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Lecture

Notes taken during lecture


Department
Chemistry
Course Code
CHM135H1
Professor
Kris Quinlan

Page:
of 3
N.B. I actually will not be scanning the slides, as there is really no need. Unless diagrams
pop up, I will only be typing up the notes I take during lecture. If you have questions or
would like clarification, email at jingwei.chen@utoronto.ca.
LECTURE 4 – Bonding and Atmospheric Chemistry
Slide No. Notes
Last
lecture
slide 24
-Ans: C
-Wavelength is inversely proportional to energy, so energy increases
in the opposite direction as wavelength
-The rightmost lines are 4->3, then 5->3, then 6->
BONDING
1-expanded octets possible – with d-orbitals being part of bonding
-1s orbitals have only two electrons
-valence electrons important for bonding
2-low ionization energy – easy to remove an electron
-electron affinity – energy associated with gaining an electron
-high electron affinity means that lots of energy are released when
electron gained
-oppositely charged ions attract each other
-surround each other – to maximize electrostatic forces
3-bringing them together lowers energy
-attraction of one hydrogen electron to nucleus of the other hydrogen
-there is an ideal distance for bonding
4-F highest – since smallest radius therefore electron close to nucleus
so attraction is very strong
-EX.
-H2 is non-polar covalent bond
-HF is polar covalent
-LiF is ionic
-Ionic bonds are between a metal and a non-metal
7-D
-There is a total dipole moment pointing to the left
8-electrons hate each other
-electron group can be a pair of electrons in a bond; 4 electrons in a
double bond, or 6 electrons in a triple bond; lone-pair
9- octahedral geometry – all bonds are 90 degrees
10 -water – angle between O-H bonds are 105 degrees
-this is a bent geometry because of the LPs
-LP on oxygen, not shared therefore more repulsive, attracted to just
one nucleus (the oxygen)
-If asked if something have a dipole moment, first determine shape
11 -5 electron groups
-lone pair tries to be as far away from the other electron groups
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-D – see-saw geometry
-Dipole moment? Yes
-The up and down dipole moments cancel out, but the total dipole
moment points left
12 -Ans: C
-4 electron groups
-N as central atom, with three F bonded to it, with a LP on the N
13 -Ans: A
-3 electron groups
-C as central atom, with three O around, each with 3 LP
-But then C does not have a full octet
-So, one of the Os LP forms a double bond between C and that O
-Resonance structures are created
GASES
3-fixed shape and volume – solids
-liquid – fixed volume, not fixed shape, not compressible
-gas – neither fixed volume nor shape
-mass – s, l, g
-volume – s, l, g
-volume – l, g
-pressure – g
-atmospheric pressure is the pressure of all gases pushing down on
the Earth
4-C
-760mmHg + 25cmHg ( 10 mm/cm) = 1010 mmHg
6-273 K = 0 degrees Celsius
-PV = nRT – this formula applies to both the initial and final values
-NRT remains the same so P1V1 = P2V2
-Subbing in the values: (150 atm)(49.0L) = (1.02 atm)V2
-V2 = 2.10 x 10^3 L
7-Ans: D
-Next Ans: C
8-EX.1.
-V = nRT/P
-n = 1.00G of KclO3 (mol/122.55g)(3mol O2/ 2mol KCl3) = 0.0122
mol
-V = (0.0122mol)(0.08206 Latm/molK)(298 K) / 1.00 atm = 0.299 L
-EX. 2.
-Want g/mol – g have, mol need
-n = PV/RT = (365 mmHg)(1 atm/ 760 mmHg)(1.500 L) / (0.08206
Latm/molK)(295.65 K) = 0.0290 mol
-0.9847 g / 0.6290 mol = 34.0 g/mol
9-Ptotal = ntotal times RT divided by V
-1.00g H2 (mol/2.016g) = 0.496 mol
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-5.00g He (mol/4.003g) = 1.25 mol
-total n = 1.745 mol
-Ptotal = (1.745 mol)(0.08206 Latm/mol K)(293K) / 5.0 L = 8.39
atm
10 -kinetic energy distribution – within sample, not all molecules have
the same kinetic energy
-lighter molecules have higher speeds (see Textbook’s Table 9.5)
12 -Ans1: A
-Ans2: C
13 - no interaction between molecules
NEXT
LECTURE
1-these are empirical equations – values of gas are measured, then
fitted to equation
-P = nRT/V = (4.89 mol)(0.08296Latm/molK)(299K)/ 1.98 L = 60.6
atm
-This number is 35% too high
-To use the other equation, the one on the slide
-P = (4.89 mol)(0.08206 Latm/mol K)(299K) / (1.98 –
4.89molx0.0427L/mol) minus (4.89 mol)^2(3.59 atmL^2/mol^2) /
1.98 L^2 = 45.8 atm (2% too high)
-Use van der Waals only when a and b given, otherwise use the ideal
gas law
2-Ans1: B
-If doubled the volume, pressure should be 1 atm, but not
-Less than 1 atm, must mean something else is going on
-There is more interaction between A and B; they spend more time
together than bouncing off the walls of the container
-Ans2: C
-Kinetic energy affected by temperature
-Kinetic energy affects average speed
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