Sodium hydride, NaH, is available commercially as a gray-white powder. It melts at \(800^{\circ} \mathrm{C}\) with decomposition. It reacts explosively with water and ignites spontaneously on standing in moist air. (a) Write a Lewis structure for the hydride ion and for sodium hydride. Is your Lewis structure consistent with the fact that this compound is a high- melting solid? Explain. (b) When sodium hydride is added very slowly to water, it dissolves with the evolution of a gas. The resulting solution is basic to litmus. What is the gas evolved? Why has the solution become basic? (c) Write an equation for the reaction between sodium hydride and 1-butyne, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\). Use curved arrows to show the flow of electrons in this reaction.

To draw the Lewis structure of the hydride ion (H-), we write the symbol of the element and then surround it with dots representing its valence electrons. Since hydrogen has only one electron and needs one more to achieve a stable electron configuration, the hydride ion has two valence electrons. So, the Lewis structure is: ``` H: ``` For sodium hydride (NaH), we will combine the Lewis structures of the sodium cation (Na+) and the hydride ion (H-). Since sodium has lost one electron, it does not have any valence electrons in its Lewis structure. Thus, the Lewis structure for NaH is: ``` Na H: ``` Now, let's discuss whether this simple Lewis structure is consistent with the fact that NaH is a high-melting solid.

Ionic compounds like NaH usually have high melting points due to the strong electrostatic forces between the positively charged cations and the negatively charged anions. In sodium hydride, the Na+ and H- ions are held together by these strong ionic bonds, resulting in a high melting point. So, the Lewis structure drawn above is consistent with the fact that NaH is a high-melting solid.

When sodium hydride is added to water, a gas is evolved and the resulting solution is basic. The gas evolved is hydrogen gas, which occurs due to the reaction between the hydride ion and water: \(\mathrm{H}^{-} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{H}_{2} + \mathrm{OH}^{-}\) The hydroxide ion (OH-) formed in this reaction is responsible for the basic nature of the resultant solution.

Finally, we need to write an equation for the reaction between sodium hydride and 1-butyne, as well as show the flow of electrons with curved arrows. 1-butyne has the structural formula \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{C} \equiv \mathrm{CH}\). The reaction between NaH and 1-butyne will occur at the triple bond between the carbon atoms. The equation for the reaction is: \(\mathrm{NaH} + \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{C} \equiv \mathrm{CH} \rightarrow \mathrm{NaOH} + \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{C}:\mathrm{H}\) The flow of electrons can be shown using curved arrows as follows: 1. The electron pair from the H- ion in NaH is transferred to the triple bond between the carbon atoms in 1-butyne. 2. One pair of electrons from the triple bond shifts towards the terminal carbon atom, resulting in a double bond between the carbon atoms. 3. The terminal carbon atom then forms a single bond with the hydrogen atom, resulting in the formation of an alkene.