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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?

Short Answer

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(a) The balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\) is: $$2\mathrm{NaAlH}_{4}\rightarrow2\mathrm{NaH} + \mathrm{Al} + 3\mathrm{H}_{2}$$ (b) In \(\mathrm{NaAlH}_{4}\), Hydrogen is the most electronegative element, and Sodium is the least electronegative. (c) The polyatomic anion is \(\mathrm{AlH}_{4}^-\) with the Lewis structure having Al in the center, connected to four H atoms through single bonds, and a negative charge on the ion. (d) The formal charge on Hydrogen in the polyatomic anion \(\mathrm{AlH}_{4}^-\) is 0.

Step by step solution

01

Decomposition of \(\mathrm{NaAlH}_{4}\)

To find the balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\), first, write down the products mentioned in the problem as \(\mathrm{NaH(s), Al(s)}\), and \(\mathrm{H}_{2}(g)\). Next, balance the equation by adjusting the number of moles of each species: $$2\mathrm{NaAlH}_{4}\rightarrow2\mathrm{NaH} + \mathrm{Al} + 3\mathrm{H}_{2}$$
02

Electronegativity of Elements

To find the most and least electronegative elements in \(\mathrm{NaAlH}_{4}\), compare the electronegativity values of Sodium (Na), Aluminum (Al), and Hydrogen (H). Based on the periodic table: Electronegativity of Na = 0.93 Electronegativity of Al = 1.61 Electronegativity of H = 2.20 The most electronegative element is Hydrogen while the least electronegative element is Sodium.
03

Identity and Lewis Structure of Polyatomic Ion

Since the electronegativity difference between the most and least electronegative elements (H and Na) is significant, the polyatomic anion would contain these two elements. Upon decomposing and releasing \(\mathrm{H}_{2}\), the remaining species in the compound, which creates the polyatomic anion, would be \(\mathrm{AlH}_{4}^-\). Now, let's draw the Lewis structure for \(\mathrm{AlH}_{4}^-\): 1. Count the total number of valence electrons: Al has 3 valence electrons, H has 1 valence electron, and since the ion has a negative charge, there's 1 extra electron. So, there are a total of 3 + 4(1) + 1 = 8 valence electrons. 2. Place the least electronegative element (Al) at the center and surround it with four H atoms. 3. Distribute the valence electrons: form four bonds between Al and each H atom, leaving 0 leftover electrons. 4. Check the octet rule: Aluminum and all Hydrogens have a full octet. To draw the \(\mathrm{AlH}_{4}^-\) Lewis structure, connect Al and the four H atoms with single bonds, and add the negative charge on the ion.
04

Formal Charge on Hydrogen in Polyatomic Ion

To calculate the formal charge on Hydrogen in \(\mathrm{AlH}_{4}^-\), use the formula: Formal charge = (Valence electrons of atom) - (Non-bonding electrons) - (1/2 × Bonding electrons) Each H atom in \(\mathrm{AlH}_{4}^-\) has: - 1 valence electron - 0 non-bonding electrons - 2 bonding electrons (as each H forms 1 bond with Al) Formal charge on H = ܽ1 - 0 - (1/2 × 2) = 0 Therefore, the formal charge on Hydrogen in the polyatomic anion \(\mathrm{AlH}_{4}^-\) is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Economy
The concept of the "hydrogen economy" centers around using hydrogen as a renewable and clean energy carrier. Unlike fossil fuels, hydrogen does not release carbon emissions when burned. It simply combines with oxygen to produce water, making it an environmentally friendly alternative. However, implementing a full hydrogen economy faces challenges, particularly in how hydrogen is stored, transported, and used economically.
  • Hydrogen as a fuel offers energy security because it can be produced from various domestic resources.
  • The transition involves sectors like transportation, where hydrogen fuel cells could power vehicles, and industry, where it can serve as a clean power source.
  • Efficient hydrogen storage solutions are key to this transition.
In essence, the hydrogen economy aims to reduce reliance on fossil fuels, lower greenhouse gas emissions, and enable a sustainable energy future. Without effective hydrogen storage, the potential of the hydrogen economy cannot be fully realized.
Metal Hydrides
Metal hydrides are compounds formed between metals and hydrogen. These hydrides are particularly critical in hydrogen storage solutions. They have the ability to store significant amounts of hydrogen within a compact volume.
  • Metal hydrides like \(\mathrm{NaAlH}_{4}\) are used for hydrogen storage because they can release hydrogen when needed.
  • The release of hydrogen from metal hydrides is often a reversible process, allowing for efficient storage and retrieval of hydrogen gas.
  • Light metal hydrides are advantageous because they have a high capacity for hydrogen storage by weight.
The energy stored in metal hydrides can be harnessed for various applications, making them promising candidates for the hydrogen economy. This capability helps overcome one of the major obstacles in hydrogen storage and usage.
Electronegativity
Electronegativity refers to the ability of an atom to attract electrons in a chemical bond. It is an essential concept when discussing chemical bonding and reactions.
  • In binary compounds, the element with higher electronegativity tends to attract electrons more strongly.
  • For instance, in \(\mathrm{NaAlH}_{4}\), hydrogen, being highly electronegative, plays a crucial role in bond formation.
  • Electronegativity differences dictate bond types; more significant differences lead to ionic bonds, while smaller differences favor covalent bonds.
Understanding electronegativity helps predict molecular structures and the behavior of substances. It is crucial for determining how electrons are distributed within a molecule, influencing its reactivity and interaction with other substances.
Lewis Structures
Lewis structures are diagrams that represent the bonding between atoms within a molecule. They showcase how valence electrons are distributed in the compound. Creating Lewis structures involve several steps:
  • Identify the total number of valence electrons available from all atoms.
  • Select the least electronegative atom as the central atom—typically the atom with the highest bonding capacity. For \(\mathrm{AlH}_{4}^-\), aluminum is central.
  • Arrange electrons around each atom to fulfill the octet rule, creating single, double, or triple bonds as necessary.
Drawing Lewis structures facilitates understanding of how atoms are connected and the molecule's overall shape. This representation aids in predicting physical and chemical properties, providing insights into how molecules interact with one another in various chemical reactions and conditions.

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Most popular questions from this chapter

Use Table \(8.4\) to estimate the enthalpy change for each of the following reactions: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{Cl}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12}(g)\) (the six carbon atoms form a six-membered ring with two \(\mathrm{H}\) atoms on each C atom)

The \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Writ the electron configurations of \(\mathrm{Ti}^{2+}\) and Ca. (b) Calculate th number of unpaired electrons for \(\mathrm{Ca}\) and for \(\mathrm{Ti}^{2+}\). (c) Wha charge would Ti have to be isoelectronic with \(\mathrm{Ca}^{2+}\) ?

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) \(\mathrm{OCS}\), (b) \(\mathrm{SOCl}_{2}\) ( \(\mathrm{S}\) is the central atom), (c) \(\mathrm{BrO}_{3}^{-}\), (d) \(\mathrm{HClO}_{2}\) (H is bonded to \(\left.\mathrm{O}\right)\).

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{SiF}_{4}\) and \(\mathrm{LaF}_{3}\), (b) \(\mathrm{FeCl}_{2}\) and \(\mathrm{ReCl}_{6}\), (c) \(\mathrm{PbCl}_{4}\) and \(\mathrm{RbCl}\).

Write the Lewis symbol for atoms of each of the following elements: (a) \(\mathrm{Al}\), (b) \(\mathrm{Br}\), (c) \(\mathrm{Ar}\), (d) \(\mathrm{Sr}\).

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