Chapter 9: The Basics of Chemical Bonding

Chapter 9: The Basicsof Chemical Bonding

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

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Chemical Bonds Attractive forces that hold atoms

together in complex substances Molecules and ionic compoundsWhy study? Changes in these bonding forces are

the underlying basis of chemical reactivity

During reaction: Break old bonds Form new bonds


3

Two Classes of Bonds

Covalent bonding Occurs in molecules Sharing of e’s

Ionic Bonding Occurs in ionic solid e’s transferred from 1 atom to another Simpler We will look at this first


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Ionic Bonds Result from attractive forces between

oppositely charged particles

Metal - nonmetal bonds are ionic because: Metals have

Low ionization energies Easily lose e– to be stable

Non-metals have Very exothermic electron affinities

Formation of lattice stabilizes ions

Na+ Cl –


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Ionic Compounds Formed from metal and nonmetal

Ionic Bond Attraction between + and – ions in ionic

compound. Why does this occur? Why is e

transferred? Why Na+ and not Na2+ or Na? Why Cl and not Cl2 or Cl+?

Na+ + ClNa + Cl NaCl(s)

e

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Ionic Compounds

Ionic crystals: Exist in 3-dimensional

array of cations and anions = lattice structure

Ionic chemical formulas: Always written as

empirical formula Smallest whole number

ratio of cation to anion


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Energetics Must look at energy of system to

answer these questions For any stable compound to form from

its elements Potential Energy of system must be

lowered. Net in energy Hf° < 0 (negative)

What are factors contributing to energy lowering for ionic compound? Use Hess’s Law to determine Conservation of energy Envision two paths


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Two Paths to Evaluate Energy1. Na(s) + ½Cl2(g) NaCl(s) Hf° = – 411.1 kJ/mol

2. Stepwise path Na(s) Na(g) Hf°(Na, g) = 107.8 kJ/mol½Cl2(g) Cl(g) Hf°(Cl, g) = 121.3 kJ/molNa(g) Na+(g) + e– IE(Na) =495.4 kJ.molCl(g) + e– Cl–(g) EA(Cl) = – 348.8 kJ/molNa+(g) + Cl–(g) NaCl(s) Hlattice = – 787 kJ/molNa(s) + ½Cl2 (g) NaCl(s) Hf° = – 411 kJ/mol


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Lattice Energy

Amount that PE of system decreases when 1 mole of solid salt is formed from its gas phase ions.

Energy released when ionic lattice forms.


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Lattice Energy Always HLattice= – = exothermic

HLattice gets more exothermic (larger negative value) as ions of opposite charge in crystal lattice are brought closer together as they wish to be.

Ions tightly packed with opposite charged ions next to each other.

Any in PE due to ionizing atoms is more than met by in PE from formation of crystal lattice. Even for +2 and –2 ions

overall exothermic to form ionic solids and they are stable compounds.

dqq

H Lattice


Your Turn!Assuming that the separation between

cations and anions in the lattice is nearly identical, which species would have the greatest lattice energy?

A. Sodium chlorideB. Calcium chlorideC. Calcium nitrideD. Sodium oxideE. Calcium oxide

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Why do Metals form Cations and

Nonmetals form Anions?Metal Left hand side of

Periodic Table IE small and positive

Little energy required to remove e to give cations

EA small and negative or positive Not favorable to attract

an e to it. Least expensive,

energy- wise, to form cation

Nonmetal Right hand side of

Periodic Table IE large and positive

Difficult to remove e

EA large and negative But easy to add e Exothermic—large

amount of E given off PE of system

Least expensive, energy- wise, to form anion

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Electron Configurations of Ions

How e'ic structure affects types of ions formed

Ex.Na 1s2 2s2 2p6 3s1 = [Ne] 3s1

Na+ 1s2 2s2 2p6 = [Ne] IE1 = 496 kJ/mol small not too difficult

IE2 = 4563 kJ/mol large ~10 x larger very difficult

Can remove 1st e, as doesn't cost too much Can’t remove 2nd e, as can't regain lost

energy from lattice doesn’t form.


