Download Chem 150 Answer Key Problem Introductory Quantum Chemistry 1

Chem 150
Answer Key Problem Introductory Quantum Chemistry
1. Determine which of the following statements are false and correct them.
a) Electromagnetic radiation is incapable of passing through water.
b) Electromagnetic radiation travels through a vacuum at a constant speed, regardless of the
wavelength.
c) Infrared light has higher frequency than visible light.
d) The glow from a fire place, the energy within a microwave oven and a for horn blast are all
forms of electromagnetic radiation.
a) FALSE. Electromagnetic radiation can pass through water (or air or vacuum for that matter).
Certain frequencies of the spectrum might be absorbed by water to excite vibrational,
rotational, electronic or nuclear energy levels but the majority of frequencies will pass through
water. Water appears colorless (unless when viewed through a layer of several meters, when it
will look blue)because it does not absorb a great deal of electromagnetic radiation in the visible
region. The picture changes when we dissolve some copper sulfate. You all know what happens.
The resulting solution is blue because the solvated copper ions absorb yellow light
(complementary color to blue).
b) correct
c) FALSE, infrared is less energetic than visible light because it has a smaller frequency. E=hν
d) FALSE, the fog horn gives off sound waves not electromagnetic waves.
2.
a) What is the frequency of radiation that has wavelength of 955 μm?
b) What is the wavelength of radiation that has a frequency of 5.50 x 1014 s-1?
c) Would the radiation in part a) or part b) be visible to the human eye?
a) ν = 3.14 x 1011 s-1
b) λ = 5.45 x 10-7 m = 545 nm
c) The radiation in b) would be visible (350-750 nm approximately).
3.
Energy from radiation can be used to cause rupture in chemical bonds (wanted e.g. as in
photosynthesis or in degradable plastics that decompose in sunlight or unwanted as in a
sunburn). A minimum energy of 941 kJ/mol is required to break the nitrogen-nitrogen bond in
N2. What is the longest wavelength of radiation that possesses the necessary energy to break
the bond? What type of radiation is that?
This problem will look much simpler if we translate in to a chemical equation.
941 kJ + 1 mol N2 (molecules)
→
2 mol N (atoms)
Energy of 941 kJmol-1 means 941 kJ applied to one mol of matter (in our case nitrogen
molecules). Molecules generally cannot accumulate energy from several photons. Only if a
photon of sufficient energy hits the dinitrogen the molecule will it dissociate. So let’s divide the
energy per mol by Avogadro’s number and get to the energy required to break just one
molecule.
1.56 x 10-18 J
+ 1 molecule N2
→
2 atoms N
Now since E = hν and c = λν we can calculate the wavelength λ = hc/E = 1.27 x 10-7 m = 127 nm
(UV light)
4.
It requires a photon with a minimum energy of 4.41 x 10-19 J to emit electrons from sodium
metal.
a) what is the minimum frequency of radiation necessary to emit electrons from sodium metal?
b)What is the wavelength of this light?
a) 6.6 x 1014 Hz
b) 450 nm
5.
Calculate the de Broglie wavelength of a muon, a subatomic particle that decays within a few
nanoseconds after formation traveling with a velocity of 8.85 x 105 cm s-1. The resting mass of a
muon is 206.8 electron masses.
Using the De Broglie relationship λ = h/p = h/(mv) with a mass of 206.8 me and a velocity of 8.85
x 105 cms-1 = 8.85 x 103 ms-1 we get λ = 3.97 x 10-10 m.
6.
a) According to the Bohr model an electron in the ground state of a hydrogen atom orbits the
nucleus at a specific radius of 0.53 x 10-10 m. In the quantum mechanical description of the
hydrogen atom, the most probable distance of the electron from the nucleus is 0.53 x 10-10 m.
Why are these two statements different?
b) Why is Schrődinger’s wave equation to describe the location of a particle very different from
the description obtained from classical physics?
c) In the quantum mechanical description of an electron what is the physical significance of the
square of the wave function Ψ2?
a) In Bohr’s model the electron orbits the nucleus. Both its position (e.g. radius from the
nucleus) and its energy are precisely known (violation of Heisenberg’s uncertainty principle). In
quantum mechanics the electron is in an orbital and not in an orbit. The most likely distance to
find the electron from the nucleus is at the same distance as Bohr’s orbit, but there is no
guaranty that it will be there, just a chance, and the chance of finding it at this radius is higher,
than finding it at any other radius (I say radius but I should really be speaking about finding it an
small volume element dV, an infinitely small shell of thickness dr at the distance r from the
nucleus).
b) because it describes the electron as a wave and not a particle
c) The square of the wave function is proportional to the electron charge density or the
probability of finding the electron. Did you ever wonder how Born deduced this relationship? In
wave theory the square of the amplitude of a wave corresponds to its intensity. So if the wave is
a light wave the amplitude of this light wave corresponds to ψ and the square ψ2 corresponds to
the intensity, which relates to the photon density, the number of photons present in a region of
space. So if we have matter waves (particle waves) the square of the wave function relates to
the electron charge density, the number of electrons present in a region of space.
