| 11. |
(a) The 4s and 3d orbitals are
close in energy. Therefore, pairing two orbitals in the 4s
orbital is less energetically feasible than promoting one to the higher
energy
3d orbital (and that will be unpaired in that 3d orbital) (b) For neutral metals, the 4s orbital is lower in energy than the 3d. However, for 2+ metal cations, the opposite is true, therefore, all metal 2+ cations lose their 4s electrons first. |
| |
|
| 14. |
Basically, pairing electrons raises the energy
of the configuration. You should recognise that the two
configurations with all unpaired electrons are lower in energy. |
| |
|
| 16. |
(a) Ga because I.E. increases left to
right (although this is a bad question because we've added the 3d
and
4p subshells that Ga has and Ca doesn't). Zeff
increases left to right, making it harder to remove an electron. (b) Mg because I.E. decreases top to bottom (for same Zeff, electrons are closer to the nucleus for Mg) (c) should be P because E.A. increases left to right (increasing Zeff). However, it turns out that the pairing energy on adding an extra electron to P causes it to actually have a lower E.A. (electron added to Si is not paired). Again, a bad question because no way to know if the pairing energy is enough to overcome the increase in Zeff. (d) [Ar]3d5 (see question 11(b)) |
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|
| 18. |
The more positive the ion, the harder to remove
an electron! |
| |
|
| 21. |
Fe (smaller radius for same Zeff) should be S (larger Zeff) but is actually P because of electron pairing. Again, not a fair question without being able to look up the actual numbers. Br (larger Zeff) N (larger Zeff) Cd (should be In, but addition of new 5p subshell raises the valence orbital energy more than the increase in Zeff can compensate) |
| |
|
| 22. |
Note: these are more fair questions than
those in question 21 - being asked to analyse anomalies is okay, asking
to predict anomalies is not. (a) S has a smaller Zeff (b) I is larger than Br (c) B has an added subshell (2p) that more than compensates for the larger Zeff (d) S has a 3p4 configuration (with one paired electron). The raise in energy caused by the electron pairing more than compensates for the lowering of energy caused by increased Zeff. (e) Cl larger than F |
| |
|
| 23. |
(a) Mg orbitals are lower in energy (and
therefore it has higher I.E.) than Na because of increased Zeff.
Orbitals also lower in E than Al because Al has added a new
subshell which is higher in E. P is a maximum in I.E. (minimum in
orbital energy) because it has a high Zeff and no
paired electrons in
its p orbitals. (b) The maxima are shifted because Na- will have the same electron configuration as Mg, and Si- the same as P, which are the lowest orbital energy configurations. |
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|
| 25. |
Size generally decreases because valence shell
electrons are not effective at shielding each other, so they only are
shielded by core electrons. In any given row, the number of core
electrons is
constant (new electrons are added to the valence levels), but the
nuclear charge increases. Therefore, valence orbitals see an
increasing positive charge at the nucleus (Zeff) as
you go left to right.
Increasing attraction of the electrons gives smaller orbitals and
smaller atomic size. |
| 1. |
(a) Double bond between one S and the C,
negative charge on S (two equivalent structures) (b) Double bond could be C=S or C=O. At this point, best to draw C=S with negative on the O as the less electronegative element. We will see later in the course that this is wrong. |
| |
|
| 2. |
(a) Best resonance structure: Se=C=N because
negative on N (again, this is actually wrong) (b) Best structure: C=S and C-O with negative charge on O (again, this is actually wrong) (c) C=S double bond, negative on O and other S (two structures) |
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|
| 3. |
If S obeys the octet rule, NSO must always have
more formal charges than SNO. (we'll see later that this isn't
necessarily true!) S=N-O(-) is best of all possible structures |
| |
|
| 4. |
O=N-C[triple]N is the nest structure (no formal
charges required) |
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|
| 5. |
T-shaped (from octahedral) |
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|
| 6. |
(a) AsH3 (largest central atom
- this is actually true, but the "real" reason will be discussed later!) (b) O3- (has more electrons on central atom) (c) SOF2 (F atoms smaller than Cl) (d) can't really say (e) BrO3- (has larger central atom) |
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|
| 8. |
(a) see-saw (tbp with 1 lone pair)
(b) linear (from tbp) (c) tetrahedral (d)
sq. planar (from octahedral) (e) bent (f) sq. planar (g) linear (h) tbp (i) tetrahedral (j) linear |
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|
| 9. |
(a) linear (b) tetrahedral
(c) tetrahedral (d) tetrahedral (e)
tetrahedral (f) octahedral (g) trigonal pyramidal (from tetrahedron) (h) sq.-based pyramid (from octahedron) (i) see-saw (j) trigonal pyramidal |
| |
|
| 10. |
(a) octahedral (b) tetrahedral
(c) bent (a radical!) (d) bent (radical)
(e) each S is trigonal pyramidal (f) each N is trigonal pyramidal |
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|
| 12. |
(a) tetrahedral (b) trigonal
pyramidal (c) tbp (d) bent (e) trigonal pyramidal (f) looks like adamantane, if you can find that in your organic book) |
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|
| 16. |
(a) trigonal pyramidal (b) bent
(c) see-saw (d) tbp (e) sq.
