Page 308, Quick Check 1. English chemist William Olding was the first

person to arrange the chemical elements intogroups.

2. John Newlands noticed that similar propertiesseemed to repeat every 8th element in the sameway that notes on a musical scale repeat every8th tone.

3. As Mendeleev had arranged similar elementsvertically on his table, the location of theblank spaces predicted the properties thatelements would possess upon beingdiscovered. When those elements wereeventually discovered, Mendeleev’s predictionsturned out to be extremely accurate. Thishelped his arrangement of the elements in hisperiodic table gain widespread acceptance.

Page 312, Quick Check 1.

Element Name

Element Symbol

Group Number

Period Number

Metal/Non-metal/

Metalloid silicon Si 14 3 metalloid

osmium Os 8 6 metal chromium Cr 6 4 metal

meitnerium Mt 9 7 metal antimony Sb 15 5 metalloid

iodine I 17 5 non-metal 2. a. The elements rearranged in order of least

metallic to most metallic are:fluorine, oxygen, sulphur, aluminum, gallium, chromium, zirconium, cesium.

b. The element with the greatest tendency togain an electron is fluorine.The element with the greatest tendency to losean electron is cesium.

Page 335, Quick Check Property Family

Number Family Name

a. reactive non-metalspossessing 7 valenceelectrons

17 halogens

b. reactive solids that form2+ cations duringreactions

2 alkaline earth metals

c. invisible gases that arealmost totally unreactive

18 noble gases

d. soft, very reactive silverysolids with 1 valence

1 alkali metals

2. The chemical formulas are:a. NaIb. BaSc. GaCl3

d. Rb2Se. Mg3P2

f. H2Se3. The two families of the periodic table containing

the most reactive elements are the halogensand the alkali metals. Elements in each of thesefamilies are within one electron of beingisoelectronic with the nearest noble gas andreadily either gain or lose one electronrespectively to achieve that electronarrangement.

Page 315, 6.1 Review Questions 1. The most important thing to know about the

periodic table is that elements in the samechemical family have similar chemicalproperties.

2. In 1913, data gathered by the young Britishchemist Henry Moseley, combined with thediscovery of isotopes, resulted in the elementsof the periodic table being re-ordered accordingto their atomic numbers rather than theiratomic masses.

3. Elements in the same chemical family havesimilar chemical properties because they havesimilar outer electron configurations and thevalence electrons are the electrons involved inchemical reactions.

4. a. The most metallic elements are located inthe lower left portion of the periodic table.b. The most non-metallic (or least metallic.elements are located in the upper right portionof the periodic table.

5. Element Properties Letter

and Symbol

a. chlorine found in carbohydrates and an elemental gas in 21% of the atmosphere

e. O

b. silver soft conductor that reacts explosively with water

producing H2 gas

d. Cs

c. neon less than 1 ounce of this solid radioactive

nonconductor exists on Earth

h. At

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d. cesium waxy yellow solid non-metal found in match heads, fertilizers, and

detergents

f. P

e. oxygen blue-gray metalloid used extensively in the

computer industry

g. Si

f. phosphorus

very reactive green gas used in the trenches in

World War I

a. Cl

g. silicon shiny solid that is the best conductor of heat

and electricity

b. Ag

h. astatine invisible unreactive gas used in lasers and some

electric street signs

c. Ne

6. Element Properties Family

Number unreactive gas used in electric street

signs comprising 0.93 % of the atmosphere

18

shiny multivalent solid, good conductor, forms coloured compounds

6

soft silvery solid, good conductor, reacts vigorously with water

1

gray-white metalloid predicted by Mendeleev and discovered in 1886

14

reactive metal present in bones and teeth possessing two valence electrons

2

yellow-green gaseous non-metal and the most reactive of all the elements

17

7. Properties of metals include:• solids at room temperature, except for mercury,

which is a liquid• generally shiny or lustrous when freshly cut or

polished• good conductors of heat and electricity• generally malleable, which means they can be

rolled or hammered into thin sheets• generally ductile, which means they can be

rolled or stretched into wires• generally flexible as thin sheets or wires• during chemical changes, tend to give up

electrons relatively easily to form cations8. Properties of non-metals include:

• usually gases or brittle solids at roomtemperature, except for liquid bromine

• solid non-metals can range in appearance fromdull or lustrous and translucent to opaque

• poor conductors of heat and electricity duringchemical changes,

• they tend to gain electrons from metals to formanions or share electrons with other non-

metals. 9. Except where large ionic charges result, atoms

of the main group elements will tend to give upor gain as many electrons as are necessary toacquire the valence electron configuration ofthe nearest noble gas.

