Question

Draw Lewis structures for the following: (a) \(\mathrm{SiH}_{4},\) (b) CO, (c) \(\mathrm{SF}_{2}\), (d) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{H}\) is bonded to \(\mathrm{O})\), (e) \(\mathrm{ClO}_{2}^{-},\) (f) \(\mathrm{NH}_{2} \mathrm{OH}\).

Short answer

The Lewis structures for the given compounds are: (a) SiH4: \[ \begin{array}{c} \text{H}\\ \text{|}\\ \text{H}-\text{Si}-\text{H}\\ \text{|}\\ \text{H} \end{array} \] (b) CO: \[ \begin{array}{c} \text{C}≡\text{O}\\ \text{:}\text{ }\text{:} \end{array} \] (c) SF2: \[ \begin{array}{c} \text{:}\text{ }\text{:}\\ \text{F}-\text{S}-\text{F}\\ \text{:}\text{ }\text{:} \end{array} \] (d) H2SO4: \[ \begin{array}{c} \text{O}=\text{S}=\text{O}\\ \text{|}\text{ }\text{ }\text{}\text{|}\\ \text{O}-\text{H}\text{ }\text{H}-\text{O} \end{array} \] (e) ClO2⁻: \[ \begin{array}{c} \text{:}\text{O}-\text{Cl}^{{-}}-\text{O:}\\ \text{ }\text{:}\text{ }\text{ }\text{ }\text{:} \end{array} \] (f) NH2OH: \[ \begin{array}{c} \text{ }\text{H}\\ \text{ }\text{|}\\ \text{H}-\text{N}-\text{O}-\text{H}\\ \text{ }\text{:}\text{ } \end{array} \]

Step-by-step solution

(a) Determine the Valence Electrons for SiH4)

To draw the Lewis structure of SiH4, we need to determine the number of valence electrons for each atom involved. Silicon (Si) has 4 valence electrons and Hydrogen (H) has 1 valence electron. Since there are four hydrogen atoms involved in the compound, there will be 4(1) = 4 total valence electrons for the hydrogen atoms, making a total of 8 valence electrons in the molecule.

(a) Draw the Lewis Structure for SiH4)

To draw the Lewis structure of SiH4, start with the central atom, Si, and connect each hydrogen atom to it with a single bond, representing the sharing of two electrons. This accounts for all 8 available valence electrons: \[ \begin{array}{c} \text{H} \\ \text{|} \\ \text{H}-\text{Si}-\text{H} \\ \text{|} \\ \text{H} \end{array} \]

(b) Determine the Valence Electrons for CO)

To draw the Lewis structure of CO, we need to determine the number of valence electrons for each atom involved. Carbon (C) has 4 valence electrons and Oxygen (O) has 6 valence electrons. This gives us a total of 10 valence electrons in the molecule.

(b) Draw the Lewis Structure for CO)

To draw the Lewis structure of CO, start with the carbon atom as the central atom and share a triple bond between the carbon and oxygen atoms. This accounts for 6 of the 10 available valence electrons. Then, add two lone pairs to the oxygen atom to make up the remaining 4 valence electrons: \[ \begin{array}{c} \text{C}≡\text{O} \\ \text{:}\text{ }\text{:} \end{array} \]

(c) Determine the Valence Electrons for SF2)

To draw the Lewis structure for SF2, we need to determine the number of valence electrons for each atom involved. Sulfur (S) has 6 valence electrons and Fluorine (F) has 7 valence electrons. Since there are two fluorine atoms involved in the compound, there will be 2(7) = 14 total valence electrons for the fluorine atoms, making a total of 20 valence electrons in the molecule.

(c) Draw the Lewis Structure for SF2)

To draw the Lewis structure for SF2, start with the sulfur atom as the central atom and connect each fluorine atom with a single bond. This accounts for 4 of the 20 available valence electrons. Then, add three lone pairs to each fluorine atom, and add two lone pairs of electrons on the sulfur atom to complete the structure: \[ \begin{array}{c} \text{:}\text{ }\text{:} \\ \text{F}-\text{S}-\text{F} \\ \text{:}\text{ }\text{:} \end{array} \]

(d) Determine the Valence Electrons for H2SO4)

To draw the Lewis structure for H2SO4, we need to determine the number of valence electrons for each atom involved. Sulfur (S) has 6 valence electrons, Oxygen (O) has 6 valence electrons, and Hydrogen (H) has 1 valence electron. There are 2 hydrogen atoms and 4 oxygen atoms, so there will be 2(1) = 2 total valence electrons for the hydrogen atoms and 4(6) = 24 total valence electrons for the oxygen atoms, making a total of 32 valence electrons in the molecule.

