Question:medium
$PH_4I + NaOH$ forms
$PH_4I + NaOH$ forms
Updated On: Jun 24, 2026
- $PH_3$
- $NH_3$
- $P_4O_6$
- $P_4O_10$
Show Solution
The Correct Option is A
Solution and Explanation
To determine the product formed when PH_4I reacts with NaOH, we need to consider the chemical reaction involved.
Phosphonium iodide (PH_4I) reacts with sodium hydroxide (NaOH) in the following manner:
PH_4I + NaOH \rightarrow PH_3 + NaI + H_2O
In this reaction, phosphine gas (PH_3) is produced, along with sodium iodide (NaI) and water (H_2O). Let's break down the process:
- The phosphonium ion (PH_4^+) reacts with hydroxide ion (OH^−) from NaOH.
- This reaction releases phosphine gas (PH_3) and water (H_2O). The iodide ion (I^−) combines with sodium ions (Na^+) to form sodium iodide (NaI).
Given the reaction, the product formed is phosphine gas (PH_3), hence the correct option is:
Correct Answer: PH_3
Let's briefly analyze the other options to rule them out:
- NH_3: This is ammonia, which is not a product of this reaction.
- P_4O_6: Phosphorus trioxide is not produced in this reaction as it requires different reactants and conditions.
- P_4O_10: Phosphorus pentoxide is also not produced in this reaction.
Therefore, the reaction correctly forms phosphine gas PH_3 when PH_4I reacts with NaOH.
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