04B_Covalent Bonding.pdf

Covalent Bonding

Bonding between two or more nonmetal atoms

Covalent Bond Example

H

F

hydrogen

atom

Fluorine

atom

H F

Naming Binary Covalent

Compounds



Order of Elements

Using Prefixes

Which element comes first in the

formula?

 The element with the lowest

electronegativity

value will always be written first in the

formula.

 _{Carbon monoxide is}

written CO not OC since 2.5 < 3.5

C 2.5 N 3.0 O 3.5 F 4.0 H 2.1 Si 1.8 P 2.1 S 2.5 Cl 3.0 Br 2.8 I 2.5

Using prefixes to tell how many

of each element is present.



Prefixes:

• 1 mono • 2 di

• 3 tri • 4 tetra • 5 penta • 6 hexa

P

2

O

5

Special case: when the

first

prefix

is mono, leave it off.



CO is carbon monoxide

leave off the

mono in front of first element C



SiF

is silicon tetrafluoride

not mono

silicon tetrafluoride



N

O is dinitrogen monoxide

don’t leave

off the mono on the second element!

The compounds are binary....so

use the -ide ending.

 H hydride

 _{C carbide}  N nitride

 _{P phosphide}  O oxide

 _{S sulfide}  F fluoride

 _{Cl chloride}

ide

Name to Formula is easy! Just

use the prefixes.



sulfur dichloride is

SCl

2



dinitrogen tetraoxide is

N

2

O

4

(in smog)

Covalent Naming Practice



Formula to Name

•

CO

•

N

O

•

PH



Name to Formula

• carbon

tetrachloride • sulfur dioxide • diboron

Covalent Lewis Dot Structures



Valence Electrons (VE)

Covalence Number (CN)

Covalent bonds are formed using

the valence electrons of the atoms



Step 1: Find the total

number of valence

electrons for the formula

given by taking the

Valence Electron Count (VE)

SF

VE

1 S x 6

2 F x 7

6

14

----20

SF

has

Covalence Number - the number

of bonds an element prefers

 Most of the time (about 95%) an element will form the same number of bonds. This gives a clue on how the molecule is put together.

 Covalence Numbers

• 1 for H (always for H), F, Cl, Br,and I

• 2 for O, S, and Se

• 3 for N and P

Valence Electron Count (VE)

SF

VE

1 S x 6

2 F x 7

6

14

----20

S wants

more

bonds,

put S in

center

CN

2

Bonding the Molecule

 Two electrons are shared to form a covalent bond. They are

represented by a line (-).

 _{All atoms in a}

formula must be bonded together.

F - S - F

The Octet Rule

 Each atom in the

molecule must have 8 electrons around it, except for H

which only has one bond.

 _{Pairs of dots called}

lone pairs are

placed around each atom up to 8

F - S - F

The Octet Rule

 Each atom in the

molecule must have 8 electrons around it, except for H

which only has one bond.

 _{Pairs of dots called}

lone pairs are

placed around each atom up to 8

F - S - F

Is this the correct dot structure?

 The total valence

electrons used must exactly equal what was calculated in the table. VE = 20

 Each atom will also have its covalence number in CHEM 2A. S has 2 bonds and F has 1 bond.

F - S - F

count all dots =16

counts 2 per bond = 4

Is this the correct dot structure?

 Each atom in the

molecule must have 8 electrons around it (except H).

 WHEN VE, CN,

AND OCTET RULE ARE ALL OBEYED YOU HAVE A

CORRECT

STRUCTURE.

F - S - F

each F has 2e from

one bond + 6e

Double Bonds

 Use when there are not enough

electrons to give all atoms their octet. If there are not

enough more must be shared.

 Every 2 electrons short means one double bond.

VE

CN

CH

O

1 C x 4

2 H x 1

1 O x 6

Double Bonds

 First bond all the

atoms to C because it has the highest

covalence number.

 Notice it still does not have 4 bonds.

VE

CN

CH

O

1 C x 4

2 H x 1

1 O x 6

4

2

6

---12

4

1

2

H - C - H

Double Bonds

 With 3 bond, 6e have been used.

 6 more VE remain to give an octet

 That leaves C without an octet.

