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Midterm Test

Midterm Test. Tuesday, October 10th (This Tuesday) at 7:30pm. Practice is test on website. Section A, M.W.F. 8:00 – 8:50 am, Room D634. 90 minute test. Chapters: 1, 2, 7, 8. Review. Chapter 1. Units. SI System. Dimensional Analysis. Scientific notation. Prefixes and suffixes.

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Midterm Test

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  1. MidtermTest Tuesday, October 10th (This Tuesday) at 7:30pm Practice is test on website. Section A, M.W.F. 8:00 – 8:50 am, Room D634. 90 minute test Chapters: 1, 2, 7, 8

  2. Review Chapter 1 Units SI System Dimensional Analysis Scientific notation Prefixes and suffixes Significant figures Propagation through addition and multiplication Precision/Acuracy

  3. Chapter 2 Dalton’s atomic theory of matter Elemental Forms Chemical & physical properties Subatomic particles - protons, neutrons, and electrons Models of the atom – Rutherford's model Atomic number and mass number - # of n’s, p’s, & e’s Isotopes - calculating average atomic mass and percent abundance Elements - names and symbols Avogadro’s number and the mole Periodic table (groups and periods) – Common properties in each group

  4. Chapter 7 Properties of waves - wavelength, frequency, amplitude, speed Electromagnetic spectrum - speed of light Planck’s equation and Planck’s constant Wave-particle duality for light , electrons, etc.) Atomic line spectra: Balmer and Rydberg Series Ground vs. excited states Heisenberg uncertainty principle Bohr and Schrödinger models of the atom Quantum numbers (n, l, ml) Shells (n), subshells (s,p,d,f) Shapes and properties of atomic orbitals –#nodes, # lobes

  5. Chapter 8 Diamagnetic vs. paramagnetic vs. ferromagnetic substances Electron spin Pauli exclusion principle Orbital box diagrams Electron configurations Aufbau order and its exceptions. Predicting ions using electron configurations Core vs. valence electrons Effective nuclear charge Periodic trends (atomic radius, ionization energy, electron affinity, ionic radius)

  6. Chapter 9 Bonding and Molecular Structure: Fundamental Concepts Countless arrangements of atoms are possible!! Bonding behaviour: What does it depend on? How do atoms form bonds? What kind of bonds exist? How do we predict bonding behaviour?

  7. 2s22p2 1s2 2s22p6 1s2 Valence Electrons Given an electron configuraton one can group the electrons into the core and valence e’s Ex) C 6 e’s core valence Atoms can form bonds by sharing or exchanging valence e’s. The core e’s don’t participate, just like the noble gas is unreactive. Then main objective for each atom is to achieve a noble gas configuration for its valence electrons Ex) Ne Has a full valence shell with 8 e’s C needs 4 more e’s to achieve a full shell. How?

  8. • N • C • • • • • • • Electron Dot Structures (G. N. Lewis) The first row elements will either want to achieve He or Ne configurations. Those who attain Ne configurations need 8 e’s in their valence shell The number of valence electrons can neatly be depicted by arranging them around the atomic symbol Ex) Ex) C has 4 valence e’s N has 5 valence e’s

  9. Electron Dot Structures Electrons placed around the four sides of the atom symbol individually or in pairs depending on the # of valence e’s. Initially on e’s is placed on each side and are paired up when all four sides are occupied, after which they are paired until the valence shell is full.

  10. Octet Rule Notice that all noble gas configurations have an outer shell with ns2.. . np6. Ne = 2s22p6 Kr = 4s23d104p6 Rn = 6s25d104f146p6 Ar = 3s23p6 Xe = 5s24d105p6 Ignoring the d and f e’s the valence shell contains 8 e’s. The d and f subshell contributions can be ignored if they are full Therefore the valence e’s in groups 13(3A) to 18(8A) can be described in terms of ns and np e’s In each case atoms want to have 8 electrons hence the Octet rule.

