Basic Concepts Basic Concepts of Chemical of Chemical

BondingBonding

Chapter 8Chapter 8

Three Types of Chemical BondsThree Types of Chemical Bonds

• Ionic bondIonic bond– Transfer of electronsTransfer of electrons– Between metal and nonmetal ionsBetween metal and nonmetal ions

• Metallic bondMetallic bond

– Bonding electrons relatively free to Bonding electrons relatively free to movemove

• Covalent bondCovalent bond– Sharing of electronsSharing of electrons– Between nonmetal atomsBetween nonmetal atoms

H F FH

Polar covalent bond ≡ a covalent bond with greater electron density around one of the two atoms

electron richregion

electron poorregion e- riche- poor

+ -

Electronegativity: the ability of an atom in a molecule to attract electrons to itself

Fig 8.6 Electronegativities of the Elements

Figure 8.7

Polar Covalent Bonds

The greater the difference in electronegativity, the more polar is the bond.

Most polar Least polar

Table 8.3

Fig 8.9

1. Sum up all valence electrons. Add 1 for each negative Add 1 for each negative charge. Subtract 1 for each positive charge.charge. Subtract 1 for each positive charge.

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

3. Complete octets of atoms connected to central atom.

4. Place remaining electrons on central atom.

5. If not enough electrons to give central atom an octet, try multiple bonds.

Writing Lewis Structures (p 314)

Write the Lewis structure of nitrogen trifluoride (NF3).

Step 2 – N is less electronegative than F, put N in center

F N F

F

Step 1 – 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

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

Step 2 – C is less electronegative than O, put C in center

O C O

O

Step 1 – 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

Formal Charges

Formal charge - 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.

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

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 best Lewis structure…

…is the one with the fewest charges

…puts a negative charge on the most electronegative atom.

Formal Charges

Draw Lewis structure for ozone, O3

or

But this is at odds with the true, observed structure of ozone, in which…

…both O−O bonds are the same length:

Resonance

Just as green is a synthesis of blue and yellow…

…ozone is a synthesis of these two resonance structures.

Resonance structure - 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?

e.g., ozone

2- 2-2-

Resonance

The organic compound benzene, C6H6, has two resonance structures:

It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.

Exceptions to the Octet Rule

Too few electrons

H HBeBe – 2e-

2H – (2)1e-

4e-

BeH2

BF3

B – 3e-

3F – (3)7e-

24e-

F B F

F

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

Total = 24

Exceptions to the Octet Rule

Odd number of electrons

N – 5e-

O – 6e-

11e-

NO N O

Too many electrons (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

Average Bond Enthalpies

Bond Type

Bond Enthalpy

(kJ/mol)

C‒C 348

C=C 614

C≡C 839

C‒N 293

C=N 615

C≡N 891

Bond Enthalpies

Single bond < Double Bond < Triple Bond

ChemicalBonding

These are average bond enthalpies, not absolute bond enthalpies The C−H bonds in methane, CH4, will be a bit different than

the C−H bond in chloroform, CHCl3

Table 8.4 Average bond Enthalpies (kJ/mol)

Fig 8.14 Estimating Enthalpies of Reaction

Hrxn = (bond enthalpies of bonds broken) -

(bond enthalpies of bonds formed)

In this example:

• one C-H bond and one Cl-Cl bond are broken

• one C-Cl and one H-Cl bond are formed

Fig 8.14 Estimating Enthalpies of Reaction

CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)

So,

H = [D(C−H) + D(Cl−Cl)] − [D(C−Cl) + D(H−Cl)]

= [(413 kJ) + (242 kJ)] - [(328 kJ) + (431 kJ)]

= (655 kJ) - (759 kJ)

= -104 kJ

Fig 8.14 Estimating Enthalpies of Reaction

CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)

Bond Type

Bond Length

(pm)

C‒C 154

C=C 133

C≡C 120

C‒N 143

C=N 138

C≡N 116

Lengths of Covalent Bonds

Bond Lengths

Triple bond < Double Bond < Single Bond