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Ex. Ca [Ar] 4s2 Ca2+ [Ar]

IE1 small = 590 kJ/mol not too difficult

IE2 small = 1140 kJ/mol not too difficult

IE3 large = 4940 kJ/mol too difficult Can regain by lattice energy ~2000

kJ/mole if +2, –2 charges. But 3rd e too hard to remove Can't recoup required energy through

lattice formation. doesn't happen


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Electron Configurations of Ions Stability of Noble Gas Core below

valence e's effectively limits # e's that metals lose.

Ions formed have Noble gas e configuration True for anions and cations

Ex. Cl 1s2 2s2 2p6 3s2 3p5 = [Ne]3s2 3p5

Cl 1s2 2s2 2p6 3s2 3p6 = [Ar] Adding another e

requires putting it into next higher n shell Energy cost too high


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Ex.O 1s2 2s2 2p4 O 1s2 2s2 2p5 EA1 = 141 kJ/mol

O2 1s2 2s2 2p6 = [Ne] EA2 = +844 kJ/mol

EAnet = +703 kJ/mol endothermic Energy required more than made up for by

in HLattice caused by higher 2 charge


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Electron Configurations of IonsGeneralization:

When ions form Atoms of most representative elements (s

and p block) Tend to gain or lose e's to obtain nearest

Noble gas e configuration Except He (2 e's), all noble gases have 8

e's in highest n shellOctet Rule Atoms tend to gain or lose e's until they

have achieved outer (valence) shell containing octet (8 e's)


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Octet Rule Works well with

Group IA and IIA metals

Al Non-metals

H and He can't obey Limited to 2 e's in n

= 1 shell Doesn't work with

Transition metals Post transition

metals


Your Turn!What is the correct electron configuration

for Cs and Cs+ ?

A. [Xe] 6s1 , [Xe] B. [Xe] 6s2 , [Xe] 6s1

C. [Xe] 5s1 , [Xe]D. [Xe] 6s1 , [Xe] 6s2

E. [Xe] 6p2 , [Xe] 6p1

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Post Transition MetalsEx.

Sn [Kr] 4d10 5s2 5p2 Sn2+ [Kr] 4d10 5s2

Neither has Noble Gas e configuration Have emptied 5p subshell

Sn4+ [Kr] 4d10 Does have empty n = 5


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Transition Metals 1st e's lost from ns orbital of outer shell Lose e's from highest n first, then highest ℓEx. Fe [Ar] 3d6 4s2

Fe2+ [Ar] 3d6 loses 4s e's 1st Fe3+ [Ar] 3d5 then loses 3d e

Extra stability due to exactly half-filled d subshell.

Consequences TM2+ common oxidation state as remove 2 e's

from outer ns shell Ions of larger charge result from loss of d e's.


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Transition Metals Not easy to predict which ions form and

which are stable But ions with exactly filled or half-filled

d subshells are extra stable and tend to form.

Mn2+ [Ar]3d5 Fe3+ [Ar]3d5 Zn2+ [Ar]3d10


Your Turn!What is the correct electronic

configuration for Cu and Cu2+ ?A. [Ar] 3d 9 4s2, [Ar] 3d 9

B. [Ar] 3d 10 4s1, [Ar] 3d 8 4s1

C. [Ar] 3d 10 4s1, [Ar] 3d 9

D. [Ar] 3d 9 4s2, [Ar] 3d 10 4s1

E. [K] 3d 9 4s2, [Ar] 3d 9

Filled and half-filled orbitals are particularly stable.

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Predicting Cation Configurations

Consider Bi, whose configuration is [Xe]6s2 4f14 5d10 6p3. What ions are expected?

