Question

Draw Lewis structures for the following: (a) \(\mathrm{Ba}^{2+}\) (b) \(\mathrm{Al}^{3+}\) (c) \(\mathrm{SO}_{3}^{2-}\) (d) \(\mathrm{CN}^{-}\) (e) \(\mathrm{HCO}_{3}^{-}\)

Short answer

\(\mathrm{Ba}^{2+}\) and \(\mathrm{Al}^{3+}\) ions have no valence electrons or Lewis structures; \(\mathrm{SO}_{3}^{2-}\) has a central sulfur atom bonded to three oxygen atoms and lone pairs to conform to expanded octet; \(\mathrm{CN}^{-}\) has a triple bond between C and N with additional lone pairs on N; \(\mathrm{HCO}_{3}^{-}\) has an H connected to an O, a double bond between C and an O, and single bonds to other Os with appropriate lone pairs and charges.

Step-by-step solution

Drawing Lewis Structure for \(\mathrm{Ba}^{2+}\)

Since barium is an alkaline earth metal, it loses two electrons to form a \(\mathrm{Ba}^{2+}\) ion. In its ionic form, \(\mathrm{Ba}^{2+}\) has no valence electrons and thus the Lewis structure is simply the symbol \(\mathrm{Ba}\) with a 2+ charge indicated.

Drawing Lewis Structure for \(\mathrm{Al}^{3+}\)

Aluminum, being a member of group 13, loses three electrons to form a \(\mathrm{Al}^{3+}\) ion. Similar to \(\mathrm{Ba}^{2+}\), \(\mathrm{Al}^{3+}\) has no valence electrons in its ionic form and the Lewis structure is just the symbol \(\mathrm{Al}\) with a 3+ charge.

Drawing Lewis Structure for \(\mathrm{SO}_{3}^{2-}\)

Sulfur trioxide with a 2- charge has a total of 26 valence electrons (24 from three oxygen atoms, 6 from sulfur, and an additional 2 from the charge). Draw a single bond between the sulfur atom and each oxygen atom. Then, complete the octet around each oxygen atom with lone pairs, ensuring sulfur has an expanded octet if necessary. Place the remaining electrons on sulfur and place brackets around the structure with a 2- charge.

Drawing Lewis Structure for \(\mathrm{CN}^{-}\)

The cyanide ion has 10 valence electrons (5 from carbon, 5 from nitrogen, and an additional 1 from the charge). Draw a triple bond between carbon and nitrogen to fulfill both atoms' octet, accounting for 6 electrons used, then place the remaining 4 electrons as lone pairs around nitrogen. Enclose the structure in brackets and assign a 1- charge.

Drawing Lewis Structure for \(\mathrm{HCO}_{3}^{-}\)

The bicarbonate ion has a total of 24 valence electrons (4 from hydrogen, 4 from carbon, 18 from three oxygen atoms, and an additional 1 from the charge). Connect hydrogen to one of the oxygens, which will not hold additional lone pairs, connect carbon with single bonds to two oxygens and one double bond to the remaining oxygen to fulfill the octet rule. Distribute remaining electrons as lone pairs on the oxygen atoms. Enclose in brackets with a 1- charge.

Key concepts

Chemical Bonding

Chemical bonding is the force that holds atoms together in chemical compounds. This strong attraction between atoms, ions, or molecules allows for the formation of chemical substances. Atoms bond to achieve greater stability, which often comes in the form of full or stable electron shells. There are three primary types of chemical bonds:

• Ionic bonds, which occur when one atom donates an electron to another, creating oppositely charged ions that attract each other, as seen in the case of \(\mathrm{Ba}^{2+}\) and \(\mathrm{Al}^{3+}\) ions.
• Covalent bonds, which involve sharing pairs of electrons between atoms to achieve stability, are observed in molecules like \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{CN}^{-}\), and \(\mathrm{HCO}_{3}^{-}\).
• Metallic bonds, found in metals, involve a 'sea' of shared electrons that are free to move around, giving metals their distinctive properties.

Each bond type follows specific rules and contributes to the unique properties of the resultant compound. The structure and strength of these bonds directly determine the physical and chemical properties of the substances, such as melting point, boiling point, reactivity, and electrical conductivity.

Valence Electrons

Valence electrons play a critical role in chemical bonding. They are the electrons located in the outermost shell of an atom and are responsible for the atom's chemical properties. Atoms interact and bond with other atoms using these valence electrons to achieve a full valence shell, which is typically associated with a more stable, lower-energy configuration.

In the examples of \(\mathrm{Ba}^{2+}\) and \(\mathrm{Al}^{3+}\), the original atoms lose their valence electrons, resulting in cations with full valence shells below the outermost one. Conversely, in molecules like \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{CN}^{-}\), and \(\mathrm{HCO}_{3}^{-}\), the atoms share their valence electrons to bond. When depicting valence electrons in Lewis structures, we must count the total number of valence electrons that will participate in the bonding, including any extra electrons due to negative charges, as seen with the polyatomic ions.

Octet Rule

The octet rule is a fundamental concept in chemistry that refers to the tendency of atoms to prefer to have eight electrons in their valence shell. This rule predicts the formation, configurations, and properties of different chemical compounds.

While the octet rule applies to most cases, there are exceptions. Elements in periods 3 and below of the periodic table, such as sulfur in \(\mathrm{SO}_{3}^{2-}\), can have an expanded octet because they have d orbitals that can accommodate more electrons. The step-by-step solutions for \(\mathrm{CN}^{-}\) and \(\mathrm{HCO}_{3}^{-}\) show that multiple bonds (such as double and triple bonds) may form to satisfy the octet requirement for both atoms involved in the bond. The fulfillment of the octet rule results in more chemically stable molecules. In ionic compounds like \(\mathrm{Ba}^{2+}\) and \(\mathrm{Al}^{3+}\), the cations form by losing electrons and achieving a stable noble gas configuration, implicitly adhering to the octet principle.