Lewis Symbols and Structures

D. Latimer; D. Vanderwel; J. Hollett

40 Lewis Symbols and Structures

Learning Objectives

By the end of this section, you will be able to:

Write Lewis symbols for neutral atoms and ions
Draw Lewis structures depicting the bonding in simple molecules

Thus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.

Lewis Symbols

We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:

A Lewis structure of calcium is shown. A lone pair of electrons are shown to the right of the symbol.

(Figure) shows the Lewis symbols for the elements of the third period of the periodic table.

Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table.

Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:

Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:

(Figure) demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.

Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. The total number of electrons does not change.

Lewis Structures

We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:

The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:

Two Lewis structures are shown. The left-hand structure shows two H atoms connected by a single bond. The right-hand structure shows two C l atoms connected by a single bond and each surrounded by six dots.

A single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.

The Octet Rule

The other halogen molecules (F₂, Br₂, I₂, and At₂) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.

The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl₄ (carbon tetrachloride) and silicon in SiH₄ (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:

Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH₃ (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:

Double and Triple Bonds

As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH₂O (formaldehyde) and between the two carbon atoms in C₂H₄ (ethylene):

A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN^–):

Writing Lewis Structures with the Octet Rule

For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:

For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:

Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
Place all remaining electrons on the central atom.
Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

Let us determine the Lewis structures of SiH₄, ${\text{CHO}}_{2}{}^{\text{−}},$ NO⁺, and OF₂ as examples in following this procedure:

Determine the total number of valence (outer shell) electrons in the molecule or ion.
- For a molecule, we add the number of valence electrons on each atom in the molecule:
  
  $\begin{array}{l}\\ \phantom{\rule{0.8em}{0ex}}{\text{SiH}}_{4}\\ \phantom{\rule{0.8em}{0ex}}\text{Si: 4 valence electrons/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 atom}=4\\ \underset{¯}{+\phantom{\rule{0.2em}{0ex}}\text{H: 1 valence electron/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{4 atoms}=4}\\ \\ \phantom{\rule{16.0em}{0ex}}=\text{8 valence electrons}\end{array}$
- For a negative ion, such as ${\text{CHO}}_{2}{}^{\text{−}},$ we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):
  
  $\begin{array}{l}\\ \phantom{\rule{0.5em}{0ex}}{\text{CHO}}_{2}{}^{\text{−}}\\ \phantom{\rule{0.48em}{0ex}}\text{C: 4 valence electrons/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 atom}=4\\ \phantom{\rule{0.8em}{0ex}}\text{H: 1 valence electron/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 atom}=1\\ \phantom{\rule{0.05em}{0ex}}\text{O: 6 valence electrons/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{2 atoms}=12\\ \underset{¯}{+\phantom{\rule{6.5em}{0ex}}\text{1 additional electron}=1}\\ \\ \phantom{\rule{15.50em}{0ex}}=\text{18 valence electrons}\end{array}$
- For a positive ion, such as NO⁺, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:
  
  $\begin{array}{}\\ \\ \phantom{\rule{0.1em}{0ex}}{\text{NO}}^{+}\\ \text{N: 5 valence electrons/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 atom}=5\\ \\ \phantom{\rule{0.4em}{0ex}}\text{O: 6 valence electron/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 atom}=6\\ \phantom{\rule{0.35em}{0ex}}\underset{¯}{+\phantom{\rule{0.5em}{0ex}}\text{−1 electron (positive charge)}\phantom{\rule{1.8em}{0ex}}=-1}\\ \\ \phantom{\rule{15.02em}{0ex}}=\text{10 valence electrons}\end{array}$
- Since OF₂ is a neutral molecule, we simply add the number of valence electrons:
  
