1. Chapter 4 Forces between Particles

2. Learning Objectives
Use electronic configurations to determine the number of electrons gained or lost by atoms as they achieve noble gas electronic configurations
Use the octet rule to correctly predict the ions formed during the formation of ionic compounds, and write correct formulas for binary ionic compounds containing a representative metal and a representative nonmetal
Correctly name binary ionic compounds
Determine formula weights for ionic compounds
Write correct formulas for ionic compounds containing representative metals and polyatomic ions, and correctly name binary covalent compounds and compounds containing polyatomic ions

3. 4.1 Noble Gas Configurations
Characterized by two electrons in the valence shell of helium and eight electrons in the valence-shell electrons for the other members of the group (Ne, Ar, Kr, Xe, and Rn)

4. 4.2 Ionic bonding
Atoms will gain or lose sufficient electrons to achieve an outer electron arrangement identical to that of a noble gas
Arrangement usually consists of eight electrons in the valence shell
Simple ion: Atom that has acquired a net positive or negative charge by losing or gaining electrons

5. Simple Ion Examples
Magnesium, Mg, has two valence electrons that it loses to form a simple ion with a +2 electrical charge
Ion is written as Mg2+
Oxygen, O, has six valence electrons
Tends to gain two electrons to form a simple ion with a –2 electrical charge
Ion is written as O2–

6. Simple Ion Examples (continued)
Bromine, Br, has seven valence electrons
Tends to gain one electron to form a simple ion with a –1 electrical charge
Ion is written as Br–
How many electrons are gained or lost to obtain the ions on the following figure?

7. Determining Ionic Charges for Representative Elements
Representative metals will form ions having the same positive charge as the number (Roman numeral) of the group to which they belong
Representative nonmetals will form ions with a negative charge equal to 8 minus the number (Roman numeral) of the group to which they belong
For example, strontium, Sr, a group IIA metal forms Sr2+ ions and phosphorus, P, a group VA nonmetal forms P3– ions
Hence:
Group IA elements always form +1 charged ions
Group IIA elements always form +2 charged ions
Group VIA elements always form -2 charged ions
Group VIIA elements always form -1 charged ions

8. Ionic Bond Formation
Ions with positive charges are attracted to ions with negative charges
Attractive force between such ions holds them together
Ionic bonds form between simple ions when representative metal atoms lose valence electrons and the electrons are gained by representative nonmetal atoms
Both atoms are changed into ions with noble gas configurations
Resulting ions are then attracted to each other

9. Isoelectronic Term that literally means same electronic
Used to describe atoms or ions that have identical electronic configurations

10. Example 4.4 - Ions and Noble Gas Configurations
Show how Na can achieve a noble gas configuration and become an ion by gaining or losing electrons

11. Example Solution
Electronic structure of sodium (Na) is represented below using an abbreviated configuration and a Lewis structure
First representation makes it obvious that the Ne configuration with 8 electrons in the valence shell would result if the Na atom lost the single electron located in the 3s subshell
Loss is represented by the following equation
Notice that the removal of a single negative electron from a neutral Na atom leaves the atom with 11 positive protons in the nucleus and 10 negative electrons
This gives the atom a net positive charge
Atom has become a positive ion

12. Example 4.5 - Losing and Gaining Electrons
Use the periodic table to predict the number of electrons lost or gained by atoms of the following elements during ionic bond formation
Write an equation to represent the process in each case Li
Any group IIA(2) element, represented by the symbol M

13. Example Solution
Lithium (Li), a metal, is in group IA(1); therefore it will lose one electron per atom:
Any group IIA(2) metal will lose two electrons; therefore

14. 4.3 Ionic Compounds
Substances that result when ionic bonds form between positive and negative ions
Binary ionic compound: When ionic compounds are formed by the reaction of only two elements
Cu2O
CuO

15. Figure 4.1 - Reaction of Sodium Metal and Chlorine Gas

16. Example 4.6 - Electronic Transfer Processes
Represent the electron-transfer process that takes place when the following pair of elements react ionically
Determine the formula for the resulting ionic compound
Mg and F

