A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s),\) and $\mathrm{H}_{2}(g) . \mathrm{NaAlH}_{4}$ 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, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion. (d) What is the formal charge on hydrogen in the polyatomic ion?

Step by step solution

01

Write a Balanced Equation for the Decomposition of NaAlH4

We are given that the decomposition of NaAlH4 produces NaH, Al, and H2 as products. To balance the equation, we need to ensure that the number of atoms of each element on both sides of the equation is equal. The balanced equation for the decomposition is: \[NaAlH_4 \rightarrow NaH + Al + 2H_2\]

02

Determine Electronegativity of Elements in NaAlH4

Electronegativity describes the ability of an atom to attract shared electrons. According to the periodic table, electronegativity generally increases from left to right and bottom to top. In NaAlH4, we have 3 elements: sodium (Na), aluminum (Al), and hydrogen (H). Using the trend mentioned above, we can determine the most and least electronegative elements in NaAlH4. The most electronegative element is hydrogen, and the least electronegative is sodium.

03

Predict the Identity of the Polyatomic Anion and Draw a Lewis Structure

Based on the electronegativity differences, we expect the polyatomic anion to consist of aluminum and hydrogen because they are closer in electronegativity compared to sodium. Aluminum can form a tetrahedral structure with 4 hydrogen atoms with a -1 charge, resulting in the polyatomic ion AlH4-. To draw the Lewis structure for AlH4-, follow these steps: 1. Count the total number of valence electrons: Al has 3 valence electrons, and each H has 1 valence electron, making a total of 7 valence electrons. Since AlH4- has a -1 charge, add one more electron. 2. Place the central atom (Al) in the center and surround it with 4 hydrogen atoms. 3. Distribute the remaining electrons (8) as pairs in covalent bonds between Al and each H atom. Now, we have a valid Lewis structure for the AlH4- ion.

04

Determine the Formal Charge on Hydrogen in the Polyatomic Ion

The formal charge on an atom can be calculated using the formula: Formal charge = Valence electrons - Non-bonding electrons - (1/2 * Bonding electrons) For hydrogen in the AlH4- ion, the number of valence electrons is 1, there are no non-bonding electrons, and there is 1 bonding electron (the electron shared in the covalent bond with Al). Thus, we can calculate the formal charge on hydrogen: Formal charge = 1 - 0 - (1/2 * 1) = 0 The formal charge on hydrogen in the AlH4- polyatomic ion is 0.

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