chem notes
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chem notes

Course: CH 101, Spring 2008

School: N.C. State

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Chap. 4 -Cations are produced by the loss of the valence electrons with the highest n quantum number. Consequently, first row transition metals lose their 4s electrons before they lose any 3d electrons. -Metals are characterized by low ionization energies, so they lose electrons to become cations. The charge on the cation is determined by the number of electrons that are lost. Some metals lose all of their...

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4 -Cations Chap. are produced by the loss of the valence electrons with the highest n quantum number. Consequently, first row transition metals lose their 4s electrons before they lose any 3d electrons. -Metals are characterized by low ionization energies, so they lose electrons to become cations. The charge on the cation is determined by the number of electrons that are lost. Some metals lose all of their valence electron, but others lose only some of their valence electrons. The following rules help determine which electrons are lost: 1. Monatomic cations with charges greater than +3 are very rare 2. Electrons from the highest n quantum number are lost first. This is important in determining the cations formed by transition metals. 3. Electrons from the highest l quantum number are lost first within a shell. This is important for the heavier metals in Groups 3, 4, and 5. Cations formed by metals: -Group 1A and 2A metals lose their electrons and become +1 and +2 respectively. -Group 3A metals lose all of their valence electrons to form +3 ions. Tl forms both +3 and +1 ions but not a +2 ion. The reason is that the heavier main group elements can lose only a portion of their valence shell. Tl is 6s26p1. Both valence sublevels are in the same level, so the one with the highest l quantum number is emptied first. Thus, Tl can lose the 6p and not the 6s to form the +1 ion, but it cannot lose the 6s and not the 6p to form a +2 ion. -Group 4A metals, +4 monatomic ions do not exist, so the Group 4A metals cannot lose all of their valence electrons. However, the heavier metals in the group (Sn and Pb) can lose the electrons in the sublevel with the highest l quantum number, the outermost p sublevel, to form +2 ions. -Transition Metals lose electrons in the level with the highest n quantum number. Thus, most transition elements form +2 ions. Scandium is an exception because it loses all three valence electrons to form Sc3+ (no +2 ion). Silver forms only a +1 ion and copper forms both +2 and +1 ions. In addition, several transition metals form a +3 ion in addition to a +2 ion. Anions formed by non-metals: -Nonmetals form anions by gaining the number of electrons required to fill their p sublevel -Nonmetals are electronegative, so they tend to gain electrons to become anions. The number of electrons gained equals the number required to fill their valence shell. A filled valence shell for a nonmetal contains eight electrons (two s and six p electrons). -charge on an anion = group number - eight Group 7A gains an electron to be -1, Group 6A becomes -2, Group 5A becomes -3 -Monatomic anions with charges of -4 do not exist, so the Group 4A nonmetals do not form anions. -A cation is smaller than its atom, while an anion is larger than its atom. -The size of an ion is determined by the n quantum number of the outermost electrons and the effective nuclear charge that they experience. -A loss of electrons increases Zeff and, if the valence shell is emptied, decreases the n quantum. Consequently, cations are smaller than their parent atom. -A gain of electrons decreases Zeff. Thus, anions are larger than their atoms. -The oxidation state of an atom in a compound is the charge it would have if its bonds were ionic -Oxidation states are obtained by assigning all bonding electrons to the more electronegative atom in each bond. Thus, we can conclude that the oxidation state of an atom: -is negative if it is more electronegative atom -positive if it is less electronegative -The chlorine atom is one of the more electronegative atoms, which means that its valence orbitals are lower in energy than those of any metal (such as Na) and most nonmetals (such as H), so the bonding electrons are usually assigned to Cl. Cl needs only one electron to fill is valence shell, so its oxidation state is usually -1. -In order for chlorine to achieve a positive oxidation state, the unfilled orbitals on the other atom must be lower in energy than those on chlorine. Thus, chorine has a positive oxidation state only when bound to fluorine or oxygen (the only two elements that are more electronegative -H is less electronegative than most nonmetals. Thus, if the bond between hydrogen and a nonmetal were ionic, the electron would have to transfer from the higher energy orbital on H into the orbital at lower energy on the nonmetal, which would give the hydrogen atom a +1 oxidation state -The hydrogen atom is more electronegative than any metal, so it is assigned the bonding electrons. It needs only one electron to fill is valence shell (1s sublevel), so it attains an oxidation state of -1 when it bonds to a metal. -The sum of the oxidation states of the atom in a compound must sum to zero. The sum of oxidation states in an ion must sum to the charge on the ion -The oxidation state of an atom lies between its group number and its