VE

CN

CH

O

1 C x 4

2 H x 1

1 O x 6

4

2

6

---12

4

1

2

H - C - H

Double Bonds

 To get 4 bonds on C and 2 bonds on O and have all atoms obey the octet rule, a

double bond is made.

 Now VE = 12, CN is correct for all, and octet.

VE

CN

CH

O

1 C x 4

2 H x 1

1 O x 6

4

2

6

---12

4

1

2

H - C - H

O

Triple Bonds

 With 2 bond, 4e have been used.

 12 more VE remain to give an octet

 That leaves C without an octet.

VE

CN

CPF

1 C x 4

1 P x 5

1 F x 7

4

5

7

---16

4

3

1

C - F

Triple Bonds

 To get 4 bonds on C and 3 on P and have all atoms obey the octet rule, a triple bond is made.

 Now VE, CN, and octet are all correct.

VE

CN

CPF

1 C x 4

1 P x 5

1 F x 7

4

5

7

---16

4

3

1

C - F

P

Now try these



POF



C

3

H

9

P

More Lewis Dots

VE

CN

POF

1P x 5

1O x 6

1F x 7

5

6

7

18

3

2

1

More Lewis Dots

VE

CN

3C x 4

9H x 1

1P x 5

12

9

5

26

4

1

3

C

H

P

H – C – C – C – P - H

H H H

More Lewis Dots

K

S

Should I make a table?

NO! It’s ionic.

Polyatomic ions – many atoms

ions

A group of atoms held together by

covalent bonds that has acquired

A polyatomic ion is a connected

group, don’t break it apart!

 The charge is for the entire group not for a single atom in it.

 If you need more than one of a polyatomic ion you must put it in ( ) to show it.

EXAMPLE: Ca(NO₃)₂ has 2 NO₃-1 _nitrate

Naming with polyatomic ions



Have a list of the ions on page 94 in

your text book on your notes.



Name the metal then look up the name

for the polyatomic ion. If it is a variable

charge metal you must use the charge

on the ion to figure out the charge on

the metal.

Name these examples….



Na

3

PO

3



Ca(ClO)

2



FeCO

3

The answers are…



Sodium phosphite



Calcium hypochlorite



Iron (II) carbonate

note that

there is ONE carbonate ion with

a –2 charge (not 3).

Now from name to formula

 Write out each ion with their charges.

 The total charges must equal zero in the

compound. Determine how many of each ion to use.

Try these examples…



Aluminum sulfate



Magnesium nitrite



Chromium (III) bicarbonate

The answers…



Aluminum sulfate

Al

(SO

)



Magnesium nitrite

Mg(NO

)



Chromium (III) bicarbonate

Cr(HCO

)

Polarity in Bonds and Molecules

Electronegativity

Bond Polarity – Unequal Sharing



Calculate the difference between the

electronegativity values for the atoms in

a bond to determine the polarity.



If the difference is less than 0.5, we will

call it nonpolar (not significant polarity)

Which element comes first in the

formula?

 The element with the lowest

electronegativity

value will always be written first in the

formula.

 _{Carbon monoxide is}

written CO not OC since 2.5 < 3.5

C 2.5 N 3.0 O 3.5 F 4.0 H 2.1 Si 1.8 P 2.1 S 2.5 Cl 3.0 Br 2.8 I 2.5

Which bond is most polar?



C – O



N – O

Which bond is most polar?



C – O

3.5 – 2.5 = 1.0



N – O 3.5 – 3.0 = 0.5



C – N

3.0 – 2.5 = 0.5



All three bonds are

Sharing the Electrons

 Nonpolar is equal sharing of the

electrons in a bond.

 Polar is unequal sharing, the

Diagram for Polar and Nonpolar

Covalent.

+ +

+ +₊

NONPOLAR

Equal

sharing

POLAR

VSEPR Theory



Valence Shell Electron Pair Repulsion

X = any atom E = lone pair

X = 2

X + E = 3

In-class assignment

Draw Lewis Dot, State geometry



HOBr



H

Te



NBr

Polarity of Molecules



To determine the polarity of a molecule

you must first

• Draw Lewis Structure

• Determine the geometry



Symmetrical molecules are generally

not polar

Molecular Symmetry

CF

H

O

•Both molecules contain polar bonds

Polar or Nonpolar?



CO



NO