Skip to main content
Ch.8 - Basic Concepts of Chemical Bonding
Chapter 8, Problem 90c

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, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion.

Verified step by step guidance
1
Identify the elements present in NaAlH_4: sodium (Na), aluminum (Al), and hydrogen (H).
Consider the electronegativity values: Na (0.93), Al (1.61), and H (2.20).
Determine the most electronegative element: hydrogen (H) is more electronegative than both Na and Al.
Predict the polyatomic anion: AlH_4^- is likely, as Al is less electronegative than H, allowing H to form a polyatomic anion with Al.
Draw the Lewis structure for AlH_4^-: Place Al in the center, surround it with four H atoms, and ensure the total charge is -1 by adding an extra electron to complete the octet around Al.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electronegativity

Electronegativity is a measure of an atom's ability to attract and hold onto electrons in a chemical bond. It plays a crucial role in determining the type of bond formed between atoms, whether ionic or covalent. In the context of NaAlH4, understanding the electronegativity differences between sodium, aluminum, and hydrogen helps predict the nature of the bonds and the resulting polyatomic anion.
Recommended video:
Guided course
02:10
Electronegativity Trends

Ionic and Covalent Bonds

Ionic bonds form between atoms with significantly different electronegativities, resulting in the transfer of electrons from one atom to another, creating charged ions. Covalent bonds, on the other hand, involve the sharing of electrons between atoms with similar electronegativities. NaAlH4 contains both types of bonds, which is essential for understanding its structure and the behavior of the polyatomic anion formed during hydrogen release.
Recommended video:
Guided course
01:16
Chemical Bonds Example 1

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They are useful for visualizing the arrangement of electrons and predicting the geometry of molecules. Drawing the Lewis structure for the polyatomic anion in NaAlH4 will help clarify its electron distribution and the nature of its bonds, aiding in the understanding of its stability and reactivity.
Recommended video:
Guided course
04:28
Lewis Dot Structures: Ions
Related Practice
Textbook Question

(b) Using these partial charges and the atomic radii given in Figure 7.8, estimate the dipole moment of the molecule.

Textbook Question

(c) The measured dipole moment of BrCl is 0.57 D. If you assume the bond length in BrCl is the sum of the atomic radii, what are the partial charges on the atoms in BrCl using the experimental dipole moment?

Textbook Question

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, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (b) Which element in NaAlH4 is the most electronegative? Which one is the least electronegative? Which element in NaAlH4 is the least electronegative?

Textbook Question

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, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (d) What is the formal charge on hydrogen in the polyatomic ion?

Textbook Question

Structures A, B, and C show the connectivity of the atoms in three different molecules that are isomers of C3H4O. By completing the Lewis structures of these molecules, complete the information in the following table:

Isomer A Isomer B Isomer C

Number of single bonds

Number of double bonds

Number of triple bonds

Number of nonbonding pairs

Textbook Question

The triiodide ion, I3-, exists, whereas the corresponding ion with fluorine, F3-, does not. The I3- ion has a linear structure in which two outer I atoms are each bonded to a central I atom. Although I3- is a known ion, F3- is not.

c. Which of the following statements about the existence of I3- versus the nonexistence of F3- is or are true?

i. The Lewis structure of I3- shows 12 electrons around the central I atom.

ii. Elements from the second row of the periodic table generally do not form hypervalent molecules and ions.

iii. An I atom can form a hypervalent molecule or ion more readily than an F atom because of the larger size of the I atom.