10. The formulas for the stable ions are:a. Be2+ b. Te2– c. Cs+ d. Ra2+

e. Ga3+ f. Se2- g. In3+

11. a. Properties of the alkali metals include:• all  soft,  silvery  solids• the most reactive of all metals• the oxide compounds of the alkali metals

dissolve in water to produce strongly basicsolutions

• all corrode rapidly in air to a dull grayappearance, react vigorously with water toproduce hydrogen gas

• readily form compounds with non-metals• readily lose that outer electron to form 1+

cations and so assume the electronconfiguration of previous noble gas

b. Properties of the alkaline earth metalsinclude:• silver-coloured reactive metals.• not as reactive as the alkali metals, but also

readily form compounds with non-metals• their oxides are also alkaline in solution but

unlike alkali compounds, a number of group 2compounds have a low solubility in water

• readily form 2+ cations by losing those twovalence electrons and so will achieve theidentical electron configuration of the nearestnoble gas

c. Properties of the halogens include:• the most reactive family of elements in the

periodic table• the only chemical family in which all three

states of matter are represented• elemental halogens exist as diatomic

molecules• readily form compounds with metals,

hydrogen, carbon, and other non-metals• when reacting with metals, halogens typically

gain a single electron forming 1– anionsd. Properties of the noble gases include:• colourless gases• generally unreactive• all of the noble gases, except helium, have

filled s and p sublevels and have “stable octets”

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12. Group 2 Core

Notation Group

17 Core

Notation Group

18 Core

Notation Be [He] 2s2 F [He]2s2 2p5 He 1s2

(no core notation) Mg [Ne] 3s2 Cl [Ne] 3s23p5 Ne [He] 2s22p6 Ca [Ar]4s2 Br [Ar] 4s24p5 Ar [Ne] 3s23p6 Sr [Kr] 5s2 I [Kr] 5s24d105p5 Kr [Ar] 4s2 3d10 4p6 Ba [Xe]6s2 At [Xe] 6s2 4f 14 5d106 p5 Xe [Kr] 5s24d105p6 Ra [Rn] 7s2 Rn [Xe] 6s2 4f14 5d10 6p6

Page 320, Quick Check 1. As we move left to right across a chemical

period in the periodic table, the energy levelremains the same and so the attractive forcethat the valence electrons experience from thenucleus is the predominant force operating thatinfluences the size of the atoms. This forceincreases as we move across the period and sothe atoms generally become smaller.

2. As we move down a chemical family, thenumber of energy levels present increases inatoms. Even though the nuclear charge alsoincreases, the increasing energy levels are thepredominating factor determining atomic size.As a result, the attractive force felt by thevalence electrons decreases and the atomic sizeincreases.

3. Effective shielding is most prevalent movingdown a chemical family rather than across achemical period. The evidence of this is that thesizes of atoms increase as we move down achemical family and decrease as we move left toright across a chemical period.

Page 323, Practice Problems 1. The elements ranked in order of decreasing

ionization energy are:argon > sulphur > silicon > aluminum >magnesium > sodium > rubidium > cesium

2. Members of this chemical family have thehighest IE1 in their period. 18Members of this chemical family have thelowest IE1 in their period. 1Members of this chemical period have thehighest IE1 in their family. The uppermost periodin that family.Members of this chemical period have thelowest IE1 in their family. 7

3. If 2s electrons spend more of their time nearerthe nucleus than 2p electrons, then 2s electronsare capable of partially shielding 2p electrons

from the nuclear charge. This should

make it easier to remove boron’s single 2p electron, which explains boron’s lower first ionization energy.

Page 326, 6.2 Review Questions 1. The regular and predictable changes in e

lemental properties as we move across a periodor down a chemical family in the periodic tableare known as periodic trends.

2. The quantum mechanical model tells us that theouter boundaries of an atom are not hard anddefinite, but rather are the edges of chargeclouds enclosing the regions of highestprobability of finding an atom’s outer electrons.

3. Ionization energy and electronegativity bothincrease as we move left to right across a periodand move up a chemical family in the periodictable.

4. As we move in any direction horizontally acrossa period or vertically in a chemical family, thetrend in atomic size is generally opposite to thetrends seen in ionization energy andelectronegativity.

5. The valence electrons of both lithium andfluorine are in the second energy level and as aresult, those valence electrons experience asimilar amount of electron shielding. Becausefluorine’s nucleus has six more protons, theeffective nuclear charge (Zeff) seen by fluorine’svalence electrons is much greater than thatexperienced by lithium’s single valenceelectron. As a result, fluorine atoms are smallerthan lithium atoms.

6. The largest atoms are located in the lower leftregion of the periodic table and the smallestatoms are located in the upper right region ofthe periodic table.

7. a. A cation will always be smaller than its parentneutral atom because of increased attraction ofthe outer electrons for the nucleus anddecreased repulsion of the electrons for eachother.

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b. An anion will always be larger than its parentneutral atom because of decreased attractionof the outer electrons for the nucleus andincreased repulsion of the electrons for eachother.

8. Inner or “core” electrons are effective atshielding the outer electrons from the attractiveforce of the nucleus and exerting a repulsiveforce on those outer clouds. As the extent ofthat shielding increases, (which occurs whendescending a chemical family), the atomic sizesincrease and the ionization energies decrease.

9. Where on the Periodic Table Elements

Show: Largest Atomic Radii lower left

Smallest Atomic Radii upper right Lowest Ionization

Energy lower left

Highest Ionization Energy

upper right

Lowest Electronegativity

lower left

Highest Electronegativity

upper right

10. Lithium’s electron configuration is 1s2 2s1.Lithium’s single valence electron is shieldedfrom the nuclear charge by the inner or core 1s2

electrons. It is therefore relatively easy toremove that outer electron and so we see thatthe first ionization energy is low. After thatelectron is removed however, the next electronremoved is an inner core electron, which is notshielded at all from the nuclear charge. As aresult, lithium’s second ionization energy isapproximately 14 times greater than the firstionization energy!

11. When such reactions occur, elements with lowionization energies and low electronegativitieswill tend to lose valence electrons to elementswhose ionization energies andelectronegativities are high.

