
The hydride ion ${{\text{H}}^{-}}$ is stronger base than hydroxide ion $\text{O}{{\text{H}}^{-}}$. Which of the following reactions will occur if sodium hydride $\left( \text{NaH} \right)$ is dissolved in water?
A. ${{\text{H}}^{-}}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\to {{\text{H}}_{3}}{{\text{O}}^{-}}$
B. ${{\text{H}}^{-}}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{O}{{\text{H}}^{-}}+{{\text{H}}_{2}}$
C. ${{\text{H}}^{-}}+{{\text{H}}_{2}}\text{O}\to $ No reaction
D. None of the above
Answer
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Hint: The hydride ion in saline hydrides is a strong base. Strong base is one which produces hydroxide ions $\left( \text{O}{{\text{H}}^{-}} \right)$ when dissolved in water. Ionise the ionic compound. Write the reaction of the cations and anions formed with water to form the required products.
Complete step by step answer:
Let us discuss the strong base and its reactions in water.
Sodium hydride $\left( \text{NaH} \right)$ is an ionic compound which ionizes into sodium cations $\left( \text{N}{{\text{a}}^{+}} \right)$ and hydride anions $\left( {{\text{H}}^{-}} \right)$ . The reaction is $\text{NaH}\to \text{N}{{\text{a}}^{+}}+{{\text{H}}^{-}}$.
The reactions which will take place when the ions are dissolved in water are given below-
(1) The dissolved hydride ions will react with water, when dissolved with water. The hydride ion will abstract proton $\left( {{\text{H}}^{+}} \right)$ from water $\left( {{\text{H}}_{2}}\text{O} \right)$ to form hydrogen gas together $\left( {{\text{H}}_{2}} \right)$ and hydroxide ion $\left( \text{O}{{\text{H}}^{-}} \right)$ is left is the solution. The reaction is ${{\text{H}}^{-}}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{O}{{\text{H}}^{-}}+{{\text{H}}_{2}}$.
(2) The left hydroxide ions $\left( \text{O}{{\text{H}}^{-}} \right)$ in the solution reacts with the sodium cations $\left( \text{N}{{\text{a}}^{+}} \right)$ in the solution to form the hydroxide of sodium metal as sodium hydroxide $\left( \text{NaOH} \right)$.
The products formed in the reaction are sodium hydroxide $\left( \text{NaOH} \right)$ and hydrogen gas $\left( {{\text{H}}_{2}} \right)$.
${{\text{H}}^{-}}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{O}{{\text{H}}^{-}}+{{\text{H}}_{2}}$ is reaction will occur if sodium hydride $\left( \text{NaH} \right)$ is dissolved in water.
The correct option is option ‘b’.
Note: The one simple concept that can be used here to find the correct answer is hydrides of alkaline metals react with water to form their corresponding hydroxides and evolve hydrogen gas. The reaction is $\text{NaH}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{NaOH}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\left( \text{g} \right)\uparrow $. The answer is obtained directly which is option ‘b’.
Complete step by step answer:
Let us discuss the strong base and its reactions in water.
Sodium hydride $\left( \text{NaH} \right)$ is an ionic compound which ionizes into sodium cations $\left( \text{N}{{\text{a}}^{+}} \right)$ and hydride anions $\left( {{\text{H}}^{-}} \right)$ . The reaction is $\text{NaH}\to \text{N}{{\text{a}}^{+}}+{{\text{H}}^{-}}$.
The reactions which will take place when the ions are dissolved in water are given below-
(1) The dissolved hydride ions will react with water, when dissolved with water. The hydride ion will abstract proton $\left( {{\text{H}}^{+}} \right)$ from water $\left( {{\text{H}}_{2}}\text{O} \right)$ to form hydrogen gas together $\left( {{\text{H}}_{2}} \right)$ and hydroxide ion $\left( \text{O}{{\text{H}}^{-}} \right)$ is left is the solution. The reaction is ${{\text{H}}^{-}}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{O}{{\text{H}}^{-}}+{{\text{H}}_{2}}$.
(2) The left hydroxide ions $\left( \text{O}{{\text{H}}^{-}} \right)$ in the solution reacts with the sodium cations $\left( \text{N}{{\text{a}}^{+}} \right)$ in the solution to form the hydroxide of sodium metal as sodium hydroxide $\left( \text{NaOH} \right)$.
The products formed in the reaction are sodium hydroxide $\left( \text{NaOH} \right)$ and hydrogen gas $\left( {{\text{H}}_{2}} \right)$.
${{\text{H}}^{-}}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{O}{{\text{H}}^{-}}+{{\text{H}}_{2}}$ is reaction will occur if sodium hydride $\left( \text{NaH} \right)$ is dissolved in water.
The correct option is option ‘b’.
Note: The one simple concept that can be used here to find the correct answer is hydrides of alkaline metals react with water to form their corresponding hydroxides and evolve hydrogen gas. The reaction is $\text{NaH}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\to \text{NaOH}\left( \text{aq}\text{.} \right)+{{\text{H}}_{2}}\left( \text{g} \right)\uparrow $. The answer is obtained directly which is option ‘b’.
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