Rewrite config’n: [Xe]4f14 5d10 6s2 6p3

Bi3+ and Bi5+

Consider Fe, whose configuration is: [Ar]4s2 3d6 What ions are expected?

Rewrite config’n: [Ar]3d6 4s2

Fe2+ and Fe3+


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Predicting Anion Configurations

Non-metals gain electrons to become isoelectronic with next larger noble gas

O: [He]2s22p4 + ? e– → ?

N: [He]2s22p3 + ? e– → ?

2 O2– : [He]2s2 2p6

3 N3– : [He]2s2 2p6


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Lewis Symbols Electron bookkeeping method Way to keep track of e–’s Write chemical symbol surrounded by dots for

each e–

H

Group # IA IIA IIIA IVA

Valence e–'s 1 2 3 4

e– conf'n ns1 ns2 ns2np1 ns2np2

Li

Na

Be

MgB

Al

C

Si

He


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Lewis Symbols

Group # VA VIA VIIA VIIIA

Valence e-'s 5 6 7 8

e- conf'n ns2np3 ns2np4 ns2np5 ns2np6

N O F Ne

P S Cl Ar

For the representative elements Group # = # valence e–’s

He


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Lewis Symbols

Can use to diagram e– transfer in ionic bonding

Na + Cl Na+ + Cl

+ Mg2+ + O 2Mg O


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Covalent Compounds

Form individual separate molecules Atoms bound by sharing e–’s

Do not conduct electricity Often low melting point

Covalent Bonds Shared pairs of e–’s between 2 atoms 2 H atoms come together, Why?


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Covalent Bond Attraction of valence e– of 1 atom by

nucleus of other atom Shifting of e– density As distance between nuclei ,

probability of finding either e– near either nucleus

Pulls nuclei closer together


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Covalent Bond

As nuclei get close Begin to

repel each other

Both have high positive charge

Final internuclear distance between 2 atoms in bond Balance of attractive and repulsive forces Net attraction since bond forms


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Electron Pair Bond

e–’s come together in molecular bond In sense 1s orbital of each H is filled e–’s share same region of space

their spins must be paired (1 and 1 )

Refer to as e– pair bondhttp://www.visionlearning.com/library/module_viewer.php?c3=&mid=55&l=


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Bond Dissociation Energy, D

Energy needed to break one mole of bond in covalent molecule in gas phase

Symbolized D D varies with bond type

Single, double, triple D varies with atoms involved D is always positive

Formation of bonds stabilizes structure energetically


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Bond Bond Energy

(kJ mol−1) Bond

Bond Energy (kJ mol−1)

C–C 348 C–Br 276

C═C 612 C–I 238

C≡C 960 H–H 436

C–H 412 H–F 565

C–N 305 H–Cl 431

C═N 613 H–Br 366

C≡N 890 H–I 299

C–O 360 H–N 388

C═O 743 H–O 463

C–F 484 H–S 338

C–Cl 338 H–Si 376

Chemical Potential Energy ΔH°rxn = {D (Bonds broken) } {D (Bonds formed) }


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Covalent Bond

2 quantities characterize this bondBond Length (bond distance)

Distance between 2 nuclei = rA + rB

Bond Energy Also bond strength Amount of energy released when bond

formed ( PE) or Amount of energy must put in to “break”

bond


Your Turn!Which species is most likely covalently

bonded?

A. CsClB. NaFC. CaF2

D. COE. MgBr2

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Lewis Structures Molecular formula drawn with Lewis

Symbols Method for diagramming electronic

structure of covalent bonds Uses dots to represent e–s Covalent bond

Shared pair of e–s Each atom shares e–s so has complete octet

ns2np6 Noble Gas e– configuration Except H which has complete shell with 2

e–s


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Octet Rule: When atoms form covalent bonds, tend to share sufficient e–’s so as to achieve outer shell having 8 e–’s

Indicates how all atoms in molecule are attached to one another

Accounts for ALL valence e–’s in ALL atoms in molecule

Let’s look at some examplesNoble Gases: 8 valence e–’s

Full octet ns2 np6

Stable monatomic gases Don’t form compounds


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Lewis Structures

Diatomic Gases: H and Halogens

H2

H· + ·H H:H or HH Each H has 2 e–’s through sharing Can write shared pair of e–’s as line

() : = covalent bond


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Lewis StructuresDiatomic Gases:

F2

Each F has complete octet Only need to form one bond to

complete octet Pairs of e–’s not included in covalent

bond are called Lone Pairs Same for rest of Halogens: Cl2, Br2, I2

F F+ F F F F


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Lewis Structures

Diatomic Gases: HF

Same for HCl, HBr, HI. Molecules are diatomics as need only

1e– to complete octet Separate molecules

Gas in most cases because very weak intermolecular forces

H F+ H F H F


Your Turn!How many electrons are required to

complete the octet around nitrogen, when it forms N2 ?

A. 2B. 3C. 1D. 4E. 6

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Lewis Structures Many nonmetals form more than 1 covalent bond

Needs 4 e-’sForms 4 bonds



methane ammonia water

C

H

CH H

H H

CH H

H

NH H

H

NH H

H

O H

H

O H

H

ON


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Multiple Bonds

Single Bond Bond produced by sharing one pair

of e–’s between 2 atoms

Many molecules share more than one pair of e–’s between 2 atoms Multiple bonds


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Double Bonds

2 pairs of e–’s shared between 2 atoms

Ex. CO2

Triple bond 3 pairs of e–’s shared between 2

atomsEx. N2

O OC C OO C OO

N N N N N N


Your Turn!Which species is most likely to have

multiple bonds ?A. COB. H2O

C. PH3

D. BF3

E. CH4

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Carbon Compounds Carbon-containing compounds

Exist in large variety Mostly due to multiple ways in which C can

form bonds

Functional groups Groups of atoms with similar bonding Commonly seen in C compounds

Molecules may contain more than 1 functional group


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Important Compounds of Carbon Alkanes

Hydrocarbons Only single bonds

Isomers Same molecular formula Different physical

properties Different connectivity

(structure)

HC

H

H

CC

H

H

C

H

H

H

H

H

HCC

H

H

C

C

H

H

H

H

HH

H

Ex.CH4 methaneCH3CH3 ethane CH3CH2CH3 propane

iso-butanebutane


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Hydrocarbons

Alkenes Contain at least

one double bond

Alkynes Contain at least

one triple bond

HCC

HH

H Ethylene (ethene)

HC

H

H

CC

H

H

C

HH

H

butene

HCCH acetylene (ethyne)

HC

H

H

CC

H

H

CH butyne


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Oxygen Containing Organics

Alcohols Replace H with

OH

Ketones Replace CH2 with

C=O Carbonyl group

H

CH O

H H

HCC

H

H

C

O

H

H

H

OC

H

C

H

H

H

H

H

methanol ethanol

acetone


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Carbonyl Group

Carbon in hydrocarbon with double bond to oxygen Aldehydes Ketones Carboxylic acids Amides

::


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Oxygen Containing Organics Aldehydes

At least 1 atom attached to C=O is H

Organic Acids Contains carboxyl

group COOH

Esters Alkyl group replaces

H on carboxylic acid

HCC

OH

H

H

acetaldehyde

acetic acid

methyl acetate(methylethanoate)

OCC

OH

H

H

H

OCC

OH

H

H

C

H

H H


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Nitrogen Containing Organics Amines

derivatives of NH3 with H’s replaced by alkyl groups

HNC

H

H

H

H

HCN

H

H

C

H

H

H

H

methylamine dimethylamine

Amides Contain -NH2 on carbon chain with carbonyl

acetamide (ethanamide)NCC

OH

H

H

H

H


Your Turn!How many isomers are there of butanol?A. noneB. 2C. 3D. 6E. 4

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Electronegativity and Bond Polarity

2 atoms of same element form bond Equal sharing of

e’s

2 atoms of different elements form bond Unequal sharing of

e’s


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Why?

1 atom usually attracts es more strongly than other

Result: Unbalanced distribution

of e density within bond e cloud tighter around Cl in HCl Slight + charge around H Slight – charge around Cl Not complete transfer of e from 1 atom to

another.


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Electronegativity and Bond Polarity

Leads to concept of Partial charges

+

H——Cl + on H = +0.17 on Cl = 0.17


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Polar Covalent Bond

Aka Polar bond Bond that carries partial + and –

charges at opposite ends Bond is dipole

2 poles or 2 charges involvedPolar Molecule Molecule has partial + and – charges

at opposite ends due to a polar bond

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Dipole Moment Quantitative measure of extent to which bond

is polarized. Dipole moment = Charge on either end

distance between them μ = q × r Units = debye (D) 1D = 3.34 × 10–30 C·m (Coulombs·meter) The size of the dipole moment or the

degree of polarity in the bond depends on the differences in abilities of bonded atoms to attract e’s to themselves


Table 9.3 Dipole Moments and Bond Lengths for Some

Diatomic Molecules

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Electronegativity () Relative attraction of atom for es in bond Ability of bonded atom to attract es to

itself Quantitative basis

Table of electronegativities - fig 8.5 Difference in electronegativity

= estimate of bond polarity = |1 2|

Ex. N—H Si—F + +


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Electronegativity Table


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Electronegativities Must know for 2nd row and H. (H) ~ 2.0 (F) ~ 4.0 by 0.5 for each element as go to left

H

2

Li Be B C N O F

1 1.5 2 2.5 3 3.5 4

P S Cl

2 2.5 3.0


Learning Check Calculate the bond polarity of the following

HCl

CO

CH

LiF

BeO

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(H)~2 (Cl)~3 = 3 – 2 = 1

(C)~2.5 (O)~3.5 = 3.5 – 2.5 = 1 (C)~2.5 (H)~2 = 2.5 – 2 = 0.5 (Li)~1 (F)~4 = 4 – 1 = 3

(Be)~1.5 (O)~3.5 = 1.5 – 3.5 = 2


Your Turn!Which of the following species has the

least polar bond?

A. HClB. HFC. HID. HBr

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Trends in Electronegativity from left to right

across period as Zeff from top to bottom

down group as n

Ionic and Covalent Bonding 2 extremes of bonding Actual is usually somewhere in

between.


Using Electronegativities

Difference in electronegativity Measure of ionic character of

bond

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BA


Using Electronegativities Nonpolar Covalent Bond

No difference in electronegativity Ionic Character of bond

Degree to which bond is polar > 1.7 means mostly ionic

>50% ionic More electronegative element almost completely

controls e

< 0.5 Means almost purely covalent Nonpolar; < 5% ionic

0.5 < < 1.7 polar covalent68


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Result

Elements in same region of Periodic Table i.e., 2 nonmetals Have similar ’s Bonding more covalent

Elements in different regions of Periodic Table i.e., metal and nonmetal Have different ’s Bonding predominantly ionic


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Reactivities of Elements Related to

Electronegativities Parallels between and its reactivity Tendency to undergo redox reactions


Reactivities of Elements Related to

Electronegativities Metals Low , means easy to oxidize (groups I and

IIA) High , means hard to oxidize (Pt, Ir, Rh, Au,

Pd) Reactivity across row as

Nonmetals Oxidizing power of nonmetal (how easily

reduced) parallels . Oxidizing power across row as Oxidizing power down column as

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Drawing Lewis Structures

Very useful Way of diagramming structure Used to describe structure of

molecules Can be used to make reasonable

accurate predictions of shapes of molecules


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Drawing Lewis Structures

Not all molecules obey the Octet Rule. Holds rigorously for 2nd row elements like

C, N, O, and F B and Be sometimes have less than octet

BeCl2, BCl3 2nd row can never have more than 8 e’s 3rd row and below, atoms often exceed

octet Why?

n = 3 shell can have up to 18 e’s as now have d orbitals in valence shell


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Method for Drawing Lewis Structure

1. Decide how atoms are bonded Skeletal structure = arrangement of

atoms. Central atom

Usually given first Usually least electronegative

2. Count all valence es. (All atoms)3. Place 2 es between each pair of

atoms Draw in single bonds


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Method for Drawing Lewis Structure

4. Complete octets of terminal atoms (atoms attached to central atom) by adding es in pairs

5. Place any remaining e’s on central atom in pairs

6. If central atom does not have octet Form double bonds If necessary, form triple bonds


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Ex. N2F2

2 N = 2 5e = 10 e 2F = 2 7e = 14 e

Total = 24 e

single bonds 6 e 18 e

F lone pairs 12 e 6 e

N electrons 6 e 0 e

F N N F

F N N F

F N N F

Skeletal Structure

Complete terminal atom octets

Put remaining es on central atom


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Ex. N2F2

Not enough electrons to complete N octets

Must form double bond between N’s to satisfy both octets.

F N N FF N N F


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Ex. SiF4

1 Si = 1 4e = 4 e 4F = 4 7e = 28 e

Total = 32 e


F lone pairs 24 e 0 e

Skeletal Structure

Complete terminal atom octets

F

SiF F

F

F

SiF F

F


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Ex. H2CO3

CO32 oxoanion, so C central, and

O’s around, H+ attached to two O’sO

CO OH H

1 C = 1 4e = 4 e

3 O = 3 6e = 18 e

2 H = 2 1e = 2 e

Total = 24 e

single bonds 10 e

14 e

O lone pairs 14 e

0 e

O

CO OH H

But C only has 6 e


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Ex. H2CO3 (cont.)

Too few e Must convert 1

of lone pairs on O to 2nd bond to C

Form double bond between C and O

O

CO OH H

O

CO OH H


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Ex. PCl5

PClCl

Cl

Cl

Cl

PClCl

Cl

Cl

Cl

1 P = 1 5 e = 5 e 5 Cl = 5 7e = 35 e

Total = 40 e single bonds 10 e

30 e

Cl lone pairs 30 e 0 e

P has 10 e

OK as 3rd row element Can expand its shell


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Ex. BBr3

B has only 6 e

Does not form double bond Has incomplete octet

1 B = 1 3 e = 3 e 3 Br = 3 7e = 21 e

Total = 24 e single bonds 6 e

18 e

Br lone pairs 18 e 0 e

BBr

Br

Br

BBr

Br

Br


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Ex. CO2CO O

CO O

1 C = 1 4e = 4 e 2 O = 2 6e = 12 e

Total = 16 e


O lone pairs 12 e 0 e

C only has 4 e

Must form 2 double bonds to O to complete C’s octet

3 ways you can do this

CO O CO O

CO O OO O

C OO C OO

Which of these is correct?

Need another criteria Come back to this


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Lewis Structures—Theory Lewis Structures meant to describe how

atoms share es in chemical bonds Theory

What experimental observations back up this theory?

Look at properties related to number of e pairs shared between 2 atoms


Your Turn!What is wrong with the following

structure?

A. Too few total electronsB. Too many total electronsC. Lack of octet around nitrogenD. Too many electrons around the N atomE. Structure is correct as written

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O N O


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Review of Covalent Bond Properties

Bond Length Distance between nuclei of bonded

atomsBond Energy Energy required to separate bonded

atoms into neutral particlesBond Order Number of pairs of es shared between

2 atoms Measure of amount of e density in bond


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Bond Order, Bond Length and Bond Energy

Greater Bond Order = Greater e density Nuclei held together more tightly Larger bond energy, D

Larger D means Nuclei drawn closer together Shorter bond length

As bond order , bond length , and bond energy Assumes comparing bonds between same 2

elements


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1 S = 1 6e = 6 e

4 O = 4 6e = 24 e

2 H = 2 1e = 2 e

Total = 32 e

single bonds 12 e

20 e

O lone pairs 20 e

0 e

Ex. H2SO4

O

SO OH H

O

O

SO OH H

O

• n=3, has empty d orbitals• Could expand it's octet• Could write structure with double bonds.