17. Example Solution
Magnesium (Mg) of group IIA(2) will lose two electrons per atom, whereas fluorine (F) of group VIIA(17) will gain one electron per atom
Therefore, it can be written as:

18. Example 4.6 - Solution (continued)
It is apparent that two fluorine atoms will be required to accept the electrons from one magnesium atom
From another point of view, two F– ions will be needed to balance the charge of a single Mg2+ ion
Both observations lead to the formula MgF2 for the compound
Note the use of subscripts to indicate the number of ions involved in the formula
Subscript 1, on Mg, is understood and never written

19. Binary Ionic Compound Formulas
Binary ionic compounds typically form when a metal and a nonmetal react
Metal tends to lose one or more electrons and forms a positive ion
Nonmetal tends to gain one or more electrons and forms a negative ion
Symbol for the metal is given first in the formula
Formula for a binary ionic compound represents the minimum number of each ion that will provide equal numbers of positive and negative electrical charges when combined together

20. Binary Ionic Compound Formula Examples
Sodium and fluorine
Sodium, a group IA metal, will form sodium ions with the symbol Na+
Fluorine, a group VIIA nonmetal, will form fluoride ions with the symbol F–
Minimum number of ions needed to give the same number of positive and negative charges is one of each
One Na+ provides one positive charge and one F– provides one negative charge
Correct formula that results is NaF

21. Binary Ionic Compound Formula Examples (continued 1)
Sodium and sulfur
Sodium is a group IA metal and will form sodium ions with the symbol Na+
Sulfur is a group VIA nonmetal and will form sulfide ions with the symbol S2─
Minimum number of ions required to give the same number of positive and negative charges is two Na+ ions and one S2– ion
Two Na+ ions provide two positive charges and one S2– ion provides two negative charges
Resulting formula is Na2S

22. Binary Ionic Compound Formula Examples (continued 2)
Aluminum and oxygen
Aluminum is a group IIIA metal and will form ions with the symbol Al3+
Oxygen is a group VIA nonmetal and will form ions with the symbol O2–
Minimum number of ions required to give the same number of positive and negative charges is two Al3+ ions and three O2– ions
Resulting formulas is Al2O3

23. 4.4 Naming Binary Ionic Compounds
Binary ionic compounds are named using the following pattern:
Name = Metal + nonmetal stem + -ide
Stem of the name of a nonmetal is the name of the nonmetal with the ending dropped

24. Naming Binary Ionic Compounds (continued)
Some metal atoms, especially those of transition and inner-transition elements, form more than one type of charged ion
Iron forms both Fe2+ and Fe3+ ions
Number of positive charges on the metal ion is indicated by a Roman numeral in parentheses following the metal name
Compounds FeCl2 and FeCl3 contain iron ions with 2+ and 3+ charges, respectively
Their names are iron (II) chloride and iron (III) chloride
FeCl2
FeCl3

25. Examples of Binary Ionic Compound Names
Name K2O
Name = Metal name + Nonmetal stem + -ide
Name = Potassium + Ox- + -ide = Potassium oxide
Name Mg3N2
Name = Magnesium + Nitr- + -ide = Magnesium nitride
Name BeS
Name = Beryllium + Sulf- + -ide = Beryllium sulfide
Name AlBr3
Name = Aluminum + Brom- + -ide = Aluminum bromide

26. Example 4.7 - Naming Binary Ionic Compound
Name the following binary ionic compound
Ca3N2

27. Example Solution
Elements are calcium (Ca) and nitrogen (N), hence the name calcium nitride

28. Example 4.8 - Formulas for Ionic Compounds
Write the formula for ionic compound that would form between the following simple ions
Note that the metal forms two different simple ions, and name each compound two ways
Cr3+ and S2–

29. Example 4.8 - Solution Charges on the combining ions are 3+ and 2–
Smallest combining ratio balances the charges in two Cr3+ (a total of 6+ charge) and three S2– (a total of 6– charge)
Formula is Cr2S3
Names are chromium(III) sulfide and chromic sulfide