group number minus 8. -The more electronegative the element, the more likely it is to be found in its lowest (most negative) oxidation state, and the lower its ionization energy, the more likely it is to be found in its highest oxidation state -The valence orbitals of metals are high in energy, so they seldom accept electrons to obtain negative oxidation states. Thus, the metal is always in a positive oxidation state in compounds with nonmetals. -The valence orbitals of nonmetals are much lower than those of metals, so nonmetals are usally in their lowest oxidation states when bound to metals. The more electronegative a nonmetal is, the more likely it is to assume its lowest oxidation state. However, nonmetals can achieve positive oxidation states when bound to more electronegative elements. They can obtain oxidation states as high as their group number when bound to highly electronegative elements (especially O and F) -The Oxidation State Rules are listed in order of priority, so they should be used in the order given; that is, a rule takes precedence over any rule below it, or any rule can be violated only to satisfy a rule above it: 1. The oxidation states of the atoms in an element are all zero: When the valence orbitals of the two atoms are the same, the bonding electrons are assumed to be shared not transferred. For example, the oxidation state of Cu in metallic Cu is zero, both F atoms in F2 are zero (the only time F is not -1), and all eight sulfur atoms in S8 are zero. 2. F is -1:The valence orbitals of F are lower than the valence orbitals on any other element, i.e., F is the most electronegative element. Consequently, it is always assigned the bonding electrons. There is only one compound in which a F atom is not -1. What is it? Hint: see the only rule that takes priority over this Rule 3. 1A metals are +1, 2A metals are +2, and Al is +3: The valence orbitals in these metals are very high in energy, i.e., these metals have low ionization energies. They also become isoelectronic with a noble gas when they form the ions. 4. H is +1: The 1s orbital on H is lower in energy than the valence orbitals of most metals and higher in energy than those of most nonmetals. Therefore, H is +1 except when Rule 1 forces it to be 0 or Rule 3 (metals with higher energy valence orbitals) force it to be -1. 5. O is -2: Oxygen is the second most electronegative atom so it is usually assigned the bonding electrons. However, it is not -2 when it is elemental O2 (Rule 1) or bonds to F (Rule 2). In addition, it can be -1 if Rules 3 and 4 force it. Compounds in which the oxidation state of oxygen is -1 are called peroxides. Peroxides contain O22-, which has an O-O bond. Hydrogen peroxide (H2O2) is a common peroxide 6. 7A elements are -1: Halogens are electronegative so they tend to fill their valence shell to attain -1 oxidation states. However, they can attain positive oxidation states when bound to more electronegative atoms, such as oxygen or more electronegative halogens, e.g., Cl is +7 in ClO41-, Br is + 5 in BrO31-, and I is +3 in IF3 -The likely formula of a compound can be predicted by assigning likely oxidation states to each of the atoms 1. Assign oxidation states to the two elements. If the elements are not listed in the oxidation state rules, assign +Group number to the less electronegative element and -(8 - Group number) to the more electronegative element. 2. Determine the lowest common multiple (LCM) of the two oxidation states (charges) assigned in Step 1. The total positive and negative charges provided by the two ions must equal the LCM 3. The number of atoms of an element in the formula is equal to the LCM divided by the oxidation state of the atom as determined in step 1. This assures that the sum of the oxidation states is zero. -One or both of the ions in an ionic compound can be polyatomic ions. Polyatomic ions consist of several atoms bound together by non-ionic bonds. Most polyatomic ions are oxoanions, anions that consist of a central atom surrounded by oxygen atoms. The central atom is often in a high oxidation state because it is surrounded by the very electronegative oxygen atoms. -Ions that have charges of -2 and -3 pick up protons to produce protonated anions, which are named by placing hydrogen (or dihydrogen) and a space in front of the name of the ion. An older, but still common, method for naming some of these ions is to replace the 'hydrogen and a space' with simply 'bi' with no space. Thus, HS 1- is either the hydrogen sulfide ion or the bisulfide ion. -The oxidation state of the metal is given as a Roman numeral in parenthesis when naming inorganic compounds that contain a metal that can attain more than one oxidation state. -Monatomic anions are named by changing the ending of the name of the nonmetal to ide. For example: Cl1- is the chloride ion, O2- is the oxide ion, and P3- is the phosphide ion -If a metal attains the same oxidation state in all of its compounds, then its cation name is the name of the metal. The common metals that have the same oxidation state in all of their compounds are Group 1A and Group 2A metals, Al, Sc, Zn, and Ag. Thus, Sc3+ is the scandium ion and Ag1+ is the silver ion. -If the metal can adopt different oxidations in their compounds (most transition metals and the larger Group 3A and 4A metals), then the oxidation state is indicated with Roman numerals in parentheses after the metal. There is no space between the name of the metal and the parenthesis. For