planar
(f) trigonal pyramidal (g) trigonal planar (h) bent (i) tetrahedral (j) octahedral (k) sq.-based pyramidal (l) T-shaped (m) tetrahedral (n) trigonal planar |
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|
| 17. |
The following are polar: a, b, c, f, h, k, l,
m, n |
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|
| 20. |
Without numerical information, it is not fair
to rank all of these in increasing bond polarity. However, you
should be able to come up with the following: (a) N more negative (b) O more negative (c) C more negative (but hard to tell without numbers) (d) O more negative (e) Br more negative (hard to tell without numbers) (f) Cl more negative Overall polarity is, in increasing order, CI, SCl, PBr, CN, NO, OCl |
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|
| 21. |
(a) N cannot form 5 bonds (not big enough, not
enough orbitals) (b) Same as (a) (remember that OF4 requires 4 bonds AND a lone pair) |
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|
| 22. |
oops - shouldn't have assigned this yet! (it's
for section 7) |
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|
| 24. |
Should have T-shaped form based on a tbp (2 lone
pairs). CF3 should be equatorial because it is the
biggest group and there's more room for it in the equatorial position. |
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|
| 25. |
Structure for OCCCO is linear because no lone
pairs on central C. When there's N as central atom, need a lone
pair
on the N, therefore it becomes bent. Turns out that N5+
has the smallest bond angle, but we don't really care about that. |
| 1. |
(a) D3d
(b) D3h (c) Cs
(d) C2h |
| |
|
| 2. |
(a) D2h
(b) Cs (c) 1,1-isomer: C2v;
trans-1,2: C2h; cis-1,2:
C2v |
| |
|
| 3. |
(a) Dooh
(b) and (c) Coov (d) Cs
(e) Coov |
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|
| 4. |
(a) D2h
(b) C2v (c) C2h
(d) Cs |
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|
| 5. |
(a) C2h (b) D6h
(c) Cs (d) C3v
(e) C2 (f) C2v
(g) D4d (h) D3h
(i) D3 (j) who cares? (D2d) |
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|
| 6. |
(a) D3d
(b) D2d (c) Td
(d) %*#(@ stupid question (e) D2h
(f) the following are C2v: 1,2,3 / 1,2,4; the 1,3,5-isomer is D3h (g) whatever (h) D2d |
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|
| 8. |
from 3-12: (a) C3v
(b) C3v (c) C2v
(O in trigonal plane) (d) C2v
(e) C3v (f) Td from 3-16: (a) C3v (b) C2v (c) C2v (d) D3h (e) D4h (f) C3v (g) D3h (h) C2v (i) Td (j) Oh (k) C4v (l) C2v (m) C3v (n) C2v |
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|
| 9. |
(a) (2) Dooh
(3) D3h (4) Td
(5) D3h (6) Oh
(7) D5h (8) D4d (b) (2/2) Dooh (3/1) C2v (3/2) C2v (3/3) D3h (4/1) C3v (4/2) C2v (4/3) C3v (4/4) Td (5/1) C2v (5/2) C2v (5/3) D3h (6/1) C4v |
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|
| 12. |
(a) E, C2, i,
σh |
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|
| 13. |
(a) E, three different C2,
i, three different σ (sort of like σh, σv,
and σv') |
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|
| 17. |
Problem 7: (e), (i) are chiral Problem 6: none Problem 7: don't bother |
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|
| 20. |
(I) C2 (II) C2
(III) Ci |
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|
| 21. |
draw CH3 in a C3v
orientation, then two more H's out the top in a plane so that one of
the σv mirror planes bisect them |
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|
| 23. |
You're brave if you want to tackle this one!
