CH 101 1st Edition Lecture 9Outline of Last Lecture I. Review of Periodic TrendsII. Ionization EnergyIII. Electronegativity Outline of Current Lecture I. Electron configuration of ionsII. Oxidation statesA. The Six Rules of OxidationIII. Predicting Oxidation Formulas IV. Possible Oxidation States for Unknown Elements X, Y and ZCurrent LectureV. Electron configuration of ions- You already learned how to find the electron configuration of an element, but why about ions? To find the electron configuration of ions you take the configuration of the element and either subtracting (in the case of a cation) electrons from the orbitals in this order p, s, d, f , etc., or adding (in the case of a anion) electrons starting with the lowest orbital .If all of the valence electrons are subtracted or the orbital is filled you just write the noble gas that has the same amount of electrons (isoelectronic).Isoelectronic – When two atoms have the same electron configuration they are isoelectronic. For example, Na1+ and Ne are isoelectronic because they both have 10 electrons. Ex) What is the electron configuration s of the ion Zr2+ ?Zr = [Kr] 5s2 4d2<- First find the electron configuration of the elementThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Zr2+ = [Kr] 5s24d2<- Zr2+ is a cation that tells you Zr is losing 2 electrons, take the two electrons from the orbitals in the given order p, s, d, f, since there is no p orbital in the noble gas electron configuration of Zr, you take electrons from the s orbital.Zr2+ = [Kr] 4d2<- This is the correct electron configuration of the ion Zr2+.Ex) What is the electron configuration of the ion Se2-?Se = [Ar] 4s2 3d10 4p4 <- Find the electron configuration of the element first Se2- = [Ar] 4s2 3d10 4p4 <- Look at the ion and determine that it is an anion, Se is gaining 2 electrons. Find the lowest orbital which is p and add 2 electron to it.Se2- = [Ar] 4s2 3d10 4p6<- This configuration is the same at the noble gas [Kr], you can find this by going to Se on the periodic table and then counting two to the right for the two electron it gained, you will see that you land on Kr.[Kr] <- The electron configuration of the ion Se2-VI. Oxidation states- Typically the elements in the main groups of the periodic table can have oxidation numbers that range from their group number to their group number minus 8. For example the oxidation state range for S is +6 to -2 because Sulfur is in main group 6 and 6 – 8 = -2. - When it comes to transition metals, they generally all have an oxidation number of 2+. However, the exceptions are that some groups tend to be 1+ and others 3+, just know that transition metals do not gain electrons, the give them away.B. The Six Rules of Oxidation- The six oxidation rules guide you toward finding the oxidation state by providing information on the pieces of the pieces of the molecule that make up the oxidation state.Rule 1: The oxidation state of an element is always = 0. F = 0, H2 = 0, O8 = 0.Rule 2: The oxidation number of Fluorine (F2) is always = -1.Rule 3: The oxidation number of any element in Group 1 is always = +1. K = +1, Li = +1.The oxidation number of any element in Group 2 is always = +2. Ca = +2, Ba = +2.The oxidation number of Aluminum (Al) is always = +3.Rule 4: The oxidation number of Hydrogen changes depending on who it bonds with, a metal or a nonmetal. When bonded to a metal, the oxidation number of H is -1. When bonded to nonmetals its +1.Rule 5: The oxidation number of Oxygen (O) is always = -2Rule 6: The oxidation number of the elements in Group 17 (halogens) is most commonly = -1, This rule is only used once all the other rules have been applied to the compound.Ex) What is the oxidation state of Bromine in the compound SrBr2?SrBr2, __ + __ = 0 <-Start by setting up an equation (oxidation state of Sr) + (oxidation state of Br2) = 0, and set the equation equal to zero since the charge of the compound is 0 or neutral. +2 + __ = 0 <- See that Look at rule 3, it states that Group 2 will always have an oxidation state of +2. See that Sr is in Group 2 and solve the equation.+2 + -2 = 0 <- Solve for Br by seeing that 2 + (-2) = 0Br2 = -2 meaning thatBr = -1<- This is the oxidation state for Br in the compound SrBr2Ex) What is the oxidation state of Cr in the compound Cr2O72- ?Cr2O72-, __ + __ = -2 <- The equation is not equal to 0 here because Cr2O72- is an anion with a charge of -2.__ + (-2(7)) = -2<- Use rule 5 to tell you that the oxidation number for oxygen is -2. Multiply this number by the number of oxygen atoms in the compound to find the oxidation number for O7.__ + -14 = -2 <- What + (-14) equals -2? You could find this by adding -2 and 14 to get 12.+12 + -14 = -2 <- Now we know they oxidation number for Cr2 is +12 but what about for just oneCr atom?Cr2 = +12 meaning that Cr = +6 <- The oxidation number for Cr in the compoundCr2O72- is +6.VII. Predicting Formulas- You can predict formulas when you are looking at a compound made up of a metal and a nonmetal (an ionic compound). All you do is state the oxidation number of the group that the element is in and then “criss-cross” the oxidation number with the other element’s oxidation number and then simplify if you can.Ex) Predict the formulafor the compound formed between Ca + S?Ca2+S2-<- By looking at the periodic table you know that the oxidation state or charge of Ca is +2 and S is -2. Note that the cation always comes before the anion in a compound.Ca2S2<- Now bring the 2+ that goes with Ca and bring it down to S and vice versa. CaS<- The denominators are equal and can therefore be simplified. *Remember that an element can have many different oxidation states, just because the oxidation state of Cl is 1- in the compound LiCl, doesn’t mean it is 1- in the compound NaClO, in fact Cl has an oxidation state of 1+ in this compound.VIII. Possible Oxidation States for Unknown Elements X, Y and Z- In this electron configuration chart, you see that element Z have two electron in its outer shell, element Y has 5, and element X has 6. The electron the element is willingto give away to create a positive oxidation state are circled. The spaces where the element is wants to gain electrons to create a negative oxidation state are squared. Using what you know about valence electron, electronegativity, and the periodic