Name:__________________________
Chem 202
Section: _____________
Quiz 4
9/26/2008
1) 1. Draw molecular orbital diagrams (labeling all orbitals) for H22-, He22+ and H2+.
Using your drawings, give the bond order for each, order them in terms of
increasing stability, and explain why H2+ has the highest electron affinity of the
three molecules. (4)
Answer:
H22He22+
H2 +
Labeling: Axes must be labeled E for energy, atomic orbitals must be labeled
as 1s, orbitals must be shown to add together (i.e., with dashes), molecular
orbitals can be labeled with sigma notation or bonding/antibonding.
Bond order of H22-= (2-2)/2 = 0
Bond order of He22+ = (2-0)/2 = 1
Bond order of H2+ = (1-0)/2 = !
Stability - H22-<H2+<He22+
When the cationic hydrogen gains an electron, its bond order goes up. Since the electron
will go into the bonding/sigma orbital, a stronger bond will form. Thus, it really wants
(has an affinity for an electron). The cationic helium does not want an electron since that
will decrease its bond order/stability and the anionic hydrogen has no more room to add
an electron since both its molecular orbitals are filled.
2) Discuss the differences between the sp, sp2, and sp3 hybridizations in terms of the
atomic orbitals that they are comprised of. Give examples of molecules that exhibit each
type of hybridization, what the remaining p orbital(s) in the sp and sp2 hybrid orbitals are
used for, and how the hybridization tells you how many sigma and pi orbitals are present.
(3)
Answer: sp3 hybrids (1 s orbital and three p orbitals) have four effective pairs, sp2
hybrids (1 s orbital and 2 p orbitals) have three effective pairs, and sp hybrids (1 s
orbital and 1 p orbital) have two effective pairs. Examples of each are: CH4 or NH3
(sp3), ethylene (sp2), and CO2 or N2 (sp). If other examples are used, check to make
sure of the correct hybridization. The remaining p orbitals in the sp2 and sp hybrids
are used to make pi bonds. The total of the exponents of each hybrid designation
gives a combination of the number of sigma bonds and lone pairs. For each p orbital
that does not participate (i.e., sp2 and sp), a pi bond or two pi bonds will be present. If
the number of sigma bonds is less than 4, either lone pairs or pi bonds are present.
Name:__________________________
Chem 202
Section: _____________
Quiz 4
9/26/2008
1) It takes 436 kJ to break apart a mole of H2.
a) Will a light of wavelength !=274.3 nm have enough energy to break apart one
hydrogen molecule? (Show your work!) (2)
b) Why, in terms of what we have learned this year, is light able to break apart
molecules? (2)
Answer: E=hc/! =(436,000 J/mol) * (1 mol/6.02e23 molecules)
b) Energy from light can promote one (or more) electrons from one energy level to the
next. In this case, the “energy levels” are bonding and antibonding orbitals. The electron
is promoted from the bonding orbital to the antibonding orbital, which
a) causes the bond order to go to zero (or -1), thus breaking the bond.
b) causes the electrons to move from between the nuclei to opposite sides of the
nuclei, causing the positive nuclei to repel each other.
Buzz words: Energy levels, excite from bonding to antibonding orbitals, bond order,
between the nuclei (bonding zone), opposite sides of nuclei (antibonding zone)
a) and b) should be worth the same. So either both 1.5 or both 2.
2) Calculate the atomic radius of a neutral chlorine and its most stable ion. (both are in
the ground state) (3)
Answer:
Answer:
Clorine=Cl
n=3
r=
(3)(8.85x10 !12 C 2 J !1m !1 )(6.626x10 !34 Js) 2
(" )(9.11x10 !31 kg)(1.6x10 !19 C) 2 (4)
r = 3.98x10 !11 m
Cloride Ion=Cl-1
n=3
(3)(8.85x10 !12 C 2 J !1m !1 )(6.626x10 !34 Js) 2
r=
(" )(9.11x10 !31 kg)(1.6x10 !19 C) 2 (3.5)
r = 4.55x10 !11 m
Also accepted for full credit(not technically correct since n=valence shell not # of
valence electrons)
Chlorine = Cl
Chlorine Ion = Cl-
3) We learned about hybridized orbitals and how different molecules hybridize their
orbtitals in order to form bonds. Draw and label the hybridized orbital for CN-. Also label
where the ! and " bonds are. (You can either draw the bonds formed between orbitals, or
show which orbitals the bonds form between.)(3)
sp Hyrbidized orbital
Pi Bond
CC
N
Pi Bond
The sigma bond is not labeled but is located between the C and N nuclei.
Make sure that sp orbital doesn’t look just like a p orbital, note asymmetric difference in shape between
normal p and sp orbital.
Name:__________________________
Chem 202
Section: _____________
Quiz 4
9/26/20
1) The bond energy of Br2 is 192.807 kJ/mol. Br2 is known to be a light sensitive
compound and special precautions are made to not expose Br2 to light. We know
from quantum mechanics that if a precise wavelength of light corresponds to a
change in energy, an electron can be promoted to a higher energy state. Calculate
the wavelength of light needed to break a Br-Br bond. (3)
192.807kJ/mol*1000 J/kJ * (1mol/6.022e23) =hc/!
!=620.86 nm
2) Bill Nye the science guy has come up with a new hybrid version of helium and has
gotten it to form a diatomic molecule with normal helium. However, the diatomic
molecule is very unstable. Give the most ‘likely’ molecular diagram (MO) for the
diatomic helium and the bond order. What can Bill Nye do to make the molecule more
stable? (4)
A:
Bond order= (# bonding - # anti-bonding)/2
BO=(2-1)/2=0.5
Bill Nye can make the molecule more stable by getting the two hybrid versions of the
helium atom to bind and make the diatomic helium. This would give a bond order of 1
which is more stable than a bond order of 0.5
3) We know that we can constructive or destructive interference when two orbitals
come together to form a bond. For the following pairs of orbitals indicate whether
they will have constructive or destructive interference.
Answer
a) No, opposite signs
b) No, constructive on bottom, destructive on top.
c) Yes, same sign, constructive on top and bottom.