12. When such reactions occur, elements with highionization energies and high electronegativitieswill tend to gain valence electrons fromelements whose ionization energies andelectronegativities are low. (The results of the reactions described in questions 11 and 12 will be the formation of cations by atoms losing valence electrons and the formation of anions by atoms gaining valence electrons.)

13. When two non-metal atoms, each with relativelyhigh ionization energies and electronegativities

react, as both attract their valence electrons strongly and neither have a tendency to lose them, they are likely to form bonds by sharing valence electrons rather than by transferring them.

14. Electron configuration for nickel: [Ar] 3d84s2

Electron configuration for zinc: [Ar] 3d104s2

Notice that when the electron configurationsare written in order of increasing sublevel size,rather than increasing energy, we see that zinchas a filled 3d sublevel with 2 more electrons inthe inner 3d cloud than nickel. These 2additional electrons increase the effectiveshielding of zinc’s 2 outer electrons and thusreduce the attractive force they feel from thenucleus. As a result, zinc atoms are larger thannickel atoms.

Page 329, Quick Check 1. a. 2.46 × 109

b. 1.63 × 10–24

c. water

Page 331, Quick Check 1. Clockwise or counterclockwise2. The nuclear spin change of the proton

in the nucleus of a hydrogen atom in adistinct environment within a moleculecauses the different “signals” on an NMRspectrum

3. A methane molecule would only have onedistinct line on its NMR spectrum because thereis only one distinct nuclear spin in the hydrogenatoms as all four hydrogen atoms have the samechemical environment.

Page 334, Quick Check 1. O-H stretch in phenols(wavenumber ~3400 cm-1)2. N–N < N=N < NΞN

N–N will absorb infrared photons with thelongest wavelength

Page 336, Quick Check 1. a.2. b.3. [a] > [b]4. ~ 375 nm

Page 338, Quick Check

1. H: ↑1s

He: ↑↓1s !" He: ↑↓

Li: ↑↓1s

↑2s

Li: ↑↓1s

↑↓2s Be: ↑↓!"

↑↓!"

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B: ↑↓1s

↑↓2s

↑

2p B: ↑↓!"↑↓!"

↑!"

2. (1) 3 peaks show the 3 orbitals (1s, 2s, 2p)(2) heights of peaks are proportional to thenumber of electrons in nth energy levels (2electrons in 1s and 2 electrons in 2s have peaksof the same height; 1 electron in 2p has a peakhalf the height of the 1s and 2s peaks/orbitals)(3) orbital diagrams clearly show sublevels (px,py, pz) and electron spins where PES does not

3. The first peak represents the 1s orbital. Thesepeaks increase in energy from elements 1 to 5due to the increase in protons of the elements,and thus an increase in nuclear charge (Zeff).

4. The heights of the final peaks represent theelectrons in the highest orbital present in thatelement. The heights alternate from short (firstelectron in s orbitals) to tall (second electron in sorbitals) and finally back to short in the 2porbital of Boron.

Page 340, Practice Problems 1. a. Mg b. P c. S2.

Page 341, Quick Check 1. See practice problem 1a., b., c..

2. F = kQ1Q2

r 2 . Q1 is the effective nuclear charge, Zeff

= 10 – 2 = +8 for neon and 8 – 2 = +6 for oxygen, while Q2 = -1 for both as this is the charge on any valence electron. The valence electrons are in principal quantum level 2 for both neon and oxygen, however due to the differences in Zeff, r is smaller for neon. The result of the larger produce of Q1 × Q2 and the smaller r value for neon means a much stronger force holding the valence shell electrons according to Coulomb’s Law. The same argument holds for the third peak in each.

3. The first and second peaks are the same heightbecause for all three elements (Mg, P, S), thereare 1s2, 2s2 electrons. The third peak heightsvary because the presence of electrons in thenext 2p orbital, vary. (i.e. C only has one 2pelectron, O has three 2p electrons, and Ne hassix 2p electrons. Thus the heights of the fifthpeak will vary to show the number of electronswithin the 3p orbital.)

Page 342, 6.3 Review Questions 1.

Region of EM Spectrum :

Spectroscopic Technique:

Photons Used To:

Radio Waves NMR (nuclear magnetic resonance)

Measures energy changes causing nuclear spin to study molecular structure.

Infrared IR (infrared) spectroscopy

Increases forces of attraction between nuclei and electrons and increases forces of repulsion between nuclei

UV/Visible UV/Vis spectroscopy EMR (electromagnetic radiation)

Remove valence electrons (photoionization)

X-rays PES (photoelectron spectroscopy)

Removes core electrons

2. As the atomic number/number of protonsincreases, the effective nuclear charge increases.Because all of the valence electrons currentlybeing compared to are all in the second energylevel, the increasing first ionization energy isdue to the increasing positive charge on thenucleus (increasing Q values of Coulomb’s Law).Coulomb’s Law that as the force of attractionincreases, the distance between an electron andthe nucleus decreases, making it harder toremove the valence electro, increasing the firstionization energies when going across thesecond period.