7.
Give the values for n, l, and ml for a) each orbital of the 2p subshell, b) each orbital in the 5 d
subshell.
a) for a 2p orbital n equals 2 (the orbital is in the 2nd principle shell) and l = 1 (it is a p orbital
(shape!)) ml can be –1, 0 and 1 (there is a px, a py and a pz orbital)
b) For a 5d orbital n is 5 (the orbital is in the 5th principle shell) and l = 2 (it is a d-orbital) ml
depends on the value for l and can only be -2, –1, 0 and +1 and +2 (there are always just five dorbitals no matter if they are 3d, 4d or 15d orbitals)
8.
What is the maximum number of electrons in an atom that can have the following quantum
numbers: a) n = 2, ms = -1/2; b) n = 5, l = 3; c) n = 4, l = 3, ml = -3; d) n = 4, l = 1, ml = 1.
a) n=2, ms -1/2 the 2nd principle shell consists of 2 subshells, the 2s and the 2p subshells that can
hold 8 electrons (Octave). The one s orbital can only hold 2 electrons, the three p orbitals in the
p subshell can hold 6 electrons. Let’s just write down all the different possibilities and remember
that no two electrons might posses the same four quantum numbers.
n = 2, l = 0, ml = 0, ms = +1/2
n = 2, l = 0, ml = 0, ms = -1/2
n = 2, l = 1, ml = 0, ms = +1/2
n = 2, l = 1, ml = 0, ms = -1/2
n = 2, l = 1, ml = 1, ms = +1/2
n = 2, l = 1, ml = 1, ms = -1/2
n = 2, l = 1, ml = -1, ms = +1/2
n = 2, l = 1, ml = -1, ms = -1/2
Only the bold sets of quantum numbers satisfy the requirement of having n = 2 and
ms = -1/2 so there are four electrons that can have these two quantum numbers:
n = 2 and ms = -1/2
b) n = 5, l = 3 this means there are 14 possible combinations for ml and ms (there are 14 f block
members)
ml = -3 and ms = +1/2
ml = -3 and ms = -1/2
ml = -2 and ms = +1/2
ml = -2 and ms = -1/2
ml = -1 and ms = +1/2
ml = -1 and ms = -1/2
ml = 0 and ms = +1/2
ml = 0 and ms = -1/2
ml = +1 and ms = +1/2
ml = +1 and ms = -1/2
ml = +2 and ms = +1/2
ml = +2 and ms = -1/2
ml = +3 and ms = +1/2
ml = +3 and ms = -1/2
c) Since 3 quantum numbers are fixed we can only have two electrons that have the same n (are
in the same principle shell), the same l (are in the same subshell) and the same ml (are in the
same orbital). Those two electrons must differ in their fourth quantum number ms
n = 4, l = 3, ml = +3 and ms = +1/2
n = 4, l = 3, ml = +3 and ms = -1/2
d) Similar picture as above: 2 electrons and they must have opposite spin (ms = -1/2, ms = +1/2)
9.
a) What are valence electrons? b) What are unpaired electrons? c) How many valence electrons
does a P atom possess and how many of them are unpaired?
a) valence electrons are electrons in the outermost electronic shell (closed shell electrons, the
electrons that occupy the completely filled inner shells are referred to as inner electrons or core
electrons) Only s and p electrons are referred to as valence electrons but not d or f-electrons!
b) Unpaired electrons have the same or parallel spin.
c) Phosphorus has 5 valence electrons in the electronic configuration:([Ne] 3s2 3p3). Three of
these electrons are unpaired. (You might find it helpful to draw the electronic configuration of
phosphorus with an orbital diagram to answer the question about the number of unpaired
electrons.)
10.
Represent the electronic configuration of the following elements with an orbital diagram: F, S,
Ar, Ti, and Ni.
The orbital diagrams below represent the electronic configuration of F, S, Ar, Ti and Ni
respectively.
11.
Draw a 2pz orbital. Make sure to include a labelled axis or coordinate system on your drawing
and show phasing either as + and – or shaded and not shaded.
