Problem 112
Question
Which of the following reactions written here are correct? (1) \(\mathrm{B}+\mathrm{NaOH} \longrightarrow 2 \mathrm{Na}_{3} \mathrm{BO}_{3}+\mathrm{H}_{2}\) (2) \(\mathrm{P}_{4}+\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaH}_{2} \mathrm{PO}_{2}+\mathrm{PH}_{3}\) (3) \(\mathrm{S}+\mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{Na}_{2} \mathrm{~S}+\mathrm{H}_{2} \mathrm{O}\) (a) 1 only (b) 3 only (c) 2 and 3 (d) 1,2 and 3
Step-by-Step Solution
Verified Answer
Option (c) 2 and 3 are correct reactions.
1Step 1: Analyze Reaction (1)
The given reaction is \( \mathrm{B} + \mathrm{NaOH} \longrightarrow 2 \mathrm{Na}_3\mathrm{BO}_3 + \mathrm{H}_2 \). - Boron generally does not react with NaOH to produce sodium borate directly.- Also, the balancing of the equation does not hold as boron is not acting as a reducing agent to release hydrogen as described.- Therefore, this reaction is not possible as written.
2Step 2: Analyze Reaction (2)
The given reaction is \( \mathrm{P}_4 + \mathrm{NaOH} + \mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{NaH}_2\mathrm{PO}_2 + \mathrm{PH}_3 \).- Phosphorus reacts with NaOH in presence of water to produce phosphine (\( \mathrm{PH}_3 \)) and sodium hypophosphite (\( \mathrm{NaH}_2\mathrm{PO}_2 \)). - Generally, the equation is balanced with proper stoichiometry, and this reaction is possible.- Therefore, this reaction is correctly written.
3Step 3: Analyze Reaction (3)
The given reaction is \( \mathrm{S} + \mathrm{NaOH} \longrightarrow \mathrm{Na}_2\mathrm{~S}_2\mathrm{O}_3 + \mathrm{Na}_2\mathrm{S} + \mathrm{H}_2\mathrm{O} \).- Sulfur reacts with NaOH in the presence of water to form sodium thiosulfate (\( \mathrm{Na}_2\mathrm{~S}_2\mathrm{O}_3 \)) and sodium sulfide (\( \mathrm{Na}_2\mathrm{S} \)).- The equation is properly balanced for sulfur, oxygen, and sodium.- Thus, this reaction is possible, making it correctly written.
Key Concepts
Boron and Sodium Hydroxide ReactionPhosphorus and Sodium Hydroxide ReactionSulfur and Sodium Hydroxide Reaction
Boron and Sodium Hydroxide Reaction
The reaction between boron (B) and sodium hydroxide (NaOH) is not straightforward. Boron typically interacts poorly with NaOH under normal conditions. This reaction proposes the formation of sodium borate (Na3BO3) and hydrogen gas (H2), but there's a catch. For a reaction to occur, boron must act as a reducing agent, releasing hydrogen. However, boron does not easily provide electrons to reduce water in NaOH to hydrogen gas. This makes the proposed reaction impossible as written. Furthermore, the balancing of the equation in its given form is incorrect, as the stoichiometry of reactants and products does not align. Therefore, this reaction between boron and sodium hydroxide needs more than a simple symbolic equation to properly illustrate what could happen with this element interacting with a strong base like NaOH.
Phosphorus and Sodium Hydroxide Reaction
In the presence of water, phosphorus (P4) reacts with sodium hydroxide (NaOH) differently. This specific reaction produces phosphine (PH3) and sodium hypophosphite (NaH2PO2). Phosphorus shows lively chemistry with NaOH under such conditions. What's interesting here is that this reaction is both balanced and plausible. The stoichiometric relationship maintains the integrity of atoms on both sides of the reaction equation. Phosphine, a gaseous product, is noted for its distinctive smell akin to garlic, although the primary focus is usually how it exemplifies a reduction-oxidation process. Here, phosphorus partially oxidizes while protons reduce to form hydrogen-containing phosphine. Documenting reactions like these provides students with valuable examples of oxidative-reduction patterns.
Sulfur and Sodium Hydroxide Reaction
When sulfur (S) encounters sodium hydroxide (NaOH) in the presence of water, it undergoes a transformation resulting in compounds like sodium thiosulfate (Na2S2O3) and sodium sulfide (Na2S). Sulfur's interaction with NaOH is fascinating for its production of varying sulfur-oxygen compounds. This reaction is balanced and thermodynamically feasible, providing a valid example of a redox process. In this scenario, sulfur can undergo different oxidation states, demonstrating the dynamic chemistry sulfur can engage in when reacting with bases like NaOH. The balanced equation supports both reaction feasibility and teaching value, as it showcases the potential change in sulfur's oxidation states and its ability to combine into complex compounds. This emphasizes how sulfur and sodium hydroxide reactions are vital in understanding elemental chemistry.
Other exercises in this chapter
Problem 109
If \(\mathrm{NaOH}\) is added to an aqueous solution of \(\mathrm{Zn}^{2+}\) ions, a white precipitate appears and on adding excess \(\mathrm{NaOH}\), the preci
View solution Problem 111
\(2 \mathrm{~g}\) of aluminium is treated separately with excess of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and excess of \(\mathrm{NaOH}\). The ratio of the
View solution Problem 113
Match the following: List I List II 1\. Sorel's cement (i) \(\mathrm{MgCl}_{2}\) 2\. Albite (ii) \(\mathrm{MgO}\) 3\. A salt of carnalite (iii) \(\mathrm{NaAlSi
View solution Problem 114
Aluminium is more reactive than iron because its standard reduction potential is higher. Still aluminium is less easily corrdoed than iron because (a) Al reacts
View solution