O

SO OH H

O


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How Do We Know Which is Accurate?Experimental evidence

In this case bond lengths from X-ray data S—O bonds (no H attached) are shorter 142

pm S—O—H, S—O longer 157 pm Indicates that 2 bonds are shorter than

other 2 Structure with S=O for 2 O’s without H’s is

more accurate Preferred Lewis Structure

Even though it seems to violate octet rule unnecessarily


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Need Criteria to Predict When This Occurs

Covalent bond = shared e pair If es shared equally

Each atom “owns” ½ of e pair or 1 e When “break” bond

½ of these es going to each atom in bond So for S

have 4 bonds Have 4 es 2 es less than 6 es, S has as atom

In bookkeeping sense S has 2+ charge if it obeys octet rule


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Formal Charge (FC)

Apparent charge on atom Bookkeeping method Does not represent real chargesFC = # valence e # lone pair e ½ (#

bonding e)FC = # valence e [# bonds to an atom

+ # unshared e ] Indicate Formal charges by placing

them in circles around atoms


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FC = #valence e [#bonds to atom + # unshared e ]

Structure 1FCS = 6 (4 + 0) = 2

FCH = 1 (1 + 0) = 0

FCO(s) = 6 (1 + 6) = 1FCO(d) = 6 (2 + 4) = 0

Structure 2FCS = 6 (6 + 0) = 0

FCH = 1 (1 + 0) = 0

FCO(s) = 6 (2 + 4) = 0

FCO(d) = 6 (2 + 4) = 0

O

SO OH H

O

O

SO OH H

O

+2

-1

-1


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H2SO4

No formal charges on any atom in Structure 2

Conclusion: When several Lewis Structures are possible Those with smallest formal charges

Most stable Preferred

Most Stable Lewis Structure1.Least number of FC's best2.All FC 13.Any negative FC on most electronegative

element


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Can Use FC to Explain B Chemistry BCl3

Why doesn’t a double bond form here? FCB = 3 – 0 – 3 = 0

FCCl = 7 – 6 – 1 = 0 All FC's = 0 so stable, doesn't need to

form double bond


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CO2 Which Structure is Best Use FC to determine

which structure is best

FCC = 4 (4 + 0) = 0

FCO(s) = 6 (1 + 6) = 1

FCO(d) = 6 (2 + 4) = 0

FCO(t) = 6 – (3 + 2) = +1

Central structure Best All FC’s = 0

CO O

C OO

CO O

1

-1

+1

+1


Your Turn!What is the formal charge on Xe for the

following ?

A. +2, +4B. +2, +3C. +4, 0D. +4, +2

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O Xe O

O

O

O Xe O

O

O


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1 N = 1 5e = 5 e

3 O = 3 6e = 18 e

1 charge = 1 e

Total = 24 e

single bonds 6 e

18 e

O lone pairs 18 e

0 e

Resonance: When Single Lewis Structure Fails

NO

O

O

NO

O

O

N = 3 O = 3.5


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Ex. NO3

Lewis structure predicts 1 bond shorter than other 2

Experimental observation: All 3 N—O bond lengths are same All shorter than N—O single bonds

Have to modify Lewis Structure e can't distinguish O atoms

Can write 2 or more possible structures simply by moving where e are Changing placement of e

NO

O

O


99

What are Resonance Structures? Multiple Lewis Structures for single molecule

No single Lewis structure is correct Structure not accurately represented by any 1

