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Chemical Bonding I:Lewis Theory

Review Questions

G9.1 Bonding theories are central to chemistry because they explain how atoms bond together to

form molecules. Bonding theories explain why some combinations of atoms are stable andothers are not.

Chemical bonds form because they lower the potential energy between the charged particlesthat compose the atom. Bonds involve the attraction and repulsion of charged particles.

9.3 The three types of bonds are ionic bonds, which occur between metals and nonmetals and arecharacterized by the transfer of electrons; covalent bonds, which occur between nonmetalsand are characterized by the sharing of electrons; and metallic bonds, which occur betweenmetals and are characterized by electrons being pooled.

In a Lewis structure, the valence electrons of main-group elements are represented as dotssurrounding the symbol for the element. The valence electrons can be determined from thegroup that they are in on the periodic table.

Bonds are formed when atoms attain a stable electron configuration. Since the stable config-uration usually has eight electrons in the outermost shell; this is known as the octet rule.

In Lewis theory, a chemical bond is the sharing or transfer of electrons to attain stable elec-tron configurations for the bonding atoms. If electrons are transferred, the bond is an ionicbond. If the electrons are shared, the bond is a covalent bond.

9.7 In Lewis theory, we represent ionic bonding by moving electron dots from the metal to the non-metal and then allowing the resultant ions to form a crystalline lattice composed of alternatingcations and anions. The cation loses its valence electron(s) and is left with an octet in the previ-ous principal energy level; the anion gains electron(s) to form an octet. The Lewis structure of theanion is usually written within brackets with the charge in the upper right-hand corner, outsidethe brackets. The positive and negative charges attract one another, resulting in the compound.

9.8 To use Lewis theory to determine the formula of an ionic compound, determine the numberof valence electrons lost by the metal to form an octet, and the number of electrons gained bythe nonmetal to form an octet. Consider the ionic compound formed between sodium andsulfur. The Lewis structures for sodium and sulfur are as follows:

Na. '?:

Sodium must lose one electron to achieve an octet, while sulfur must gain two electrons toachieve an octet. So the compound requires two sodium atoms for each sulfur atom givingthe formula

9.9 Lattice energy is the energy associated with forming a crystalline lattice of alternating cationsand anions from the gaseous ions. Since the cations are positively charged and the anions arenegatively charged there is a lowering of potential — as described by Coulomb's law — whenthe ions come together to form a lattice. That energy is emitted as heat when the lattice forms.

289

292 Chapter 9 Chemical Bonding I: Lewis Theory

would have if all bonding electrons were shared equally between the bonded atoms. Formal charge canbe calculated simply by taking the number of valence electrons in the atom and subtracting the number ofelectrons that it "owns" in a Lewis structure. An atom in a Lewis structure "owns" all of its lone pair elec-trons and 1/2 of its bonding electrons.

Formal charge = number of valence electrons - (number of lone pair electrons +1/2 number of bond-ing electrons).

The concept of formal charge is useful because it can help us distinguish between competing skeletal struc-tures or competing resonance structures.

9.29 The octet rule has some exceptions because not all atoms always have eight electrons surrounding them.The three major categories are 1) odd octets—electron species, molecules, or ions with an odd number ofelectron, for example, NO; 2) incomplete octets—molecules or ions with fewer than eight electrons aroundan atom, for example, BF3; and 3) expanded octets—molecules or ions with more than eight electronsaround an atom, for example,

9.30 Elements in the third row of the periodic table and beyond often exhibit expanded octets. Elements in thefirst or second row of the periodic table can never have expanded octets.

9.31 The bond energy of a chemical bond is the energy required to break 1 mole of the bond in the gas phase.Since breaking bonds is endothermic and forming bonds is exothermic we can calculate the overall enthalpychange as a sum of the enthalpy changes associated with breaking the required bonds in the reactants andforming the required bonds in the products.

9.32 A reaction is exothermic when weak bonds break and strong bonds form. A reaction is endothermic whenstrong bonds break and weak bonds form.

9.33 When metal atoms bond together to form a solid, each metal atom donates one or more electrons to anelectron sea.

9.34 Metals conduct electricity because the electrons in a metal are free to move. The movement or flow of elec-trons in response to an electric potential is an electric current. Metals are also excellent conductors of heatbecause of the highly-mobile electrons, which help to disperse thermal energy throughout the metal.