example, Fe2+ is the iron(II) ion, while Fe3+ is the iron(III) ion -Binary ionic compounds are named as the name of the cation followed by the name of the anion. Some examples: -NaCl = sodium chloride (Group 1A are always +1) -Mg3N2 = magnesium nitride (Group 2A are always +2) -FeO is iron(II) oxide because iron also forms Fe2O3, which is iron(III) oxide -Naming Oxoanions of the Elements of Groups 4, 5, and 6: -Suffixes are used to indicate the oxidation state of the central atom in most oxoanions: -ate implies that the central atom is in its highest oxidation state (group number) -ite tells us that the oxidation state of the central is lower than the highest oxidation state by two because the ion has one less oxygen atom than the corresponding ion that ends in -ate. -Removing an oxygen atom from an oxoanion reduces the oxidation state of the central atom by two, but it does not change the charge on the ion. Some examples: -Group 4 CO32- is the carbonate ion because Group 4A Carbon is in the +4 oxidation state -Group 5 PO43- is the phosphate ion because Group 5A phosphorus is in the +5 oxidation state. However, NO31- is the nitrate ion. Removing a single oxygen atom produces NO21-, the nitrite ion. -Group 6 SO42- is the sulfate ion because Group 6A sulfur is in +6 oxidation state. Removal of one oxygen atom produces the sulfite ion, SO32-. -The formula of a polyatomic ion can often be inferred from its name. As an example, we determine the formula of the dichromate ion. We obtain the following from the name 1. chrom chromium, a 6B metal 2. -ate chromium is in its highest oxidation state, which is +6 for a 6B metal 3. di there are two chromium atoms in the formula -The ion is an anion, so there must be enough oxygen atoms at -2 each to cancel two chromium atoms at +6 each and to produce an anion. Six oxygen atoms would produce a compound (no charge), not an ion, so a minimum of seven oxygen atoms is required. Two chromium atom at +6 each and seven oxygen atoms at -2 each and would produce a charge of 2(+6) + 7(-2) = 12 -14 = -2. Thus, the dichromate ion is Cr2O72-A covalent bond results when atoms share one or more pairs of electrons. The electrons that are shared in a covalent bond are called bonding electrons or bonding pairs. -The bond energy (D) is the energy required to break a mole of bonds in the gas phase -The bond length is the separation between two bound atoms at the position of minimum energy. -The equilibrium distance between two bound atoms in a molecule is called the bond length of the bond. The H-H bond, length, which is 0.74 , is a very short bond because the two atoms are so small that they must get very close to interact. The I-I bond length, which is 2.7 , is very long because the two atoms are so large. Most bond lengths lie between 1 and 3 . -The energy of two bound atoms is lower than that of the separated atoms because the bonding electrons on each atom interact with both nuclei. Thus, energy is released when atoms bond. In the case of the H-H bond, 436 kJ is released when two moles of hydrogen atoms combine to form one mole H-H bonds. This means that the minimum energy of two interacting hydrogen atoms is -436 kJ/mol. Separating the two bound atoms requires breaking the bond, which requires an input of the same amount of energy that was released when the bond formed. The energy required to break or dissociate a bond in the gas phase is called the bond or dissociation energy and given the symbol D. The H-H bond energy in H2 is D = 436 kJ/mol is an average bond energy as most bond energies lie between 100 and 1000 kJ/mol. The I-I bond is very weak with a bond energy of only 151 kJ/mol -If the electronegativities of two atoms in a bond are different, then the bond is polar. Polar bonds have bond dipoles, which are represented with arrows pointing from the negative pole toward the positive with a line is through the positive end. -The presence of a positive and a negative pole in a covalent bond produces a bond dipole. -A covalent bond with a bond dipole is said to be a polar covalent bond. -A covalent bond with no bond dipole is a purely covalent bond. -Bond dipoles are frequently represented with arrows pointing from the positive pole toward the negative pole. A line is then drawn through the arrow so as to make a plus sign at the positive end Bond types vary continuously from purely covalent to polar covalent to ionic. There is no clear distinction between a polar covalent bond and an ionic one. -covalent bonds: Bonds with 0.4 have less than 5% ionic character and behave like nonpolar bonds. Consequently, we will refer to them as covalent bonds. The C-H and PH bonds are covalent bonds. -ionic bonds: Bonds with 1.8 are over 50% ionic, so we will refer to them as ionic bonds. KBr and NaCl are ionic bonds. -polar covalent bonds: Bonds that lie between the covalent and ionic limits given above will be called polar covalent bonds. The PO and HCl bonds are polar covalent bonds. -Assume that metal-nonmetal bonds are ionic and nonmetal-nonmetal bonds are covalent, but it is not always true Naming Binary Compounds: -Binary compounds are compounds that contain only two different The elements. name of a binary covalent compound consists of the name of the less electronegative element followed by the name of the more electronegative element with its ending changed to -ide. The number of atoms of each element present in a molecule is given by a Greek prefix except that the prefix 'mono' is not used for the