Chapter 4 Forces between Particles

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Presentation transcript:

Chapter 4 Forces between Particles

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

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)

4.2 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

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–

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?

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

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

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

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

Example 4.4 - 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

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

Example 4.5 - 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

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

Figure 4.1 - Reaction of Sodium Metal and Chlorine Gas

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

Example 4.6 - 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:

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

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

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

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

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

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

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

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

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

Example 4.7 - Solution Elements are calcium (Ca) and nitrogen (N), hence the name calcium nitride

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–

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

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

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

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

Example 4.9 - 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

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

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

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

Table 4.7 - Some Common Polyatomic Ions

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

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

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

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

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–

Example 4.18 - 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