Question

To address energy and environmental issues, there is great interest in powering vehicles with hydrogen rather than gasoline. One of the most attractive aspects of the "hydrogen economy" is the fact that in principle the only emission would be water. However, two daunting obstacles must be overcome before this vision can become a reality. First, an economical method of producing hydrogen must be found. Second, a safe, lightweight, and compact way of storing hydrogen must be found. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. One of the most attractive hydrides is \(\mathrm{NaAlH}_{4}\), which can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(\mathrm{s})\), Al(s), and \(\mathrm{H}_{2}(\mathrm{~g})\). NaAlH \(_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4} .\) (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, what do you think is the identity of the polyatomic anion? Draw a Lewis structure for this ion.

Short answer

The balanced equation for the decomposition of NaAlH4 is: \[NaAlH_4 \to NaH + Al + 2H_2\] In NaAlH4, hydrogen (H) is the most electronegative element, and sodium (Na) is the least electronegative element. The polyatomic anion in NaAlH4 is the AlH4- ion. Its Lewis structure has aluminum (Al) in the center, surrounded by 4 hydrogen (H) atoms connected by single bonds and the negative charge on the aluminum atom: H | H–Al–H | H

Step-by-step solution

(a) Write a balanced equation for the decomposition of NaAlH4

From the information given, we know that NaAlH4 decomposes into NaH(s), Al(s), and H2(g). So, we can write the unbalanced equation as: \[NaAlH_4 \to NaH + Al + H_2\] Now, we need to balance the equation. It is already balanced as written: \[NaAlH_4 \to NaH + Al + 2H_2\] This is the balanced equation for the decomposition of NaAlH4.

(b) Identifying the most and least electronegative elements in NaAlH4

To determine the most and least electronegative elements, we should refer to their electronegativities on the periodic table: 1. Sodium (Na) has an electronegativity of 0.93. 2. Aluminum (Al) has an electronegativity of 1.61. 3. Hydrogen (H) has an electronegativity of 2.20. From the values above, hydrogen (H) is the most electronegative element, and sodium (Na) is the least electronegative element in NaAlH4.

(c) Identifying the polyatomic anion and drawing its Lewis structure

Based on electronegativity differences, we can infer that the polyatomic anion in NaAlH4 is the AlH4- ion. Now, we will draw its Lewis structure. First, we calculate the total number of valence electrons: - Aluminum (Al) has 3 valence electrons. - Hydrogen (H) has 1 valence electron (but there are 4 hydrogen atoms, so we multiply by 4). Total valence electrons = 3 + 4 = 7 electrons Considering the negative charge on the ion, there is an extra electron. So, the total number of valence electrons is 8. Now, we can draw the Lewis structure of the AlH4- ion: 1. Place the least electronegative atom (Al) in the center and surround it with the hydrogen atoms (H). 2. Connect the hydrogen atoms to aluminum with single covalent bonds (electron pairs). 3. Distribute the remaining electrons (lone pairs) to satisfy the octet rule for each atom. The Lewis structure for the AlH4- ion has aluminum in the center, surrounded by 4 hydrogen atoms connected by single bonds. The octet rule is satisfied for all the atoms, and the negative charge is placed on the aluminum atom. H | H–Al–H | H

Key concepts

The Hydrogen Economy

The hydrogen economy aims to revolutionize how we produce, store, and consume energy, moving us away from fossil fuels to hydrogen as a clean energy source. This transition is appealing because the primary emission from hydrogen-powered devices is water, significantly reducing greenhouse gas emissions. To achieve this, we face two major challenges: finding an economical way to produce hydrogen and developing effective storage solutions.

To address these challenges, research focuses on using light metal hydrides like sodium alanate (\(\text{NaAlH}_4\)). These compounds store hydrogen efficiently due to their high weight percentage of hydrogen. Understanding the decomposition process of these hydrides is crucial for their use in the hydrogen economy.

In essence, by focusing on hydrogen production and storage, we aim to create a sustainable energy infrastructure that could drastically reduce our reliance on fossil fuels.

NaAlH4 Decomposition

Sodium alanate, or \(\text{NaAlH}_4\), is a promising material for hydrogen storage. Its decomposition releases hydrogen gas along with solid sodium hydride (\(\text{NaH}\)) and aluminum (\(\text{Al}\)). The balanced chemical equation for this decomposition is:

\[ \text{NaAlH}_4 \to \text{NaH} + \text{Al} + 2\text{H}_2 \]

This reaction is integral to storing and releasing hydrogen safely and effectively.

The advantages of using \(\text{NaAlH}_4\) include its ability to release 5.6% of its mass as hydrogen, making it a high-density hydrogen storage medium. However, achieving this decomposition process efficiently at a reasonable temperature remains a focus of ongoing research. Catalysts and precise control of reaction conditions are explored to optimize this process for practical use.

Through further research, reliable and efficient hydrogen release from \(\text{NaAlH}_4\) could significantly impact the practicality of the hydrogen economy.

Electronegativity in NaAlH4

Electronegativity refers to an atom's ability to attract and hold onto electrons. Within the \(\text{NaAlH}_4\) molecule, electronegativity differences help us understand its molecular interactions.

Hydrogen (\(\text{H}\)) is the most electronegative element in \(\text{NaAlH}_4\), with a value of 2.20, allowing it to form polar interactions. Sodium (\(\text{Na}\)), having the least electronegativity at 0.93, tends to lose electrons, forming ionic bonds. Aluminum (\(\text{Al}\)) falls in between with a value of 1.61.

These differences in electronegativity contribute to the formation of the polyatomic anion AlH4−.

Understanding these electronegativities is essential as it helps predict molecular stability and reactivity—key factors in hydrogen storage materials. The differences also dictate the binding nature, as larger differences often indicate stronger ionic character, crucial for efficient hydrogen storage and release.

Lewis Structure of AlH4-

A Lewis structure represents the valence electrons in a molecule or ion, essential for visualizing bonding and molecular geometry. For the AlH4− ion, the Lewis structure can easily be drawn by following these steps:

- Determine the total number of valence electrons: Aluminum (\(\text{Al}\)) provides 3 electrons, and each of the four hydrogens (\(\text{H}\)) provides 1 electron, totaling 7. Accounting for the negative charge adds an additional electron, making it 8 in total.

- Place aluminum as the central atom due to its lowest electronegativity and surround it with hydrogen atoms, each connected by a single bond.

- Each hydrogen shares a pair of electrons with aluminum, fulfilling hydrogen’s requirement for stability with two electrons in its outer shell.

- The extra electron from the negative charge is included in forming complete electron pairs around aluminum, satisfying the octet rule for aluminum.

This structure helps us understand the reactivity and bonding environment of AlH4−, which is crucial for its role in hydrogen storage applications. It illustrates the distinct role each element plays within the ion and gives insight into the electron sharing and formal charge distribution critical for predicting chemical behavior.