Review Lecture Worksheet
(the answers are written in white font, switch all font to black to view them)
Rank the 3rd period p- block elements (In order of atomic number that is: Na, Mg, Al, Si, P, S, Cl,
ArAr) in terms of electron affinity, electronegativity and ionization energy. Explain all.
Explanation: of general trend. As you go across the periodic table toward the noble gasses,
effective nuclear charge increases. This causes all three of the trends listed above to increase as
well.
Ionization Energy: Low-High: Na, Al, Mg, Si, S, P, Cl
Explanation of exceptions: because Mg and P have stable filled and half-filled subshells
respectively they have an abnormally high ionization energy. Causing them to “switch”
orders with the Al and S respectively.
Electron affinity: Low-High: Mg, Na, Al, P, Si, S, Cl
Explanation of exceptions: because Mg and P have stable filled and half-filled subshells
respectively they have an abnormally low electron affinities. Causing them to “switch”
orders with the Na and Si respectively.
Electronegativity: Low-High: Na, Mg, Al, Si, P, S, Cl
Explanation of exceptions: because electronegativity is something we discuss with
atoms in a bond, all the electron configurations are already stable, leaving
electronegativity to only be related to the effective nuclear charge and not to the
electron configurations.
Draw ICl4- and ICl4+. For each- What is electron geometry, molecular geometry, bond angles,
hybridization of the central atoms. What orbital does each atom use to bond with? (answer on
next page)
36 e-
34 e-
Draw the MO diagram for CO assume that 2p is higher in energy than 2p. Be sure to label all
atomic and molecular orbitals, including subscripts. Be sure to fill in all electrons necessary.
What is the bond order? Is it paramagnetic or diamagnetic? (answer on next page)
How does constructive and destructive interference relate to molecular orbital diagrams?
B.O.: ½ (8-2)=3
diamagnetic
Constructive interference forms bonding orbitals as
shown by the overlap in the pictures
Destructive interference forms antibonding orbitals as
shown by the overlap in the pictures
I’ve shown many pictures of the same thing here so you
can see different “artist renditions” and pick whichever
version is most helpful to help you understand what is
going on.
What is the molecular formula for the following? Give the hybridization of circled atoms (for
the sake of practice do it for all of them till you get the hang of it, any are fair game). How many
sigma bonds are present? How many pi bonds are present? Any guess what molecule it is?
C17H13ClN4 (hints: be careful to get the carbon at the
end of the line, up near the top. Don’t forget it also has
3 hydrogens attached).
Hybridization: most common mistake here is on the
nitrogens, don’t forget there is a lone pair that isn’t
drawn in line structures, making the =N- nitrogens sp2
hybridized and the nitrogen with three single bonds
sp3 hybridized.
Sigma bonds: 38
pi bonds present: 9
The molecule is Alprazolam, better known as xanex (I
thought it appropriate for a review sheet)
Rank the following molecules in order of bond angles: H2O, CH4, ClO3- BH3, NO2, NO2-. For each
molecule state the hybridization and what orbital each bond is formed with, as well as where the
lone pairs on the central atom is (are) located. For H2O, ClO3- and NO2 rank in order of expected
bond length and strength.
H2O, ClO3-, CH4, NO2-, NO2, BH3
Bond orders: H2O= 1
NO2=1.5
ClO3-= 5/3=1.666
As bond order increases length decreases and strength increases.
Length: ClO3-< NO2< H2O
Strength: ClO3->NO2>H2O