Ions

Chapter 4 Compounds and Their Bonds 4.1 Valence Electrons 4.2 Octet Rule and Ions

LecturePLUS Timberlake

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Chemical Bonds

Attraction between two or more atoms Interaction between valence electrons Ionic bonds Covalent bonds LecturePLUS Timberlake

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Valence Electrons Electrons in the highest (outer) electron level Have most contact with other atoms Known as valence electrons Outer shells of noble gases contain 8 valence electrons (except He = 2) Example:

Ne

2, 8

Ar

2, 8, 8 LecturePLUS Timberlake

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Electron Dot Structures Symbols of atoms with dots to represent the valenceshell electrons 1A

2A

3A

4A

5A

6A

7A

8A

H•

He: •

Li• Be•

• •

B•

•

•

••

C•

•

• •

Na• Mg•

• •

Al•

••

N•

• • •

O•

•• ••

Si •

•

••

•

: F • :Ne : ••

••

P•

••

•

••

••

S•

Timberlake • LecturePLUS • ••

••

:Cl • :Ar : ••

••

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Learning Check A.



X would be the electron dot formula for 1) Na

B.

  X 

1) B

2) K

3) Al

would be the electron dot formula 2) N

LecturePLUS Timberlake

3) P

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Solution A.



X would be the electron dot formula for 1) Na

B.

  X 

2) N

2) K would be the electron dot formula 3) P

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Octet Rule An octet in the outer shell makes atoms stable Electrons are lost, gained or shared to form an octet Unpaired valence electrons strongly influence bonding

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Formation of Ions from Metals Ionic compounds result when metals react with nonmetals Metals lose electrons to match the number of valence electrons of their nearest noble gas Positive ions form when the number of electrons are less than the number of protons

•

Group 1A metals →

ion 1+

Group 2A metals → Group 3A metals →

ion 2+ ion 3+

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Formation of Sodium Ion Sodium atom

Na • 2-8-1 11 p+ 11 e0

– e−

Sodium ion

→

Na +

2-8 ( = Ne) 11 p+ 10 e1+ LecturePLUS Timberlake

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Formation of Magnesium Ion Magnesium atom

Magnesium ion

•

Mg •

– 2e−

→

2-8-2 12 p+ 12 e0

Mg2+ 2-8 (=Ne)

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12 p+ 10 e2+

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Some Typical Ions with Positive Charges (Cations) Group 1A

Group 2A

Group 3A

H+

Mg2+

Al3+

Li+

Ca2+

Na+

Sr2+

K+

Ba2+

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Learning Check A. Number of valence electrons in aluminum 1) 1 e2) 2 e3) 3 eB.

C.

Change in electrons for octet 1) lose 3e2) gain 3 eIonic charge of aluminum 1) 32) 5LecturePLUS Timberlake

3) gain 5 e-

3) 3+ 12

Solution A. Number of valence electrons in aluminum 3) 3 eB.

Change in electrons for octet 1) lose 3e-

C.

Ionic charge of aluminum 3) 3+ LecturePLUS Timberlake

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Learning Check Give the ionic charge for each of the following: A. 12 p+ and 10 e1) 0 2) 2+ 3) 2B. 50p+ and 46 e1) 2+ 2) 4+

3) 4-

C. 15 p+ and 18e2) 3+ 2) 3-

3) 5-

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Solution Give the ionic charge for each of the following: A. 12 p+ and 10 e2) 2+ B. 50p+ and 46 e2) 4+ C. 15 p+ and 18e2) 3LecturePLUS Timberlake

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Learning Check A. Why does Ca form a Ca2+ ion?

B. Why does O form O2- ion?

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Solution A. Why does Ca form a Ca2+ ion? Loses 2 electrons to give octet 2-8-8-2 2-8-8 (like Ar) B.

Why does O form O2- ion? Gains 2 electrons to give octet 2-6 + 2e2-8 (like Ne) LecturePLUS Timberlake

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Ions from Nonmetal Ions In ionic compounds, nonmetals in 5A, 6A, and 7A gain electrons from metals

Nonmetal add electrons to achieve the octet arrangement

Nonmetal ionic charge: 3-, 2-, or 1LecturePLUS Timberlake

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Fluoride Ion unpaired electron ••

:F• ••

2-7 9 p+ 9 e0

octet ••

+ e−

1-

: F: ••

2-8 (= Ne) 9 p+ 10 e1ionic charge LecturePLUS Timberlake

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Learning Check Complete the names of the following ions: 5A 6A 7A N3− O2− F− nitride __________ fluoride P3− S2− ___________ __________

Cl− _________ Br − _________

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Solution 5A

6A

7A

N3−

O2−

F−

oxide

fluoride

S2−

Cl−

sulfide

chloride

nitride P3− phosphide

Br −

bromide LecturePLUS Timberlake

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