Please refer to your handouts. You must be able to draw any s and p orbitals!
12.
What is the uncertainty in the position of an electron whose speed is known with an uncertainty
of 1 m/s.
ΔxΔp ≥ h/4π since p = mu, ΔxΔ(mu) ≥ h/4π or
(ΔxΔu)m ≥ h/4π which
-5
yields Δx = 5.8 x 10 m. This seems like a small and insignificant number but if you consider that
atoms are approx. 10-10 m in size it means we have no idea if the electron is in the atom or not.
13.
How does Valence bond theory describe the bonding in a) HCN and b) C2H4 (ethylene)? Sketch
and label the atomic and/or hybrid orbitals that overlap. Clearly show phasing and occupancy of
each orbital. How many σ-bonds and how many π-bonds (if any) have you formed in each
molecule?
HCN
First we draw a Lewis structure of the molecule!
H
C
N
Next we determine the shape of the molecule:
Both carbon and nitrogen are surrounded by two electron groups (EGG = 2) which suggests
linear geometry:
H
C
N
linear
Select the hybridization that is consistent with the shape (sp3 for tetrahedral, sp2 for trigonal
planar and sp for linear!)
Since both carbon and nitrogen have EGG=2 (linear) we select sp hybridization for both:
Remember! sp hybridization means we algebraically combine one of the three 2p orbitals with
the one available 2s orbital:
For clarity I used shaded and not shaded to represent the phase of the orbital (instead of +/-)
+
One 2p orbital
+
+
one 1s orbital → algebraically combine → two sp hybrid orbitals
=
Little shaded loops not shown for clarity!
Now we also need to include the two unmodified 2p orbitals which will be orthogonal to the sp-hybrid
orbitals to minimize repulsion.
Next we need to find the occupancy of each orbital for both carbon and nitrogen:
Energy
hybridize
2p
2p
sp hybrid
2s
Fill the orbitals with the appropriate number of electrons:
Energy
2p
sp hybrid
Carbons four valence electrons
2p
sp hybrid
Nitrogens five valence electrons
Note that the orbital filling occurs singly until both the hybrid set and the unmodified p orbitals are filled
and not as represented in the diagram below:
Energy
2p
sp hybrid
The energy difference between the sp-hybrid orbitals and the unmodified 2p orbitals is too small to
compensate for the energy required to pair two electrons in a single orbital (spin pairing energy)
So we have one hydrogen atom with one electron in a 1s atomic orbital. We also have one carbon
atom with two singly occupied sp-hybrid orbitals and two singly occupied 2p orbitals. Finally we have
one nitrogen atom with one singly occupied sp-hybrid orbital, one doubly occupied (filled) sp-hybrid
orbital and two singly occupied unmodified (not hybridized) 2p orbitals. In pictures this looks like this:
H(left) one 1s1; C (centre), two singly occupied 2p and two singly occupied sp hybrid orbitals
N(fare right), two singly occupied 2p,one singly occupied sp hybrid and one full sp hybrid orbital
We need to connect the elements H and C and N with a chemical bond. This is accomplished by
overlapping the singly occupied 1s orbital of hydrogen with one of the two singly occupied sp hybrid
orbitals of carbon. The other singly occupied sp-hybrid-orbital of carbon overlaps with the one (and
only) singly occupied sp-hybrid orbital on nitrogen. This pairs up the electrons and forms two sigma
(σ)bonds.
As the carbon and nitrogen atoms have two sets of singly occupied 2p orbitals they can also overlap
(side on) to produce two π bonds. In pictures this looks like this:
This diagram is consistent with the Lewis structure which shows a triple bond between C and N. In
valence bond theory carbon and nitrogen are connected via one σ-and two π-bonds. The lone pair on
carbon is represented by the filled sp hybrid orbital on N which cannot engage in bonding. Only half
filled orbitals are allowed to overlap! Over and above what Lewis theory predicts, Valence Bond theory
suggests that there is no rotation about the carbon-nitrogen triple bond.
H
C
N
Ethylene:
H
H
C
H
C
H
The Lewis structure suggests trigonal planar geometry at both carbon atoms (EGG =3) which suggests
sp2-hybridization:
Algebraically combine two of the three 2p orbitals with the one 2s obital resulting in three sp2 hybrid
orbitals:
+
+
=
shaded little loops omitted for clarity
Including the one unmodified p orbital which will be orthogonal to the plane of the sp2 hybrid orbitals
we get something that looks almost like a trigonal bipyramid with the one p-orbital in axial position and
the three hybrid orbitals in equatorial position.