first atom in the formula. Thus, CO is carbon monoxide (not monocarbon monoxide), but N2F4 is dinitrogen tetrafluoride. Many compounds have common names. For example, H2O is water, NH3 is ammonia, and NO is nitric oxide -By convention, the less electronegative element is written first in the formula except if one of the elements is hydrogen. Many hydrogen containing compounds are acids, and, by convention, hydrogens at the beginning of the formula are considered to be acidic hydrogens. Ammonia is written NH 3 even though hydrogen is the less electronegative element, but dihydrogen sulfide is written as H2S because it is an acid number prefix 1 mono 2 di 3 tri 4 tetra 5 penta 6 hexa 7 hepta 8 octa 9 nona 10 deca Lewis Symbols: -The structures of covalently bound molecules and ions depend upon the distribution of the valence electrons. The distribution of valence electrons can be represented by Lewis structures. The Lewis symbol of an atom shows the valence electrons spread in four different regions and remaining unpaired until each region has at least one electron. Lewis symbols represent the bonding atom, not the isolated atom. The number of valence electrons for a main group element is simply the element's group number, so the electron distribution given in the Lewis symbols is the same for all atoms in a group. - Lewis structures almost always show eight valence electrons around each atom. An exception is hydrogen, which has only two - Lewis structures almost always show eight valence electrons around each atom. An exception is hydrogen, which has only two. -Filled or closed shells are very stable and atoms strive toward these configurations when they bond -Nonmetals can also achieve closed shells by sharing electrons. A closed shell for a nonmetal consists of eight electrons (2 s and 6 p electrons). Thus, nonmetals strive to obtain eight electrons, which is called an octet, in their valence shell. This tendency is summarized by the octet rule -nonbonding pairs of electrons are called lone pairs -The number of shared pairs in a molecule can be determined from the number of valence electrons available from the atoms. -SP = [ER - VE] SP is the number of shared pairs in the molecule ER is the number electrons required to give each atom an octet (or duet for H). It is equal to the number on non-hydrogen atoms times 8 (octet) plus the number of hydrogen atoms times 2 (duet). VE is the sum of the valence electrons on all of the atoms in the molecule -Bonding pairs are shown as lines in a Lewis structure, but lone pairs are shown as two dots -The number of bonding pairs in a bond is called its bond order. -Bonds get shorter and stronger as their bond order increases. Determining Lewis formulas To determine the Lewis structure of a molecule 1. Determine the number of shared pairs as before: SP = (ER - VE) where o ER = electrons required = eight times the number of non-hydrogen atoms + 2 times the number of hydrogens. o VE = valence electrons = sum of the valence electrons on all of the atoms in the molecule and any charge on an ion. The number of valence electrons is increased by the negative charge of an anion and decreased by positive charge on a cation. VE is the number of electrons that must be shown in the final Lewis structure. 2. Draw the skeleton of the molecule (connect all of the atoms with single lines), then determine how many more shared pairs must be added and add them. 3. Add lone pairs to atoms to assure that each non-hydrogen atom has an octet. 4. Check your structure to be certain that each atom has an octet and that the number of electrons shown in the structure is the same as the number of valence electrons determined above. Lewis structures should never show double or triple bonds to hydrogen or to a halogen. More than one single bond must be drawn to a halogen when it is the central atom as in ClO41-, but a double or triple bond should not be used. -Resonance structures are Lewis structures that differ only in the placement of the electron pairs. Curved arrows are used to show how the electrons can be moved to convert from one resonance structure into another Determining the number of resonance structures: -Determine the number of shared pairs 1. Determine the number of electrons required with no sharing (ER) 2. Determine the number of valence electrons (VE) 3. Determine the number of shared pairs (SP) Resonance structures form when the number of shared pairs exceeds the number of identical bonding regions in which they can be placed. Resonance structures for molecules or ions having a single central atom occur when the number of shared pairs exceeds the number of regions by one. In this case, the number of structures equals the number of bonding regions in which the extra shared pair can be placed bond order = (number of shared pairs)/(number of bonding regions) *where only the bonding regions in which the double bond can be placed are considered -A common misconception among students is that molecules have bond orders. They do not; bonds have bond orders -The formal charge (FC) on an atom is the charge that the atom would have if the bonds were assumed to be purely covalent. FCA = VE - (NB + BE) -The oxidation state (OX) of an atom is the charge that the atom would have if the bonds were assumed to be ionic so the bonding electrons are assigned to the more electronegative atom in the bond -The charge that is assigned to an atom in a molecule equals the number of valence electrons in the free