Chemistry 3830 – Fall 2010
Answers to Assignment #1
What is the ratio of the energy of a ground state He+ to that of a ground state Be3+ ion?
1.
The ground-state energy of any one electron system is given by E n =1 = −
Z 2 me 4 , i.e. from the Bohr equation written for
8ε 02 h 2
just a single state, and with n = 1. If we are simply comparing He+ with Be3+, all the constants fall away and we get the ratio
E(He+)/E(Be+3) = 22/42 = 0.25
2.
The ionization energy of hydrogen is 13.6 eV. What is the difference in energy between the n =1 and n = 6 levels?
Here again we use the Bohr equation: E n − E m = hν =
experimental
("Rydberg-like") constant
⎛1 1⎞
E1 − E 6 = −13.6⎜⎜ 2 − 2 ⎟⎟ = −13.2 eV
⎝1 6 ⎠
E1
=
1⎞
Z 2 me 4 ⎛ 1
; first calculating n=1 to n=∞ which gives us an
2 2 ⎜ 2 − 2 ⎟
8ε 0 h ⎝ nn nm ⎠
–13.6
eV.
Then
we
can
use
this
to
calculate
3.
What is the relationship between the possible angular momentum quantum numbers to the principal quantum number?
The rule stemming from the solution to the Schrödinger equation for the H atom is: l = 0, 1, 2, ….(n – 1). In words, the
values of the angular momentum quantum number are a family of positive integers ranging from zero to one less than the
value of the principal quantum number.
4.
How many orbitals are there in a shell of principal quantum number n? (Hint: begin with n = 1, 2 and 3 and see if you can
recognize a pattern.
The number of orbitals in any given shell is simply given by n2.
5.
Using sketches of 2s and 3p orbitals, distinguish between (a) the radial wavefunction, (b) the radial density distribution
function, and (c) the angular wavefunction.
Ψ
2
4πr Ψ
2s radial wavefunction
r→
Ψ
2
2s radial distribution function
2s angular
wavefunction
r→
2
4πr Ψ
3p radial wavefunction
r→
2
3p radial distribution function
3p angular
wavefunction
r→
6.
Give the ground state electron configuration of (a) C, (b) F, (c) Ca, (d) Ga3+, (e) Bi, (f) Pb2+, (g) Sc, (h) V3+, (i) Mn2+, (j)
Cr2+, (k) Cu, (l) Gd3+, (m) W, (n) Eu3+, (o) Eu2+, (p) Mo4+.
C: [He]2s22p2; F: [He]2s22p5; Ca: [Ar]4s2; Ga3+: [Ar]3d10; Bi: [Xe]4f145d106s26p3; Pb2+: [Xe]4f145d106s2; Sc: [Ar]3d14s2; V3+:
[Ar]3d2; Cr2+: [Ar]3d4; Cu: [Ar] 3d104s1; Gd3+: [Xe]4f7; W: [Xe]4f145d46s2; Eu3+: [Xe]4f6; Eu2+: [Xe]4f7; Mo4+: [Kr]4d2.
7.
Which atom should have the larger covalent radius, (a) potassium or calcium; (b) fluorine or chlorine? Give your
reasoning.
(a) Potassium should be larger than calcium. They are both in the same period of the PT, and radii decrease from left to
right. To prove this, we can calculate Z*. K: Z* = 19 – [10 + (8 × 0.85)] = 2.2
Ca:
Z* = 20 – [10 + (8 × 0.85) +
(0.35)] = 2.85. Thus we expect the forces operating on the valence shell in calcium to be greater than those in potassium,
leading to a smaller radius in calcium.
(b) Chlorine should be larger than fluorine. We would expect Z* to be very similar, since the two atoms are in the same
group of the periodic table. However, the valence electrons in chlorine are 3p electrons, as opposed to the 2p in fluorine.
Thus the electrons occupy a shell with a larger Bohr radius, and being further from the nucleus they will be less attracted to
the nucleus and the atom will be larger.
8.
Rationalize the first four ionization energies of Al (5.98, 18.82, 28.44, 119.96 eV) on the basis of ground-state electronic
configurations.
Al (5.98) [Ne]3s23p1; Al+ (18.82) [Ne]3s2; Al2+ (28.44) [Ne]3s1; Al3+ (153.77) [He]2s22p5. It is progressively harder to
remove an electron from the more positively charged ions. However, there is a huge jump in removing the fourth electron,
since this electron comes from a core orbital that experiences a much greater Z*.
1-7
9.
What are the names, symbols, and roles of the four quantum numbers associated with the hydrogen atom?
•
•
•
•
n = principal quantum number - controls the energy of the electrons in a hydrogen atom
l = orbital angular momentum q.n. - dictates the shape of the orbital
ml = orbital magnetic q.n. - determines the orientation of the orbital in 3-D space
ms = electron spin q.n. - determines the spin of the electron, which may be up or down
10. Sketch the orbitals of atomic hydrogen for the first three quantum shells. In which of these orbitals are the signs associated
with the wave functions arranged symmetrically?
For sketches of the orbitals, see the lecture notes on p.23. Note that the s and d orbitals are symmetric.
11. What orbitals do the following quantum number combinations describe? Are they even permissible solutions of the
Schrödinger equation? (i)
n=2 l=1
ml = -1 (ii)
n=1 l=0
ml = -4
(i) n = 2
(ii) n = 1
l=1
l=0
ml = -1 one 2p orbital (allowed)
ml = -4 a 1s orbital, not permissible with a value of ml = -4
12. Which of the following orbitals are allowed for the hydrogen atom by quantum mechanics? 5p; 5h; 6d
5p is permissible, since for n =5, l can be up to 4, which includes l = 1
5h is not permissible, since this requires l = 6
6d is permissible, since for n =6, l can be up to 5, which includes l = 2
13. Consider the following radial probability density-distribution plot and respond to the associated questions.
a) How many radial nodes are there?
b) If the total number of nodes is 3, what type of orbital is involved?
c) Which orbital would it be if there were one more node?
2
4 πr Ψ
2 (nm-1 )
4
3
2
1
0.1
0.2
0.3
0.4
0.5
0.6 0.7
radius (nm)
a) 1 at 0.1 nm
b) 2 angular nodes, one radial, ∴ 4d
c) 3 angular nodes, one radial, ∴ 5f
14. On the left-side graph below is a crude sketch of the radial component of a wavefunction for a hydrogen atom. The
principle quantum number is given.
2
4 πr Ψ
Ψ
2
n=4
r→
r→
a) Identify the orbital Ψn, l, ml =Rn,l • Yl,ml
b) Sketch roughly the radial probability density function corresponding to this wavefunction.
a)
Ψ4,0,0 =R4,0 • Y0,0 Sketch filled in below:
2-7
2
Ψ
4 πr Ψ
n=4
2
a 4s orbital, since all
three nodes are radial
r→
r→
15. On the right-side graph below is a crude sketch of the radial probability density function for a hydrogen atom. The
principle quantum number is given.
2
4 πr Ψ
Ψ
2
n=2
r→
r→
a) At the left, sketch the radial component of the wavefunction that corresponds to this probability function.
b) Identify the orbital by its common name
c) If instead of n = 2, the sketch at right belonged to an orbital with n = 3, what orbital would that be? (Of course, then the
sketch you have drawn at the left would no longer be valid!)
a) Done on sketch below.
b) Done on sketch below.
c) This would make it 3p.
2
4 πr Ψ
Ψ
2
n=2
It is a 2s orbital
r→
r→
16. What are the advantages to write the orbital (wavefunction) (a) in a complex form and (b) in a real form?
When you write the orbitals (wavefunction) in the complex form (using the imaginary number i) you are able to define the
quantum number ml, however, you cannot visualize them in terms of Cartesian coordinates (except for ml = 0). Using the
real wavefunctions, it is possible to label the orbitals with Cartisian coordinates, such as px and py. In the real form, most
orbitals cannot be described by the q.n. ml (exception: ml = 0).
17. On the left-side graph below is a crude sketch of the radial component of one wave function of a hydrogen atom.
Ψ
r→
r→
a) At the right, sketch the corresponding radial probability density distribution function for this wave function.
b) If the total number of nodes is 2, identify the orbital both by its common name and the quantum numbers which
uniquely describe it.
3-7
b) If instead the total number of nodes is 4, would the sketch on the right still be a valid ] plot? If so, identify the orbital
both by its common name and the quantum numbers which uniquely describe it.
a) Done on sketch below.
b) Done on sketch below.
c) Yes it would still be valid. Now the number of angular nodes is 3, so it must describe a 5f orbital (n = 5, l = 3).
Ψ
1 radial node, therefore 1 angular
node, so must be a 3p wavefunction
r→
r→
18. For the following molecules and ions, provide the Lewis structure and the VSEPR geometry (the central atom(s) are
underlined). Include any resonance, and show all non-zero formal charges. Indicate approximate bond angles around the
central atoms. And clearly indicate the isomers you choose if isomerism is possible. Assign point groups to the molecular
structures.
SF5Cl
BeH2
OCS
SO42–
..
:Cl : ..
:F..:
..
S
F.. :
:F
..
:..F:
: F:
..
CH3F
NO2F
IF4–
NNO
HSSH
GeH4
[NH4]+
IF5
..
:F :
H
octahedral
all angles ca. 90°
..
: : F : ..
:F..:
..
:F
.. Xe F.. :
:..F:
: F:
distorted octahedral
angles less than 90°
H
H ca.109°
H
..
:Cl :
:
H Be H
T-shaped mol geometry
angles ca. 90°
linear
180°
I
:
H
H
Xe
O
: O: .. :
tetrahedral
trigonal pyramidal
all angles exactly 109.5° angles <109.5°
..
..O
..
..S
linear
180°
..
:F
..
:..F:
..
N
.. ..
:F..:
I F.. :
..
square planar
angles exactly 90°
..
F.. :
:F :
ca.109° ..
H
:Cl :
..
:O
..
ICl3
SeCl4
KrF2
H
tetrahedral el-pair geometry tetrahedral el-pair geometry
about both S atoms
about nitrogen
trigonal pyramidal
molecular geometry
..
Cl
.. :
..
Ge
XeF6
TeF6
BrNO
SCN–
S S
tetrahedral
all angles ca. 109°
H
H
NHF2
XeO3
[BrBr2]–
OSO
..
O
..
+ ..
N F.. :
..
:O
..
..
: F : ..
:F..:
..
Te
F.. :
:F
..
:..F:
: F:
..
trigonal planar
ca. 120° (not exact)
ca.90° ..
.. :Cl :
: Cl
:
.... Se
Cl
:
:Cl :
ca.120° ..
..
octahedral
all angles ca. 90°
H
H
N+
H
see-saw
:
H
tetrahedral
all angles exactly 109.5°
4-7
..
O
..
+ ..
N F.. :
- ..
:O
..
:
..
:Br :
Br
: Br:
..
:
linear mol geometry
angles 180°
..
N
..
: Br
..
:
:
Kr
:F :
..
:
O
.. :
.. : O:
..
:F :
:O
..
+ ..
- ..
N
N
O
..
..
.. O
:
: O .. :
S
+ resonance structures
bent
angle <120°
linear mol geometry
angles 180°
.. ..
:F..:
..
:F
.. I F.. :
:..F:
:F
.. :
square pyramidal
angles < 90°
:O
..
linear
180°
tetrahedral
all angles exactly 109.5°
..
S
- ..
:S
..