3. Wavelength Range

a. frequency b. Range c.Spectroscopy

700 nm to 400 nm

See i. Visible Light

UV/Vis

400 nm to 10 nm

See ii. Ultraviolet UV/Vis

10 to 0.01 nm

See iii. X-ray PES

i. 4.29 × 1014 s–1 to 7.50 × 1014 s–1 ii. 7.50 × 1014 s–1 to 3.0 × 1016 s–1

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iii. 3.0 × 1016 s–1 to 3 × 1019 s–1

4. Fluorine5. a. Mg b. Ga

c. Kr d. Al6. a. Phosphorus

d. 9.57 MJ/mol, The 3p3 and 3s2 electrons can beemitted.

7. a. Neb. A – 1s; B – 2s; C – 2pc. Peak A is far to the left because of the closedistance that 1s electrons have to the nucleus. Itrequires more energy to remove 1s electrons.d. Peak C is three times higher than peak Bbecause it represents the Binding Energyrequired to remove three times as manyelectrons as in peak B. (Peak B represents thetwo electrons in 2s, Peak C represents the sixelectrons in 2p).

Page 348, Quick Check 1. When metal atoms react with non-metal atoms,

metal atoms tend to lose one or more valenceelectrons to non-metal atoms forming metalcations and non-metal anions.

2. Three types of chemical bonds based on thedifferent elements involved:

Atoms Involved in Chemical Bond

Type of Chemical Bond

metal bonded to non-metal

ionic bond

non-metal bonded to non-metal

covalent bond

metal bonded to metal

metallic bond

3. Ionic bond formation is typically associated withmetals from groups 1 and 2 reacting with non-

metals from groups 16 and 17 of the periodic table.

Page 352, Practice Problems 1. a. BaBr2 b. BeO c. Sr3N2

d. MgCl2 e. FrF2.

a. RbF As ∆EN = 3.2 bond is ionic

c. KBr As ∆EN = 2.0 bond is ionic

b. RaCl2 As ∆EN = 2.1 bond is ionic

d. Na2O As ∆EN = 2.6 bond is ionic

3. a. Na3N ∆EN = 2.1b. SrBr2 ∆EN = 1.8c. LiCl ∆EN = 2.0d. CsF ∆EN = 3.3e. Rb2O ∆EN = 2.7Compounds in order of increasing ionic bondcharacter: SrBr2 < LiCl < Na3N < Rb2O < CsF

Page 352, Practice Problems 1. a. A compound with an ionic bond:

NaCl (∆EN = 2.1) b. A compound with a polar covalent bond:

AlN (∆EN = 1.5)c. A compound with a non-polar covalent bond

NCl3 (∆EN = 0)

2. a. H2O ∆EN = 1.4b. PCl3 ∆EN = 0.9c. CI4 ∆EN = 0d. SiO2 ∆EN = 1.7e. AlN ∆EN = 1.5Compounds in order of most equal to mostunequal electron sharing: CI4 PCl3 H2O AlNSiO2

3. Elements Present

Formula ∆EN Value Nature of Bonds

Atom Possessing

Greater Electron Density

C and S CS2 0 covalent neither B and Cl BCl3 1.0 polar

covalent chlorine

Al and O Al2O3 2.0 ionic oxygen N and I NI3 0.5 polar

covalent nitrogen

Ca and F CaF2 3.0 ionic fluorine

Page 355, 6.4 Review Questions 1. a. The energy associated with the bonded

atoms must be less than when the atoms areapart.b. This tells us that the attractive forces existingbetween the bonded atoms exceed therepulsive forces.

2. An ionic crystal lattice is the three-dimensionalsymmetrical arrangement of cations and anionsin a solid ionic crystal. The vast number ofinterionic forces present in such a crystal mustbe overcome to melt an ionic compound andthis explains the high melting points of suchcompounds.

3. A crystal lattice shows us that no neutralindependent molecules exist in ioniccompounds. The formulas that we write simply

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represent the smallest whole number ratios of cations to anions that exist in ionic compounds.

4. a. The attractive forces associated with Ionicbonds are the electrostatic forces betweenpositively charged cations and negativelycharged anions.b. The attractive forces associated withcovalent bonds are the electrostatic forcesbetween negatively charged electrons andadjacent positively charged nuclei.

5. a. Similarities between ionic and covalentbonds include:• Both involve valence electron clouds.• Both involve electrostatic attractions between

oppositely charged species.• The formation of both bonds begins with the

valence electrons of two atoms experiencingthe attractive force of adjacent positive nuclei.

• Ionic and covalent bonds can both be strongb. Differences between ionic and covalentbonds include:• Ionic bonds involve the transfer of valence

electrons forming ions whereas covalentbonds involve the sharing of valenceelectrons with no ion formation.

• Ionic bonds form only between metals andnon-metals whereas covalent bonds usually(but not always) form between two non-metals.

• Ionic bonding does not result in moleculeformation whereas covalent bonding usuallydoes.

• The attractive forces associated with covalentbonds are the electrostatic forces betweennegatively charged electrons and adjacentpositively charged nuclei.

• The attractive forces associated with Ionicbonds are the electrostatic forces betweenpositively charged cations and negativelycharged anions.

6. Elements Compoun

d Formula ∆EN

Value Nature of

Bonds Present a. rubidium

andoxygen

Rb2O 2.7 ionic

b. strontiumandbromine

SrBr2 1.8 ionic

c. carbonandsulphur

CS2 0 covalent

d. silicon andchlorine

SiCl4 1.2 polar covalent

7. ∆EN for MgS = 1.3∆EN for H2O = 1.4We see that the bonds in water actually possessslightly more ionic character than those inmagnesium sulphide even though the formercompound contains two non-metals and thelatter compound contains a metal and a non-

metal. 8. The formula for glucose and other molecular

compounds does not represent a ratio as doionic formulas. Rather, they represent the actualnumber of atoms of each element existing in anindividual molecule of the compound.