atom minus the number of valence electrons assigned to the atom in the molecule. The number of valence electrons in the free atom (VE) is given by the atom's group number. The number of electrons assigned to the atom in the molecule is the number of nonbonding electrons (NB) plus those bonding electrons (BE) that are assigned to it -Lewis structures that minimize formal charge are preferred -The best Lewis structure is the one in which all formal charges are closest to zero and negative formal charge resides on the more electronegative atoms 6.1-1. Electron Regions -Atoms obeying the octet rule can have only two, three, or four electron groups VSEPR -The geometry of electron groups around a central atom depends only on the number of groups -The transition from a two- to a three-dimensional image of a molecule can be accomplished with the valence- shell electron-pair repulsion(VSEPR) model. VSEPR is based on the premise that the regions of negative charge around an atom repel one another and thus adopt positions that minimize their repulsion -The three possible orientations of electron groups around an atom that obeys the octet rule are: number of groups 2 3 4 Angle 180 degrees(same plane) 120 degrees(same plane) 109 degrees(not same plane) -If all of the electron groups around a central atom are not identical, the predicted bond angles are only approximate -The angles apply only to situations where all four electron groups are the same, which is not all that common. Thus, the angles between electron groups around an atom that obeys the octet rule will be 180o or close to 120o or 109o, but it will equal 120o or 109o only when all groups are the same. -Deviation from the predicted angles can be caused by differences in the size of the bound atoms because large atoms tend to move apart to avoid 'bumping' one another. Deviation also results because the interactions between lone pairs and bonding pairs are different. Bonding pairs have a more diffuse charge density than do lone pairs, so other electron regions tend to move away from lone pairs. The following gives the relative strengths of the interactions: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair -bonding pairs move away from the lone pairs by moving closer to one another. The deviation from the predicted angles increases with the number of lone pairs. Use the following to predict relative bond angles: 1. Determine the number of electron groups around the atom where the angle forms. 2. If there are no lone pairs and the atoms are nearly the same size, the angle will be 180o, 120o, or 109o. 3. If there are lone pairs, the angles decrease from the values predicted in Step 2. The deviation is greater for two lone pairs than for one -The locations occupied by the lone pairs are not used when describing the shape of a molecule -The shapes shown in Figure 6.2 show the orientations that can be adopted by the electron groups surrounding a central atom that obeys the octet rule. However, we can determine the positions of only the atoms, not the lone pairs, so a molecular shape describes the shape adopted by only the atoms not the electron groups. The lone pairs help establish what that shape is, but the name of the shape applies only to that taken by the atoms. 1. -Determine the number of electron groups around the central atom. 2. Determine which of the shapes shown in Figure 6.2 applies. 3. Name the molecular shape adopted by the atoms. -Molecules, such as PF5, that have five electron regions around a central atom adopt a trigonal bipyramidal structure -Molecules, such as SF6, that have six groups assume an octahedral structure -Rules for Mixing Use the following rules when mixing orbitals: 1. The number of orbitals produced must always equal the number of atomic orbitals used to construct them. 2. Regions in which the atomic orbitals have the same sign (shading) add constructively, which makes the produced orbital larger in that region, but regions where the orbitals have opposite phases add destructively, which makes the combined orbital smaller in that region Sigma Bonds -In all of the examples shown in Figure 6.7, the overlap region lies on the line between the two atoms, which is called the internuclear axis. Bonds in which the bonding electron density falls on the internuclear axis are called sigma bonds. Pi Bonds -The electron density in a pi bond lies above and below the internuclear axis, not on it. Thus, Sigma bonds result from end-on overlap of s or p orbitals and place electron density on the internuclear axis, while pi bonds are produced from side-on overlap of p and or d orbitals and place a nodal plane through the internuclear axis -All bonds contain one and only one sigma bond. Double bonds contain one Sigma and one Pi bond, and triple bonds contain one Sigma and two Pi bonds. The bond order of a bond is simply the sum of the number of Sigma and Pi bonds that it contains. -Orbitals produced by combining two or more atomic orbitals on the same atom are called hybrid orbitals, and the process by which they are formed is called hybridization -The hybrid orbitals that adopt the geometries for two, three, and four electron groups can be constructed from combinations of s and p orbitals. Hybrid orbitals are used to hold lone pairs and to form Sigma bonds. The remaining p orbitals (those not used to construct the hybrid orbitals) are used to form Pi bonds. Bonding versus Antibonding Orbitals -In MO theory, atomic orbitals (AO's) are combined to form molecular orbitals (MO's). The number of MO's created is always equal the number of AO's used to create them. Thus, combining two AO's produces two MO's, which differ in the way in which the AO's are combined. 