  11. Cl + Cl • Cl • • • • • • • • • • • • • • • • • • • • • • • • • Cl • • Octet Rule An element can achieve octet status in one of three ways: 1. It can gain valence electrons to make an anion: Ex) Cl [Ne]3s23p5 Cl-[Ne]3s23p6=[Ar] 2. It can lose valence electrons to make an cation: K+ [Ar] Ex) K [Ar]4s1 3. It can share valence electrons with another atom to make a covalent bond. This will generally involve two atoms of similar electronegativity (ability to attract electrons) Ex) Cl2

  12. Octet Rule Nonmetals normally gain electrons to obtain a complete octet. Ex) Cl to Cl- O to O-2 S to S2- P to P3- Metals normally lose electrons to obtain a complete octet. Ex) Na to Na+ Ca to Ca2+ Al to Al3+ Atomic charge = # protons – # electrons Ex) Al3+ has 13 protons and 10 electrons q = 13 – 10 = +3 Elements in groups 1, 2 and 3, ionic charge is group #: Ca: 2(A) → q = +2 Li: 1(A) → q = +1 Ex) Al: 3(A) → q = +3 Elements in groups 15, 16 and 17, ionic charge is group # -18: Ex) As: 15 → q = -3 S: 16 → q = -2 F: 17 → q = -1

  13. Cl + Na - + Na Ionic Compounds Recall that opposite charges attract each other. A cation and an anion, experience an electrostatic force, given by: e =unit of charge k =8.988 × 109N m2/C2 Coulonb’s Constant This force pulls them together to make an ionic bond. q1 q2 r r Cl -

  14. The energy released to form an ionic bond: Exercise Compute the force that a sodium cation and chloride anion experience when 10.00 nm apart q1 = +1 q2 = -1 F = ke2q1q2/r2 r = 10 nm = 1.000*10-8 m F = (8.988 × 109N m2/C2)(1.6022 *10-19 C)2(1)(-1) (1.000*10-8 m)2 F = -2.307*10-12 N

  15. Compute the energy of formation of the ionic bond in NaCl, where the bond length is 279 pm q1 = +1 q2 = -1 E = ke2q1q2/r r = 279 pm = 2.79*10-10 m E = (8.988 × 109N m2/C2)(1.6022 *10-19 C)2(1)(-1) (2.79*10-10 m) E = -8.27*10-19 N m = J How energy for one mole of NaCl bonds? E(total) = E(bond) *(# of bonds) = (-8.27*10-19 J/bonds)(6.022*1023 bonds/mol) = 498,000 J/mol = 498 kJ/mol

  16. LiF formation Li loses all e’s in its valence shell Large reduction in atomic radius: Li – 152 pm Li+ - 78 pm Vol = 1/7th Ionization Energy consumed As an atomic gas. F gains electron its valence shell Expansion in the radius because effective Z* is decreased due to extra electron. F – 71 pm F- - 133 pm Vol = 6x Electron Affinity Energy released As an atomic gas.

  17. LiF formation Energy is released to form the ionic bond

  18. Ionic Materials Lattice-3-D pattern of ions - Minimize repulsive forces -Maximize attractive forces - No net charge - Large lattice energy released - High melting point - Brittle

  19. Formation of ionic materials ElectronAffinity Whether element from ionic compounds depends on the balance between: 1) Ionization energy 2) Electron affinity 3) Lattice energy 4) Phase transition energies 5) Bond energies Dissociation energy Ionization energy Lattice energy vaporization Formationenergy

  20. Covalent Compounds Covalent bond - valence electrons are shared between two or more atoms. i.e. bonding electrons are “co-valent”. The combination of electrons forms one entity. In Ionic systems - the cation and anion are separate entities The wavefunctions for each electron constructively interfere and form a combined wavefunction, called a molecular orbital 1s 1s(1) 1s(2) Molecule Atom 2 Nucleus 1 Nucleus 2 Atom 1 Electrons are shared to complete the valence shell of each atom

  21. Covalent Compounds Covalent bonds form if energy is released when atoms bond: The negative energy of reaction means that the product (H2) is more stable than the reactants (2 × H). 2 electrons are shared completing the 1s orbital for each H Bond dissociation energy -Energy is required to break the bond. The lowest energy point is at the average bond length.

  22. Lewis structure diagrams Cl • Cl - • Cl Cl Cl + Cl • Cl • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Cl • • The exact solutions of molecular quantum mechanics are highly complex The Lewis dot diagrams of atoms can be combined to depict bonding in molecules These Lewis diagramsreflect the underlying quantum mechanics and serve primarily as a bookkeeping device for the valence electrons in the molecule The guiding principle behinds Lewis diagrams is that each atom in molecules achieve noble gas electron configuration by sharing electrons This is known as the octet rule because the majority of noble gases have 8 valence e-, except for H which only requires two e’s to complete its valence. The electrons of a chemical bond are represented by a dash There can also be non-bonding electrons, which are written as dots ●●

  23. Steps for Drawing Lewis Electron Dot Structures • Determine the central atom. The rest are terminal atoms. 2. Determine the total number of valence electrons. 3. Use one pair of electrons to make a single bond between each pair of bonded atoms. 4. Use any remaining electrons as lone pairs around each terminal atom (except H) so that each terminal atom has a complete octet, if possible. 5. Allow for any deficit or excess of e’s only for the central atom. 6. Check the central atom for too few or too many electrons. 7. If there are too few electrons, increase the bond order of one or more bonds by sharing non-bonded electrons. (i.e. make double or triple bonds, as necessary) 8. Calculate formal charge for all atoms, and indicate any which are not zero.

  24. F F F F C • • C • • • F F C F F • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • F • Ex) CF4 Central Terminal 4 v.e’s 7 v.e’s

  25. Tips When finished review it to verify that correct number of atoms and electrons were used and that the octet rule is obeyed Remember that Lewis dot structures do not show a molecule’s shape. If there is more than one acceptable solution, the true electron distribution is a hybrid of the possible distributions. This is called resonance. If it is impossible to avoid having one atom with too few or too many electrons, make sure it is the central atom. Elements in the 1st or 2nd period can never have more than eight electrons under any circumstance. Molecules with odd numbers of electrons form free radicals and cannot fully obey the rules.

  26. Tips Larger molecules are treated as a sequence of central atom problems. The central atom can be generally be chosen using a few rules: 1. The central atom is never H or F. 2. If you have many atoms of one element and one atom of another, the lone atom is the central one. 3. Given a choice, the central atom is not O. 4. Given a choice, C is central. Ex) Choose the central atom for the following molecules: (a) BBr3 (b) CH3Cl (c) CH2O (d) POCl3

  27. H H H • • • • H H H • • C C • • • • • • • • • - - H H H H N H H H H H C C H • • H H N • • • • • • • Draw Lewis dot structures for the following molecules (a) NH3 (b) C2H6 (c) C2H4 (d) C2H2 a) b)

  28. H H H H • • • • • • • • H H H H • • • • C C C C • • • • • • • • • • • • • • • • • • • H H H H C • C • • • • H C C H C • • • C • • • • H • C C H H H Draw Lewis dot structures for the following molecules (a) NH3 (b) C2H6 (c) C2H4 (d) C2H2 c) d)

  29. O Cl • • • • • • • • • • • • • • We have miscounted the e’s Ex) ClO- The bonding e’s have been double counted!!! [ ]q The bonding e’s are shared and therefore count only as half an electron for each atom Extra e added to complete Valence on O. #e = core e’s + ½(bond e’s) + non-bond e’s #p = 17 #p = 8 #e = 18 #e = 9 # e’s Cl = 10 + ½ (2) + 6 = 17 10 core 8 valence 2 core 8 valence qCl = 17 – 17 = 0 # e’s O = 2+ ½ (2) + 6 = 9 qCl = #p - #e = 17 – 18 = -1 qO = #p -#e = 8 – 10 = -2 qCl = 8 – 9 = -1 Therefore the –ve charge is on O Total charge = -3 ???? It needs it to complete its valence

