The correct answer is option 2. Disproportionation reaction.
Let's break down the reaction of white phosphorus (\(P_4\)) with \(NaOH\) in detail: The chemical reaction is as follows: \[ P_4 + 3NaOH \rightarrow NaH_2P_2 + PH_3 \]
1. White Phosphorus (\(P_4\)): White phosphorus exists as \(P_4\) molecules, where four phosphorus atoms are arranged in a tetrahedral structure. \(P_4\) is a reactive form of phosphorus and can undergo various reactions.
2. Reaction with Sodium Hydroxide (\(NaOH\)): White phosphorus reacts with sodium hydroxide (\(NaOH\)). The balanced chemical equation indicates the stoichiometric proportions of reactants and products.
3. Products Formed: The reaction produces two main products: \(NaH_2P_2\) (sodium hypophosphite) and \(PH_3\) (phosphine gas).
4. Sodium Hypophosphite (\(NaH_2P_2\)): Sodium hypophosphite is formed as a result of the oxidation of phosphorus. The oxidation state of phosphorus in \(NaH_2P_2\) is lower compared to its state in \(P_4\). The hypophosphite ion (\(H_2PO_2^-\)) is a reducing agent in this context.
5. Phosphine Gas (\(PH_3\)): Phosphine (\(PH_3\)) is formed as a result of the reduction of phosphorus. Phosphine is a gas that is liberated in the reaction. The phosphorus in \(PH_3\) has a lower oxidation state compared to \(P_4\).
Overall Reaction: \[ P_4 + 3NaOH \rightarrow NaH_2P_2 + PH_3 \]
Type of Reaction: Disproportionation Reaction: In a disproportionation reaction, an element undergoes both oxidation and reduction in the same reaction. In this case, phosphorus in \(P_4\) is oxidized to form \(NaH_2P_2\) (higher oxidation state) and reduced to form \(PH_3\) (lower oxidation state). The reaction represents the disproportionation of phosphorus in the context of oxidation states.
Summary: The reaction of white phosphorus with sodium hydroxide is an example of a disproportionation reaction, where phosphorus undergoes both oxidation and reduction simultaneously, resulting in the formation of sodium hypophosphite (\(NaH_2P_2\)) and phosphine gas (\(PH_3\)). |