1. Bonding MO's A combination of two AO's that have the same phase in the region between the two nuclei results in a bonding MO. Bonding MO's are characterized by an accumulation of electron density between the two atoms. 2. Antibonding MO's A combination of two AO's of opposite phase in the region between the two nuclei results in an antibonding MO. Antibonding MO's are characterized by an annihilation of electron density between the two atoms. That is; antibonding interactions contain nodal planes perpendicular to the bonding axis. Antibonding MO's are designated with a "*". For example, the Sigma* and Pi*, (pronounced "sigma star" and "pi star") are the antibonding combinations that contain nodal planes perpendicular to the bonding axis. -Bonding MO's increase the electron density between the nuclei, while antibonding MO's contain nodal planes perpendicular to the internuclear axis. -Combining two s-orbitals results in electron density on the bonding axis, so both combinations are classified as Sigma. The combination of orbitals of the same phase increases electron density on the bonding axis, so it is the bonding Sigma orbital. The combination of opposite phases produces a nodal plane perpendicular to the bonding axis, so it is the antibonding Sigma* orbital. -Head-on combination of two p orbitals results in electron density on the bonding axis, so both combinations are classified as Sigma. The combination of orbitals of the same phase increases electron density on the bonding axis, so it is the bonding Sigma orbital. The combination of opposite phases produces a nodal plane perpendicular to the bonding axis, so it is the antibonding Sigma* orbital. -Antibonding orbitals are very high in energy -Each molecular orbital is characterized by an energy level, and the electrons in a molecule fill the molecular energy levels in the same manner that they fill atomic orbitals. That is; the electrons fill the molecular energy levels at lowest energy while obeying both Hund's rule and the Pauli exclusion principle -Many of the properties of a molecule are dictated by the nature of and energy difference between occupied MO that is highest in energy and the unoccupied MO that is lowest in energy. These two orbitals are referred to as the HOMO and LUMO. HOMO; The Highest Occupied MO LUMO; The Lowest Unoccupied MO -The energy changes resulting from the combinations of two s-orbitals are shown in an MO diagram like the one shown in Figure 6.15. Three important characteristics of these diagrams are: The energy of bonding interactions is lower than that of the atomic orbitals by delta E. The energy of antibonding orbitals is higher than that of the atomic orbitals by (deltaE*) deltaE* > deltaE -Bond order = Number of bonding pairs - Number of antibonding pairs -MO's can be bonding, nonbonding, or antibonding 1. -The number of MO's equals the number of atomic orbitals used to construct them. 2. Each MO contains one more nodal plane than the MO that it is immediately beneath it in energy. There are no nodal planes in the lowest energy MO, and there is a nodal plane between each pair of atoms in the highest energy MO. 3. The nodal planes are placed symmetrically even if it means placing them on an atom. 4. Nodal planes cannot be placed on adjacent atoms. 5. Nodal planes are not placed on terminal atoms. The bonding character of an MO spread over several atoms depends upon the relative number of bonding and antibonding interactions. In the following discussion, we use B = number of bonding interactions and A = number of antibonding interactions. B > A: The MO is bonding, and its energy is lower than the energy of the AO's used to construct it. B < A: The MO is antibonding, and its energy is higher than the energy of the AO's used to construct it. B = A: The MO is nonbonding, and its energy is close to the energy of the AO's used to construct it. Four-atom Example: Butadiene

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Chapter 4: Analysis of Financial Statements I. Liquidity Ratios Use 2002 statements for Allied Food Products for all ratio examples. A. Current Ratio = CA / CL = 1,000 / 310 = 3.2 times B. Quick Ratio = CA Inv. / CL = 1,000 615 / 310 = 1.2 timesI
N.C. State - BUS - 320
Chapter 15 I. CashWorking Capital ManagementA. Why do firms hold cash? 1. 2. 3. 4. Transaction Motive day to day cash transactions Precautionary Motive in case of less predictable CF's Speculative Motive take advantage of bargain purchases Com
N.C. State - BUS - 320
Chapter 5: The Financial Markets I. Financial markets bring together those wanting to borrow money with those having surplus funds. They serve an intermediaries that facilitate the movement of funds between lenders and borrowers as well as between bu
N.C. State - BUS - 320
Chapter 14 I.Distributions to ShareholdersHow does dividend policy affect stock price? 3 Basic Views: 1) Dividend policy is irrelevant; investors can design their own dividend policy. If they are not receiving enough, sell shares. If receiving to
N.C. State - BUS - 320
Chapter 11Basics of Capital BudgetingCapital Budgeting: Planning expenditures for assets whose cash flows extend beyond one year.Project S Year Cash Flows 0 -$1,000 1 500 2 400 3 300 4 100Project L Cash Flows -$1,000 100 300 400 600Note: 1.
