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RVCC Fall 2009 CHEM 103 – General Chemistry I. Chapter 8: Ionic and Covalent Bonding. Chemistry: The Molecular Science, 3 rd Ed. by Moore, Stanitski, and Jurs. Bonding – What holds atoms together?.

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chapter 8 ionic and covalent bonding

RVCC Fall 2009

CHEM 103 – General Chemistry I

Chapter 8:Ionic and Covalent Bonding

Chemistry: The Molecular Science, 3rd Ed.

by Moore, Stanitski, and Jurs

bonding what holds atoms together
Bonding – What holds atoms together?

Octet rule: To form bonds, atoms gain, lose, or share e- to achieve a valence shell of 8 (or isoelectronic with a noble gas).

  • Ionic bond –

an electrostatic attraction between a cation and an anion that forms when electrons transfer from one atom to another.

  • Covalent (Molecular) bond –

the net attractive force that results from the sharing of electrons between atoms.

ionic bonds

:

.

:

Cl

:

Ionic Bonds

An ionic bond is formed by the transfer of electrons from one atom (metal with low EA) to another (nonmetal with high EA). The resultant ions are held together by electrostatic attraction.

Na

.

[Ne]3s1 [Ne]3s23p5

Na+ Cl-

[Ne] [Ne]3s23p6 = [Ar]

Each atom has satisfied the octet rule.

ionic bonds1

:

F

.

:

:

:

F

Mg

.

.

.

:

:

:

:

:

:

:

:

:

:

Ionic Bonds

MgF2

-

-

F

Mg

2+

F

slide5

Ionic Compounds - Properties

  • crystalline
  • high melting point
  • high boiling point
  • soluble in water
  • electrolytes

Crystal Lattice

  • hard
  • brittle
covalent bonding g n lewis 1916

Attraction Stable bond Repulsion

Covalent Bonding - G.N. Lewis (1916)

Some atoms share e- to form bonds. When two nonmetals bond, they often share electrons since they have similar attractions (EA) for them. This sharing of valence electrons is called the covalent bond.

  • Number of bonds = Number shared e- pairs.
covalent bonds
Covalent Bonds

H2

.

.

:

+

H

H

H

H

single covalent bonds
Single Covalent Bonds

Lewis structures:

show ALL valence electrons

dot = 1 e- line = 1 pair of e-

single bond- one shared pair of e-

H − H

H

H

single covalent bonds1

H

H – C – H

H

..

..

..

..

..

..

..

F – N – F

F

..

..

..

..

..

H – O – H

..

..

..

..

H – F

H – F

Single Covalent Bonds

# of e- shared

to form an octet

(8-A group#)

# of e- shared

Group # of to form an octet Example

valence e- (8 - A group#)

4A 4 4 C in CH4

5A 5 3 N in NF3

6A 6 2 O in H2O

7A 7 1 F in HF

..

lewis structures

:

:

:

:

H

Cl

Lewis Structures

Lewis electron-dot formulas or Lewis structures.

bonding pair

lone pairs

:

:

:

bonding pair

H

Cl

:

lone pair

  • An electron pair is either a bonding pair (shared between two atoms) or a lone pair (an electron pair that is not shared).
multiple bonds

H

H

:

:

:

:

C

C

:

:

H

H

Multiple Bonds

In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared.

  • It is possible to share more than one pair. A double bond involves the sharing of two pairs between atoms.

C has octet.

H OK with 2.

or

multiple bonds1

:::

H

H

C

C

or

:

:

Multiple Bonds

Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms.

Elements that form multiple bonds: C, O , N, S

the procedure
The Procedure
  • Using the molecular formula, count the total number of valence electrons available (bonding + lone pairs).
    • Valence electrons for each atom corresponds to group #
    • Adjust for charge (add electron for each minus, delete electron for each plus)

2(1) + 6

3(1) + 6 - 1

the procedure1
The Procedure
  • Make a skeleton by connecting the atoms with single bonds only. When connecting atoms, remember…
    • Put the least electronegative atom in the center. (Usually the first listed in the chemical formula.)
    • Hydrogen is ALWAYS a terminal atom. More electronegative atoms are terminal (F, O…)
    • Make the structure symmetric.