The malleability of metals and the ductility of metals are also accounted for by the electron sea model. Sincethere are no localized or specific "bonds" in a metal, it can be deformed relatively easily by forcing the metalions to slide past one another. The electron sea can easily accommodate these deformations by flowing intothe new shape.

Problems by Topic

Valence Electrons and Dot Structures•

• N *9.35 N : Is22s22p3 • The electrons included in the Lewis structure are 2s22p3.

* Ne *Ne : Is22s22p6 * •• * The electrons included in the Lewis structure are 2s22p6.

(a) Al: Is22s22p63s23pl

•• Al •

(b) Na+: Is22s22p6

(c) Cl: Is22s22p63s23;>5

Na+

Chapter 9 Chemical Bonding I: Lewis Theory 293

(d) Cl~: Is22s22p63s23p6

9.38 (a) : Is22s22p63s23p6

2-

(b) Mg: Is22s22p63s2

(c) Mg2+: Is22s22p6

2+Mg

(d) P: Is22s22p63s23p3

*•• p .•

Ionic Lewis Structures and Lattice Energy

9.39 (a) NaF: Draw the Lewis structures for Na and F based on their valence electrons. Na: 3s1 F: 2s22p5

• •Na» J F •

••

Sodium must lose one electron and be left with the octet from the previous shell, while fluorine needsto gain one electron to get an octet.

Na • c •• r •-

(b) CaO: Draw the Lewis structures for Ca and O based on their valence electrons. Ca: 4s2 O: 2s22p4

• •ca: : o .•

Calcium must lose two electrons and be left with two Is electrons from the previous shell, while oxy-gen needs to gain two electrons to get an octet.

2 +Ca

2 -

(c) SrBr2: Draw the Lewis structures for Sr and Br based on their valence electrons. Sr: 5s2 Br: 4s24p5

••Sr I 5 Br •

••Strontium must lose two electrons and be left with the octet from the previous shell, while bromineneeds to gain one electron to get an octet.

2 +Sr 2 ! Brl.


Simple Covalent Lewis Structures, Electronegativity, and Bond Polarity

9.49 (a) Hydrogen: Write the Lewis structure of each atom based on the number of valence electrons.

H- »H

When the two hydrogen atoms share their electrons, they each get a duet, which is a stable configu-ration for hydrogen.

H H

(b) The halogens: Write the Lewis structure of each atom based on the number of valence electrons.

• Y • • Y •, A A (

If the two halogens pair together they can each achieve an octet, which is a stable configuration. So,the halogens are predicted to exist as diatomic molecules.

:" — ":

(c) Oxygen: Write the Lewis structure of each atom based on the number of valence electrons.

: o • • o :In order to achieve a stable octet on each oxygen, the oxygen atoms will need to share two electronpairs. So, oxygen is predicted to exist as a diatomic molecule with a double bond.

•• ••

(d) Nitrogen: Write the Lewis structure of each atom based on the number of valence electrons.

N • N• •

In order to achieve a stable octet on each nitrogen, the nitrogen atoms will need to share three elec-tron pairs. So, nitrogen is predicted to exist as a diatomic molecule with a triple bond.

N; N

9.50 Write the Lewis structure for N and H based on the number of valence electrons.

H

• •N H

H N H • N H H N H

If nitrogen combines with three hydrogen atoms, the nitrogen will achieve a stable octet and each hydrogenwill have a duet of electrons. This is a stable configuration. If the nitrogen were to combine with only twohydrogen atoms, the nitrogen could only achieve a seven electron configuration, which is not stable. Also, ifthe nitrogen were to combine with four hydrogen atoms, the nitrogen would have a nine electron configura-tion, which is not stable. So, Lewis theory predicts that nitrogen will combine with three hydrogen atoms.

PH3: Write the Lewis structure for each atom based on the number of valence electrons.

••• P • . H


Phosphorus will share an electron pair with each hydrogen in order to achieve a stable octet.

H -

(b) SC12: Write the Lewis structure for each atom based on the number of valence electrons.

•• ••: s . ; ci •• ••The sulfur will share an electron pair with each chlorine in order to achieve a stable octet.

.. ".: s—ci.A:"• CI •• •

(c) HI: Write the Lewis structure for each atom based on the number of valence electrons.• •H. . ,:••

The iodine will share an electron pair with hydrogen in order to achieve a stable octet.• •

H—i :••

(d) CtLi: Write the Lewis structure for each atom based on the number of valence electrons.

The carbon will share an electron pair with each hydrogen in order to achieve a stable octet.

H

H-i-H

9.52 (a) NFs: Write the Lewis structure for each atom based on the number of valence electrons.

The nitrogen will share an electron pair with each fluorine in order to achieve a stable octet.

: F — n — f l• • I ••

• J: •• F •


The carbon will share three electron pairs with oxygen in order to achieve a stable octet.The oxygen atom is more electronegative than the carbon atom, so the oxygen will have a partial negativecharge and the carbon will have a partial positive charge.

• r- r> •• c u •

To estimate the percent ionic character, determine the difference in electronegativity between carbonand oxygen.From Figure 9.8 we find the electronegativity of C is 2.5 and O is 3.5. The electronegativity difference (AEN)is AEN = 3.5 - 2.5 = 1.0.From Figure 9.10, we can estimate a percent ionic character of 25%.

9.58 BrF: Write the Lewis structure for each atom based on the number of valence electrons.

The bromine and fluorine will share an electron pair to achieve a stable octet.The fluorine atom is more electronegative than the bromine atom, so the fluorine will have a partial nega-tive charge and the bromine will have a partial positive charge.

F :

To estimate the percent ionic character, determine the difference in electronegativity between bromineand fluorine.From Figure 9.8 we find the electronegativity of Br = 2.8 and F = 4.0. The electronegativity difference (AEN)is AEN = 4.0 - 2.8 = 1.2.From Figure 9.10, we can estimate a percent ionic character of 30%.

valent Lewis Structures, Resonance, and Formal Charge

9.59; (a) €14: Write the correct skeletal structure for the molecule.

Calculate the total number of electrons for the Lewis structure by summing the number ofvalence electrons of each atom in the molecule.

(number of valence e " for C) + 4(number of valence e ~ for I) = 4 + 4(7) = 32Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom.

All 32 valence electrons are used.

If any atom lacks an octet, form double or triple bonds as necessary to give them octets.All atoms have octets; the structure is complete.


(b) N2O: Write the correct skeletal structure for the molecule.N is the less electronegative, so it is central.

N - N - O


2(number of valence e " for N) + (number of valence e " for O) = 2(5) + 6 = 16Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom.

All 16 valence electrons are used.If any atom lacks an octet, form double or triple bonds as necessary.

I N ^ N — oj

All atoms have octets; the structure is complete.(c) SiH :̂ Write the correct skeletal structure for the molecule.

H is always terminal, so Si is the central atom.

H

H - Si - H

H


(number of valence e ~ for Si) + 4(number of valence e ~ for H) = 4 + 4(1) = 8Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms andfinally to lone pairs on the central atom.

H

H - Si - H

All 8 valence electrons are used.If any atom lacks an octet, form double or triple bonds as necessary to give them octets.All atoms have octets; the structure is complete.

(d) C12CO: Write the correct skeletal structure for the molecule.C is the least electronegative, so it is the central atom.

O

•ciCalculate the total number of electrons for the Lewis structure by summing the number ofvalence electrons of each atom in the molecule.

(number of valence e " for C) + 2(number of valpnre e ~ for CI) + (number of valencee~ for O) = 4 + 2(7) + 6 = 24


If any atom lack an octet, form double or triple bonds as necessary to give them octets.Lastly, write the Lewis structure in brackets with the charge of the ion in the upper right-hand corner.

o :2-

(a) N2H2: Write the correct skeletal structure for the molecule.

H - N - N - H


2(number of valence e ~ for N) + 2(number of valence e " for H) = 2(5) + 2(1) = 12Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms andfinally to lone pairs on the central atom.

• • ••H - N - N - H

••


• • ••

H - N=N - H

All atoms have octets (duets for H); the structure is complete.

(b) N2HLi: Write the correct skeletal structure for the molecule.

N— N

Calculate the total number of electrons for the Lewis structure by summing the valence elec-trons of each atom in the molecule.

2(number of valence e " for N) + 4(number of valence e ~ for H) = 2(5) + 4(1) = 14Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms andfinally to lone pairs on the central atom.

*^N— N

All 14 valence electrons are used.If any atom lacks an octet, form double or triple bonds as necessary to give them octets.All atoms have octets (duets for H) structure is complete.

(c) C2H2: Write the correct skeletal structure for the molecule.

H - Q - C - H


2(number of valence e "for C) + 2(number of valence e~ for H) = 2(4) + 2(1) = 10


Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms andfinally to lone pairs on the central atom.

H - C - C* - H«•


H - C==C - H

All atoms have octets (duets for H); the structure is complete.

(d) C2H,j: Write the correct skeletal structure for the molecule.

H H

\Calculate the total number of electrons for the Lewis structure by summing the number ofvalence electrons of each atom in the molecule.

2(number of valence e " for C) + 4(number of valence e ~ for H) = 2(4) + 4(1) = 12Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms andfinally to lone pairs on the central atom.

H H


H H

/=C\H H

All atoms have octets (duets for H) structure is complete.

9.62 (a) H3COCH3: Write the correct skeletal structure for the molecule.

H H

- H

iCalculate the total number of electrons for the Lewis structure by summing the number ofvalence electrons of each atom in the molecule.

2(number of valence e " for C) + (number of valence e " for O) + 6(number of valencee'forH) = 2(4) + 6 + 6(1) = 20


9.66

H H

I •• |C=S H - S

I II

Calculate the formal charge on each atom in structure I by finding the number of valence electronsand subtracting the number of lone pair electrons and one-half the number of bonding electrons.

Hjeft Htop C S

number o f valence electrons 1 1 4 6- number o f lone pair electrons 0 0 0 4- l/2(number o f bonding electrons) 1 1 4 2

Formal charge 0 0 0 0The sum of the formal charges is 0, which is the overall charge of the molecule.Calculate the formal charge on each atom in structure II by finding the number of valence electronsand subtracting the number of lone pair electrons and one-half the number of bonding electrons.

Hleft Htop S C

number o f valence electrons 1 1 6 4- number o f lone pair electrons 0 0 0 4- l/2(number o f bonding electrons) 1 1 4 2

Formal charge 0 0 + 2 - 2The sum of the formal charges is 0, which is the overall charge of the molecule.Structure I is the better Lewis structure because it has the least amount of formal charge on each atom.

»"

-S C H H C S-

IICalculate the formal charge on each atom in structure I by finding the number of valence electronsand subtracting the number of lone pair electrons and one-half the number of bonding electrons.

T T T T T T T T C f

**left ^-nop Bright ^ibottom ^ ^~number o f valence electrons 1 1 1 1 6 4- number o f lone pair electrons 0 0 0 0 0 4- l/2(number o f bonding electrons) 1 1 1 1 4 2

Formal charge 0 0 0 0 + 2 - 2The sum of the formal charges is 0, which is the overall charge of the molecule.Calculate the formal charge on each atom in structure II by finding the number of valence elec-trons and subtracting the number of lone pair electrons and one-half the number of bondingelectrons.

number o f valence electrons 1 1 1 1 4 6- number o f lone pair electrons 0 0 0 0 0 4- l/2(number o f bonding electrons) 1 1 1 1 4 2

Formal charge 0 0 0 0 0 0The sum of the formal charges is 0, which is the overall charge of the molecule.Structure II is the better Lewis structure because it has the least amount of formal charge oneach atom.


(b) NO: Write the correct skeletal structure for the molecule.

N 0


(number of valence e ~ for N) + (number of valence e ~ for O) = 5 +6 = 11Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms, and finally to lonepairs the central atom.

(c) C102:

(a)

All 11 valence electrons are used.N has an incomplete octet. It has 7 electrons because we have an odd number of valenceelectrons.

Write the correct skeletal structure for the molecule.Cl is less electronegative so it is central.

O Cl 0


(number of valence e " for Cl) + 2(number of valence e ~ for O) = 7 +2(6) = 19Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs of the central atom.

-o :••••o= = CI=

orAll 19 valence electrons are used.

Cl will have either an incomplete octet or an expanded octet. Because Cl brings 7 elec-trons, there is an odd number of electrons in the structure.

Write the correct skeletal structure for the ion.

O

O -O

Calculate the total number of electrons for the Lewis structure by summing the number ofvalence electrons of each atom in the ion and adding 3 for the 3 - charge.