N.C. State - BUS - 320
Chapter 1: Overview of Financial Management I. Three main forms of Business Organization A. Sole proprietorship 1. Unincorporated business owned by one individual. 2. Advantages: easily and inexpensively formed, subject to few government regulations,
N.C. State - BUS - 320
Chapter 3: Financial Statements, Cash Flows, and Taxes I. The Balance Sheet Allied Food Products: December 31 Balance SheetsASSETS20052004 $80 315 415 $810 870LIABILITIES &amp; EQUITY A/P Notes Payable Accruals Total Current Liab. Long Term Debt
The University of Akron - PHYS - 292
Chapter 39 p. 1CHAPTER 39 Quantum Mechanics Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc
The University of Akron - PHYS - 292
Chapter 40 p. 1CHAPTER 40 Quantum Mechanics of Atoms Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E
Northeastern - BIO - 101
Student Questions on Recent Lectures[thanks to students submitting questions!]Chapter 8 Q. Why do the axons cross over to the opposite side of the spinal cord in fig. 8.1 and 8.6a? A. Both pain and tactile information end up in the opposite cereb
The University of Akron - PHYS - 292
Chapter 41p. 1CHAPTER 41 Molecules and Solids Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf
The University of Akron - PHYS - 292
Ch. 42 Page 1CHAPTER 42 Nuclear Physics and Radioactivity Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm.1.To find the rest mass of an particle, we subt
UCF - EUH - 2000
Hellenic Age- Graco meant foreigner, name for themselves- Hellas Hellas- the conditions Semi-arid, not much rain No major navigable rivers, little rain Sparse woodlands Pasture land rare, not a lot of cattle, horses Mountainous &quot;wine-dark sea&quot;, tempe
The University of Akron - PHYS - 292
Ch. 43 Page 1CHAPTER 43 Nuclear Energy; Effects and Uses of Radiation Note: 1. A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2 /4p(8.85 1012 C2/N m2)= 2.30 1028 J m = 1.44 MeV fm.We find the product nucleus by bal
UCF - EUH - 2000
Hellenistic Key Terms Hellas- Greece Hellenic- Greek Hellenistic- Greek-like Macedonia- north of Greece, divided into low lands &amp; high hinterlands, good resources- iron &amp; timber &amp; open pasture land, beset on all sides by enemiesIllyrians, Thracians (
The University of Akron - PHYS - 292
Ch. 44 Page 1CHAPTER 44 Elementary Particles Note: A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.631034 J s)(3108 m/s)(109 nm/m)/(1.601019 J/eV); E = (1.24103 eV nm)/ = (1.241012 MeV m)/. 7.29 Ge
The University of Akron - PHYS - 292
Ch. 45Page 1CHAPTER 45 Astrophysics and Cosmology Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm. 1. Using the definition of the parsec, we find the equiv
The University of Akron - PHYS - 292
Chapter 1 CHAPTER 1 - Introduction, Measurement, Estimating 1. (a) Assuming one significant figure, we have 10 billion yr = 10 109 yr = 1 1010 yr. 10 yr)(3 107 s/yr) = (b) (1 10 3 1017 s. (a) (b) (c) (d) (e) (f) (g) (a) (b) (c) (d) (e) (f) (a) (b) (c
UCF - EUH - 2000
The Dark Ages: Premise The centuries following the demise of the Roman Empire New order needed in Roman vacuum New govts form, they do not bring stability The Roman Catholic Church steps in Rise of Islam, break w/ Constantinople lead to a focus north
The University of Akron - PHYS - 292
Chapter 2CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which givest = 5.0 h. 88 km/h.2. 3.We find the average speed from average speed = d/t = (280 km)/(3.2 h) =
UCF - EUH - 2000
Frederick Barbarossa/Holy Roman Empire Domesday Book Angevin Dynasty Henry, Richard, John, Edward of England (know Scandinavians- Vikings) Premise -monarchs attempt to centralize gov't -result of personal ambition, NOT nationalism or ethnic identity
UCF - THE - 2000
As I walked into the theater, I was pleasantly surprised when I saw the set onstage. It was extravagant, to say the least. It being a smaller theater, I wasn't expecting what I saw. The seats I had reserved happened to be the best seats in the house!
UCF - THE - 2000
Chapters 1-6, intro Scantrons, 50 on each side 4078967365 ogg- caveman middle east- feel that acting is an offence to Allah some feel acting creates false representation of man shadow puppets, shadow becomes character- so to not offend Allah goes to
UCF - EUH - 2000
Western civ overview History is about change Interaction, geography, etc, change the way people live, think, &amp; behave Summer, Babylon, mid east: beginnings of national organization Professionalism Beginnings of global trade (even before empires) Reli
UCF - EUH - 2000
Sugar Europe gets a taste during the Crusades Levant returns to Muslim hands, &amp; with them the cane fields Cyprus- set up sugar plantations Cyprus uses slaves from Byzantine slave market 1453: Constantinople falls to the Turks, cutting off the slave i
UCF - EUH - 2000
The end of the renaissance Niccolo machiavelli Florentine author Believer in the Florentine republic Effect on usa: checks &amp; balances The dour realist Anxiety for italys future Appeal for a strong leader: the Prince (1513) Wrote it to Lorenzo medichi
UCF - EUH - 2000
Cultural Comparison Through ComedyOlivia LoucksWestern CivilizationWednesdays 6pmThe cultures of ancient times and today have many similarities and differences that can be shown through comedy. Through analyzing a work of comedy from ancient tim
UCF - FRE - 2200
Voila! La manire nouvelle et moderne d'emporter l'picerie!Les couturiers comme Stella McCartney et Consuelo Castiglioni de Marni ont dessine les cabas trs lgants!Il y a beaucoup de clbrits qui utilisent les cabas comme Julia Roberts, Keira Knight
UCF - FRE - 2200
Porte-feuille #3 Allez sur le site web: (you can copy and paste it) http:/www.radio-canada.ca/audio-video/index.shtml#idMedia=1529681&amp;urlMedia=http:/www.radiocanada.ca/Medianet/2007/CBFT/5Sur5200711171700_1.asx&amp;dureeMedia=451 Regardez le reportage et
UCF - FRE - 2200
Bonjour, Aujoud'hui nous allons parler de La Cte d' Ivoire. La Cte d'Ivoire est un beau et charment addition la communaut francophone. Elle se situe l'ouest de l'Afrique. La Cte d'Ivoire est aussi de la meme que Nouveau Mexique avec neuf fois la po
UCF - FRE - 2200
9/6/07 Les jeunes en Guadeloupe, Vous avez dit que les jeunes nord-americains ont beaucoup d'argent a depenser et qu'ils consomment beaucoup. Quand on compare avec d'autres pays, c'est vrai. Mais, moi meme, je fais d'economies beaucoup. Chaque mois,
UCF - FRE - 2200
++Alpha Blondy (Cote d'Ivoire)Alpha Blondy est un chanteur de reggae ivoirien. Il est ne Seydou Kone, Dimbokro, Cote d'Ivoire, le 1er janvier 1953. Il etait le premier de neuf enfants dans sa famille. En 1973, il allait aux Etats-Unis pour etud
UCF - FRE - 2200
Le week-end dernier, je suis rentree a la maison. Vendredi, je suis allee a une fete d'anniversaire pour mon ami John. Je lui ai donne vingt dollars. Apres la fete, je suis rentree a la maison avec mon copain Joe et mes meilleurs amis Jane et Chris.