  30. F F B • F • • • • • • • • • • • • • • • • • • • • • • • Formal Charge Formal charge (Qf) is the charge on an atom assuming that every bond is completely covalent. All bonding electrons are shared equally Qf = #p’s – [ (# core e’s) + ½ (# bonded e’s) + (# non-bonded e’s)] Qf = (# valence e’s) – [ ½ (# bonded e’s) + (# non-bonded e’s)] As a general rule, we want to keep the formal charge on each atom as close to 0 as possible (without giving any atom more than a complete octet). Ex) BF3 Qf(B) = 3 –[½ (6) + 0] = 0 Qf(F) = 7 –[½ (2) + 6] = 0 Note: valence of B is incomplete!!

  31. • B • • • F - - B F F F F F • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Let’s try another arrangement taking an extra set of bonding electrons from one of the F’s -1 Qf(B) = 3 –[ ½ (8) + 0] = -1 0 0 Qf(F) = 7 –[ ½ (2) + 6] = 0 or Qf(F) = 7 –[ ½ (4) + 4] = +1 +1 Total q = -2 + 0 + 1 + 1 = 0 Therefore the previous arrangement is preferred Large charge separation!!! ]0 0 0 0 0

  32. O O O O O O N • • • ]- N O N • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • O O • Consider the anion NO3-. 0 Incomplete valence on two O’s Lets try to fill them using the remaining 2 e’s on N 0 -1 0 Charge separation is the same As above 0 -1 0 Total charge is -1 All valences are complete 0

  33. ]- ]- N O N O O O -1 -1 -1 O = N = O O = N _ O O _ N = O • • • • • • • • • • • • • • • • • • • • • • • • • • • • O O O O O Resonance This arrangement is also correct? This one too!!! When there are several permutations that are equivalently correct, the actual situation is an average between them. This phenomenon is known are resonance, which can be depicted using bidirectional arrows

  34. .. .. .. .. .. .. .. .. : : : : O _ O = O O _ O _ O .. .. .. .. .. O = O _ O .. .. .. .. .. O = O =O .. .. . C C C . . C C C C C C . . . C C C C C C C C C Ex) O3 X -1 -1 +1 -1 -1 +1 X Ex) Benzene, C6H6 +2 ?

  35. F- F F Li+ d Bond polarity and electronegativity In a symmetric molecule such as H2, the concept of covalence is unambiguous: the electrons are evenly shared by the two atoms. But in LiF, we have complete electron transfer: Li+ and :F-, i.e. ionic bonding There is a continuum of behaviour between these two extremes Ionic bonds are completely polarized towards the opposite charged ions. As the electronegativity difference decreases, e’s are more likely to be shared, but unequally: polar covalent bonds In polar covalent bonds, there is a bond dipole, which is indicated by a vector There are also partial positive and negative charges, indicated by δ+ and δ-

  36. Pauling Electronegativity We use Pauling electronegativity values to determine bond polarity. Electronegativity is the ability of an atom in a molecule to attract electrons to itself. • A difference in Pauling electronegativity between elements of more than 2 units is enough to cause ionic bonding General trend in element electronegativity

  37. .. .. .. .. : H _C = S _ O O _ C N : : .. .. O = C = O .. .. .. H Evaluating Lewis diagrams - Revisited How do you chose between several possible Lewis diagrams? • Most important is achieving the Octet rule: any structure that obeys the octet rule is better than any structure that does not • 2. Any structure that minimizes the sum of the absolute values of the Formal Charge is better • 3. The diagram that associates negative Formal Charge with more electronegative elements and positive Formal Charge with electronegative elements is to be preferred over others Let us now apply this rule to some more complex cases: 1) OCN- (cyanate) 2) CO2 (carbon dioxide) 3) H2CSO (sulfine) -1 +1 -1