4(number of valence e " for O) + (number of valence e ~ for P) + 3 = 4(6) + 5 + 3 = 32Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom.

:o •

: o—P—o ••*

o •• •


All 32 valence electrons are used.Lastly, write the Lewis structure in brackets with the charge of the ion in the upper right-hand corner.

-i 3-

O

All atoms have octets (duets for H); the structure is complete.Calculate the formal charge on each atom by finding the number of valence electrons andsubtracting the number of lone pair electrons and one-half the number of bonding electrons.

! <

:o —

• c•

Oleft Of0p6 66 61 1

•••

mm

) :•

3-

Bright ^bottom6 66 61 1

p

504

number of valence electrons- number of lone pair electrons- l/2(number of bonding electrons)

Formal charge -1 -1 -1 -1 +1The sum of the formal charges is -3, which is the overall charge of the ion. However, we canwrite a resonance structure with a double bond to an oxygen because P can expand its octet.This leads to lower formal charges on P and O.

•

*•O l

• (•

•

i n •

) !•

—: o—p-

• •

-o

o * 0 '

-|3~

? O-••

•o'

Using the leftmost structure, calculate the formal charge on each atom by finding the num-ber of valence electrons and subtracting the number of lone pair electrons and one-half thenumber of bonding electrons.


Formal charge 0 -1 -1 -1 0The sum of the formal charges is -3, which is the overall charge of the ion. These resonanceforms would all have the lower formal charges associated with the double bonded O and P.

Oleft642

Otop

661

Oright661

Obottom I6 56 01 5

(b) CN Write the correct akeletal structure for the ion.



(number of valence e " for C) + (number of valence e ~ for N) + 1 = 4 + 5 + 1 = 10Distribute the electrons among the atoms, giving octets \ to as many atoms as possible.Begin with the bonding electrons, then proceed to lone pairs on terminal atoms, and finallyto lone pairs on the central atom.

• •:c — N :


Lastly, write the Lewis structure in brackets with the charge of the ion in the upper right-hand corner.

All atoms have octets; the structure is complete.Calculate the formal charge on each atom by finding the number of valence electrons andsubtracting the number of lone pair electrons and one-half the number of bonding electrons.

C Nnumber of valence electrons 4 5- number of lone pair electrons 2 2- l/2(number of bonding electrons) 3 3

Formal charge -1 0The sum of the formal charges is - 1, which is the overall charge of the ion.

(c) SC»32 ~: Write the correct skeletal structure for the ion.

O

o — s^— oCalculate the total number of electrons for the Lewis structure by summing the valence elec-trons of each atom in the ion and adding 2 for the 2 - charge.

3(number of valence e ~ for O) + (number of valence e ~ for S) + 2 = 3(6) + 6 + 2 = 26Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom.

: o — s — o :•• •• ••All 26 valence electrons are used.



., 2-

o o ;

Calculate the formal charge on each atom by finding the number of valence electrons andsubtracting the number of lone pair electrons and one-half the number of bonding electrons.

2-

Otop Oright661

661

661

623


Formal charge -1 -1 -1 +1The sum of the formal charges is -2, which is the overall charge of the ion. However, we canwrite a resonance structure with a double bond to an oxygen because S can expand its octet.This leads to a lower formal charge.

-o

2-

• O

-o

2-••

o

o

2-

Using the leftmost resonance form, calculate the formal charge on each atom by finding thenumber of valence electrons and subtracting the number of lone pair electrons and one-halfthe number of bonding electrons.


Formal charge 0 - 1 - 1 0The sum of the formal charges is -2, which is the overall charge of the ion. These resonanceforms would all have the lower formal charge on the double bonded O and S.

(d) C1O2~ : Write the correct skeletal structure for the ion.

Oleft642

0,0p661

Origh6

61

t S624

O Cl O


2(number of valence e ~ for O) + (number of valence e " for Cl) + 1 = 2(6) + 7 + 1 = 20


Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms andfinally to lone pairs on the central atom.