UCF - SYG - 2000
Cohabitation Remaining single Marriage without children Homosexual relationships Family rights for homosexuals Adoption Several states ban adoption by homosexuals Marriage Legal in 1 state (Mass) No explicit prohibition- 4 states and Washington D.C.
UCF - THE - 2000
I walked into the theater with the stage set as a restaurant. It was very elaborate and was excited to see this show. I had actually planned on auditioning for the show, but with my foot surgery this past summer, that didn't happen. I had gotten the
UCF - EUH - 2000
Works Cited&quot;Lysistrata, by Aristophanes.&quot; About.Com. 2008. 26 Feb. 2008 &lt;http:/ancienthistory.about.com/library/bl/bl_text_aristophaneslysistrata1.htm&gt;.Anchorman: the Legend of Ron Burgundy. Dir. Adam McKay. Perf. Will Ferrell, Christina Applegat
UCF - EUH - 2000
Black Death might not have been the Bubonic Plague Believed spread by fleas on rats Spread to Iceland, so maybe spread by person to person Starts in Italy, arriving from East Waves of plague traveled back and forth across Europe Some places lose up t
The University of Akron - PHYS - 292
Chapter 3CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45 D 2y D 2 = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =
Radford - ENG - 102
Maura Cashen Dr.Ren ENGL 102 7 February 2008 Does Water Have a Future? Everyone has a passion. For some, this passion is out of the ordinary like Sandor Ellix Katz who has a passion for food. You may say that a passion for food is not unique; however
Radford - MUSIC - 121
NAME: _Maura Cashen_ MUSC 121 Test #2 February 29, 2008 DIRECTIONS: I. Save file as MUSC121_ (Insert Last Name Here). For example, if I were taking this test I would save it as MUSC121_Isley. Please be SURE to also list your name at the top of the pa
The University of Akron - PHYS - 292
Chapter 41CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ~ 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: ?F = ma; 255 N = m(2.20 m/s2), which
The University of Akron - PHYS - 292
Chapter 5CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F kFN = 0; y-component
Radford - MUSIC - 121
NAME: _Maura Cashen _ MUSC 121 Test #3 March 31, 2008 DIRECTIONS: I. Save file as MUSC121_ (Insert Last Name Here). For example, if I were taking this test I would save it as MUSC121_Isley. Please be SURE to also list your name at the top of the page
The University of Akron - PHYS - 292
Chapter 6CHAPTER 6 - Gravitation and Newton's Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.67 1011 N m2/kg2)(5.98 1024 kg
Radford - ENG - 102
Maura Cashen English 102 February 7, 2008 Stuff Essay &quot;Every women's best friend&quot; Most women wear one of the same cosmetics everyday day. Even women, who claim they are not into wearing make-up, probably dap some of this on themselves before walking
The University of Akron - PHYS - 292
Chapter 7CHAPTER 7 - Work and Energy 1. 2. 3. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force,
The University of Akron - PHYS - 292
Chapter 8CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ! kxf2 0; 35.0 J = ! (82.0 N/m)xf2 0, which gives xf = 2. 3. 4. For t
Oklahoma Christian - ECONOMICS - 2113
Macroeconomics Homework Chapter 8:1-16 1. Determine whether each of the following statements is true or false a. Some people who are officially unemployed are not in the labor forcefalse b. Some people in the labor force are not working- true c. Ever
The University of Akron - PHYS - 292
Chapter 9CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled gases from F = ?p/?t = (?m/?t)v = (1200 kg/s)(50,000 m/s) = 6.0 107 N. An equal, but opposite, force will be exerted on the rocket: 6.0 107 N, up. For the mome
Oklahoma Christian - ECONOMICS - 2113
Homework Chapter 7:1-10 1. Identify the component of aggregate expenditure to which each of the following belongs a. A U.S. resident's purchase of a new automobile manufactured in JapanNet Exports b. A household's purchase of one hour of legal advice