  $\begin{array}{l}\\ \phantom{\rule{0.8em}{0ex}}{\text{OF}}_{\text{2}}\\ \phantom{\rule{1.25em}{0ex}}\text{O: 6 valence electrons/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 atom}=6\\ \phantom{\rule{0.1em}{0ex}}\underset{¯}{+\phantom{\rule{0.2em}{0ex}}\text{F: 7 valence electrons/atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{2 atoms}=14}\\ \phantom{\rule{16.28em}{0ex}}=\text{20 valence electrons}\end{array}$
Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)

When several arrangements of atoms are possible, as for ${\text{CHO}}_{2}{}^{\text{−}},$ we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In ${\text{CHO}}_{2}{}^{\text{−}},$ the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl₃, S in SO₂, and Cl in ${\text{ClO}}_{4}{}^{\text{−}}.$ An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
- There are no remaining electrons on SiH₄, so it is unchanged:
Place all remaining electrons on the central atom.
- For SiH₄, ${\text{CHO}}_{2}{}^{\text{−}},$ and NO⁺, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
- For OF₂, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:
Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
- SiH₄: Si already has an octet, so nothing needs to be done.
- ${\text{CHO}}_{2}{}^{\text{−}}:$ We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:
- NO⁺: For this ion, we added eight valence electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:
  
  This still does not produce an octet, so we must move another pair, forming a triple bond:
- In OF₂, each atom has an octet as drawn, so nothing changes.

Writing Lewis Structures

NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H₃CCH₃), acetylene (HCCH), and ammonia (NH₃). What are the Lewis structures of these molecules?

Solution

Calculate the number of valence electrons.
HCN: (1 $×$ 1) + (4 $×$ 1) + (5 $×$ 1) = 10
H₃CCH₃: (1 $×$ 3) + (2 $×$ 4) + (1 $×$ 3) = 14
HCCH: (1 $×$ 1) + (2 $×$ 4) + (1 $×$ 1) = 10
NH₃: (5 $×$ 1) + (3 $×$ 1) = 8
Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
Where needed, distribute electrons to the terminal atoms:

HCN: six electrons placed on N
H₃CCH₃: no electrons remain
HCCH: no terminal atoms capable of accepting electrons
NH₃: no terminal atoms capable of accepting electrons
Where needed, place remaining electrons on the central atom:

HCN: no electrons remain
H₃CCH₃: no electrons remain
HCCH: four electrons placed on carbon
NH₃: two electrons placed on nitrogen
Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:
HCN: form two more C–N bonds
H₃CCH₃: all atoms have the correct number of electrons
HCCH: form a triple bond between the two carbon atoms
NH₃: all atoms have the correct number of electrons

Check Your Learning

Both carbon monoxide, CO, and carbon dioxide, CO₂, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO₂ has been implicated in global climate change. What are the Lewis structures of these two molecules?

Answer:

Fullerene Chemistry

Carbon soot has been known to man since prehistoric times, but it was not until fairly recently that the molecular structure of the main component of soot was discovered. In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley ((Figure)), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C₆₀ buckminsterfullerene molecule ((Figure)). An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C_60. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.

Richard Smalley (1943–2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. Upon his death in 2005, the US Senate honored him as the “Father of Nanotechnology.” (credit: United States Department of Energy)

A photo of Richard Smalley is shown.

Exceptions to the Octet Rule

Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:

Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.

Odd-electron Molecules

We call molecules that contain an odd number of electrons free radicals. Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.

To draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes:

Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from N) + 6 (from O) = 11. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.
Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N–O single bond:
N–O
Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell:
Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.
Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)

Electron-deficient Molecules

We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13, outer atoms that are hydrogen, or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH₂, and boron trifluoride, BF₃, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in BF₃, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B–F single bonds. This suggests the best Lewis structure has three B–F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the B–F bonds are slightly shorter than what is actually expected for B–F single bonds, indicating that some double bond character is found in the actual molecule.

Two Lewis structures are shown. The left shows a beryllium atom single bonded to two hydrogen atoms. The right shows a boron atom single bonded to three fluorine atoms, each with three lone pairs of electrons.