4 pairs 4 pairs

2 pairs left 1 pair left

the procedure2
The Procedure
  • Put the left over electrons as lone pairs, preferably on the more electronegative atoms Is the octet rule satisfied?
    • If YES, then you’re done…
the procedure3
The Procedure
  • If you are electron-deficient(not enough electrons to complete an octet), then some atoms must share more than two electrons. “If you have a lone pair, make those two atoms share.”
    • Ex. C2H4
the procedure4
The Procedure
  • If you have excess electrons, at least one atom must have an expanded valence
    • Must be element from third period or lower
    • Usually the central atom
    • e.g. SF4
slide20

General Rule

  • The total number of valence electrons on an atom (from bonds & lone pairs) cannot exceed that atom’s maximum valence.
    • First period: 2 electrons (s)
    • Second period: 8 electrons (s,p)
    • Third period & below: prefer to have 8, but can expand when necessary (s,p,d)
writing lewis dot formulas
Writing Lewis Dot Formulas

20 e- total or 10 pairs

SCl2

8 left

0 left

:

:

:

:

:

Cl

S

Cl

:

:

:

writing lewis dot formulas1

O

:

:

:

:

:

:

C

:

:

:

Cl

Cl

Writing Lewis Dot Formulas

24 e- total 12 pairs

COCl2

9 left

0 left

Note that the carbon has only 6 electrons.

writing lewis dot formulas2

9 e- left

O

:

:

:

:

:

:

C

:

:

:

Cl

Cl

Writing Lewis Dot Formulas

COCl2

12 pairs

0 e- left

To fulfill the octet rule…

“If you have a lone pair, make those two atoms share!”

writing lewis dot formulas3

:

:

:

:

:

:

Writing Lewis Dot Formulas

COCl2

24 e- total

18 e- left

:

:

O

0 e- left

C

Cl

Cl

Note that the octet rule is now obeyed.

slide25

Writing Lewis Dot Formulas

Practice

N2 SF4

O2 ClO3-

HCN ClO2-1

PO4-3 NO3-1

slide26

Question

no

no

no

no

no

  • First evaluate the total valence electrons:
  • 24 e-
  • 26 e-, wrong
  • 24 e-, looks OK
  • 24 e-, one F has too many
  • 24 e-, N not enough
  • 24 e-, but least electronegative has to be in the center
  • 24 e-, no bond between two N.
exceptions to the octet rule
Exceptions to the Octet Rule

Although many molecules obey the octet rule, there are exceptions where the central atom has less or more than eight electrons.

Incomplete octet – B, H

Boron has 3 valence electrons

BF3

.. ..

:F – B – F:

.. ..

:F:

..

exceptions to the octet rule1

:

:

:

F :

F :

: F

:

:

:

:

: F :

: F :

:

Exceptions to the Octet Rule

If a nonmetal is in the third period or greater it can accommodate as many as twelve electrons as the central atom.

PF5

P

exceptions to the octet rule2

:

:

:

:

F :

F :

: F

: F

:

:

:

:

Exceptions to the Octet Rule

In sulfur tetrafluoride, SF4, the sulfur atom must accommodate two extra lone pairs for a total of 5 electron pair (10 electrons)

:

S

formal charge and lewis structures
Formal Charge and Lewis Structures

In certain instances, more than one feasible Lewis structure can be illustrated for a molecule. For example,

:

:

H

C

N

H

N

C

or

The concept of “formal charge” can help us decide which structure is correct.

formal charge and lewis structures1
Formal Charge and Lewis Structures

formal charge = valence e- before bonding– valence e- after bonding

= valence e- - [1/2 bonding e- + lone pair e-]

:

:

H

C

N

H

N

C

or

H: 1-½(2) = 0

H: 1-½(2) = 0

C: 4 - ½(8) = 0

C: 4 – (½(6)+2) = -1

N: 5 – (½(8) + 2) = 0

N: 5 – (½(8)) = +1

formal charge and lewis structures2

formal charges

:

:

H

C

N

H

N

C

or

0

0

0

0

+1

-1

Formal Charge and Lewis Structures
  • Smaller formal charges are more favorable
  • More electronegative (or higher EA) atom should have negative formal charges
  • Like charges should not be on adjacent atoms
  • Net formal charge should be the overall charge on the molecule/ion.
practice
Practice

Determine the most stable structure for dinitrogen oxide. (All structures have 16 valence electrons.)

N=N=O N-N≡O N ≡ N - O

-1 +1 0 -2 +1 +1 0 +1 -1

formal charge= valence e- - [1/2 bonding e- + lone pair e-]

practice formal charge
Practice - Formal Charge
  • Which structure is correct?

:

O

N

Cl

O

N

Cl

or

0

0

0

-1

0

+1

delocalized bonding resonance

:

:

:

:

:

:

O

O

:

:

:

:

O

O

O

O

:

:

Delocalized Bonding: Resonance

The structure of ozone, O3, can be represented by two different Lewis electron-dot formulas.

The bond lengths for the above structures are:

O – O 132 pm O = O 112 pm

However, experiments show that both bonds are identical.

delocalized bonding resonance1

O

O

O

Delocalized Bonding: Resonance

According to theory, one pair of bonding electrons is spread (delocalized) over the region of all three atoms.

