Ionic Compounds PDF

Ionic Compounds CHAPTER 3 Compounds A compound is two or more elements chemically combined. Elements combine chemically by: gaining, losing, or s h a r i n g electrons. The number of valence electrons determines how many electrons the elements will gain, lose, or share. Two Main Types of Co...

Ionic Compounds CHAPTER 3 Compounds A compound is two or more elements chemically combined. Elements combine chemically by: gaining, losing, or s h a r i n g electrons. The number of valence electrons determines how many electrons the elements will gain, lose, or share. Two Main Types of Compounds Ionic Compounds are formed through the exchange of e–... One element/atom loses e–, another gains e–. Molecular Compounds are formed by elements/atoms sharing e–. Main-Group Ions For both ionic and molecular compounds, each element arrives at a core configuration of valence electrons in the compound. This is a total of 8 e– in its outermost s and p orbitals, also called an octet. For ionic compounds, the number of valence electrons in the neutral atom determines whether the element will gain or lose e–, and how many it will gain or lose to achieve its octet. * Important to know this. Main-Group Ions Example: Cl’s electron configuration is: 1s2 2s2 2p6 3s2 3p5. It has seven highest-level (valence) electrons (3s2 3p5). To achieve an octet (8 valence e–), Cl needs to gain only 1 e–, and will become an anion with a charge of –1. Cl: 3s2 3p5 + e– → Cl–1: 3s2 3p6 Main-Group Ions Sodium’s electron configuration is: 1s2 2s2 2p6 3s1. It has one valence electron (3s1). To achieve an octet, sodium could gain 7 e–, filling the 3s and 3p orbitals. Na: 3s1 + 7e– → Na–7: 3s2 3p6 This is too large number of electrons to be gained... Main-Group Ions Another way to achieve an octet: sodium can lose its 1 valence e–, leaving behind a filled set of 2s and 2p orbitals. Na: 2s2 2p6 3s1 → Na+1: 2s2 2p6 + e– Sodium becomes a cation with a charge of +1. Ionic Compounds In forming the ionic compound of sodium and chlorine, there is an exchange of an e–. sodium will lose 1 e–...achieving octet!...and... chlorine will gain this e–... achieving octet! Once the ions are formed, ALL the positively-charged sodium ions (Na+) are attracted to... ALL the negatively-charged chloride ions (Cl–). Ionic Compounds However, there is no molecule (single unit) of: one sodium ion and one chloride ion. The ratio of one sodium ion to one chlorine ion is shown as its chemical formula: NaCl. The positive ion is always written first in the formula. Ionic Compounds In this chemistry, Na lost one e– and Cl gained one e– In forming an ionic compound, a metal will lose e– and become a cation; a nonmetal will gain e–and become an anion. The total number of electrons lost by the metal MUST EQUAL the total number of electrons gained by the nonmetal. * Important to know this. Ionic Compounds Magnesium (Mg) is in the second group of the periodic table, therefore it has two valence electrons (3s2). When magnesium forms an ion, it will lose these two e– to become a +2 cation. Mg → Mg+2 + 2e– It is preferable for magnesium to lose two e– rather than gain six e–. Ionic Compounds Nitrogen (N) is in the third group of the p block (fifth group of the main group elements). It has five valence electrons (2s2 2p3). When nitrogen forms an ion, it will gain three e– to complete its octet,...becoming a –3 anion. N + 3e– → N–3 It is preferable for nitrogen to gain three e – rather than lose five e –. Ionic Compounds In forming an ionic compound, the total number of electrons lost must equal the total number of electrons gained. Since Mg can only lose 2 e–, and N must gain 3 e–, the ratio cannot be 1:1... Ionic Compounds In fact, three Mg must lose a total of six e–, so that two N can gain these six e–. Mg → Mg+2 + 2e– 3Mg → 3Mg+2 + 6e– N + 3e– → N–3 2N + 6e– → 2N–3 NOTE: We can multiply the whole ‘chemical equation’ by whole numbers!...Just like a math equation! Ionic Compounds The periodic table can be used to quickly determine the common charge on the elements forming an ionic compound. Metals and semimetals with one, two, or three valence e – will lose these e – to become +1, +2, or +3 cations, respectively. Nonmetals and semimetals with five, six, or seven valence e – will gain three, two, or one e –, respectively. They form –3, –2, and –1 anions. * Important to know this. Ionic Compounds TEST QUESTION: What are the common charges for the elements potassium (K), barium (Ba), phosphorus (P), and iodine (I)? (Answer before moving on to the next slide.) Ionic Compounds TEST QUESTION: What is the formula for the ionic compound formed from lithium (Li) and sulfur (S)? (Answer before moving to next slide.) TEST QUESTION: What is the formula for the ionic compound formed from aluminum (Al) and oxygen (O)? What is