•• ft

• o — ct—o :


.. •• ..I: o — ci —o :


I:"- -a--o:

Oleft Onght Clnumber o f valence electrons 6 6 7- number o f lone pair electrons 6 6 4- l/2(number o f bonding electrons) 1 1 2

Formal charge -1 -1 +1The sum of the formal charges is -1, which is the overall charge of the ion. However,we can write a resonance structure with a double bond to an oxygen because Cl canexpand its octet. This leads to a lower formal charge.

o :

Using the leftmost resonance form, calculate the formal charge on each atom by finding thenumber of valence electrons and subtracting the number of lone pair electrons and one-halfthe number of bonding electrons.

Oleft Oright Clnumber o f valence electrons 6 6 7- number o f lone pair electrons 4 6 4- 1 /2(number o f bonding electrons) 2 1 3

Formal charge 0 - 1 0The sum of the formal charges is -1, which is the overall charge of the ion. Theseresonance forms would all have the lower formal charge on the double bondedO and Cl.

9.72 (a) SO42 ": Write the correct skeletal structure for the ion.

O

-O

O


2(number of valence e ~ for O) + (number of valence e " for Br) + 1 = 2(6) + 7+1=20Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom.

.. •• ••tO Br O :.. •• ••



o :

Oright Brnumber o f valence electrons 6 6 7- number o f lone pair electrons 6 6 4- l/2(number o f bonding electrons) 1 1 2

Formal charge -I- -1 +1The sum of the formal charges is - 1, which is the overall charge of the ion.However, we can write a resonance structure with a double bond to an oxygenbecause Br can expand its octet. This leads to a lower formal charge.

0:^Br - O :•• •• ••

Using the leftmost resonance form, calculate the formal charge on each atom byfinding the number of valence electrons and subtracting the number of lone pairelectrons and one-half the number of bonding electrons.

Oleft Oright Br

number of valence electron 6 6- number o f lone pair electrons 4 6 4- l/2(number o f bonding electrons) 2 1 3

Formal charge 0 - 1 0The sum of the formal charges is -1, which is the overall charge of the ion. These reso-nance forms would both have the lower formal charges on the double bonded O and Br.

(a) PF5: Write the correct skeletal structure for the molecule.

F

F


(number of valence e " for P) + 5(number of valence e ~ for F) = 5 +5(7) = 40


Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom. Arrange additional electrons around the central atom, giving it anexpanded octet of up to 12 electrons.

• F • .»

. •• I ./ .. *: F—P •• •F ; "'••

(b) I3 : Write the correct skeletal structure for the ion.

I I I


3(number of valence e ~ for I) + 1 = 3(7) + 1 = 22Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom. Arrange additional electrons around the central atom, giving it anexpanded octet of up to 12 electrons.

• • •• ••I I I I •


(c) SF^. Write the correct skeletal structure for the molecule.

F


(number of valence e " for S) + 4(number of valence e ~ for F) = 6 + 4(7) = 34Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms aspossible. Begin with the bonding electrons, and then proceed to lone pairs on terminal atomsand finally to lone pairs on the central atom. Arrange additional electrons around the cen-tral atom, giving it an expanded octet of up to 12 electrons.

I F — s — F I• • •

• F •• r •


(d) Gep4: Write the correct skeletal structure for the molecule.

Ge-


(number of valence e ~ for Ge) + 4(number of valence e ~ for F) = 4 + 4(7) = 32Distribute the electrons among the atoms, giving octets to as many atoms as possible. Begin withthe bonding electrons, then proceed to lone pairs on terminal atoms and finally to lone pairs onthe central atom.

F •

9.74 (a) C1F5: Write the correct skeletal structure for the molecule.

-cr

Calculate the total number of electrons for the Lewis structure by summing the valence elec-trons of each atom in the molecule.

(number of valence e " for Cl) + 5(number of valence e ~~ for F) = 7 +5(7) = 42Distribute the electrons among the atoms, giving octets to as many atoms as possible. Beginwith the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom. Arrange additional electrons around the central atom, giving it anexpanded octet of up to 12 electrons.

I F -

F• •

(b) AsF6 : Write the correct skeletal structure for the ion.


(d) XeC>4: This is a covalent compound between two nonmetals.Write the correct skeletal structure for the molecule.

0

Xe—O

Calculate the total number of electrons for the Lewis structure by summing the valence electrons ofeach atom in the molecule.

(number of valence e ~ for Xe) + 4(number of valence e ~ for O) = 8 + 4(6) = 32Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms as possi-ble. Begin with the bonding electrons, then proceed to lone pairs on terminal atoms and finally to lonepairs on the central atom.