An atom like the boron atom in BF₃, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH₃ reacts with BF₃ because the lone pair on nitrogen can be shared with the boron atom:

Hypervalent Molecules

Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Elements in the third and higher periods (n ≥ 3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules. (Figure) shows the Lewis structures for two hypervalent molecules, PCl₅ and SF_6.

In PCl₅, the central atom phosphorus shares five pairs of electrons. In SF₆, sulfur shares six pairs of electrons.

Two Lewis structures are shown. The left shows a phosphorus atom single bonded to five chlorine atoms, each with three lone pairs of electrons. The right shows a sulfur atom single bonded to six fluorine atoms, each with three lone pairs of electrons.

In some hypervalent molecules, such as IF₅ and XeF₄, some of the electrons in the outer shell of the central atom are lone pairs:

Two Lewis structures are shown. The left shows an iodine atom with one lone pair single bonded to five fluorine atoms, each with three lone pairs of electrons. The right diagram shows a xenon atom with two lone pairs of electrons single bonded to four fluorine atoms, each with three lone pairs of electrons.

When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom.

Writing Lewis Structures:

Octet Rule Violations Xenon is a noble gas, but it forms a number of stable compounds. We examined XeF₄ earlier. What are the Lewis structures of XeF₂ and XeF₆?

Solution

We can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply.

Calculate the number of valence electrons:
XeF₂: 8 + (2 $×$ 7) = 22
XeF₆: 8 + (6 $×$ 7) = 50
Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom:
Distribute the remaining electrons.
XeF₂: We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving each F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell d orbitals and can accommodate more than eight electrons. The Lewis structure of XeF₂ shows two bonding pairs and three lone pairs of electrons around the Xe atom:

XeF₆: We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom:

Check Your Learning

The halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens BrCl₃ and ${\text{ICl}}_{4}{}^{\text{−}}.$

Answer:

Two Lewis structures are shown. The left depicts a bromine atom with two lone pairs of electrons single bonded to three chlorine atoms, each with three lone pairs of electrons. The right shows an iodine atom, with two lone pairs of electrons, single boned to four chlorine atoms, each with three lone pairs of electrons. This structure is surrounded by brackets and has a superscripted negative sign.

Key Concepts and Summary

Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient molecules, and hypervalent molecules.

Chemistry End of Chapter Exercises

Instructions

Complete the following exercises by providing answers to each question. There are a total of 20 questions in this section. Please be aware that certain questions may contain subsections

1. Write the Lewis symbols for each of the following ions:

(a) As^3–

(b) I^–

(c) Be²⁺

(d) O^2–

(e) Ga³⁺

(f) Li⁺

(g) N^3–

Answer(s):

(a) eight electrons:

A Lewis dot diagram shows the symbol for arsenic, A s, surrounded by eight dots and a superscripted three negative sign.

(b) eight electrons:

A Lewis dot diagram shows the symbol for iodine, I, surrounded by eight dots and a superscripted negative sign.

(c) no electrons Be²⁺
(d) eight electrons:

A Lewis dot diagram shows the symbol for oxygen, O, surrounded by eight dots and a superscripted two negative sign.

(e) no electrons Ga³⁺
(f) no electrons Li⁺
(g) eight electrons:

A Lewis dot diagram shows the symbol for nitrogen, N, surrounded by eight dots and a superscripted three negative sign.

2. Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements:

(a) Cl

(b) Na

(c) Mg

(d) Ca

(e) K

(f) Br

(g) Sr

(h) F

3. Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed:

(a) MgS

(b) Al₂O₃

(c) GaCl₃

(d) K₂O

(e) Li₃N

(f) KF

Answer(s):

(a)

Two Lewis structures are shown. The left shows the symbol M g with a superscripted two positive sign while the right shows the symbol S surrounded by eight dots and a superscripted two negative sign.