9. The melting of molecular covalent compoundsdoes not involve the breaking of covalentchemical bonds within the molecules, but ratherovercoming the relatively weak attractive forcesbetween those molecules in the solid phase.

10. Diamond is a type of covalent substance isknown as a network covalent solid. Rather thanconsisting of individual molecules as molecularcovalent compounds do, these substances areheld together by covalent bonds that extendthroughout the entire sample. This means thatthe “molecule” is literally as big as the sampleitself. To melt such a substance, all of thecovalent bonds within this giant molecule mustbe broken and this accounts for the very highmelting point.

11. The bond in HCl is a polar covalent bond (∆EN =0.9) whereas the bond in N2 must be purecovalent (∆EN = 0). This means that electrondensity in the HCl molecule is concentrated onthe chlorine side of the molecule making thatend of the molecule somewhat negative andthe hydrogen end of the molecule somewhatpositive. We might expect therefore, that HClmolecules would attract each other morestrongly than N2 molecules where the electrondensity is evenly distributed throughout themolecule. (The melting point of HCl is –114.2°Cand the melting point of N2 is – 210°C.)

Page 359, Quick Check 1. The dots in a Lewis dot symbol for an atom or

ion represent the number of electrons in thevalence shell – the number of electrons in theouter s and p sublevels.

2. For the elements in Group 1, 2, and 13 the dotstells us the number of electrons the each atomloses when forming a cation – this also equalsthe magnitude of the resulting positive chargeon that cation.

3. In a correctly drawn Lewis symbol for a non-metal, the number of unpaired dots shown

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represents either the number of electrons that atom must gain when forming an anion, or the number the electrons must share to complete its octet when forming covalent bonds.

Page 359, Practice Problems

Page 362, Quick Check 1. Because the elements Be, B, and Al each possess

fewer than 4 Valence electrons (2, 3, and 3respectively), they are unable to achieve avalence a valence octet of four electrons pairsby sharing all of their valence electrons whenforming covalent compounds.

2. The skeleton “S–H–H” Is incorrect because itshows a central hydrogen with two bonds.Hydrogen normally forms only one bond bysharing its single electron.

3. The total number of valence electrons in each ofthe following molecules are:a. H2Se (2 × 1) + (6) = 8 valence electronsb. CCl4 (4) + (4 × 7) = 32 valence electronsc. NF3 (5) + (3 × 7) = 26 valence electronsd. PCl5 (5) + (5 × 7) = 40 valence electronse. SF6 (6) + (6 × 7) = 48 valence electrons

Page 363, Practice Problems 1.

2.

There are 3 bonding pairs and 1 lone pair of electrons.

3.

b. The central boronhas an incompleteoctet of 6 electrons

Page 366, Practice Problems 1. 2. 3.

Page 370, 6.5 Review Questions 1.

Incomplete Valence Octet

Valence Octet

Expanded Valence

Octet H C Si Be N P B O S Al F Cl

2. Helium and neon are both invisible andunreactive gases whose similar behaviourqualifies each of them as members of the familywhich incudes other unreactive gases. They dohave a different number of outer electrons, butthe outer electron cloud of each is a particularlystable configuration.Helium only has 2 electrons, and those arerepresented by the to dots in its Lewis structure.That single pair of outer electrons fills the 1s sublevel and completes the 1st energy level, whichexplains why helium is unreactive.Neon has 8 outer electrons represented as 4pairs of dots. Those 8 electrons fill the 2s and 2psublevels, complete the 2nd energy level andgive neon a stable valence octet consisting of 4pairs of electrons. As a result, neon is alsounreactive.

3. Element Pair Probable

Central Atom

Probable Peripheral

Atom a. Phosphorus and Chlorine Phosphorus Chlorineb. Nitrogen and Oxygen Nitrogen Oxygen c. Carbon and Sulphur Carbon Sulphur d. Nitrogen and Hydrogen Nitrogen Hydrogen e. Oxygen and Fluorine Oxygen Fluorine

4.

5. OF2 H2S PCl3 CCl2F2

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6. OH– AlH4

– CN–

7. SF6 PCl5 ICl4

– SeBr4

8. CS2 SCO O3 NO3

–

9.

Page 476, Quick Check 1. VSEPR theory states electrons will position

themselves so far away from each other as possible to minimize repulsion forces.

2. BP-BP<BP-LP<LP-LP3. Each has a different number of lone-pairs so a

different molecular shape is formed.

Page 478, Practice Problems 1.

Lewis Structure AXmEn Notation

Molecular Shape (Name and Diagram)

a.

AX5

Trigonal Bipyramidal

b.

AX4E2

Square Planar

2. Chemical Formula

Lewis Structure

AXmEn Notation

Molecular Shape (Name and Diagram)

a.

CCl4 AX4

Tetrahedral

b.

PF3 AX3E

Trigonal Pyramidal c.

SCl2 AX2E2

Bent or Angular

Page 482, Practice Problems 1.