  • In fact, the actual bond length is 127.8 pm (in between 132 and 112pm).
  • The actual molecule is a hybrid or composite structure and not different structures that change back and forth… although, we often represent it that way.
delocalized bonding resonance2

O

H O N

O

O

H O N

O

O

H O N

O

Delocalized Bonding: Resonance

Lewis resonance structures, have the same atoms in the same positions. Only an electron pair position is different.

resonance structures
Resonance Structures

Which pair does NOT represent resonance structures?

O=S O O S=O

S=S O S O=S

slide39

-2

O

O C

O

-2

O

O C

O

Resonance Structures

Draw resonance structure(s) for the following:

-2

O

O C

O

slide40
“All covalent bonds are created equal but some are more equal than others.”

(We assumed equal sharing when we calculated formal charge.)

electronegativity
Electronegativity…

…is a measure of the ability of an atom in a molecule to draw bonding electrons to itself when bonded.

Periodic Trend - Electronegativity

increases

across a period

decreases

down a group

slide42

Electronegativity

Notice, there are NO values for

EN for the noble gases.

types of bonds
Types of Bonds

0.9 3.0

Na+ Cl-

2.5 2.1

C - H

1.6 1.6

Zn Zn

  • Ionic:
    • ΔEN >1.8
    • electron transfer
  • Covalent:
    • ΔEN <1.8
    • electron sharing
  • Metallic:
    • electron-sea model or band theory
covalent bonds en 1 8
Covalent Bonds (EN<1.8)
  • Non-polarcovalent- ΔEN = 0 – 0.5
  • Examples: H-H, Cl-Cl, C-H bonds
  • Polarcovalent - ΔEN = 0.5 – 1.8
  • Examples: H-O, C-Cl, C-O bonds
bond polarity
Bond Polarity

In HCl we have a partial negative charge on the chlorine (denoted d-)and a partial positive charge on the hydrogen (denoted d+)

The bond is polar covalent.

d+

d-

:

H :Cl:

:

bond polarity1
Bond Polarity

Arrange the following bonds from the most to the least polar:

HH, HCl, HF, HI, HBr

Compare the electronegativity of Cl, F, I and Br:

Least Most

Electronegative Electronegative

I Br Cl F

Determine polarity:

HH HI HBr HCl HF

non polar most polar

practice1
Practice

Which of the following bonds in each pair are more polar?

C-S or C-O

Cl-Cl or O=O

N-H or C-H

bond length
Bond Length

Bond length (or bond distance) is the distance between the nuclei of two bonded atoms (the sum of atomic radii).

slide51

Bond Length

The size of atoms that form the bond determine the length.

C - N 147 pm

C - C 154 pm

C - P 187 pm (P, period 3)

bond length multiple bonds
Bond Length – Multiple Bonds

As the electron density between atoms increases the bond

lengths decrease; the atoms are pulled together more strongly.

decrease

bond enthalpy

Bond length Bond enthalpy

C

C

154 pm

346 kJ/mol

C

C

134 pm

602 kJ/mol

C

C

120 pm

835 kJ/mol

Bond Enthalpy

Bond Enthalpy – the enthalpy change that occurs when the bond between two bonded atoms in the gas phase is broken and the atoms are separated completely at constant pressure.

As the electron density between two atoms increases, the bond

gets shorter and stronger.

NaCl

lattice energy

-786 kJ/mol

MgO

lattice energy

-3791 kJ/mol

bond enthalpy1
Bond Enthalpy

Bond Enthalpies can be used to calculate the standard enthalpies of reaction (gas phase, STP)

DHº = ∑[(moles of bonds) × D(bonds broken)] -

∑[(moles of bonds) × D(bonds formed)]

DHº - standard enthalpy of reaction

bond enthalpy2
Bond Enthalpy

Estimate the DHº for the following reaction:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

:

:

:

:

:

+ 2 :O = O:  : O = C = O: + 2H – O - H

:

4 C – H bonds 1 O = O bond 2 C = O bond 2 H – O bond

per molecule per molecule per molecule per molecule

slide56

:

:

:

:

:

+ 2 :O = O:  : O = C = O: + 2H – O - H

:

4 C – H bonds 1 O = O bond 2 C = O bonds 2 H – O bonds

per molecule per molecule per molecule per molecule

4 C – H bonds 2 O = O bonds 2 C = O bonds 4 H – O bonds

DHº = ∑[(moles of bonds) × D(bonds broken)] -

∑[(moles of bonds) × D(bonds formed)]

= [4 × D(C-H) + 2 × D (O=O)] – [2 × D (C=O) + 4 × D(H-O)] =

[4 × 416 + 2 × 498] – [2 × 803 + 4 × 467] = -814 kJ