What is the occupancy in each of these orbitals:
Energy
hybridize
2p
sp2 hybrid
2p
2s
Fill the orbitals with the appropriate number of electrons:
Energy
2p
2
sp hybrid
Carbons four valence electrons
Note that again the orbital filling occurs singly until both the hybrid set and the unmodified p orbital are
filled and not as represented in the diagram below:
Energy
2p
sp hybrid
Again the energy difference between the sp-hybrid orbitals and the unmodified 2p orbital is too small to
compensate for the energy required to pair two electrons in a single orbital (spin pairing energy)
Now let’s make some bonds!
We have 4 hydrogen atoms which have one electron each in a spherical 1s orbital:
We also have two carbon atoms with three singly occupied sp2 hybrid orbitals and one singly occupied
unmodified 2p orbital.
We can make a carbon-carbon sigma(σ) bond by overlapping two singly occupies sp2 hybrid orbitals
head on! This pairs up the two electrons in the σ bond. We also get a π bond because the two
unmodified and singly occupied 2p orbitals overlap side on. This is the C=C double bond!
Finally we have four more singly occupies sp2 hybrid orbitals on the two carbon atoms that can
overlap head on with four singly occupied 1s orbitals of hydrogen to make four carbon-hydrogen
σ-bonds.
This suggests that ethylene (5 sigma and one pi-bonds) should be planar and there is no rotation
about the C=C double bond.
14.
Given below is the molecular orbital diagram for diatomic species of the first period. Assume the
species H2- was postulated to exist, a dihydrogen anion.
a) How many electrons does this species possess?
b) Show how the electrons occupy the molecular orbital diagram below.
c) What is the bond order in H2- ?
d) Would you expect a bond length longer or shorter as that in dihydrogen?
e) Is the species dia- or paramagnetic?
f)
Answer:
Should H2- exist?
a) There are three electrons altogether in H2- , one from each hydrogen atom and one to
account for the charge (see diagram below on the left)
b) see diagram below on the right, c) Bond order is (2-1)/2 = 0.5
d) longer, e) paramagnetic, f) should exist although much less stable than dihydrogen
15. Given below is an empty an unlabeled molecular orbital diagram of homonuclear diatomic species
of the second period in the periodic table for atomic number Z equal or greater 8.
a) Label the atomic as well as the molecular orbitals.
b) How many of the molecular orbitals are bonding and how many are anti bonding?
c) How does MO theory describe the bonding in fluorine gas? Fill the MO diagram with the
appropriate number of electrons. What is the bond order in F2?
d) Is F2 diamagnetic or paramagnetic?
a) see MO diagram
b) 5 molecular orbitals are boning and 5 are antibonding (denoted with *)
c) Bondorder =(# of e- in bonding MO’s - # of e- in anti-bonding MO’s)/2 = (10-8)/2 =1
d) F2 is diamagnetic. All electrons are paired!
s* 2p
F2
p* 2p
p* 2p
2p
2p
p2p
p2p
s2p
2s
s* 2s
2s
s2s
1s
s* 1s
1s
s1s
16. Why does the conductivity of a semiconductor such as silicon increase with temperature?
In band theory silicon posessesa filled valence band and an empty conduction band. The two are
separated by only a small band gap which allows thermal excitation of electrons from the valence band
in the conduction band. This “mobility” of electrons is enhanced as temperature increases and more and
more electrons have sufficient energy to enter the conduction band.
Formulas and constants
pV = nRT
piVi(Ti)-1 = pfVf(Tf)-1
Ptotal = p1 + p2 + p3 + .... + pn
[p + n2a/V2][V – nb] = nRT
Vmol = 22.4 L mol-1 at STP (1 atm = 101.325 kPa and 273.15 K)
R = 8.3145 L kpa mol-1 K-1
Fundamental wave equation
c = λν
Plank’s equation
E = hν
Heisenberg’s uncertainty principle
ΔxΔp ≥ h/4π
De Broglie’s wave-particle dualism
λ = h/p
Impulse
p = mv
Kinetic energy
Ekin = 0.5mv2
Planks constant
h = 6.626 10-34 Js
Avogadro’s number
Na = 6.022 x 1023 mol-1
Unified atomic mass unit
1 u = 1.660 x 10-27 kg
Mass of electron
me = 9.109 x 10-31 kg
Mass of proton
mp = 1.673 x 10-27 kg
Mass of neutron
mn = 1.675 x 10-27 kg
Speed of light
c = 3 x 108 ms-1
End of formula sheet. The orbital pictures on the next page will not be on the formula sheet!