Symmetric Molecules Asymmetric Molecules

AXmEn Notation

Shape of Molecule

AXmEn Notation

Shape of Molecule

AX2 Linear AX2E Bent or Angular

AX3 triangular planar

AX2E2 bent or angular

AX4 tetrahedral AX3E trigonal pyramidal

AX5 trigonal bipyramidal

AX4E see-saw

AX6 octahedral AX3E2 t-shaped

AX2E3 linear AX5E square pyramidal

AX4E2 square planar

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2.

3. In the previous Practice Problems numbers 1 and 2

Molecular Formula

Polar or Non-Polar Molecule?

1. a. SOF4 polar b. XeF4 non-polar

2. a. CCl4 non-polar b. PF3 polar c. SCl2 polar

Page 391, Practice Problems 1. Formula Shape Hybridization Polar?

CH4 Tetrahedral sp3 non-polar H3O+ Trigonal

Pyramid sp3 polar

OF2 Bent sp3 polar SO3 Trigonal

Planar sp2 non-polar

HCN Linear sp polar SF6 Octahedral sp3d2 non-polar

SOF4 Trigonal Bipyramidal

sp3d polar

Page 394, Practice Problems – Assigning and Using Formal Charges 1.

Structure One: Carbon: F.C. = Valence e-‘s – [dots + lines] = 4 – [0 + 4] = 0

1st and 2nd: Oxygen: F.C. = 6 – [4 + 2] = 0

Structure Two: Carbon: F.C. = 4 – [0 + 4] = 0 1st Oxygen: F.C. = 6 – [6 + 1] = -1 2nd Oxygen: F.C. = 6 – [2 + 3] = +1

The First Species has 0 F.C. on all atoms, the second species has two non-zero formal charges assigned hence the first species is more stable and more likely to form according to formal charge assignment. (Note that both structures obey the stable octet rule all the way around, so the first species is undoubtedly the more likely to form.

2. a. b. c.  

d. e. f.  

Structure 1: Oxygen: F.C. = 6 – [4 + 2] = 0 Carbon: F.C. = 4 – [0 + 4] = 0

Sulfur: F.C. = 6 – [4 + 2] = 0

Structure 2: O: F.C. = 6 – [6 + 1] = -1, C: 4 – [0 + 4] = 0, S: 6 – [2 + 3] = +1

Structure 3: O: 6 – [2 + 3] = +1 C: 4 – [0 + 4] = 0, S: 6 – [6 + 1] = -1

Structure 4: O: 6 – [0 + 4] = +2, C: 4 – [4 + 2] = -2, S: 6 – [4 + 2] = 0

Structure 5: O: 6 – [0 + 4] = +2, C: 4 – [6 + 1] = -3, S: 6 – [2 + 3] = +1

Structure 6: O: 6 – [0 + 4] = +2, C: 4 – [2 + 3] = -1, S: 6 – [6 + 1] = -1

All structures’ atoms obey the stable octet rule. Structure 1 is MOST stable and Structure 5 is LEAST stable.

Lewis Structure

AXmEn Notation

Molecular Shape

(Name and Diagram)

Polar Molecule

? (Yes / No)

a.

AX3 Trigonal Planar

Yes

b.

AX3E

Trigonal Pyramidal

Yes

c.

AX5

Trigonal Bipyramidal

No

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3. With no expanded octet:Upper O’s:  

6 − 6 +2

2⎛⎝

⎞⎠ = 6 – 6 +1( ) = −1

Lower O:  

6 – 4 +4

2⎛⎝

⎞⎠ = 6 – 4 + 2( ) = 0

Central S:  

6 – 0 +8

2⎛⎝

⎞⎠ = 6 – 0 + 4( ) = −2

With an expanded octet:

O’s:   6 − 4 −4

2= 0

S’s:   6 − 0 −12

2= 0

Formal charge indicates the structure with an expanded octet and three double bonds is most likely. Recent studies show that structures whose

atoms all obey the stable octet rule may, in fact, be more stable than those with minimal formal charge.

Page 396, 6.6 Review Questions 1. a. The letters stand for: Valence Shell Electron

Pair Repulsion.b. VSEPR theory allows us to use two-dimensional Lewis diagrams for molecules toquite accurately predict the three-dimensionalshapes of those molecules.

2. As lone-pair electrons are attracted to only oneatomic nucleus, they are held less tightly thanbonding electron groups. Their electron cloudstherefore occupy more space and exert morerepulsive force on bonding electron groupsthan those groups exert on each other.

3. Both BF3 and CH2O have the same shapebecause both of them are “AX3” molecules. Eventhough the peripheral atoms in eachmolecule possess different numbers of non-

bonding electrons and one of the “AX” bonds on CH2O is a double bond, the shapes are identical

4. AXmEn Notation Molecular Shape AXmEn Notation Molecular Shape

AX3 angular AX4E T-shapedAX2E3 trigonal bipyramidal AX2E octahedral AX4 trigonal pyramidal AX3E2 square pyramidal

AX3E trigonal planar AX6 square planar AX2E2 tetrahedral AX5E angular AX5 linear AX4E2 see-saw

5. a. The X – A – X bond angles in ammonia will be smaller than in methane (107o vs. 109.5o) because ammonia isan AX3E molecule and methane is an AX4 molecule. The lone pair on the central nitrogen in ammonia occupiesmore space than the bonded pairs on the central carbon in methane and will therefore force the bondedelectron pairs in ammonia closer together.b. Methane is a symmetrical AX4 molecule and is non-polar. The intermolecular forces acting are thereforeweak London dispersion forces. Ammonia is an asymmetrical AX3E polar molecule that exhibits much strongerhydrogen bonding.

