1

Bonding: General Concepts8※ Types of chemical bonds

Covalent bonding

Ex. H2E

0

458

(kJ/mol)

0.074 r (nm)

Repulsions of nucleus and es

Zero interaction at long distance

H H

r

H-H bond length

Two es shared by two Hs： covalent bonding

Ionic bondingEx. Na+ Cl: ionic forces operate

Coulomb attraction:

rQQ

E 2119 )nmJ1031.2(

charges

Polar covalent bondEx. H–F

Unequal sharing of e

Charge distribution occurs

H F

Stronger attraction for e

Fractional charge

A dipole

2

※ Electronegativity

The ability of an atom in a molecule to attract shared es

Linus Pauling (1901-1995)

A–B Expected BE =2

BEBE BBAA

expectedB-AactualB-A )BE()BE(

The greater EN difference, the greater value

Define EN(A) – EN(B) =Assign EN(F) = 4.0

102.0

Fluorine atom with the highest EN

EN for all elements can be determined

EN difference for A–B Bond type0 covalent

polar covalent

large ionic (A+, B)

A B

Higher EN

F

Increasing EN

DecreasingEN periodic table

3

※ Bond polarity and dipole moment

A dipole has a dipole moment ()

Will orient in an electric field

= Q·R

charge (c)

distance (m) Unit: debye (D)1D = 3.336 × 1030 cm

For polyatomic molecules

OH

H

net dipole

C OO

The center of all positive atoms and the center of all negative atoms do not merge

: net dipole = zero

※ Ions: Electron configurations and sizes

Ionic compoundsIn general A M+

non-metal main group metal

Non-metal: gains e

Metal: loses ereach noble gas configuration

Ex. Ca: [Ar]4s2

O: [He]2s22p4Ca2+: [Ar]O2: [He]2s22p6 = [Ne]

4

◎ Sizes of ionsdetermines the structure and stability of ionic compounds

Negative ions: larger than parent (gain e)Positive ions: smaller than parent (lose e)

Trend: Down a group larger

Ex. Li+ Be2+ O2 F

60 31 140 136 (pm)Na+

95

◎ The energy

M+(g) + X(g) M–X(s) energy released = lattice E

OverallEx.

Formation of an ionic compound energy depends on many factors difficult to predict qualitatively

Lattice E = )( 21

rQQ

k

Li(s) + 1/2 F2(g) LiF(s)

Li(g)

F(g)

Li+(g)ionization E

F(g)electron affinity

+

latticeenergy1/2 BE of F2

= 78.9160.7

520.5

328.0

1049.0

616.9 Unit:kJ/mol

At 298 K

LEtheo

= 966Born-Habercycle

5

※ Partial ionic character of covalent bonds

In the gas phase: no true ionic bond

Polar covalent bond

percent ioniccharacter

%100moment dipole calculatedmomentdipolemeasured

X+ Y

Ex. NaCl(g): 75%

In general: > 50% ionic solid

Ambiguity: NH4+Cl, Na2SO4

covalent bonded group

Definition of ionic compoundconducts electric current when melted

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※ The covalent bond: A model

Ex. CH4 C + 4H requires 1652 kJ

CH4 is more stable than the discreet atomsby 1652 kJ

Consider forces exist between C and 4H – bonding

Consider localized C–H bonds

To break each bond = = 413 kJ 4

1652

Ex. CH3Cl C + 3H + Cl requires 1578 kJ

3 C–H bonds, 1 C–Cl bond

1578 – 3(413) = 339 kJ/mol

Strength of C–Cl bond

Assumption: bond strength does not vary with structures

★ The covalent bond modelConsidering electrons are localized between two atoms

7

※ Covalent bond energiesand chemical reactions

Fact 1 HCH4(g) CH3(g) + H(g) 435 kJ/molCH3(g) CH2(g) + H(g) 453 CH2(g) CH(g) + H(g) 425 CH (g) C(g) + H(g) 339

total = 1652Average: 413

BEs are structurally dependent

Average BEs are used

Bond types

es sharedsingle 2double 4triple 6

Ex. BE (kJ/mol) bond length (Å)C–C 347 1.54C=C 614 1.34CC 839 1.20

8

◎ Bond energy and enthalpy

Ex. H2(g) + F2(g) 2HF(g)

2H 2F 2H, 2F

identical

Bond E difference = H

H = D (bonds broken) – D (bonds formed)

Bond E (+)

H = DH-H + DF-F – 2DH-F

= 432 + 154 – 2(565)= –544 kJ

Calculated from Hf: H = 2(–271) = –542 kJHf

o(HF)

※ The localized electron bonding model

(Valence bonding model)

A covalent bond is formed between two atomssharing electrons using atomic orbitals

Pairs of es localized on an atom: lone pairsPairs of es localized between atoms: bonding pairs

9

※ Lewis structures

Describe the arrangement of valence electrons

G. N. Lewis (18751946) observed:In most stable compounds the atoms achieve noble gas electron configuration

Molecules with covalent bonds

Duet ruleH, He

H:H He:

Octet ruleC, N, O, F: second row nonmetals

F F F Ne

Ex. H2O

H· H·

1 1 6

O

Total: 8 es

O HH

10

Ex. CO2

:C:

4 6 6

O

Total: 16

C OO

O

Does not obey octet rule

Try and error

C OO = C OO

C N

Ex. CN One extra charge

4 + 5 + 1 = 10 e

Not good

C N Not good

C N Not good

C N Good

C N

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Ex. N2

10 e isoelectronic with CN

N N

Ex. NH3

8 e

CH4

8 e

NO+

5 + 6 1 = 10 e

N HH

H

N HH

H

C HH

H

H

C HH

H

H

N O+

CF4

32 eC FF

F

F

C FF

F

F

※ Exceptions to the octet rule

◎ Less than 8

B

F

FF

Only 6 e

How about

B

F

FF

Fact: B is very reactive toward electron rich molecules

H NH

H

+ B

F

FFN BH F

H

H

F

F

12

◎ More than 8

S

F

F

FF

FF

12 e

Classical explanationuse of low lying empty d orbitals

(3d for sulfur)

PCl55 + (5 × 7) = 40 e

P

Cl

Cl

ClCl

Cl

10 e

Allow the central atom to expand its valency

Ex. XeO3

8 + (3 × 6) = 26 e Xe

O

O O

obeys

BeCl22 + (2 × 7) = 16 e

Be ClCl

Only 4 e

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※ Resonance

Ex. NO3

N

O

O O

Predict: a short double bondtwo long single bonds

Fact: Equal length (between a single and a double)

A new theory must be formulated

Actually there are three possible structures:

N

O

O ON

O

O ON

O

O O

We can consider the real structure as the avg. of the three– a resonance hybrid

Any single structure can not represent the real structure Each Lewis structure is called resonance structure

(exists only on paper) Represented as

N

O

O ON

O

O ON

O

O O

★ double headed arrow (not )

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Only allow electrons to move(nuclei stay at the same position)

Must be proper Lewis structuresEx. Carbon can not have five bonds

◎ Odd-electron molecule

Nitric oxide NO5 + 6 = 11

N O N O N O

Not good A radical with unpaired e

＊Keep the number of unpaired e the lowest

C CH

H H

HC C

H

H H

HNot a viable resonance structure

NO2

5 + 2(6) = 17

O N O ONO O N O ONO

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◎ Formal charge

Q: Structure of SO42 (32 e)

or

S

O

O

OO

2

S

O

O

OO

2

S

O

O

OO

2What about

10 e 12 e

How to decide?

Method: Estimate the charge

Oxidation state: O 2S +6

Over estimated

Only useful for bookkeeping

The concept of formal charge

Compare: number of e in a neutral atom·····Anumber of e in a bonded atom····B

When A = B no chargeA > B (+)A < B ()

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Define Formal charge = A – B

Assign lone pair es to the atom completelyDivide shared es equally

Ex.

S

O

O

OO

2 For OA = 6B = 6 + 2/2 = 7

lone pair es

shared es

Formal charge on O = 6 – 7 = –1

For S: A = 6 B = 0 + 8/2 = 4

lone pair esshared es

Formal charge on S = 6 – 4 = +2

S

O

O

OO

21

1

1

12 Overall charge

= (+2) + 4(1)

S O

S

O

O

OO

2 Single bonded O: 1Double bonded O: 6 – [4 + 4/2] = 0S: 6 – (10/2) = +1

S

O

O

OO

20

11

1

1

ok

17

S

O

O

OO

20

1 1

0

6 – (12/2) = 0

A better structure due to minimum formal charge

There are resonance structures

S

O

O

OO1 1

S

O

O

OO

1

1

S

O

O

OO

1

1etc.

Another school of thought:Keeping the octet rule ismore important S

O

O

OO1

1

1

1

2

Note: they are all resonance structures

Guidelines For 2nd row elements: never exceed octet rule Formal charges as low as possible Negative charge on more electronegative atoms

B

F

FF

1

1

Not good

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※ Molecular structures: The VSEPR model

To predict approximate 3-D structureA simple model:

Valence shell electron pair repulsion (VSEPR) model(useful for nonmetals)

The main postulation:minimizing electron pair repulsions

Bonding and nonbonding pairs

Ex. BeCl2Cl–Be–Cl : linear

(two pairs)

As far away as possible

BF3

B

F

FF

120o

C

H

HHH

109.5o

NHH

H

O

H

H

C

H

H

HH

90o

Trigonal planar(three pairs)

CH4

Tetrahedral(four pairs)

Cf.

Square planar(not good)

NH3

Based on the position of atoms: trigonal pyramid

H2OA bent structure(four pairs)

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Bond angles

C

H

HHH

109.5oN

HHH

107o

O

H

H

104.5o

lonepairs

0 1 2

Explanationlone pair is closer to the nucleus exerts large repulsion towards the other electron pairs

NHH

H

squeezed

Five pairs: PCl5

P

Cl

Cl

ClCl

Cl Trigonal bipyramidal

Six pairs: PCl6

P

Cl

Cl

Cl

Cl Cl

ClOctahedral

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Ex. XeF4

Xe

F

F

FF

Six pairs: octahedral arrangement

XeF

F F

F as far away as possible

Square planar shape

Ex. I3

I I I

Five pairs: trigonal bipyramidal arrangement

III

90o, 90o, 180o

I

I

I

120, 120, 120

I

I

I

90, 90, 120

Best(no 90o∠ repulsions)

linear

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Treat double or triple bonds as one effective pair

N

O

O O

Ex.

3 pairs Bent structure

SO2S

OOS

O O

In fact ∠O-S-O ~120o with little distortion long pair is satisfied with 120o∠