sos chem 121 final dec 2011
TRANSCRIPT
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Chemistry 121 Review Package
Table of Contents: p1 Extra Questions
p7 - Answer Key to Extra Questions p17 Review Session Key
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1. Consider the structure to the right:
State the shape and hybridization of all non-H atoms.
2. Draw the Lewis structures, state the shape, and draw the perspective diagram for XeO3.
3. Sketch R3s(r), R23s(r), and 4 2R2
3s(r) vs. r.
4. Rank in order of increasing ionization energy: Rb+, Ca+, K+.
5. Rank the following in order from smallest to largest: Mg+, Na+, Ne, O2-.
6. Rank the Lattice energies from lowest to highest between NaCl, NaF and NaI
7. Rank in order of increasing ionization energy: O, N-, Ne+, F+, P
8. Rank in order of increasing electron affinity: Ne, Fr, B, Li.
9. Rank in order of increasing size: F, Ne+, O-, B2-, C3-.
10. Classify each of the bonds in the following molecules as Polar, Non-Polar, or Ionic, and draw a Lewis Structure for all of the non-ionic structures.
a) CSCl2
b) MgO
c) NOF3
d) [NO]+
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11. Which of the following bonds is most polar?
a) C-N
b) Li-N
c) B-N
d) Rb-N
e) Na-N
12. Which of the following contain double bonds?
a) N2
b) NOBr3
c) MgO
c) CO2
13.
a) MgO2
b) N2H4
c) N3
14. Draw all chemically reasonable resonance structures of [BN2]3- (connectivity N-B-N).
15. Arrange the following in order of increasing melting point:
a) NaCl, H2O, CH3CH2OH, PCl5
b) AsBr3, Kr, SiS2, PBr5, PCl5
16. A factory was recently built in an area which subsequently experienced rising levels of acid rain.
Which chemical is the factory most likely emitting into the atmosphere? a) PCl5 b) H2S c) SO3 d) ClO2
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17. 3pz vs. y.
18. for the 3dxy
19. freshly prepared sample of B4+ discovered how to excite the lone electron in each ion to the 5f orbital, and you want to record the emissions spectrum of this unique species. Unfortunately, although the lab seemed deserted, the pplaying an awful, secretive prank. a) Given this list of photon wavelengths (nm) emissions, determine which should not exist: 162, 75.0, 51.3, 26.3, 19.5, 17.4, 4.9, 4.1, 3.9, 3.8 b) At the end of the experiment, assuming the sample is exposed to no other light sources, will all the electrons end up in the ground state (ie. the 1s orbital)?
20. Write out an equation for the following reactions (this question just asks you for the reactions
optional than the other questions): a) Potassium with excess oxygen?
b) Lithium with limited oxygen?
c) Sodium Oxide with water?
d) Calcium with Chlorine gas? e) Be with carbon dioxide? f) Gallium (Ga) with oxygen? g) Boron undergoing a thermite reaction? h) Aluminum with water?
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i) The Hall-Heroult process? j) Silicon with water? k) Formation of Silica (at 1000°C)? l) Germanium (Ge) with Bromine gas? m) Germanium (Ge) with limited Oxygen at room temperature? n) The formation of Phosphorous acid? o) Arsenic (As) with limited bromide gas? p) Production of sulfur trioxide? q) The formation of sulfurous acid? r) The formation of sulfuric acid?
21. Balance these reactions
a) _ NaBr(aq) + _ H2O(l) _ Br2(g) + _ H2(g) + _ NaOH(aq) b) I2(g) + H2O(l) HOI(aq) + HI(aq)
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22. Rationalize why elements in the first 2 rows of the periodic table cannot expand their octet
23. What is the hybridization of the carbon atom(s) in each of the following species? CCl2O SiCl2(CH3)2 CO2
24. Draw a diagram depicting the orbital cross sections in BF3
25. Draw an orbital overlap diagram for ammonia
26. Which will have a stronger pi bond, CO2 or CS2?
27. Will the molecule H2Si=Si=SiH2 be planar or twisted?
28. Draw out energy level diagrams for B2 and F2 and explain any differences between them. Also, calculate the bond order of each compound.
29. 1 2 3 4 MOs in a molecule have 0, 1, 2, and 3 nodal planes respectively, then rank these orbitals in terms of increasing antibonding nature.
30. Why are the noble gases found as single atoms? (Hint: Draw out the MO for Ne!)
31. Look back at the N2 MO diagram we drew during the session. How would the diagram look if the molecule was N2
+?
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1. Consider the structure to the right:
State the shape and hybridization of all non-H atoms. 1,2,3,15 3 4,5,6,7,8,9,10,11 trigonal planar so sp2
2. Draw the Lewis structures, state the shape, and draw the perspective diagrams for XeO3.
tetrahedral - trigonal pyramidal
3. Sketch R3s(r), R23s(r), and 4 2R2
3s(r) vs. r. Please not that the beginning of the graph should contact the y-axis at some point; the function I graphed were just not conducive to showing it on a reasonable scale.
Xe
O
O O
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4. Rank in order of increasing ionization energy: Rb+, Ca+, K+.
Ca+, Rb+, K +
5. Rank the following in order from smallest to largest: Mg+, Na+, Ne, O2-.
Na+, Ne, O2-, Mg+
6. Rank the Lattice energies from lowest to highest between NaCl, NaF and NaI Ans: NaF>NaCl>NaI
7. Rank in order of increasing ionization energy: O, N-, Ne+, F+, P
N-<P<O<F+<Ne+
Reasons:
- Ne^+ already lost one electron. The usually noble gas does not want to lose another electron, therefore, its ionization energy is high
- F^+ is highly electronegative, and therefore would rather get electrons than lost another one. However , because Ne is a noble gas, it still has a higher ionization energy
- N^- has gained an electron has a low ionization energy
- P has valence electrons in the n = 3 shell. The valence electrons are therefore not held very tightly, and are easy to remove.
8. Rank in order of increasing electron affinity: Ne, Fr, B, Li.
Ne< F r<L i<B
Reasons:
- Ne is already a very stable noble gas and therefore does not want an extra electron
- F r is the least electronegative atom and therefore has the a very low attraction for an extra electron
- B has a greater # of protons than L i, therefore adding an extra electron into the same shell in both elements is more favourable for B
9. Rank in order of increasing size: F, Ne+, O-, B2-, C3-.
Ne<F<O<C<B
Reasons:
- Ne^+ and F are isoelectronic, but Ne has more protons, therefore has more pull for the electrons, and is therefore smaller than F
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- O ^- and F are isoelectronic, but O^- has 1 less proton than F , therefore the electrons are not held as close and O^- is larger than F
- C^3- and F are isoelectronic, but for the same reason as O ^2-, F is smaller . C^3- has more electrons for the number of protons that it has, which is why it is bigger than O^2-
- B^2- is the largest because it has the least number of protons, and there is therefore less pull for the electrons that it has.
10. Classify each of the bonds in the following molecules as Polar, Non-Polar, or Ionic, and draw a Lewis Structure for all of the non-ionic structures.
a) CSCl2
covalent, polar
b) MgO
ionic
c) NOF3
polar covalent
d) [NO]+
polar covalent
11. Which of the following bonds is most polar?
a) C-N
b) Li-N
c) B-N
d) Rb-N
e) Na-N
12. Which of the following contain double bonds?
a) N2
b) NOBr3
c) MgO
c) CO2
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13.
a) MgO2
b) N2H4
c) N3
14. Draw all chemically reasonable resonance structures of [BN2]3- (connectivity N-B-N).
15. Arrange the following in order of increasing melting point:
a) NaCl, H2O, CH3CH2OH, PCl5
PCl5 < C H3C H2O H < H2O < NaC l b) AsBr3, Kr, SiS2, PBr5, PCl5
PCl5 < C H3C H2O H < H2O < NaC l
16. A factory was recently built in an area which subsequently experienced rising levels of acid rain. Which chemical is the factory most likely emitting into the atmosphere?
a) PCl5 b) H2S c) SO3 d) ClO2
17. 3pz vs. y. no graph here; 3pz exists along z-axis.
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18. for the 3dxy
19. morning, you get to the lab extra early to obtain an emissions spectrum for a freshly prepared sample of B4+ discovered how to excite the lone electron in each ion to the 5f orbital, and you want to record the emissions spectrum of this unique species. Unfortunately, although the lab seemed deserted,
playing an awful, secretive prank. a) Given this list of photon wavelengths (nm) emissions, determine which should not exist: 162, 75.0, 51.3, 26.3, 19.5, 17.4, 4.9, 4.1, 3.9, 3.8 F irst, draw out the possible transitions. Starting from the 5f, the 3d and 4d orbitals are accessible, but no 5 to 2 or 1 transitions are possible. These correspond to wavelengths of 17.4 and 3.8nm. Next, from the 4d we can access the 3p or 2p, but not the 1s orbital, so 4 to 1 transition is forbidden the wavelength for this is 3.9nm. F rom the 3d we can access the 2p, and from the 2p we can access the 1p, so all other transitions are allow between principal electrons shells. b) At the end of the experiment, assuming the sample is exposed to no other light sources, will all the electrons end up in the ground state (ie. the 1s orbital)? No, since if they transition from the 3p to the 2s they cannot decay further , nor can they be excited because there is no light energy being provided to give them energy to get out of
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explained in (a).)
20. Write out an equation for the following reactions: a) Potassium with excess oxygen? Ans: 2K(s) + O2(g) 2O2(s) b) Lithium with limited oxygen?
Ans: 2Li(s) + (1/2)O2(g) 2O(s) c) Sodium Oxide with water?
Ans: Na2O(s) + H2O(l) (aq) d) Calcium with Chlorine gas? Ans: Ca(s) + Cl2 (g) 2(s) e) Be with carbon dioxide? Ans: 2Be(s) + CO2(g) (s) + C(s) f) Gallium (Ga) with oxygen? Ans: 4Ga(s) + 3O2(g) 2Ga2O3(s) g) Boron undergoing a thermite reaction? Ans: 2B(s) + Fe2O3(g) B2O3(s) +2Fe(l) + Energy h) Aluminum with water? Ans: Al(s) + H2O(l) No Reaction! i) The Hall-Heroult process? Ans: 2Al2O3(l) +3C(s) 4Al(l)+ 3CO2(g) j) Silicon with water? Ans: Si(s) + H2O(l,g) No Reaction k) Formation of Silica (at 1000°C)? Ans: Si(s) + O2(g) SiO2(g)
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l) Germanium (Ge) with Bromine gas? Ans: Ge(s) + 2Br2(g) GeBr4(l) m) Germanium (Ge) with limited Oxygen at room temperature? Ans: Ge(s) + 2O2(g) No Reaction n) The formation of Phosphorous acid? Ans: P4O6(s) + 6H2O(l) 4H3PO3(aq) o) Arsenic (As) with limited bromide gas? Ans: As4(s) + 6Cl2(g) 4AsCl3(l) p) Production of sulfur trioxide? Ans: 2SO2(g) + O2(g) 2SO3(g) q) The formation of sulfurous acid? Ans: SO2(g) + H2O H2SO3(aq) r) The formation of sulfuric acid? Ans: SO3(g) + H2O H2SO4(aq)
21. Balance these reactions a) _ NaBr(aq) + _ H2O(l) _ Br2(g) + _ H2(g) + _ NaOH(aq) Ans: 2 NaBr(aq) + 2 H2O(l) 1 Br2(g) + 1 H2(g) + 2 NaOH(aq) b) I2(g) + H2O(l) HOI(aq) + HI(aq) Ans: 1 I2(g) + 1 H2O(l) 1 HOI(aq) +1 HI(aq)
22. Rationalize why elements in the first 2 rows of the periodic table cannot expand their octet No d-orbitals to hybridize
23. What is the hybridization of the carbon atom(s) in each of the following species? CCl2O sp2 SiCl2(CH3)2 sp3 CO2 - sp
24. Draw a diagram depicting the orbital cross sections in BF3
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25. Draw an orbital overlap diagram for ammonia
26. Which will have a stronger pi bond, CO2 or CS2? C O2, because it is formed from the 2p orbitals. Similar orbitals have better overlap than dissimilar orbitals.
27. Will the molecule H2Si=Si=SiH2 be planar or twisted? Twisted
28. Draw out energy level diagrams for B2 and F2 and explain any differences between them. Also, calculate the bond order of each compound.
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29. 1 2 3 4 MOs in a molecule have 0, 1, 2, and 3 nodal planes respectively, then rank these orbitals in terms of increasing antibonding nature.
1 , 2 , 3 , 4
30. Why are the noble gases found as single atoms? (Hint: Draw out the MO for Ne!)
BO = 0, therefore Ne2 does not exist
BO = 1 for both molecules Differences:
- F2 has more electrons than B2 since atomic fluorine has more electrons than boron - The energies of the sigma-2p and pi-2p orbitals are reversed for F2, due to mixing
between 2s and 2p orbitals of boron (F2 has same relative energies of orbitals as O2)
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31. Look back at the N2 MO diagram we drew during the session. How would the diagram look if the molecule was N2
+?
1.
1s 1s
1s
1s
E
2p 2p
2s 2s
1s
1s
N2 N N+
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Per iodic T rends:
1) The 4 blocks: s, p, d, and f 2) The special names for elemental categories.
alkali metals, alkaline earth, pnictogens, chalgogens, halogens, noble gases, (bottom) lanthanides, actinides.
3) Zeff increases in which direction? left to right 4) Ionization energy increases from: bottom to top, left to right. SO if i were to ask you a
wrong, because of the exception in the trend due to the p orbitals being higher in energy examine the graph at the end of Ch9 if you need to.
5) Electron Affinity increases from: bottom to top, left to right (again) except for noble gases. Why??? because they have full valence octets.
6) Atomic radius increases from: r ight to left, top to bottom (opposite I E).
1. Rank the following in terms of increasing ionization energy, then size, and then identify any special groups they belong to in their elemental states. S2-, Cl-, Ar, Ca2+, Ge4+, U
size: Ge4+, Ca2+, A r , C l-, S2-, U ionization energy: U , S2-, C l-, A r , Ca2+, Ge4+. naming: do it quickly
2. Which of the following statements is INCORRECT? a) The radius of F- is smaller than that of Cl-. b) The radius of C l+ is greater than that of A l. c) The radius of Cl is smaller than that of P. d) The radius of I is larger than B.
Lattice energy describes the energy required to break one mole of crystal into the ions in gas phase (at infinite separation). Nbecause ions are packed in crystal lattices. However, certain properties still matter; size, and charge. Smaller atoms will take more energy to separate because they will influence each other more. Larger charges also have a large influence because they will more strongly attract each other. The online package will have more practice with all of these trends if you want it. Intermolecular Forces: Different types of interactions exist between molecules. The relative strength of these bonds is:
London Dispersion Forces < Dipole-Dipole Interactions <Hydrogen Bonds < Ionic < Network Covalent
London Forces, also known as dispersion forces, are caused by the movement of electrons to
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induce a dipole within individual molecules. This results in an attraction between molecules. Polarizability is how easily electrons are moved in a molecule. Larger molecules are more polarizable and have higher melting points. Dipole-Dipole Interactions arise from the attraction between species that have a permanent dipole moment. Electronegativity difference increases dipole moment, which increases melting point. Hydrogen Bonds are special cases of dipole-dipole interactions
3. Which of the following pure substances will exhibit the strongest hydrogen bonding? a) CH3CH2NH2 b) PCl3 c) CH3OCH3 d) CH3CHO e) BH3
4. Arrange the following in order of increasing boiling point: CH3OH, NaF, Ar, MgO, and H2O A r (noble gas), methanol, water (better H-bonding than methanol), NaF , then MgO (stronger attractive forces due to larger charge on the ions). bp NaF = 1695; bp MgO 3600.
L ewis/Resonance Structures and VSEPR Theory:
First: Hybridization - Why do we need it?
Consider ethylene. The shape at each carbon is ______________. How can we achieve this shape using s and p orbitals? We MUST hybridize, and in doing so, we can rationalize almost any bonding geometry.
In this case, we combine the outer s-orbital of each carbon with 2 of its p-orbitals to get 3 planar hybrid orbitals, as show below. These 3 sp2 orbitals
- -bond between the carbons.
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The wavefunctions of atomic orbitals can be added or subtracted to give new wavefunctions called hybrid orbitals. Only orbitals in the same shell can be mixed together. - How do we recognize it?
We can determine the hybridization by determining the SHAPE of the atom.
*Note that links are not the same as bonds. Double and triple bonds count as one link.
5. Label the structures with the hybridizations of the non-hydrogen atoms in the structures below.
6. Your experiments indicate that PCl3 has a dipole moment, but AlCl3 does not. Propose a molecular geometry for each molecule.
AlC l3 is tr igonal planar (dipoles cancel), PC l3 is
N O W H O W C A N W E M A K E T H ESE? P4(s) + 6C l2(g) 4PC l3(l) 2A l(s) + 3C l2(g) 2A lC l3(s) but is this how this molecule prefers to exist? No!
7. Draw Lewis structures for the following two molecules:
Number of L inks* + Lone Pairs
Hybridization VSEPR Parent Shape Bond Angles
2 sp Linear 180o 3 sp2 Trigonal Planar 120o 4 sp3 Tetrahedral 109.5o
5 sp3d Trigonal Bipyramidal 120 and 90o
6 sp3d2 Octahedral 90o
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a) C-N-O (all atoms have full octets you determine the charge) b) H3PO4 does it have a dipole moment?
(b) does have a dipole moment, since the
resonance; also, sp2 orbitals are more electronegative than sp3 orbitals Notice that we can draw a resonance structure of this, with a negative formal
charge on the O and a positive on the P perfectly reasonable. F rom this, we can see that that O withdraws through resonance A ND E N . The others only withdraw through E N . A LSO : what is the hybridization of each oxygen? The sp2 orbitals have more s-character and are more electron withdrawing as they are closer to the nucleus. How can we make (b)? P4O10(s) + 6H2O(l) 4H3PO4(aq)
8. Deprotonated tetrazole is used in solid phase DNA synthesis. Draw all reasonable resonance structures of this anion.
9. Consider the structure to the right: How many atoms can lie in the same plane? i] 5 ii] 10 iii] 15 iv] 16 v] 18
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Quantum Mechanics: There are several equations you should have memorized by now for the quantum mechanics section (Ch7), which I will reproduce on this power point deal.
E k = E light E binding - 0
h/4
of subatomic particles at the same time.) E k = � mu2
3-D One Electron Systems: There exist 3 quantum numbers for electrons in 3D boxes: n, l, and ml. Each combination that
an atomic orbital. The way we describe the electron in 3D space, we use spherical polar coordinates Math 200) . These help express a wavefunction which can be separated into a product of two functions: the radial wave function Rn,l(r) and the angular wave function Yl, ml ). R describes the size of the orbital and Y describes the shape of the orbital. Only certain values for the quantum numbers describe an orbital and give solutions to
n can be any integer >0. l can be anything up to n-1. ml can be anything from l to + l.
10. Sketch (a) the orbital, then (b) from = 0 to
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Spectroscopy of One-E lectron Species The energy associated with an orbital is given by where Z is the
atomic number and n is the principle quantum number. Notice that for a one-electron system, the orbital energy depends only on the principal shell number, not on the subshell this is due to the degenerate energies of all but the s-orbitals. When light shines through a sample, atoms absorb certain wavelengths of light which correspond to the specific energy differences between orbitals. These energy differences can be calculated
using . The wavelength of light required to provide
notice that a higher
wavelength corresponds to a lower energy. The absorption of energy causes an electron to jump to a higher energy orbital. The energy that
passes through the sample and is then detected by our eyes or by a detector. In an absorption spectrum, energies will be recorded for transitions between all states, for example, 1 2, 1 3, 1 4. On an energy level diagram, in contrast, only the specific energy of each orbital is shown. Selection rules The above spectra were determined using only changes in the principal quantum number n. If we want to assign transitions to specific pairs of orbitals (1s, 2s, etc.), we must take into account the other quantum numbers. When we do, certain transitions are not allowed, and these are determined by the following rules: - Only non-zero changes in n are allowed - l can only change by +1 or -1
11. A sample of hydrogen ions in the ground state is irradiated with ultraviolet light of wavelength 95nm. Which emission below is not possible?
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Now draw all possible transitions; the electrons must be in a 5p orbital if it was excited from g.s. hydrogen atoms. Notice the transition from 4 to 1 is impossible and use that to calculate a wavelength emission.
C rystal Structures Simple Cubic Atoms per cell: 1 Cubic holes per cell: 1 Tetrahedral holes per cell: 0 Octahedral holes per cell: 0 Body-Centred Cubic Atoms per cell: 2 Cubic holes per cell: 0 Tetrahedral holes per cell: 0 Octahedral holes per cell: 3 Face-C entred Cubic
Atoms per cell: 4 Cubic holes per cell: 0 Tetrahedral holes per cell: 8 Octahedral holes per cell: 4 *In FCC if half of the tetrahedral holes are filled, there is a 1:1 ratio of ions
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Coordination number is the number of other ions that each ion touches Stoichiometric ratio is inversely proportional to the coordination number, but proportional to the ratio of ions in the unit cell to the number ions in holes Ions in each position of a cell contribute a different fraction to the cell: Vertex ions contribute 1/8 of an ion Edge ions contribute ¼ Face ions contribute ½ Body ions contribute 1 ion To determine the number of ions in a cell, count the number of ions in each position and multiply by the fraction that they contribute. The sum of the contributions from each position is the number of ions in the cell.
8. The oxygen atoms in MgAl2O4 form a face centred cubic cell with Mg atoms in tetrahedral holes and Al atoms in octahedral holes. What fraction of each type of hole is filled? 14 Oxygen atoms in the diagram of the F C C in cell, but these are all shared: the corners are shared with 8 cells, so each is 1/8 belonging to a given unit cell. Each facial atom is split between to cells, so each is 1/2 belonging to this cell. thus, we have 4 atoms in our cell, which makes this question simple; we must have 2 A l atoms and 1 Mg. This means the tetrahedral holes are 1/8 filled with Mg and the octahedral holes are � filled with A l.
9. A graduate student found a bottle of unknown crystals of a metal on a dusty shelf in the lab, but
the bottle label had faded over time and was totally unreadable. He analyzed the crystals and
found them arranged in a simple cubic packing structure. The density of the compound was
determined to be 9.142 g/cm3 and the atomic radius was measured to be 168pm. Identify the
metal.
Work backwards starting from the atomic radius; r = 1.68x10-8cm. The side length is 2r = 3.36x10-8cm, and (2r)3 = Vol. of cell = 3.80x10-23 cm3/cell. Vol. x N mol-1 = 22.84cm3/mol. Units of molar mass are g/mol, so we need to use the density in g/cm3 to get us there, dimensionally speaking. 22.84cm3/mol x 9.142 g/cm3 = 209 g/mol This is Po.
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Chapter 3 1. Oxidation States: Oxidation States show what molecules are losing electrons and gaining electrons.
If losing electrons, atoms become positive
If gaining electrons, atoms become negative
Rules:
1) The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. If the entire compound is charged it will most likely be indicated. 2) Oxidation problems have at least one a cation and one anion. 3) Approach each question like a simple math questions as will be explained below:
1. What is the oxidation state of Mg 2+? Ans: 2+ 2. What is the oxidation state of X, in X2O6
-‐ ? Ans: 2X + 6O = -‐1 2X + 6(-‐2) = -‐1 2X = -‐1 -‐ 6(-‐2) 2X = -‐1 (-‐12) 2X = -‐1 + 12 2X = 11 X = 11/2
S-block Metals Ionic Compounds are formed through the reaction of 2 species:
One species loses electrons
One species gains electrons
Please Note: Elements in the same column react the same way. The trick to mastering these reactions lies in writing them out on flash cards and review
alittle every day.
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1. A lkali Metals:
Alkali Metals are elements in group 1
They have 1 valence electron and therefore, are really reactive as they want to get rid of their lone valence electron to have a full octet.
Cannot use Oxidation states in these reactions Reactions that you have to know for Alkali Metals!
Reaction Conditions
2Na(s) + (1/2)O2(g) 2O(s)
Oxygen is limited, Sodium Oxide is produced
2Na(s) + O2(g) 2O2(s)
Oxygen is in excess, Sodium Peroxide is produced
2Na(s) + 2H2O (l) (aq) + H2 (g) + Energy
Na2O(s) + H2O(l) (aq)
Na2O2(s) + 2H2O(l) (aq) + H2O2(aq)
2. A lkaline Earth Metals:
Alkaline Earth Metals are elements in group 2 They have 2 valence electrons and are therefore, highly reactive. of +2 in reactions:
Reaction Conditions
Mg(s) + (1/2)O2(g) (s)
Magnesium Oxide Produced
Mg(s) + 2H2O (l) 2 (aq) + H2 (g)
Magnesium oxide coating must be removed by boiling water or using dilute HCl
Mg(s) + Cl2 (g) 2(s)
2Mg(s) + CO2(g) (s) + C(s)
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Chapter 6 1. Group 13 Elements-‐ Aluminum
Alumina or Aluminum Oxide (Al2O3) is found as a mineral on earth. The Hall-Heroult process takes alumina and makes Aluminum: 2Al2O3(l) +3C(s) -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 4Al(l)+ 3CO2(g) Na3AlF6
Please note that for the reactions below, Aluminum has a +3 oxidation state. Generally oxidation states are thought of as the number of electron a species can donate or receive.
Remember that oxygen has an oxidation state of -‐2 Reactions that you have to know:
Reactions Conditions
4Al(s) + 3O2(g) 2Al2O3(s) Pure Aluminum is very reactive with oxygen and therefore forms Alumina
2Al(s) + Fe2O3(g) Al2O3(s) +2Fe(l) + Energy This is a Thermite Reaction! It releases a lot of energy!
Al(s) + H2O(l) No Reaction! This is because Aluminum has a thin coat that protects aluminum from getting in contact with water
2Al(s) + 3H2O(l) Al2O3(s) +3H2(g) Once HCl or NaOH is used, the protective coat can be dissolved
2Al(s) + 3Cl2(g) 2AlCl3(s) (dimer) Al2Cl6 is a dimer as AlCl3 does not have a full
octet
Al2Cl6 + 2N(CH3)3 2(AlCl3-N(CH3)3)
Questions:
1. Draw out 2AlCl3(s)
2. Group 14 Elements-‐ Silicon
Like Aluminum, elemental silicon does not exist naturally on earth
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In order to produce elemental Silicon: SiO2(s) +2C(s) -‐-‐-‐ Si(s)+ 2CO(g) occurs at 3000°C Please note that Silicon has an oxidation state of +4 and reacts accordingly in the reactions below
Reactions that you have to know:
Reactions Conditions
Si(s) + H2O(l,g) No Reaction At room temperature, no rxn due to protective coating
Si(s) + O2(g) No Reaction At room temperature, no rxn due to protective coating
Si(s) + O2(g) SiO2(g) At 1000°C, Silica is formed
Si(s) + 2Cl2(g) SiCl4(l) At 300°C, Silicon can react with Halogens
3. Group 15 Elements-‐ Phosphorous
phosphorous as very reactive and unstable.
Phosphorous has an oxidation state of +3 and +5 Covalently bonded phosphorous atoms like to form rings of P4 molecules White Phosphorous> Red Phosphorous> Black Phosphorous Most Reactive -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ Least Reactive
Draw White Phosphorous
Draw Red Phosphorous
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Reactions that you have to know: Reactions Conditions
2Ca3(PO4)2(s) + 10C(s) + 6SiO2(s) P4(l) +10CO(g)
+6CaSiO3(s)
Produces White Phosphorous
P4(s) + 3O2(g) P4O6(s) Reacts under limited Oxygen
P4(s) + 5O2(g) P4O10(s) Reacts under excess Oxygen
P4O6(s) + 6H2O(l) 4H3PO3(aq) Phosphorous Acid Forms
P4O10(s) + 6H2O(l) 4H3PO4(aq) Phosphoric Acid Forms
P4(s) + 6Cl2(g) 4PCl3(l) Reacts under limited chlorine gas
P4(s) + 10Cl2(g) 4PCl5(s) Reacts under excess chlorine gas
4. Group 16 Elements-‐ Sulfur The Frasch process obtains Pure Sulfur: 1. 2H2S(g) + 3O2(g) 2SO2(s) + 2H2O(g) 2. 2H2S(g) + SO2(g) 3S(l) + 2H2O(g) (300°C) 3. Cannot use Oxidation states in the reactions belo Reactions that you have to know:
Reactions Conditions
S(s) + O2(g) SO2(g) Formation of Sulfur dioxide
2SO2(g) + O2(g) 2SO3(g) Sulfur dioxide reacts with oxygen to produce sulfur trioxide
SO2(g) + H2O H2SO3(aq) Sulfurous Acid produced
SO3(g) + H2O H2SO4(aq) Sulfuric Acid produced
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5. Group 17 Elements-‐ Chlorine This row of elements have a very high affinity for electrons and therefore, they are never found in the elemental form e.g. Cl but found in the diatomic form e.g. Cl2(g)
Reactions that you have to know: Reactions Conditions
2NaCl(aq) + 2H2O(l) Cl2(g) + H2(g) +
2NaOH(aq)
Cl2(g) + H2O(l) HOCl(aq) + HCl(aq) A strong acid (Hydrochloric acid) and a weak acid (Hypochlorous acid) is produced. This reaction has equilibrium arrows and through Le
backwards HOCl(aq) + HCl(aq) Cl2(g) + H2O (l)
2NaClO3(aq) + SO2(aq) 2ClO2(g) +
Na2SO4(aq)
Formation of ClO2(g) (Chlorine dioxide)is a very dangerous chemical
6. Group 18 Elements-‐ No Reactions!
Electronic Configurations:
1. Ground State: The ground state is the lowest energy arrangement of an electron or electrons within the orbitals of an atom or ion.
e.g: Li= 1s2 2s1 The Pauli Exclusion principle:
No two electrons in a single atom can have the same four quantum numbers; if n, l, and ml are the same, ms must be different such that the electrons have opposite spins, and so on.
ms can be both +1/2 and -‐1/2
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orbitals, electrons occupy them singly with the same spin before being paired within the same p orbital
To rationalize this; remember that electrons are negative and therefore repel one another
Aufbau principle:
Electrons fill the orbitals from lowest energy to highest energy Think of the ground state as you rarely have only one electron in the 3s orbital and electrons in the 3p orbitals
Exceptions to regular orbital filling:
1. Chromium and Copper-‐ The 4s orbital is lower in energy than the 3d orbital. Generally, the 4s orbital is filled up before the 3d orbital. However, in the case of chromium and copper, it is more favorable place an electron from the 4s orbital to the 3d orbital.
Questions:
a) Draw the orbital diagrams for a) Chromium and b) Copper
See your CHiRP for this
2. When transition metals are ionized (lose electrons), they first lose electrons from their p orbitals, then s orbitals then d orbitals
a) What is the energy configuration of Copper (I) ion?
b) What is the energy configuration of Chromium (I) ion?