We've covered this mostly in the tutorial. |
| |
|
| 24. |
(a) D3d
(b) Ci (c) Ci
(d) C3 |
| 2. |
First, find the mass of the unit cell: In a NaCl lattice there are (8 corners * 1/8) + (6 faces * 1/2) = 4 total anions Because a 1:1 salt, must be 4 cations (check: 12 edges * 1/4 + 1 in centre = 4 total cations) 4 anions weigh: 4*(79.904 g/mol) / (6.022 x 1023 mol-1) = 5.307 x 10-22 4 cations weigh: 4*(6.941 g/mol) / (6.022 x 1023 mol-1) = 0.461 x 10-22 total mass = 5.769 x 10-22 g if d = density = m/V, then V = m/d = 5.769 x 10-22 g / 3.464 g/cm3 = 1.665 x 10-22 cm3 because the unit cell is cubic, all lengths are the same, so V = l*w*h = l3 therefore, l = unit cell length = V1/3 = 5.502 x 10-8 cm (= 5.502 x 10-10 m) = 550.2 pm Bearing in mind that the unit cell distance two full Li+ radii + 2 Br- radii, the interionic distance is l/2 = 275.1 pm According to Appendix B1 (using C.N. 6 numbers), Li+ is 90 pm, Br- is 182 pm; sum = 272 pm (pretty close!) |
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|
| 3. |
Both the cation and anion in CsCl have C.N. = 8,
whereas in the CaF2 lattice, only the Ca2+ has
C.N. = 8. This is to ensure that the charge balances, because the
CaF2 lattice has 2+ cations with 1- anions. It also
turns out that, because CsCl is based on a cubic lattice and CaF2
on a close-packed lattice, that Cs+ is in a bigger hole than
Ca2+. |
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|
| 4. |
We already did the first, which is a NaCl
lattice (see question 2) The second diagram is a CsCl lattice. # of cations = 1 (because it's in the centre) # of anions = 8 corners * 1/8 each = 1 total anion Therefore, molecular formula is M1X1 The third diagram is fluorite # of cations = 4 * 1 each (all internal) = 4 total cations # of anions = (8 corner * 1/8) + (12 edge * 1/4) + (6 face * 1/2) + (1 internal) = 8 total anions Therefore, this is a M1X2 salt |
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|
| 10. |
We must calculate the ΔHL from
the data given, then use a Born-Haber cycle to determine the missing
parameter. I choose to use the Born-Meyer equation because we
have all the information we need for that. ΔHL = (NazAzB/d)*(e2/4πεo)(1-d*/d)A = -772 kJ/mol (because zA = -1, zB = +1, d* = 0.3 (I used the book value), d = 2.81, A = 1.748) Therefore, "positive half" = 119.5 (half bond enthalpy) + 413 (ΔHf) + 496 (ionisation E) + 109 (enthapy of vaporisation) = 1137.5 kJ/mol [NOTE: I have used the conversion factor 1 eV = 96.4853 kJ/mol. This is on the back cover of the text.] The "negative half" = -772 (ΔHL) - ΔHEA (which we are trying to solve for!) therefore, ΔHEA = 1137.5-772 = 366 kJ/mol (remember, by tradition, ΔHEA are reported as positive numbers) |
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|
| 11. |
CaO has higher charges on its ions than KF.
Likewise, looking at the ionic radii, CaO is slightly smaller
than KF. Both of these factors (higher charge and closer
together) mean that CaO is a stronger lattice (larger bonding), which
accounts for its higher melting point and harder structure. MgO also has 2+ and 2- cations, and all anions are held in place by 8 surrounding counterions. While CaF2 has a 2+ cation, it has only a 1- anion, which is held in place by only 4 "bonds" to neighbouring cations. |
| |
|
| 14. |
As in question 10, ΔHL must be
calculated. If rK+ = 1.51 and rBr- = 1.82, d = 3.33 then ΔHL = -661.8 kJ/mol (because zA = -1, zB = +1, d* = 0.3 (I used the book value), A = 1.748) Negative half = -661.8 - 324.7 (E.A., appendix B3) = -986.5 Positive half = ΔHf + 1/2 * 29.8 + 190.2 + 79 + 418.8 (I.E. of K, Appendix B2) = ΔHf + 607.8 Therefore, ΔHf = 607.8-985.5 = -283.6 kJ/mol |
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|
| 15. |
NOTE: it occured to me while making these
answers, that the question does not make sense! LATTICE TYPE IS SUPPOSED TO BE "NaCl", NOT RUTILE As in question 11, ΔHL must be calculated. If rMg2+ = 0.86 and rO2- = 1.26, d = 2.12 (using the numbers for 6-coordinate ions) then ΔHL = -3934 kJ/mol (because zA = -2, zB = +2, d* = 0.3 (I used the book value), A = 1.748) Negative half = -3934 - 141 (1st E.A.) - = -4075 kJ/mol Positive half = ΔHf + 247 + 744 (2nd EA) + 737.8 (1st I.E.) + 1451 + 37 = ΔHf + 3216.8 Therefore, ΔHf = 3216.8 - 4075 = -858 kJ/mol |
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|
| 17. |
(a) By keeping the effective nuclear charge the
same (i.e. same atom), adding electrons increases inter-electron
repulsion in the valence shell, thereby increasing the size of the
anion as the electrons try to space themselves further apart.