6. AXmEn Category

AX2 AX3 AX4 AX5 AX2E3 AX6 AX4E2

X–A–X Bond Angle

180o 120o 109.5o 120o 180o 180o 90o 90o

7. The molecules of each compound areasymmetric AX3E molecules. If we consider the∆EN values as we move up the family frombottom to top, we see the following:

Compound SbH3 AsH3 PH3 NH3

∆EN Value 0.2 0.1 0.0 0.9

Even though asymmetry exists in each of the molecules, because the ∆EN values for the first three compounds are so small, all of the bonds in those molecules are nonpolar and so the molecules themselves are nonpolar. As a result, the intermolecular forces acting between them are London dispersion forces. However, the

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bonds in ammonia are quite polar and so the molecule itself is also polar. As nitrogen is bonded to hydrogen in a polar molecule, hydrogen bonds exist between the ammonia molecules. We might therefore expect ammonia to have the highest boiling point in the group. However, when we check Figure 6.5.9, we notice that SbH3 actually has a higher boiling point than NH3. This must mean that the London dispersion forces in SbH3 (a larger and heavier molecule), are more significant than the hydrogen bonds in ammonia.

8. Because hydrogen bonds exist betweenammonia molecules but only London dispersionforces exist between diatomic molecules ofnitrogen and hydrogen, the ammonia has ahigher boiling point and thus liquefies(condenses) at a higher temperature thannitrogen and hydrogen which remain in thegaseous phase upon cooling.

9. As a result of hydrogen bonding, water remainsliquid until 100oC. This is very important giventhe fact that most of our Earth and most of ourbodies are composed of and require liquidwater to survive. The three-dimensionalstructure of proteins and the base-pairing in thedouble helix of DNA molecules depend onhydrogen bonds. Hydrogen bond formation inice make it less dense than liquid water which iscrucial to aquatic life when bodies of waterfreeze.

10. I2 is a relatively large and massive diatomicmolecule and the strength of London dispersionforces increase as the size of the moleculesinvolved increases. This is because largeelectron clouds are more loosely held thansmaller clouds and thus more easily deformedor polarized by a nearby dipole than compacttightly-held clouds. In addition, largemolecules with more surface area have electronclouds that are spread out and so aremore easily distorted by neighboring dipoles. Asa result, the dispersion forces are strong enoughto keep the molecules of I2 attached to eachother even at room temperature.

11. Although the ionic bonds holding a crystallattice together are strong, when the surface ofthat lattice is in contact with water, each ion onthat surface will attract the oppositely chargedend of polar water molecules near them. Thatattraction between an ion and a polar moleculeis called an ion-dipole force. These attractiveforces soon overcome those existing betweenthe ions themselves and so the crystal structurebegins to break down and the ionic compound

dissolves. As the ions move away from the lattice surface, they immediately become surrounded or enclosed in what chemists call a hydration shell. Ion-dipole forces are the primary force responsible for the solubility of ionic compounds in water.

12. AXmEn

Notation Shape of Molecule

(Name and Diagram) Type of

Intermolecular Force Acting

Between Molecules

AX4

Tetrahedral

dipole-dipole

AX3

Trigonal Planar

dipole-dipole

AX6

Octahedral

London dispersion

AX5E

Square Pyramidal

dipole-dipole

13. i. a.

b. Bent (V-Shaped) – sp3

c. Dipole movement existsd. 2 sigma / 0 pie. S=0, Each Cl=0

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iii. a.

b. Square Planar – sp3d2

c. No dipole movement(all bond dipoles cancel)d. 4 sigma / 0 pie. All atoms FC=0vi. a.

b. Octahedral – sp3d2

c. Very slight dipole movementd. 6 sigma / 0 pie. F all=0, I=+1, O=-1

v. a.

b. See saw – sp3dc. Dipole movementd. F all=0, I=+1e. 4 sigma / 0 pif. F all=0, I=+1

iv. a.

b. Trigonal Planar – sp2

c. No dipole movementd. 3 sigma / 1 pie. C=0, Double bonded O=0, Single bonded O=-1

Page 405, Practice Problems

1. Be2

Energy ↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

Be Be2

Be

2. CO

Energy σ*2p

π*2p

2p2↑↑ 2p4↑↓↑↑

↑↓ ↑↓ π2p

↑↓ σ2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

C CO O

3. NO

Energy σ*2p

↑ π*2p

2p3↑↑↑ 2p4↑↓↑↑

↑↓ ↑↓ π2p

↑↓ σ2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

N NO O

Bond Order = 0 Bond Order = 3 Bond Order = 2.5

Note: CO and NO may also be shown with the π2p orbitals at a lower energy than the σ2p due to the large 2s-2p interactions that occur in the carbon and nitrogen atoms involved in these molecules (see page 5). These interactions are less pronounced in the oxygen atom. Consequently CO may be shown either as above or as in review question 5.