30. 4.5 Ionic Compound Structure
Stable form of an ionic compound is not a molecule, but a crystal in which many ions of opposite charge occupy lattice sites in a rigid three-dimensional arrangement called a crystal lattice
Lattice site: Individual location occupied by a particle in a crystal lattice

31. Ionic Compound Formulas and Weights
Formulas for ionic compounds represent only the simplest combining ratio of the ions in the compounds, not the precise numbers of atoms of each element found in a crystal lattice
Formula weight: Sum of the atomic weights of the atoms shown in the formula of an ionic compound
Similar to molecular weight
One mole of an ionic compound contains Avogadro's number (6.022 x 1023) of the simplest combining ratio of ions in the compounds

32. Example 4.9 - Comparing Molecular and Ionic Compounds
Carbon dioxide, CO2, is a molecular compound, whereas magnesium chloride, MgCl2, is an ionic compound
Molecular weight of CO2 is 44.0 u
Formula weight of MgCl2 is 95.3 u
Determine the mass in grams of 1.00 mol of each compound

33. Example Solution
1.00 mol of a molecular compound has a mass in grams equal to the molecular weight of the compound
Thus, 1.00 mol CO2 = 44.0 g CO2
For ionic compounds, 1.00 mol of compound has a mass in grams equal to the formula weight of the compound
Thus, 1.00 mol MgCl2 = 95.3 g MgCl2

34. Naming Binary Covalent Compounds
Similar to naming binary ionic compounds
Rules
Give the name of the less electronegative element
Give the stem of the name of the more electronegative element and the suffice –ide
Indicate the number of each type of atom in the molecule by means of Greek prefixes
Prefix mono- is not used when it appears at the beginning of the name

35. Examples of Naming Binary Covalent Compounds
Name = Sulfur + di- + ox + -ide = Sulfur dioxide
XeF6
Name = Xenon + hexa- + fluor + -ide = Xenon hexafluoride
H2O
Name = Di- + hydrogen + mono- + ox + -ide = Dihydrogen monoxide
Known as water
Final o of mono- was dropped for ease of pronunciation

36. Ionic Compounds Containing Polyatomic Ions
Rules for writing formulas and names for ionic compounds containing polyatomic ions are essentially the same as those used for writing formulas and names for binary ionic compounds
Formula
Metal is written first, the positive and negative charges must add up to zero, and parentheses are used around the polyatomic ion if more than one is used
Name
Positive metal ion is given first followed by the name of the negative polyatomic ion name

37. Table 4.7 - Some Common Polyatomic Ions

38. Examples of Ionic Compounds Containing Polyatomic Ions
Compound containing K+ and ClO3–
KClO3
Compound containing Ca2+ and ClO3–
Ca(ClO3)2
Compound containing Ca2+ and PO43–
Ca3(PO4)2

39. Naming Ionic Compounds Containing Polyatomic Anions
Names of ionic compounds that contain a polyatomic anion are obtained using the following pattern:
Name = Name of metal + Name of polyatomic anion
Examples
KClO3 is named potassium chlorate
Ca(ClO3)2 is named calcium chlorate
Ca3(PO4)2 is named calcium phosphate
CaHPO4 is named calcium hydrogen phosphate

40. Example 4.17 - Naming Binary Covalent Compounds
Name the following binary covalent compounds:
N2O5
CS2

41. Example 4.17 - Solution N2O5 is assigned the name dinitrogen pentoxide
The a is dropped from the penta- prefix for oxygen to avoid the pronunciation problem created by having two vowels next to each other in a name (dinitrogen pentaoxide)
CS2 is named carbon disulfide

42. Example 4.18 - Writing Compound Formulas and Names
Write formulas and names for compounds composed of ions of the following metals and polyatomic ions indicated:
Na and NO3–
K and HPO42–

43. Example Solution
Sodium (Na) is a group IA(1) metal and forms Na+ ions
Electrical neutrality requires a combining ratio of one Na+ for one NO3–
Formula is NaNO3
Name is given by the metal plus the polyatomic ion name, sodium nitrate
Potassium (K) is a group IA(1) metal and forms K+ ions
Required combining ratio of 2:1 gives the formula K2HPO4
Name is potassium hydrogen phosphate