Likewise, oxidising the atom leads to decreased inter-electron
repulsion, so the cation gets smaller. (b) Anions have the same number of electrons, therefore the same degree of interelectron repulsion. However, F- has a larger Zeff and therefore the electrons are pulled closer to the nucleus (making F- smaller) |
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|
| 18. |
As in questions 10, 14, and 15, Using rPb2+ = 1.33 and rS2- = 1.70, d = 3.03 (using the numbers for 6-coordinate ions) ΔHL = (NazAzB/d)*(e2/4πεo)(1-d*/d)A = -2888 kJ/mol (because zA = -2, zB = +2, d* = 0.3 (I used the book value), A = 1.748) Using the Born-Haber cycle: Positive half = 98 + 196 + 715.5 (1st IE) + 15.03*96.4853 (2nd IE) + 535 (includes 1st EA, 2nd EA, and atomisation from S8 solid) = 2995 Negative half = ΔHL (already used 1st EA above!) Therefore, ΔHL = -2995 kJ/mol |
| |
|
| 19. |
This is essentially describing lattice defects.
If Zn is in the lattice instead of Zn2+, there are two
extra
electrons in the lattice. These extra 2 electrons can serve as
charge
carriers in an n-doped semiconductor (n = "negative"
because
there are extra negative electrons to carry the charge). Likewise, with neutral S replacing S2-, there are 2 fewer electrons than there should be, leading to extra positive charges in the lattice. This leads to p-type semiconductors (electrons can move into the positive p holes left by the missing anion electrons). |
| 14. |
This is the Inert Pair Effect - as atomic number
increases, the energy of the ns2 orbitals (which
spend
more time near the nucleus than the np2 elctrons)
decreases
faster than the energy of the np2 elctrons. As
the
nucleus gets larger (more positive), the electrons all move faster.
Eventually, they reach relativistic speeds and the mass of the
electrons increases, causing an even greater lowering of energy of the ns2
electrons (because they spend more time near the nucleus) than for the np2
elctrons. Therefore, Pb2+ is more stable than Pb4+.
For Sn, the Sn4+ is more stable but Sn2+ is
still possible. For Ge and Si, the 2+ oxidation state is almost
never seen. |
| |
|
| 21. |
N2O: Coov NO: Coov NO2: C2v N2O3: Cs N2O4: D2h N2O5: D2h (as drawn; if bent, C2v) NO+: Coov NO2+: Dooh NO2-: C2v NO3-: D3h N2O22-: C2h (NO43-: Td) HNO2: Cs HNO3: Cs (as drawn, assuming free rotation about the N-OH bond would give C2v) |
| |
|
| 24. |
The Lewis structure of I3-
has 2 bonding pairs and 3 lone pairs around the central I. This
is
a base tbp geometry. According to VSEPR, the lone pairs will take
the
larger equatorial positions, leaving the other I's at the axial
positions
which are oriented at a 180o angle. The Lewis structure of I3+ has 2 fewer electrons, therefore is based on a tetrahedral base geometry. The two I-I bonds are therefore at an angle somewhere around 109.5o. |
| |
|
| 30. |
XeOF2: two lone pairs, Xe=O and two
Xe-F bonds gives a base tbp with 2 vertices missing (both in equatorial
plane because
they are roomier positions). The O, F, F are in a T-shape.
The
highest possible geometry is if the F's are trans to each other, giving
C2v
symmetry. XeOF4: one Xe=O and four Xe-F bonds requires 1 lone pair on the Xe. This is a base octahedral geometry missing one vertex (to give a square-based pyramid). The highest possible symmetry puts the Xe=O on the axis opposite the lone pair, to give C4v symmetry. XeO2F2: two Xe=O and 2 Xe-F requires 1 lone pair on Xe. This is a tbp missing one (equatorial) vertex. Assuming an "all-trans" orientation, the symmetry is C2v. XeO3F2: three Xe=O and 2 Xe-F requires no lone pairs! Therefore, this is tbp. Assuming the highest symmetry, all the O's are in the trigonal plane and the F's are on the axial positions. This gives D3h symmetry. |
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Last updated, February 12, 2010
Page maintained by Craig
MacKinnon