(b)

Two Lewis structures are shown. The left shows the symbol A l with a superscripted three positive sign while the right shows the symbol O surrounded by eight dots and a superscripted two negative sign.

(c)

Two Lewis structures are shown. The left shows the symbol G a with a superscripted three positive sign while the right shows the symbol C l surrounded by eight dots and a superscripted negative sign.

(d)

Two Lewis structures are shown. The left shows the symbol K with a superscripted positive sign while the right shows the symbol O surrounded by eight dots and a superscripted two negative sign.

(e)

Two Lewis structures are shown. The left shows the symbol L i with a superscripted positive sign while the right shows the symbol N surrounded by eight dots and a superscripted three negative sign.

(f)

Two Lewis structures are shown. The left shows the symbol K with a superscripted positive sign while the right shows the symbol F surrounded by eight dots and a superscripted negative sign.

4. In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:

(a)

Two Lewis structures are shown side-by-side, each surrounded by brackets. The left structure shows the symbol M with a superscripted two positive sign. The right shows the symbol X surrounded by four lone pairs of electrons with a superscripted two negative sign outside of the brackets.

(b)

Two Lewis structures are shown side-by-side, each surrounded by brackets. The left structure shows the symbol M with a superscripted three positive sign. The right structure shows the symbol X surrounded by four lone pairs of electrons with a superscripted negative sign and a subscripted three both outside of the brackets.

(c)

Two Lewis structures are shown side-by-side, each surrounded by brackets. The left structure shows the symbol M with a superscripted positive sign and a subscripted two outside of the brackets. The right structure shows the symbol X surrounded by four lone pairs of electrons with a superscripted two negative sign outside of the brackets.

(d)

Two Lewis structures are shown side-by-side, each surrounded by brackets. The left structure shows the symbol M with a superscripted three positive sign and a subscripted two outside of the brackets. The right structure shows the symbol X surrounded by four lone pairs of electrons with a superscripted two negative sign and subscripted three both outside of the brackets.

5. Write the Lewis structure for the diatomic molecule P₂, an unstable form of phosphorus found in high-temperature phosphorus vapor.

A Lewis diagram shows two phosphorus atoms triple bonded together each with one lone electron pair.

6. Write Lewis structures for the following:

(a) H₂

(b) HBr

(c) PCl₃

(d) SF₂

(e) H₂CCH₂

(f) HNNH

(g) H₂CNH

(h) NO^–

(i) N₂

(j) CO

(k) CN^–

7. Write Lewis structures for the following:

(a) O₂

(b) H₂CO

(c) AsF₃

(d) ClNO

(e) SiCl₄

(f) H₃O⁺

(g) ${\text{NH}}_{4}{}^{+}$

(h) ${\text{BF}}_{4}{}^{\text{−}}$

(i) HCCH

(j) ClCN

(k) ${\text{C}}_{2}{}^{\text{2+}}$

Answer(s):

(a)

A Lewis structure shows two oxygen atoms double bonded together, and each has two lone pairs of electrons.

In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule.
(b)

A Lewis structure shows a carbon atom that is single bonded to two hydrogen atoms and double bonded to an oxygen atom. The oxygen atom has two lone pairs of electrons.

(c)

A Lewis structure shows an arsenic atom single bonded to three fluorine atoms. Each fluorine atom has a lone pair of electrons.

(d)

A Lewis structure shows a nitrogen atom with a lone pair of electrons single bonded to a chlorine atom that has three lone pairs of electrons. The nitrogen is also double bonded to an oxygen which has two lone pairs of electrons.

(e)

A Lewis structure shows a silicon atom that is single bonded to four chlorine atoms. Each chlorine atom has three lone pairs of electrons.

(f)

A Lewis structure shows an oxygen atom with a lone pair of electrons single bonded to three hydrogen atoms. The structure is surrounded by brackets with a superscripted positive sign.