the formula for the ionic compound formed from calcium (Ca) and sulfur (S)? (Answer before moving to next slide.) Polyatomic Ions Polyatomic ions (many-atom ions) consist of a number of atoms joined together to form a single unit with a charge. Most polyatomic ions have a negative charge. There is one common polyatomic ion with a positive charge: Ammonium ion: NH4+1 Common Polyatomic Ions ClO–1 hypochlorite PO3–3 phosphite ClO2–1 chlorite PO4–3 phosphate ClO3–1 chlorate MnO4–1 permanganate ClO4–1 perchlorate CrO4–2 chromate NO2–1 nitrite Cr2O7–2 dichromate NO3–1 nitrate C2H3O2–1 acetate SO3–2 sulfite CN–1 cyanide SO4–2 sulfate OH–1 hydroxide CO3–2 carbonate HCO3–1 bicarbonate * Important to know this. Common Polyatomic Ions ClO–1 hypochlorite PO3–3 phosphite ClO2–1 chlorite PO4–3 phosphate ClO3–1 chlorate MnO4–1 permanganate ClO4–1 perchlorate CrO4–2 chromate NO2–1 nitrite Cr2O7–2 dichromate NO3–1 nitrate C2H3O2–1 acetate SO3–2 sulfite CN–1 cyanide SO4–2 sulfate OH–1 hydroxide CO3–2 carbonate HCO3–1 bicarbonate You will need to know the name, formula, and charge for the bolded polyatomic ions. * Important to know this. Polyatomic Ions In determining the formula of an ionic compound containing a polyatomic ion, the polyatomic ion is taken as a whole. The formula for the compound of calcium (Ca) and phosphate polyatomic ion (PO4–3) is: * Important to know this. TEST QUESTION: What is the formula for the ionic compound formed from sodium (Na) and sulfate ion (SO4–2)? (Answer before moving on to the next slide.) Transition Metal Ions Transition metals can have different charges: iron (Fe) typically can be a +2 or a +3 ion (Fe+2, Fe+3) copper (Cu) is usually +2, but can be +1 (Cu+1, Cu+2) zinc (Zn) is always +2 (Zn+2) The charge on a transition metal cannot be predicted from the periodic table. However, the charge on a transition metal can be determined from the formula of the ionic compound. * Important to know this. Transition Metal Ions For the compound FeO, the charge on O can be found from the periodic table: Transition Metal Ions TEST QUESTION: What are the charges on the ions in V2S5? (Answer before moving to next slide.) TEST QUESTION: What are the charges on the ions in Mn3(PO4)4? (Answer before moving to next slide.) Naming Ionic Compounds The names of ionic compounds are written with the positive ion (cation) first. This is usually a metal (and called by the element’s name) but may be ammonium polyatomic ion (NH4+1). The second half is the negative ion (anion)... If the anion is a single nonmetal it is given an -ide ending. If it is a polyatomic anion, the name of the polyatomic anion is used. * Important to know this. Names of Monatomic Anions carbon – carbide C–4 nitrogen – nitride N–3 oxygen – oxide O–2 fluorine – fluoride F–1 phosphorus – phosphide P–3 sulfur – sulfide S–2 chlorine – chloride Cl–1 bromine – bromide Br–1 iodine – iodide I–1 * Important to know this. Naming Ionic Compounds TEST QUESTION: Name the following ionic compounds: NaCl, Mg3N2, Ca3(PO4)2. (Answer before moving to next slide.) Naming Ionic Compounds One less oxygen than the -ate ion is the -ite ion. ClO3–1 chlorate ClO2–1 chlorite NO3–1 nitrate NO2–1 nitrite SO4–2 sulfate SO3–2 sulfite PO4–3 phosphate PO3–3 phosphite Note: The charge remains the same! * Important to know this. Naming Ionic Compounds TEST QUESTION: Name the following compounds: KNO3 and KNO2. (Answer before moving to next slide.) KNO3: KNO2: Naming Ionic Compounds Since transition metals can have multiple charges, compounds with DIFFERENT RATIOS can be formed using the same elements. Two compounds of iron and oxygen can be formed... BOTH with the name iron oxide: FeO and Fe2O3. To differentiate between these compounds, the charge on the transition metal is stated as a Roman numeral in the name: FeO: iron(II) oxide Fe2+ O2– Fe2O3: iron(III) oxide Fe3+ O2– * Important to know this. Naming Ionic Compounds TEST QUESTION: Name the following compound: V2S5. (Answer before moving to next slide.) Naming Ionic Compounds TEST QUESTION: Name the following compound: Mn3(PO4)4. (Answer before moving to next slide.) Naming Ionic Compounds TEST QUESTION: Name the following compound:Co2(CO3)3. (Answer before moving to next slide.) Naming Ionic Compounds Roman numerals for the charge are not used for main-group metals since their charges do not vary. Al2S3 is aluminum sulfide, not aluminum(III) sulfide; because: Al will always be +3 in a compound. Some transition metals only have one charge so they are often written without the Roman numeral... zinc is always Zn+2...and silver is always Ag+1. ZnS: zinc sulfide AgNO 3: silver nitrate Naming Ionic Compounds TEST QUESTION: Write the formulas for the following compounds: (Answer before moving to next slide.) potassium carbonate: copper(II) phosphite: iron(III) bromide: