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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