There are some deposits of nitrates and phosphates in the earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under laboratory conditions, but microbes do it easily. Ammonia, forms a large number of complexes with transition metals. Hybridization easily explains the ease of sigma donation capability of \(NH_3\) and \(PH_3\).

Phosphine is an inflammable gas, it is prepared from white phosphorus.

White phosphorus on reacting with \(NaOH\) gives \(NaH_2P_2\) and \(PH_3\) as products. This is an example of

Disproportionation reaction

The correct answer is 2. Disproportionation reaction.

Let's break down the reaction of white phosphorus (\(P_4\)) with \(NaOH\) in detail:
The chemical reaction is as follows:
\[ P_4 + 3NaOH \rightarrow NaH_2P_2 + PH_3 \]

Explanation:

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 (\(PH_3\)).