Lewis Structure Actual structure = "average" of all possible

structures Double headed arrow between resonance

structures used to denote resonance

NO

O

O

NO

O

O

NO

O

O

1

+1

1

+1+1

1

11

1


100

Resonance Structures

Lewis structures assume e is localized between 2 atoms

In cases where need resonance structures, e are delocalized Smeared out over all atoms Can move around entire molecule to give

equivalent bond distances

Resonance Hybrid Way to depict

resonance delocalizationN

O

O

O


101

Ex. CO32

CO

O

O

CO

O

O

CO

O

O

1 N = 1 5e = 5 e

3 O = 3 6e = 18 e

1 charge = 1 e

Total = 24 e

single bonds 6 e

18 e

O lone pairs 18 e

0 e

C = 2.5 O = 3.5

C only has 6 e so form double bond


102

Three Equivalent Resonance Structures

All have same net formal charges on C and O’s FC = 1 on singly bonded O’s FC = O on doubly bond O and C

1 1

1

1

1

1

CO

O

OC

O

O

OC

O

O

O


103

Resonance Structures Not Always Equivalent

2 or more Lewis Structures for same compound may or may not represent e distributions of equal energy

How Do We Determine Which are Good Contributors?

1. All Octets are satisfied2. All atoms have as many bonds as possible3a. FC 13b. Any negative charges are on

electronegative atoms.4. As little charge separation as possible.


104

Drawing Good Resonance Structures

1. All must be valid Lewis Structures2. Only e are shifted

Usually double or triple bond and lone pair Nuclei can't be moved Bond angles must remain the same

3. Number of unpaired electrons, if any, must remain the same

4. Major contributors are the ones with lowest E (see above)

5. Resonance stabilization is most important when serves to delocalize charge onto 2 or more atoms


105

1 C = 1 4e = 4 e

1 N = 1 5e = 5 e

1 O = 1 6e = 6 e

1 charge = 1 e

Total = 16 e

single bonds 4 e

12 e

O lone pairs 12 e

0 e

Ex. NCO

N C O

N C O

N = 3 C = 2.5 O = 3.5

N C O

N C O

OCN


106

Ex. NCO

FCN = 5 – 2 – 3 = 0FCC = 4 – 0 – 4 = 0FCO = 6 – 6 – 1 = –1

FCN = 5 – 4 – 2 = –1FCC = 4 – 0 – 4 = 0FCO = 6 – 4 – 2 = 0 FCN = 5 – 6 – 1 = –2FCC = 4 – 0 – 4 = 0FCO = 6 – 2 – 3 = +1

OK

Best

NotAccept-able

2 +1

1

1

N C O

N C O

OCN


107

Resonance Stabilization

Actual structure is more stable than either single resonance structure

For Benzene The extra stability is ~146 kJ/mol Resonance Energy

Extra stabilization energy from resonance


Your Turn!Which of the structures below exhibit

resonance?

A. NO2

B. H2O

C. N3-

D. N2O (nitrogen is central atom)

E. CH3CH2CH

108


109

Coordinate Covalent Bonds

Ammonia Normal covalent bonds One electron from each atom shared

between the two

3 H + N N

H

H H


110

Coordinate Covalent Bond

Ammonium Ion H+ has no electrons N has lone pair Can still get 2es shared between

them

N

H

H H+ H+ N

H

H H

H

+


111

Coordinate Covalent Bond Both electrons of shared pair come from

just one of two atoms Once bond formed, acts like any other

covalent bond Can't tell where electrons came from

after bond is formed Useful in understanding chemical

reactions


112

Coordinate Covalent Bond

Especially boron (electron deficient molecule) reacts with nitrogen compounds that contain lone pair of electrons

N

H

H H+ B

Cl

Cl ClN

H

H B

H

Cl

Cl

Cl


113

Learning Check

Ex. What is the best Lewis Structure for HClO4?

XeF4

I3–

BrF5

Documents

Chapter 9: The Basics of Chemical Bonding