(d) Draw the Lewis Structure for H2SO4)

To draw the Lewis structure for H2SO4 (with H bonded to O), start with the sulfur atom as the central atom and connect each oxygen atom to it with a double bond. This accounts for 12 valence electrons. Add one lone pair of electrons to each oxygen atom in the existing double bonds, and then connect the two remaining oxygen atoms to the two hydrogen atoms with single bonds. Add two more lone pairs to these oxygen atoms to make up the remaining valence electrons: \[ \begin{array}{c} \text{O}=\text{S}=\text{O} \\ \text{|}{ }\text{ }\ \ \text{|} \\ \text{O}-\text{H}\text{ }\text{H}-\text{O} \end{array} \]

(e) Determine the Valence Electrons for ClO2⁻)

To draw the Lewis structure for ClO2⁻, we need to determine the number of valence electrons for each atom involved and remembering there is an extra electron due to the negative charge. Chlorine (Cl) has 7 valence electrons, and Oxygen (O) has 6 valence electrons. Since there are two oxygen atoms involved in the compound, there will be 2(6) = 12 total valence electrons for the oxygen atoms, making a total of 20 valence electrons in the molecule.

(e) Draw the Lewis Structure for ClO2⁻)

To draw the Lewis structure for ClO2⁻, begin with the chlorine atom as the central atom and connect each oxygen atom to it with a single bond. This accounts for 4 of the 20 available valence electrons. Add three lone pairs to each oxygen atom and two lone pairs to the chlorine atom, and still, there is one electron left. Place this extra electron on the central chlorine atom, indicating the negative charge on the structure as well: \[ \begin{array}{c} \text{:}\text{O}-\text{Cl}^{{-}}-\text{O:} \\ \text{ }\ \text{:}\text{ }\text{ }\text{ }\text{:} \end{array} \]

(f) Determine the Valence Electrons for NH2OH)

To draw the Lewis structure for NH2OH, we need to determine the number of valence electrons for each atom involved. Nitrogen (N) has 5 valence electrons, Oxygen (O) has 6 valence electrons, and Hydrogen (H) has 1 valence electron. Since there are two hydrogen atoms involved in the compound, there will be 2(1) = 2 total valence electrons for the hydrogen atoms, making a total of 14 valence electrons in the molecule.

(f) Draw the Lewis Structure for NH2OH)

To draw the Lewis structure for NH2OH, begin with the nitrogen atom as the central atom and connect each hydrogen atom to it with a single bond. This accounts for 4 of the 14 available valence electrons. Then connect the nitrogen atom to the oxygen atom with a single bond using 2 more electrons, and place two lone pairs on the oxygen atom. Finally, connect the oxygen atom to the last hydrogen atom with a single bond and add a lone pair to the central nitrogen atom, which accounts for the remaining valence electrons: \[ \begin{array}{c} \text{ }\text{H} \\ \text{ }\text{|} \\ \text{H}-\text{N}-\text{O}-\text{H} \\ \text{ }\ \text{:}\text{ } \end{array} \]

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a pivotal role in chemical bonding. They are the electrons available for forming bonds with other atoms. To understand how atoms combine to form molecules, it is crucial to identify the number of valence electrons an atom possesses. This concept is central when drawing Lewis structures, as it guides us in determining how atoms bond together.

As we've seen in the exercise for the molecule SiH₄, silicon has four valence electrons and hydrogen has one, leading to a total of eight valence electrons that need to be accounted for in the Lewis structure. When developing content to explain valence electrons, emphasizing the recurring ‘octet rule’—the tendency of atoms to prefer eight electrons in their valence shell—is essential. Completing or having eight valence electrons gives stability to the molecule, similar to the electron configuration of noble gases.

However, some elements like hydrogen are stable with a 'duet' rule—having two electrons in their valence shell—while others may have an expanded octet, like sulfur in SO₄²⁻. It's important to recognize exceptions to the rules and teach students how to identify them.

Molecular Chemistry

Molecular chemistry focuses on the study of the chemical properties and reactions of molecules. When discussing molecular chemistry with students, the importance of understanding Lewis structures is vital. These structures are a simple way to represent molecules visually, illustrating how atoms are arranged and bonded in a molecule and the presence of lone pair electrons.

For instance, in our exercise, H₂SO₄ requires careful consideration of the placement of atoms and electrons to ensure that each atom satisfies its need for electrons (observing the octet rule wherever applicable). Understanding the Lewis structures helps predict molecule geometry, reactivity, polarity, and other chemical properties. Building a strong foundation in recognizing and drawing these structures can significantly enhance a student’s grasp of molecular behavior and interactions.

In our lesson content, using models and visual aids can be an exceptional way to break down complex molecules for students, making molecular chemistry much more approachable.

Chemical Bonding

Chemical bonding is fundamental to molecular chemistry, as it explains how and why substances combine to form new materials. Bonds are forces maintaining the atoms together in molecules, and they come in different types, such as ionic, covalent, and metallic bonds. For the purpose of our Lewis structures exercise, we focus on covalent bonding, where atoms share valence electrons to achieve stability.

An understanding of electron sharing is crucial in drawing the correct Lewis structure. For instance, CO has a triple bond, where three pairs of electrons are shared between carbon and oxygen, whereas in SF₂, sulfur and each fluorine atom share one pair of electrons through a single bond. Additionally, some atoms like chlorine in ClO₂⁻ carry a charge, which influences bonding and structure.

While designing educational material on chemical bonding, it is beneficial to incorporate interactive bonding exercises that allow students to practice identifying the types of bonds and predict the resulting molecule's stability and properties. Relating chemical bonding to real-world examples can further elucidate its relevance and captivate students' interest in the topic.