Chemical Bonding I:The Covalent Bond

Chapter 9

2

Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons thatparticpate in chemical bonding.

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

3

Lewis Dot Symbols for the Representative Elements &Noble Gases

4

Li + F Li+ F -

The Ionic Bond

1s22s11s22s22p5 1s2 1s22s22p6

[He] [Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -

LiF

Ionic bond: the electrostatic force that holds ions together in an ionic compound.

形成”離子對”

5

Ionic bond: Other example

1. LiO2

2. Mg3N2

6

Lattice energy increases as Q increases and/or

as r decreases.

Compound Lattice Energy (kJ/mol)

MgF2

MgO

LiF

LiCl

2957

3938

1036

853

Q: +2,-1

Q: +2,-2

r F- < r Cl-

Electrostatic (Lattice) Energy

E = kQ+Q-r

Q+ is the charge on the cation

Q- is the charge on the anionr is the distance between the ions

Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

E is the potential energy形成離子固體後 , 其穩定性如何 ?

7

Born-Haber Cycle for Determining Lattice Energy

Hoverall = H1 + H2 + H3 + H4 + H5° °°°°°

Lattice energy: + 1017 kJ

82+ leading to large attractive force than 1+

Chemical formula v.s lattice energy

MgCl2(1st ionic energy:738+ 2nd ionic energy: 1450 = 2188KJ/mol)

NaCl (2nd inoic energy: 4560KJ/mol) >> Lattice Energy so the NaCl2 is not formed!

9

A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.

Why should two atoms share electrons?

F F+

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairslone pairs

lone pairslone pairs

single covalent bond

single covalent bond

Covalent compound.

10

8e-

H HO+ + OH H O HHor

2e- 2e-

Lewis structure of water

Double bond – two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e-8e-

double bondsdouble bonds

Triple bond – two atoms share three pairs of electrons

N N8e- 8e-

N N

triple bondtriple bond

or

Octet rule: suitable for 2nd period(2s+ 2p[Ne])

Ex: C2H4

Ex: C2H2

11

Lengths of Covalent Bonds

Bond Lengths

Triple bond < Double Bond < Single Bond

12

Compare the difference between ionic compound and covalent compound

1. Covalent compound: gas, liquid or Low melting point solid2. Low solubility in water(can’t conductive)3. Example: NaCl(solid) and CCl4 (liquid)

13

H F FH

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms

electron richregion

electron poorregion e- riche- poor

+ -

14

Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.

Electron Affinity - measurable, Cl is highest

Electronegativity - relative, F is highest

X (g) + e- X-(g)

The attraction of single atom to its electrons

The attraction of multi atoms to its co-bonded electrons

15

The Electronegativities of Common Elements

當形成鍵結的 2原子之電負度差異很大時，易形成離子鍵。反之 , 當電負度差異小則易形成極性共價鍵。而電負度相近時，則形成非極性共價鍵。

16

Variation of Electronegativity with Atomic Number

過渡金屬有例外 ~

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Covalent

share e-

Polar Covalent

partial transfer of e-

Ionic

transfer e-

Increasing difference in electronegativity

Classification of bonds by difference in electronegativity

Difference Bond Type

0 Covalent

2 Ionic

0 < and <2 Polar Covalent

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Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the NN bond in H2NNH2.

Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic

H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Polar Covalent

N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent

19

Oxidation number(氧化數 ) and Electronegativity

Ex: NH3 (N 3- / H+ )

Actually, electronegativity is the way to define the oxidation number of a compound: the numbers of charge been transferred

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1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.

2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.

3. Complete an octet for all atoms except hydrogen

4. If structure contains too many electrons, form double and triple bonds on central atom as needed.

Writing Lewis Structures

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Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F, put N in center

F N F

F

Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)

5 + (3 x 7) = 26 valence electrons

Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

22

Write the Lewis structure of the carbonate ion (CO32-).