(g)

A Lewis structure shows a nitrogen atom single bonded to four hydrogen atoms. The structure is surrounded by brackets with a superscripted positive sign.

(h)

A Lewis structure shows a boron atom single bonded to four fluorine atoms. Each fluorine atom has three lone pairs of electrons. The structure is surrounded by brackets with a superscripted negative sign.

(i)

A Lewis structure shows two carbon atoms that are triple bonded together. Each carbon is also single bonded to a hydrogen atom.

(j)

A Lewis structure shows a carbon atom that is triple bonded to a nitrogen atom that has one lone pair of electrons. The carbon is also single bonded to a chlorine atom that has three lone pairs of electrons.

(k)

A Lewis structure shows two carbon atoms joined with a triple bond. A superscripted 2 positive sign lies to the right of the second carbon.

8. Write Lewis structures for the following:

(a) ClF₃

(b) PCl₅

(c) BF₃

(d) ${\text{PF}}_{6}{}^{\text{−}}$

9. Write Lewis structures for the following:

(a) SeF₆

(b) XeF₄

(c) ${\text{SeCl}}_{3}{}^{+}$

(d) Cl₂BBCl₂ (contains a B–B bond)

Answer(s):

(a) SeF₆:

A Lewis structure shows a selenium atom single bonded to six fluorine atoms, each with three lone pairs of electrons.

(b) XeF₄:

A Lewis structure shows a xenon atom with two lone pairs of electrons. It is single bonded to four fluorine atoms each with three lone pairs of electrons.

(c) ${\text{SeCl}}_{3}{}^{\text{+}}\text{:}$

A Lewis structure shows a selenium atom with one lone pair of electrons single bonded to three chlorine atoms each with three lone pairs of electrons. The whole structure is surrounded by brackets.

(d) Cl₂BBCl₂:

A Lewis structure shows two boron atoms that are single bonded together. Each is also single bonded to two chlorine atoms that both have three lone pairs of electrons.

10. Write Lewis structures for:

(a) ${\text{PO}}_{4}{}^{\text{3−}}$

(b) ${\text{ICl}}_{4}{}^{\text{−}}$

(c) ${\text{SO}}_{3}{}^{\text{2−}}$

(d) HONO

11. Correct the following statement: “The bonds in solid PbCl₂ are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl₂ are located on the Cl^– ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.”

Answer(s): Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb²⁺ ion has a 6s² valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.

12. Write Lewis structures for the following molecules or ions:

(a) SbH₃

(b) XeF₂

(c) Se₈ (a cyclic molecule with a ring of eight Se atoms)

13. Methanol, H₃COH, is used as the fuel in some race cars. Ethanol, C₂H₅OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO₂ and H₂O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.

Two reactions are shown using Lewis structures. The top reaction shows a carbon atom, single bonded to three hydrogen atoms and single bonded to an oxygen atom with two lone pairs of electrons. The oxygen atom is also bonded to a hydrogen atom. This is followed by a plus sign and the number one point five, followed by two oxygen atoms bonded together with a double bond and each with two lone pairs of electrons. A right-facing arrow leads to a carbon atom that is double bonded to two oxygen atoms, each of which has two lone pairs of electrons. This structure is followed by a plus sign, a number two, and a structure made up of an oxygen with two lone pairs of electrons single bonded to two hydrogen atoms. The bottom reaction shows a carbon atom, single bonded to three hydrogen atoms and single bonded to another carbon atom. The second carbon atom is single bonded to two hydrogen atoms and one oxygen atom with two lone pairs of electrons. The oxygen atom is also bonded to a hydrogen atom. This is followed by a plus sign and the number three, followed by two oxygen atoms bonded together with a double bond. Each oxygen atom has two lone pairs of electrons. A right-facing arrow leads to a number two and a carbon atom that is double bonded to two oxygen atoms, each of which has two lone pairs of electrons. This structure is followed by a plus sign, a number three, and a structure made up of an oxygen with two lone pairs of electrons single bonded to two hydrogen atoms.

14. Many planets in our solar system contain organic chemicals including methane (CH₄) and traces of ethylene (C₂H₄), ethane (C₂H₆), propyne (H₃CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules.

15. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl₂CO. Write the Lewis structures for carbon tetrachloride and phosgene.

Two Lewis structures are shown. The left depicts a carbon atom single bonded to four chlorine atoms, each with three lone pairs of electrons. The right shows a carbon atom double bonded to an oxygen atom that has two lone pairs of electrons. The carbon atom is also single bonded to two chlorine atoms, each of which has three lone pairs of electrons.

16. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom:

(a) 1s²2s²2p⁵

(b) 1s²2s²2p⁶3s²

(c) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰

(d) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴

(e) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p¹

17. The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.

(a) the amino acid serine:

(b) urea:

A Lewis structure is shown. A nitrogen atom is single bonded to two hydrogen atoms and a carbon atom. The carbon atom is single bonded to an oxygen atom and another nitrogen atom. That nitrogen atom is then single bonded to two hydrogen atoms.

(c) pyruvic acid:

A Lewis structure is shown. A carbon atom is single bonded to three hydrogen atoms and another carbon atom. The second carbon atom is single bonded to an oxygen atom and a third carbon atom. This carbon is then single bonded to two oxygen atoms, one of which is single bonded to a hydrogen atom.

(d) uracil:

(e) carbonic acid:

A Lewis structure is shown. A carbon atom is single bonded to three oxygen atoms. Two of those oxygen atoms are each single bonded to a hydrogen atom.

(a)

(b)

A Lewis structure is shown. A nitrogen atom is single bonded to two hydrogen atoms and a carbon atom. The carbon atom is single bonded to an oxygen atom and one nitrogen atom. That nitrogen atom is then single bonded to two hydrogen atoms. The oxygen atom has two lone pairs of electron dots, and the nitrogen atoms have one lone pair of electron dots each.

(c)

A Lewis structure is shown. A carbon atom is single bonded to three hydrogen atoms and a carbon atom. The carbon atom is single bonded to an oxygen atom and a third carbon atom. This carbon is then single bonded to two oxygen atoms, one of which is single bonded to a hydrogen atom. Each oxygen atom has two lone pairs of electron dots.

(d)

(e)

A Lewis structure is shown. A carbon atom is single bonded to three oxygen atoms. Two of those oxygen atoms are each single bonded to a hydrogen atom. Each oxygen atom has two lone pairs of electron dots.

18. A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.

19. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.

A Lewis structure is shown. A carbon atom is single bonded to three hydrogen atoms and another carbon atom. The second carbon atom is double bonded to another carbon atom and single bonded to a hydrogen atom. The last carbon is single bonded to two hydrogen atoms.

20. Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules.

21. How are single, double, and triple bonds similar? How do they differ?

Answer(s): Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.

Glossary

double bond: covalent bond in which two pairs of electrons are shared between two atoms

free radical: molecule that contains an odd number of electrons

hypervalent molecule: molecule containing at least one main group element that has more than eight electrons in its valence shell

Lewis structure: diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion

Lewis symbol: symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion

lone pair: two (a pair of) valence electrons that are not used to form a covalent bond

octet rule: guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond

single bond: bond in which a single pair of electrons is shared between two atoms

triple bond: bond in which three pairs of electrons are shared between two atoms

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Lewis Symbols

Lewis Structures

The Octet Rule

Double and Triple Bonds

Writing Lewis Structures with the Octet Rule

Writing Lewis Structures

Check Your Learning

Exceptions to the Octet Rule

Odd-electron Molecules

Electron-deficient Molecules

Hypervalent Molecules

Writing Lewis Structures:

Check Your Learning

Instructions

Glossary

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