Chem 222
Intro to Inorganic Chemistry
Problem Set 1 Answer Key
Summer 2011
Wednesday, May 11, 2011
1. Draw and label the 1s, 3s, 3p, and 3d orbitals, placing them correctly on cartesian axes (x,y,z).
1s and 3s:
3p: lobes can be signs/or shading
z
+
+
x
–
+
–
–
y
pz
px
py
3d: lobes can be signs/or shading
z
z
+
+
–
y
y
x
–
y
–
x
x
–
+
dyz
z
y
+
–
–
x
+
dxz
+
–
z
x
+
–
–
+
dxy
y
+
–
dz2
z
dx2–y2
2. (RC&O 2.20) Which atom should have the larger covalent radius, fluorine or chlorine? Give
your reasoning.
Fluorine has electron configuration 1s22s22p5, while chlorine has electron configuration 1s22s22p53s23p5.
Chlorine should have a larger covalent radius because its “outermost” electrons are in the 3p orbitals,
for which the radial distribution function extends much further out from the nucleus than do the 2p
orbitals. “The larger the value of n, the larger the orbital”.
Note that this trumps the fact that the Zeff felt by the outermost electrons of Cl is higher than that
experienced in F, as Zeff does not take into account the distance between the charges.
3. (a) (RC&O 1.5) Identify the orbital that has n = 5 and l = 1.
For the principle quantum number n = 5, the angular momentum quantum number, l, can have values
4,3,2,1,0. The number l=0 always corresponds to the s orbital; the l=1 always corresponds to the p
orbitals. Thus, the above quantum numbers correspond to a 5p orbital.
(b) (RC&O 1.6) Idenitfy the orbital that has n = 6 and l = 0.
See explanation for 3(a): this is the 6s orbital.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
4. (H&S 1.23) Write down the six sets of quantum numbers that describe the electrons in a
degenerate set of 5p atomic orbitals. Which pairs of sets of quantum numbers refer to spinpaired electrons?
For n=5, the p orbitals correspond to l=1, and ml= 1,0,-1. The six sets of quantum numbers that describe
the degenerate set of 5p orbitals are:
n=5, l=1, ml= 1, ms=+1/2; n=5, l=1, ml= 1, ms=–1/2: these electrons are spin-paired
n=5, l=1, ml= 0, ms=+1/2; n=5, l=1, ml= 0, ms=–1/2: these electrons are spin-paired
n=5, l=1, ml= –1, ms=+1/2; n=5, l=1, ml= –1, ms=–1/2: these electrons are spin-paired
5. (H&S 1.24) For a neutral atom, X, arrange the following atomic orbitals in an approximate
order of their relative energies (not all orbitals are listed): 2s, 3s, 6s, 4p, 3p, 3d, 6p, 1s.
E = 1s< 2s< 3s< 3p< 3d< 4p< 6s <6p
6. (H&S 1.25) Using the concepts of shielding and penetration, explain why a ground state
configuration of 1s22s1 for a Li atom is energetically preferred over 1s22p1.
Ground state configurations are always the lowest energy arrangement possible (Aufbau principle).
Because the 2s orbital penetrates the 1s orbital better than the 2p orbital does (i.e. 2s has a higher
probability close to the nucleus than 2p), a 2s1 electron feels a greater Zeff than does a 2p1 electron. A
2p1 electron experiences more shielding from the nucleus by the filled 1s orbital than does a 2s1
electron. Because the 2s1 electron feels a greater Zeff, this is a lower energy configuration.
See the radial distribution functions in Fig. 1.12 from the Introduction notes to see why this is the case.
7. (a) (H&S 1.26) For each of the following atoms, write down a ground state electronic
configuration and indicate which electrons are core and which are valence: (i) Na, (ii) F, (iii) N,
(iv) Sc.
(i) Na: [1s22s22p6] (core) 3s1 (valence)
(ii) F: [1s2] (core) 2s22p5 (valence)
(iii) N: [1s2] (core) 2s22p3 (valence)
(iv) Sc: [1s22s22p63s23p6] (core) 4s23d1 (valence)
(b) (H&S 1.27) Draw energy level diagrams (see class notes on electron configuration) to
represent the ground state electronic configurations of the atoms in (a).
(i)
(ii)
3s
(iii)
2p
2p
E
E
2p
E
2s
2s
1s
1s
2s
1s
(iv)
Chem 222
Intro to Inorganic Chemistry
Summer 2011
3d
4s
3p
E
3s
2p
2s
1s
8. (RC&O 1.12) Write noble-gas core ground state electron configurations for (i) calcium, (ii)
chromium, (iii) lead.
(i) Ca: [Ar] 4s2
(ii) Cr: [Ar] 4s13d5 (recall extra stability derived from half-filled shells)
(iii) Pb: [Xe] 6s24f145d106p2
9. (H&S 1.30) Draw energy level diagrams to show the ground state electronic configurations of
only the valence electrons in an atom of (i) fluorine, (ii) aluminum, (iii) magnesium.
(i) F ([He] 2s22p5)
(ii) Al ([Ne] 3s23p1)
2p
E
2s
(iii) Mg ([Ne] 3s2)
3p
E
3s
E
3s
10. (H&S 1.31) The ground state electronic configuration of a group 16 element is of the type
[X]ns2np4 where X is a group 18 element. How are the outer four electrons arranged, and what
rules are you using to work out this arrangement (i.e. explain your reasoning)?
The four electrons in the degenerate np orbitals are arranged in this way:
Hund’s rule states that for a set of degenerate orbitals electrons will not spin pair until each orbital in the
set contains one electron, and these single electrons will have parallel spins. Thus the first three
electrons have parallel spins and singly occupy the three p-orbitals. The fourth electron takes the
opposite spin and pairs with another electron in one of the three orbitals (impossible to say which one).
(Recall this is dictated by the Aubau principle, which says that orbitals are filled in the order of lowest
Chem 222
Intro to Inorganic Chemistry
Summer 2011
energy: the energetic cost of spin-pairing is lower than the energetic cost of placing the fourth electron
in the (n+1)s orbital. Another way to explain this is in terms of shielding effects: the Zeff felt by an np
electron is sufficiently higher than the Zeff felt by an (n+1)s electron to outweigh the extra pairing
energy required.)
11. (RC&O 2.22) Suggest a reason why the covalent radius of hafnium (144 pm) is almost identical
to that of zirconium (145 pm), the element above it in the periodic table.
This corresponds to a contraction in radius that is observed for elements in the sixth period (relative to
that predicted by Schrödinger’s wave equation). For these heavier atoms, relativistic effects on the
innermost core electrons result in a decreased orbital size, particularly for the s and p orbitals.
In addition to this relativistic phenomenon, the 6th period transition metals experience the “lanthanoid
contraction”: the electrons in Hf’s filled 4f orbitals are extremely poor at shielding the 6s and 5d
electrons, so the Zeff felt by these electrons is large enough to reduce the covalent radius of Hf to
approximately the same value as that of Zr.
12. (RC&O 2.25) Using Slater’s rules, calculate the effective nuclear charge on one of the 3d
electrons compared to that on one of the 4s electrons for an atom of manganese.
Mn has electron configuration 1s22s22p63s23p64s23d5 or [1s2] [2s2, 2p6] [3s23p6] [3d5] [4s2]
For one of the five 3d electrons, the screening or shielding, S, predicted by Slater’s rules is:
S = 2(0) + 4(0.35) + 8(1.00) + 8(1.00) + 2(1.00) = 0 + 1.4 + 8 + 8 + 2 = 19.4
Therefore Zeff = Z – S = 25 –19.4 = 5.6 for a 3d electron
For one of the 4s electrons:
S = 1(0.35) + 5(0.85) + + 8(1.00) + 8(1.00) + 2(1.00) = .35 + 4.25 + 8 + 8 + 2 = 22.6
Therefore Zeff = Z–S = 25 – 22.6 = 2.4 for a 4s electron.
NOTE: this explains in an approximate way why the 4s electrons are the first to leave when the 1st row
transition metals are ionized.
13. (RC&O 2.26) Using Slater’s rules, calculate the effective nuclear charge on a 3p electron in (a)
aluminum and (b) chlorine. Explain how your results relate to: (i) the relative atomic radii of
the two atoms and (ii) the relative first ionization energies of the two atoms.
(a) Al has electron configuration 1s22s22p63s23p1 or [1s2] [2s2, 2p6] [3s23p1]
For the 3p electron:
S = 2(0.35) + 8(0.85) + 2(1.00) = .70 + 6.80 + 2.00 = 9.50
Therefore Zeff = Z – S = 13 – 9.50 = 3.50
(b) Cl has electron configuration 1s22s22p63s23p5 or [1s2] [2s2, 2p6] [3s23p5]
For a 3p electron:
S = 6(0.35) + 8(0.85) + 2(1.00) = 2.10 + 6.80 + 2.00 = 10.90
Therefore Zeff = Z – S = 17 – 10.90 = 6.10
Chem 222
Intro to Inorganic Chemistry
Summer 2011
NOTE: the outermost (3p) electrons in chlorine experience a higher Zeff than the 3p electron in aluminum.
This is consistent with the smaller atomic radius of Cl (rcov = 99 pm, vs 130 pm for Al). It is also
consistent with the much higher 1st ionization energy for Cl (1251 kJ/mol, vs 577.5 kJ/mol for Al).
14. (RC&O 2.27) Which element should have the higher first ionization energy, silicon or
phosphorus? Give your reasoning.
Because phosphorus is further to the right in the third period than silicon, its 3p electrons should
experience a slightly higher Zeff. That means phosphorus should have the higher 1st ionization energy.
(Check a textbook or other reference – does it?)
15. (RC&O 2. 32) Which element, sodium or magnesium, should have an electron affinity closer to
zero? Give your reasoning.
Electron affinities (M + e-  M-) for elements are mostly negative (i.e. adding an electron to the neutral
element is an exothermic process). If an element has less electron affinity (i.e. the incoming electron
feels a lower Zeff), the exothermicity is reduced, which makes the value closer to zero. For Na, the
incoming electron will occupy the 3s orbital, where it will experience a relatively high Zeff because it is
not experiencing much shielding by the other 3s electron. For Mg the 3s orbital is full and is relatively
good at shielding the incoming 3p electron, which will experience a slightly lower Zeff. Thus, despite
the general increase in electron affinities across a period, Mg should exhibit a comparable or slightly
lower electron affinity relative to Na: i.e. its EA1 will be closer to zero.
16. (H&S 2.15) Using the electronegativity data in Table 2.2 determine which of the following
covalent single bonds is polar and (if appropriate) in which direction the dipole moment will
act. (i) N-H; (ii) F-Br; (iii) C-H; (iv) P-Cl; (v) N-Br.
Note: Pauling electronegativity values are well documented and constant, so you could have also
gotten these values from your first year text or the lecture notes.
(i) PN = 3.0, PH = 2.2: the N-H bond is polar in the sense N––H+, thus the dipole vector along this
bond will point toward the N.
(ii) PF = 4.0, PBr = 3.0: the F-Br bond is polar in the sense F––Br+, thus the dipole vector along this
bond will point toward the F.
(iii) PC = 2.6, PH = 2.2: while the C-H bond is slightly polar in the sense C––H+ and the (small) dipole
vector along this bond will point toward the C, this is small enough (<0.5) that it is considered non
polar.
(iv) PP = 2.2, PCl = 3.2: the P-Cl bond is polar in the sense P+–Cl–, thus the dipole vector along this
bond will point toward the Cl.
(v) PN = 3.0, PBr = 3.0: the N-Br bond is not polar.
Note that the polar vector is described in the IUPAC manner (pointing towards the negative pole), which
is different from the description in H and S textbook.
Definitions/Concepts: atomic orbital, degenerate, shielding, penetration, boundary surface, core
electrons, valence electrons, Aufbau principle, Pauli exclusion principle, nodal plane, Hund’s rule,
effective nuclear charge, shielding, 1st ionization energy, electron affinity, electronegativity, bond polarity,
dipole moment.