, except in peroxides.
• Hydrogen is usually +1 when bonded to nonmetals.

Recognizing these states helps predict and understand how atoms behave in chemical reactions.

Phosphorus Reactions

Phosphorus is a fascinating element due to its ability to exhibit multiple oxidation states. This enables phosphorus to engage in diverse reactions, including oxidation and reduction. In the exercise, phosphorus in the reactants starts in its elemental form with an oxidation state of 0. As the reaction progresses, phosphorus ends up creating two compounds:

• In "KH₂PO₂", phosphorus's oxidation state changes to +1, indicating that it has lost electrons and is oxidized.
• In "PH₃", phosphorus obtains an oxidation state of -3, showing it has gained electrons, so it is reduced.

Thus, phosphorus demonstrates its dual nature, undergoing both oxidation and reduction in the same chemical reaction. This kind of reaction, where an element is both oxidized and reduced, is known as a disproportionation reaction.

Chemical Equation Balancing

Chemical equation balancing ensures the law of conservation of mass is adhered to, meaning the number of atoms for each element remains the same on both sides of the equation. Balancing equations involves adjusting the coefficients, which are the numbers placed before the molecules or atoms involved in the reaction. Here are some steps to consider when balancing a chemical equation:

• List each element that appears in the reaction.
• Write the quantity of each element in the reactants and products.
• Adjust the coefficients to make the quantities of each element equal on both sides of the equation.
• Start by balancing elements that appear in only one reactant and one product, then move to those that appear in multiple compounds.

For example, in the given exercise reaction, four phosphorus atoms are needed because phosphorus participates in creating two separate compound types in different amounts on the products' side. Balancing ensures that the chemical reaction is accurately represented, reflecting how elements and compounds actually interact in reality.