• •• •• O •

-O• •

O

The structure as shown is an appropriate Lewis structure. However, this structure would leave a for-mal charge on Xe. It can be drawn with all double bonds which would eliminate all the formal charge.

(a) BaCO3: Ba2+

-O• •

Determine the cation and anion.Ba2+ C03

2-Write the Lewis structure for the barium cation based on the valence electrons.

Ba

Ba

5s2 Ba,2+ 5s10

Ba2+

Ba must lose two electrons and be left with the octet from the previous shell.Write the Lewis structure for the covalent anion.Write the correct skeletal structure for the ion.

O

Calculate the total number of electrons for the Lewis structure by summing the number of valenceelectrons of each atom in the ion and adding two for the 2 - charge.

(number of valence e ~ for C) + 3(number of valence e ~ for O) = 4 + 3(6) + 2 = 24


Distribute the electrons among the atoms, giving octets to as many atoms as possible. Begin with thebonding electrons, then proceed to lone pairs on terminal atoms and finally to lone pairs on the cen-tral atom.

• ••

-O •

If any atom lacks an octet, form double or triple bonds as necessary.

• •• O •


• •

o

o

The double bond can be between the C and any of the oxygen atoms, so there are resonance structures.

2-• •

• o •••o=c — o«

2-

< >

• •• o •

• •

•• ••

2

<^->

1 o I

(b) Ca(OH)2: Ca2+

Determine the cation and anion.Ca2+ OH-

Write the Lewis structure for the calcium canon based on the valence electrons.Ca4s2 Ca2+4s°

• O_l_

Ca. ca24

Ca must lose two electrons and be left with the octet from the previous shell.Write the Lewis structure for the covalent anion.Write the correct skeletal structure for the ion.

O H

Calculate the total number of electrons for the Lewis structure by summing the valence electrons ofeach atom in the ion and adding one for the 1 - charge.

(number of valence e " for H) + (number of valence e~ for O)+1 = 1 + 6 + 1 = 8


Distribute the electrons among the atoms, giving octets (or duets for H) to as many atoms as possi-ble. Begin with the bonding electrons, and then proceed to lone pairs on terminal atoms and finallyto lone pairs of the central atom.

: o-


H

(c) KNO3: K+

O

••O

Determine the cation and anion.K+ NO3~

Write the Lewis structure for the potassium cation based on the valence electrons.K 4s1 K+ 4s°

K * K+

K must lose one electron and be left with the octet from the previous shell.Write the Lewis structure for the covalent anion.Write the correct skeletal structure for the ion.

Calculate the total number of electrons for the Lewis structure by summing the valence electrons ofeach atom in the ion and adding one for the 1 - charge.

(number of valence e " for N) + (number of valence e ~ for O) = 5 + 3(6) +1 = 24Distribute the electrons among the atoms, giving octets to as many atoms as possible. Begin withthe bonding electrons, then proceed to lone pairs on terminal atoms and finally to lone pairs on thecentral atom.

•••• O-••

••

-o

If any atom lacks an octet, form double or triple bonds as necessary.

• •• O •

o


Lastly, write the Lewis structure in brackets with the charge of the ion in the upper right-hand comer.

• o

The double bond can be between the N and any of the oxygen atoms, so there are resonance structures.

o

N O •

(d) LilO: Li+

• •

: «••Determine the cation and anion.

Li+ 10-Write the Lewis structure for the lithium cation based on the valence electrons.

+ °Li 2s1 Li+ 2s1

Li* Li+

Li must lose one electron and be left with the octet from the previous shell.Write the Lewis structure for the covalent anion.Write the correct skeletal structure for the ion.

——— (l

Calculate the total number of electron for the Lewis structure by summing the number of valenceelectrons of each atom in the ion and adding one for the 1 - charge.

(number of valence e " for I) + (number of valence e ~ for O) = 7 +6 + 1 = 14Distribute the electrons among the atoms, giving octets to as many atoms as possible. Begin with thebonding electrons, then proceed to lone pairs on terminal atoms and finally to lone pairs on the cen-tral atom.

••: i -•• o :


9.84 (a) RbIO2: Rb+

••- 1 -

••-o :

chemical bonding i: lewis theory - gordon state …ptfaculty.gordonstate.edu/lgoodroad/summer...

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