Step 1 – C is less electronegative than O, put C in center

O C O

O

Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e-

4 + (3 x 6) + 2 = 24 valence electrons

Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

Step 5 - Too many electrons, form double bond and re-check # of e-

2 single bonds (2x2) = 41 double bond = 4

8 lone pairs (8x2) = 16Total = 24

23

Two possible skeletal structures of formaldehyde (CH2O)

H C O HH

C OH

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

對中性分子，所有原子的形式電荷總合為 0，對於離子，形式電荷為離子的帶電荷

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H C O HC – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-

2 single bonds (2x2) = 41 double bond = 4

2 lone pairs (2x2) = 4Total = 12

formal charge on C = 4 -2 - ½ x 6 = -1

formal charge on O = 6 -2 - ½ x 6 = +1

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

-1 +1

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C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-

2 single bonds (2x2) = 41 double bond = 4

2 lone pairs (2x2) = 4Total = 12

HC O

H

formal charge on C = 4 -0 - ½ x 8 = 0

formal charge on O = 6 -4 - ½ x 4 = 0

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

0 0

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Formal Charge and Lewis Structures

1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1 HC O

H

0 0

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A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.

O O O+ -

OOO+-

O C O

O

- -O C O

O

-

-

OCO

O

-

-

What are the resonance structures of the carbonate (CO3

2-) ion?

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The Benzene (C6H6)structure The measured bonding length is 140 pm, less than C-C(154 nm) while longer than C=C(133 pm)

分子的表現行為可以看成是共振結構的加成結果，而非兩種結構的來回切換。

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Exceptions to the Octet Rule

The Incomplete Octet

H HBeBe – 2e-

2H – 2x1e-

4e-

BeH2

BF3

B – 3e-

3F – 3x7e-

24e-

F B F

F

3 single bonds (3x2) = 69 lone pairs (9x2) = 18

Total = 24

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NH3+BF3H3N-BF3

Coordinate covalent bond(配位共價鍵 ):價電子完全由一原子提供來配對

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Exceptions to the Octet Rule

Odd-Electron Molecules

N – 5e-

O – 6e-

11e-

NO N O

The Expanded Octet (central atom with principal quantum number n > 2)

SF6

S – 6e-

6F – 42e-

48e-

S

F

F

F

FF

F

6 single bonds (6x2) = 1218 lone pairs (18x2) = 36

Total = 48

(奇墊子分子 : 自由基 radical)

EX: NO2

EX: SCl2 still satisfy the Octet rule

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Expanded Octet

XeF4: leave 4 lone pairs to the Xe atom

Margin Art 9.14

Octet

Expanded Octet

2nd 週期後的元素因填入 3d軌域，價電子會超過八個

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The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond enthalpy.

H2 (g) H (g) + H (g) H° = 436.4 kJ

Cl2 (g) Cl (g)+ Cl (g) H° = 242.7 kJ

HCl (g) H (g) + Cl (g) H° = 431.9 kJ

O2 (g) O (g) + O (g) H° = 498.7 kJ O O

N2 (g) N (g) + N (g) H° = 941.4 kJ N N

Bond Enthalpy

Bond Enthalpies

Single bond < Double bond < Triple bond

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Average bond enthapy in polyatomic molecules

H2O (g) H (g) + OH (g) H° = 502 kJ

OH (g) H (g) + O (g) H° = 427 kJ

Average OH bond enthalpy = 502 + 427

2= 464 kJ

36

Bond Enthalpies (BE) and Enthalpy changes in reactions

H° = total energy input – total energy released= BE(reactants) – BE(products)

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

endothermic exothermic

37

Use bond enthalpies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g)

H° = BE(reactants) – BE(products)

Type of bonds broken

Number of bonds broken

Bond enthalpy (kJ/mol)

Enthalpy change (kJ/mol)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond enthalpy (kJ/mol)

Enthalpy change (kJ/mol)

H F 2 568.2 1136.4

H° = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ/mol

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9.25 Use the Born-Haber cycle outlined in Section 9.3 for LiF to calculate the lattice energy of NaCl. [The heat of sublimation of Na is 108 kJ/mol and (NaCl) = −411 kJ/mol. Energy needed to dissociate mole of Cl2 into Cl atoms = 121.4 kJ].

9.17 Use Lewis dot symbols to show the transfer of electrons between the following atoms to form cations and anions: (a) Na and F, (b) K and S, (c) Ba and O, (d) Al and N.

9.29 How many lone pairs are on the underlined atoms in these compounds? HBr, H2S, CH4.

9.49 Write Lewis structures for these species, including all resonance forms, and show formal charges: (a) (b) Relative positions of the atoms are as follows:

Ch9 HW

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2H2 (g) + O2 (g) 2H2O (g)

Use bond enthalpies to calculate the enthalpy change for: