CHEM101, B2
HMWK 03
Answers
2008 01 25
Schroedinger - Periodic Trends
HT
01.) a.) Determine the number of d orbitals for the case where the max. number for n = 4.
d orbitals means l=2. We can have 5 3d orbitals and 5 4d orbitals;
for a total of 10 orbitals.
b.) Determine the number of orbitals with ml =1 for the case where the max. number for n=4.
ml can be 1 for p, d, and f orbitals (l = 1, 2, 3). So we can have ml=1 orbital for the
2p, 3p, 3d, 4p. 4d, 4f series; for at total of 6 orbitals.
02.) Make a sketch of the nodes and phases of the following orbitals; must include labeling of
coordinate system:
a.) 2pz
b.) 3dxy
c.) 3dxz
03.) Show a graph of Ψ2 (electron probability, point probability, PP) in the direction of the xaxis for a.) the 3dxy and b.) for the 3dxz orbitals.
04.) Graph the PP function for the 3dx2-y2 orbital in the direction of the
a.) x - axis, b.) y - axis c.) z- axis.
05.) Show graphs of Ψ, Ψ2 and 4πr2Ψ2 functions for the 3s orbital. Line-up the graphs one on
top of the other to clearly show location of nodes.
06.) How many d orbitals are occupied in the ground state of the following elements:
a.) Ca, 0 b.) Sr, 5 c.) V, 3 d.) Cr, 5 e.) Co, 5 f.) Br, 5 g.) I 10
07.) For element 73, tantalum, show the electron configuration for the ground state
a.) in “full, standard” notation, 1s22s22p63s23p64s23d104p65s24d105p66s25d34f14
b.) “shorthand, core” notation. [Xe] 6s25d34f14
08.) Which of the following species are paramagnetic. Show the work by writing “shorthand,
orbital filling” electron configurations.
Na, S–, Cl–, Sc3+, V, Cr, Cu2+, Ga3+, As3+
para
Na : [Ne]
S- : [Ne]
Cl- : [Ne]
3s
para
3p
3s
dia
3p
Sc3+ : [Ar]
V : [Ar]
Cu2+ [Ar]
Ga
3+
[Ar]
As3- [Ar]
dia
4s
para
3d
4s
para
3d
4s
3d
4s
3d
para; somewhat uncertain
could be 3d10, then dia
dia
09.) Consider the following elements. In which of the successive ionization energies do we
encounter a great jump in the magnitude of the IE?
a.) C,
b.) Na,
c.) Mg,
d.) P
We expect great jumps in IE once we start to lose core electrons. Therefore,
a.) C, IE5
b.) Na, IE2
c.) Mg, IE3
d.) P, IE6
10.) Rank the following species in terms of atomic size; smallest first.
If there is ambiguity. (uncertainty) in any of the placements indicate this.
Li+, Be2+, O, O2–, F, S2–, Se2–, As3–
Be2+ < Li+ < F < O < O2– < S2– < Se2– < As3–
There is some uncertainty where to place F relative to Li+ . F is smaller than Li,
but the +1 charge should reduce the size of Li+ a lot.
(We can confirm this by consulting Table 9.8 in Petrucci: Li+, 59 pm ; F, 71 pm.)
11.) Compare the differences in the
a.) EA’s between Be and B,
In Be, the new electron goes into a new subshell, 2p. The existing electrons (1s and 2s)
cover up the nucleus nicely; consequently, there is no attraction for an incoming electron.;
i.e., EA ≈ 0.
In B, the new electron would go into a 2py orbital (assuming the first one is in 2px). 2px and
2py orbitals are widely separated; i.e., don’t shield each other much. Therefore, an
incoming electron is “highly” attracted by the B nucleus and the EA1 for B has a large neg.
value.
b.) IE’s between B and C,
Electrons in different 2p orbitals don’t shield each other much. Therefore, an outer electron
in C experiences greater attraction by the nucleus than an outer electron in B.
Therefore, IE1 in C is larger than in B.
c.) IE’s between P and S. Explain the difference in each case.
In S, we remove an electron from a doubly filled p orbital, while in P an electron is
removed from a singly filled p orbital. A certain amount of repulsion between the 2
electrons in the doubly filled orbital will make removal easier.
Therefore, IE1 for S smaller than for P.