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Page 407, Quick Check 1. NO- Lewis Diagram:

NO+ Lewis Diagram:

2. Molecular Orbital Diagrams of NO-, NO, NO+

NO-

Energy σ*2p

↑ ↑ π*2p

2p3↑↑↑ 2p4↑↓↑↑

↑↓ ↑↓ π2p

↑↓ σ2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

N NO- O

NO

Energy σ*2p

↑ π*2p

2p3↑↑↑ 2p4↑↓↑↑

↑↓ ↑↓ π2p

↑↓ σ2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

N NO O

NO+

Energy σ*2p

π*2p

2p3↑↑↑ 2p3↑↓↑↑

↑↓ ↑↓ π2p

↑↓ σ2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

N NO+ O

3. Bond Orders of NO-, NO, NO+

NO- Bond Order = 2

Type of bond: double bond

Relative to NO+: weaker/

longer bond

NO Bond Order = 2.5 NO+ Bond Order = 3

Type of bond: triple bond

Relative to NO-: stronger/

shorter bond

4. Bonds of Lewis dot structures are identical to those predicted by MO theory.

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Page 408, Quick Check

1.

2. Energy σ*2p

↑ ↑ π*2p

2p4↑↓↑↑ 2p4↑↓↑↑

↑↓ ↑↓ π2p

↑↓ σ2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

O O2 O

3. Magnetic Behavior of liquid oxygen can bepredicted by the molecular orbital theory(paramagnetic behavior is due to unpairedelectrons in π*2p orbital). The Lewis dot

structure shows all electrons paired. This predicts diamagnetic behavior.

4. Diamagnetic (only paired electrons present)species: (page 440) Be2, CO, (page 442) NO+

Page 409, 6.7 Review Questions

1.

Differences Molecular Orbital

Similarities Differences Atomic Orbital

-presence of bonding/antibonding orbitals-helps predict bondorder, paramagnetismand/or diamagnetism

-both are conserved (two atomicorbitals combine to form twomolecular orbitals)-both hold maximum 2 electrons,each with opposite spins-both orbitals overlap along x-axis-helps predict bond energy andbond length

-no presence of bonding/ antibondingorbitals-cannot predict bond order, paramagnetismand/or diamagnetism

2.

Differences Bonding Orbital

Similarities Differences Antibonding Orbital

-Lower in energy-more stable-higher electron densitybetween the nuclei-results from constructiveorbital addition

-both are molecular orbitals andthus can be σ or π orbitals

-Higher in energy-less stable-lower electron density between thenuclei-results from destructive orbital addition

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3. Molecular Orbital Diagrams of Li2+, Li2, Li2

-

Li2+

Energy σ*2s

2s1 ↑ ↑ 2s1

↑ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

Li Li2+ Li

Li2

Energy σ*2s

2s1 ↑ ↑ 2s1

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

Li Li2 Li

Li2-

Energy ↑ σ*2s

2s1 ↑ ↑ 2s1

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

Li Li2- Li

Bond Orders of Li2+, Li2, Li2

-

Li2+ Bond Order = 0.5

Unstable

Relative to Li2:

weaker/longer bond

Paramagnetic

Li2 Bond Order = 1

Stable

Relative to ions:

stronger/shorter bond

Diamagnetic

Li2- Bond Order = 0.5

Unstable

Relative to Li2:

weaker/longer bond

Paramagnetic

4. uncharged species: Cl2

anion: SCl-

5. COBond Order = 3 suggests that the compound contains a triple bond. π2p orbitals are filled before σ2p due to thelarge interactions between the 2s and the 2p orbitals. These interactions force the σ2s orbital to drop in energyand more noticeably, the σ2p to rise so that the π2p orbitals actually fill before the σ2p. This irregularity in fillingorder is most commonly shown in the homonuclear molecules, B2, C2, and N2 and may not be shown inheteronuclear molecules containing only one of the atoms B, C or N.

6. a. N2 Bond Order = 3; N2+ Bond Order = 2.5

O2 Bond Order = 2; O2+ Bond Order = 2.5

NO Bond Order = 2.5; NO+ Bond Order = 3b. N2 is diamagnetic; N2

+ is paramagnetic

O2 is paramagnetic; O2+

is paramagneticNO is paramagnetic; NO+

is diamagnetic

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7. Molecular Orbital Diagrams of C2-, C2, C2

+

C2-

Energy σ*2p

π*2p

2p2↑↑ 2p2↑↑

↑ σ2p

↑↓ ↑↓ π2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

C C2- C

C2

Energy σ*2p

π*2p

2p2↑↑ 2p2↑↑

σ2p

↑↓ ↑↓ π2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

C C2 C

C2+

Energy σ*2p

π*2p

2p2↑↑ 2p2↑↑

σ2p

↑↓ ↑ π2p

↑↓ σ*2s

2s2 ↑↓ ↑↓ 2s2

↑↓ σ2s

↑↓ σ*1s

1s2 ↑↓ ↑↓ 1s2

↑↓ σ1s

C C2+ C

3. Bond Order and Stability of C2-, C2, C2

+

C2- Bond Order = 2.5

Stable

C2 Bond Order = 2

Stable

C2+ Bond Order = 1.5

Stable

8. a. H2-2: (σ1s)2(σ1s*)2

Bond Order = 0 UNSTABLE b. Ar2: (σ3s)2(σ3s*)2(σ3p)2(π3p)4(π3p*)4(σ3p*)2

Bond Order = 0 UNSTABLE c. He2

2+: (σ1s)2

Bond Order = 1 STABLE d. F2

2-: (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ3p*)2

Bond Order = 0 UNSTABLE

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