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Excited States:
The Excited states are anything but the ground state There can only be one ground state but infinite excited states
a) Write an excited state for phosphorous (I) ion
b) Write an excited state for Chromium
Magnetic Properties:
Species that have one or more unpaired electrons are paramagnetic Species with NO un paired electrons are called diamagnetic (think of dice at a craps
Questions:
a) Write out the electronic configuration for Kr+, drstate whether it is paramagnetic or diamagne
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b) While we are generally taught that Ionization energy increases from left to right on the periodic table, explain why 1) The ionization energy of Be is less than B. (Group 13 vs Group 2) 2) Rationalize why Oxygen has a lower ionization energy than Nitrogen (Group 16 vs Group 15)?
Synthesis Question: In a chemical factory, solid lithium (100.00 mol) reacts with limited oxygen gas (25.00 mol) to produce Compund X (50.00 mol). Compound X reacts with water (50.00 mol) to produce Compound Y (100.00 mol). Finally, Compound Y reacts with H2SO4 (aq) to produce Mixture Z which consists of water and 2 charged species. Write the reactions that form Compound X, Compound Y and Mixture Z.
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Valence Bond Theory Bonds between atoms are represented as the overlap of two orbitals containing one unpaired electron each, and with opposite spins using either orbital diagrams or orbital cross sections. Sigma bonds are formed by orbitals that lie on the same axis (orbitals have to point towards each other)
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Pi bonds are formed by orbitals that lie above and below the axis. * There is always exactly one sigma bond (no more and no less) between two covalently bonded atoms. *Hybrid orbitals are never involved in pi bonds. Diffuse (large) orbitals have poor overlap, and thus form very weak bonds. 2p orbitals form the best pi bonds. Orbitals of similar size have better overlap than dissimilar orbitals, and thus form stronger bonds. Questions: 1. Draw the valence bond diagram for carbon dioxide.
2. Which of the d-orbitals can form a pi bond and which can form a sigma bond with a 2pz orbital? Sigma: dz2 pi: dxz dyz
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Molecular Orbital Theory: This model of atomic bonding works thus: two atoms with thei r electrons in the atomic orbitals come close together , and the electrons wavefunctions interfere either constructively or destructively. Some key points for M O theory:
1) Number of MOs formed = number of AOs they were formed from. 2) Wave functions of similar phase reinforce each other and cause bonding interactions;
wavefunctions of opposite phase weaken each other and cause antibonding interactions. Each of these interactions gives a possible MO for the new system. Bonding orbitals contribute to bond strength, while antibonding orbitals lessen bond strength. Therefore, if you have equal bonding and antibonding interactions, you have no bond!
3) Bonding interactions are great(similar principal quantum number n) due to better orbital overlap.
4) 5) Delocalized MOs are symmetrical
1s 1s
s1
1s
s1*
1s
2s 2s
s2*
s2
1s
Li Li2
Li
E
+
- +
-‐ Same phase wave functions interfere constructively
-‐ Bonding molecular orbital (MO) formed
-‐ Bonding MO < Atomic Orbital (AO) (favourable)
-‐ Opposite phase wave functions interfere destructively
-‐ Antibonding molecular orbital (MO) formed
-‐ Antibonding MO > AO (unfavourable)
E N E R G Y is C O NSE R V E D!!!
Node: less electron density than there was before, so positive nuclei repel each other More nodes = higher energy
H O M O (highest occupied molecular orbital)
L U M O (lowest unoccupied molecular orbital)
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1. Draw the energy level diagram for N2.
2p 2p orbitals for B2, C2, and Ns, but they switch for O2 and F2. Why? Because the 2s and 2p orbitals are very close in energy and mix to disrupt the energy of the system, so the relative ordering of orbitals is perturbed. Memorize this, keeping the rationale in mind. Bond O rder: This equation makes sense since the bonding electrons provide stability for the system, while antibonding electrons provide instability to the molecule.
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2. Benzene (C6H6) has a planar ring structure with delocalized bonds. A molecular orbital ( 1
) of C6H6 is shown below (top view). The phase of the orbital is either shaded (+) or not shaded (-).
a) Draw line(s) to indicate the nodal plane(s) in 1 b) What is the total number of molecular orbitals in C6H6? 6
c) Below are 5 drawings of molecular orbitals for 6-membered ring structures. Four of them
show molecular orbitals of benzene. Circle the diagram that does not show the molecular orbital of benzene.
d) State whether 2 and 5 are higher, lower, or equal in energy to 1 . Pi-2: lower in energy Pi-5: equal in energy
1
65
4 32
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e) Calculate the average -bond order and then the average BO.
2## gelectronsantibondinctronsbondingeleBO
3206 electronsgantibondinelectronsbonding
21
63
bondsBOaverage
211
23.totalBOavg