O
.. :
+ .. : N N ..
O:
N:
linear
180°
bent
<120°
..
Comments: XeF6 is highly fluxional, the lone pair is moving from one face to another via the edge of the distorted
octahedron. For SO42- resonance structures have to be invoked (with only two formal charges).
SF5Cl
C4v
CH3F
C3v
HSSH
BeH2
OCS
SO42–
D∞h
C∞v
Td
NO2F
IF4–
NNO
C2v
D4h
C∞v
GeH4 Td
[NH4]+ Td
IF5
C4v
D2
NHF2
Cs
XeF6
XeO3 C3v
[BrBr2]– D∞h
OSO
C2v
TeF6
BrNO
SCN–
if lone pair point towards a face: C3v
ICl3
C2v
Oh
SeCl4 C2v
Cs
KrF2
D∞h
C∞v
19. Draw Lewis structures for the following molecules for which there are more than one resonance isomer. Show formal
charges, and indicate if any resonance hybrids are more important contributors than others to the true electronic structure.
The connectivity of the atoms in each of these molecules is shown by a sketch. Assign point groups to the molecular
structures.
[HCO2]-
O
H
S2O2
-
S
C
H
:O
.. :..
:O: H
- ..
:O :
+
S
..
S
:O
.. :
- ..
:O :
+
.. +
S
..
less important
-
..
N
N
S
:O :
..
:S N:
.. .. :S N:
:N
.. S :
+
:N S :
+
+
S:
:O :
equal contributions
S
S
O
O
:O:
S2N2
O
equally important
:O : ..
+
:S
.. S
.. :S N:
+
:S N:
:N
.. S :
:N
.. ..S :
-
equal contributions
:O
.. : C2v
C2h
D4h
5-7
20. Construct Lewis structures of typical resonance contributions of (a) ONC– and (b) NCO– and assign formal charges to each
atom. Which resonance contribution is likely to be the dominant one in each case?
for ONC- and NCO- : 16 valence electrons (8 valence electron pairs)
isocyanate
- .. +
:O
:
.. N
.. +
:O
.. N
:
:N
2-
neither resonance structure is
particularly good with a negative charge on C
.. O
.. :
- ..
N
..
..
O
..
2-..
N
..
+
O:
dominant
very low contribution
21. What shapes would you expect for (a) SO3, (b) SO32–, (c) IF3? Assign point groups to their molecular structures.
:O:
- .. .. .. .. .. .. ..
S
:O
:F
.. S O
.. :
.. I F
.. :
O
O
:..
:
..
:O:
: F:
..
+ resonance!
trigonal planar
T-shaped
(trigonal bipyramidal
electron group geometry)
trigonal pyramidal
D3h
C3v
C2v
22. It is possible to synthesize the ion C(CN)3–. Draw an electron-dot structure and deduce its most likely geometry. In fact, the
ion is planar. Draw one of the resonance structures that would be compatible with this finding.
:N
N:
..
normal carbanion
leads to pyramidal geometry
N
..
BUT
..
:N
-
..
N
..
N:
:N
..
-
N:
:N:
:N
..
.. N:
N
..
these resonance structures are supported by the higher electronegativity of N versus C
these resonance structures support a planar geometry
23. Show whether Ψ(t) = 8π sin(ωt) + 8π cos(ωt) is an eigenfunction of the differential operator  =
Ψ(t) = 8π sin(ωt) + 8π cos(ωt)
(d/dt) Ψ(t) =8π ω cos(ωt) - 8π ω sin(ωt)
(d2/dt2) Ψ(t) =-8π ω2 sin(ωt) - 8π ω2 cos(ωt) = - ω2 (8π sin(ωt) + 8π cos(ωt))= - ω2 Ψ(t)
6-7
d2
dt 2
24. The symmetry of the following molecules (with HC models):
C2v
C3v
Td
D2h
D3d
D3h
C2
D2d
C2v
7-7