  38. .. .. .. .. O _ C N O C _N : : : : O = C = N .. .. .. .. .. .. .. .. : : : : O _ C _ O O _ C O .. .. .. .. .. : : O C _ O .. .. .. .. .. .. : H _C _ S = O H _C = S _ O .. .. .. .. O = C = O .. .. H H Let us now apply this rule to some more complex cases: 1) OCN- (cyanate) 2) CO2 (carbon dioxide) 3) H2CSO (sulfine) -1 -1 +1 -1 -1 -2 -1 -1 -1 -1 +1 -1 +1

  39. Exercise State whether the bond is ionic or covalent and draw the dipole vectors and partial charges for: 2.1 2.8 C 3.5 2.5 C 2.1 4.0 C 0.8 4.0 I H - Br O - C H - F K - F d- d+ d+ d- d- d+ d+ d- 2.5 2.8 C 4.0 2.8 C 3.5 0.9 I 3.5 2.5 C C - Br F - I O - Na O - S d- d+ d+ d+ d- d- d+ d-

  40. %-ionic character Continuum in behaviour between pure covalent and ionic character Molecules such as H2, F2 are at 0 on this scale Common “ionic” compounds such as NaCl cover a wide range on this scale but are all over 2 The formal charge separation gives rise to ionic forces of attraction contributing to the bond energy. Therefore all bonding must be considered as having ionic and covalent contributions.

  41. Partial Charge The bond dipole can be quantified by calculating the partial charge (Q) on each atom. Partial charge, similar to formal charge, except that the electronegativities of the atoms, govern the distribution of the bonding electrons between the atoms. Recall the formula for formal charge Therefore, pure covalency assume electronegativities are the same Q(Cl) = 7 – [6 + {3.0/(2.1+3.0)}*2] = -0.2 Ex) H-Cl Q(H) = 1 – [0+ {2.1/(2.1+3.0)}*2] = 0.2

  42. Bond order: length and energy We have seen that chemical bonds between the same pairs of elements may be single, double or triple bonds Quadrupole bonds also exist between some of the d-elements How do these bonds differ? The higher the bond order, the shorter will be the bond will be. The higher the bond order the larger the bond dissociation energies • This can be illustrated by some examples:

  43. Predicting Bond Lengths Bond lengths shows the same trends as atoms size Size decreases from left to right across the Periodic Table, and increases down any group 143 154 147 136 C—C C—N C—O N—OC—Si C—P C—S Si—Si Si—P Si—S P—S 194 187 181 214 234 227 221 Increasing the bond order always shortens the bonds; however the %-shortening is not a very regular parameter, therefore not simply predicted.

  44. Lewis diagrams and Molecular Shape There is a direct link between Lewis diagrams and molecular shape The Valence Shell Electron Pair Repulsion theory states a molecule’s shape is determined by the electron pairs that surround central atoms These electron pairs(EP) include both bond pairs (BP) and lone pairs (LP) Four electron pairs define the tetrahedral shape family, as in SiCl4 We define five shape families based on the need to accommodate mutually repelling electron pairs around a central atom The balloons illustrate these “natural” shapes

  45. The 5 Shape Families The five shape families that electron pairs develop are illustrated above Linear – Trigonal-planar – Tetrahedral – Trigonal-bipyramidal – Octahedral In this case each molecule has all its electron pairs as bond pairs around the central atom Not all the shape-determining electrons pairs need be bond pairs

  46. X .. .. X X X .. .. X General Shape Family Scheme (1) 0 LP 2 BP 0 LP 3 BP 1 LP 2 BP 2 LP 1 BP 2 EP’s 3 EP’s 1 LP 1 BP

  47. Lone pairs and shape: 4 Electron Pairs When there are four electron pairs, the shape family is tetrahedral Four possibilities exist: with 0, 1, 2 or 3 lone pairs 0 LP – tetrahedral shape 2 LP – bent shape 1 LP – trigonal-pyramidal shape 3 LP – linear shape Linear Hydrogen Fluoride, HF 1 bond pair 3 lone pairs

  48. .. : : : X X X X .. .. General Shape Family Scheme 4 BP 0 LP 3 BP 1 LP 2 BP 2 LP 1 BP 3 LP

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