Chemistry 1000 (Spring 2006)
Problem Set #4: Chapter 8
Answers to Practice Problems
1.
(a)
(b)
(c)
(d)
(e)
Explain why each of the following statements is true.
Fluorine is the most electronegative element. (two reasons)
1.
Fluorine is a small element which only has two core electrons. As such, its
valence electrons (in the n = 2 shell) are relatively close to the nucleus. The
smaller the distance between oppositely charged objects (e.g. an electron and a
nucleus), the stronger the force of attraction between them.
2.
The valence electrons of fluorine feel a very high effective nuclear charge. There
is some shielding due to the two core electrons and seven valence electrons, but
the valence electrons still feel a relatively large fraction of the +9 charge of the
nucleus (about +5) – higher than for any other element.
3.
Fluorine is one electron short of having a full valence shell. As such, it would
really “like” to gain an electron to reach octet status.
Hydrogen has a greater electron affinity than helium.
Electron configurations:
H = 1s1
H-1 = 1s2
He = 1s2
He-1 = 1s22s1
Electron affinity is the energy released when gaining an electron.
When gaining an electron, hydrogen goes from having an incomplete valence shell (1s1)
to a full valence shell (1s2). Thus, H-1 is relatively stable and can form.
If helium were to gain an electron, it would go from having a full valence shell (1s2) to
having an incomplete valence shell (1s22s1). Thus, He-1 does not form because it is much
less stable than He.
Since H-1 can form and He-1 cannot, hydrogen has a greater electron affinity than helium.
Potassium has a greater second ionization energy than calcium.
Second ionization energy is the energy required to remove an electron from X+1 where X
is the element of interest.
For calcium, this is the energy required to form Ca2+ from Ca+. Ca2+ has a complete
valence shell (complete octet) whereas Ca+ does not. As such, Ca2+ is more stable and
easy to form from Ca+.
For potassium, this is the energy required to form K2+ from K+. K+ has a complete
valence shell (complete octet) whereas K2+ does not. As such, K+ is more stable and very
difficult to convert to K2+.
The atomic radius of Br -1 is larger than the atomic radius of Br.
Br and Br -1 have the same number of protons, but Br -1 has one more electron. This extra
electron partially shields the other electrons in the valence shell from the positive charge
of the nucleus. As such, the effective nuclear charge felt by the valence electrons of Br -1
is smaller than that felt by the valence electrons of Br, and they can move farther away
from the nucleus (increasing the atomic radius).
The atomic radius of nitrogen is smaller than the atomic radius of boron.
Nitrogen has two more protons in its nucleus than boron does. It also has two more
electrons, but these do not completely shield the valence electrons from the increased
positive charge of the nucleus. As such, the effective nuclear charge felt by the valence
electrons of nitrogen is higher than that felt by the valence electrons of boron. As such,
the valence electrons of nitrogen are pulled more tightly to the nucleus (decreasing the
atomic radius).
2.
(a)
(b)
(c)
3.
(a)
(b)
4.
(a)
(b)
Rank the following sets of atoms/ions according to the number of valence electrons (from
least to most).
S2-, Sb, Se, Sn2+
Sn2+ (2)
Sb (5)
Se (6)
S2- (8)
F-, Fe2+, Fe3+, Fr
Fr (1)
Fe3+ (5)
Fe2+ (6)
F- (8)
C, Ca, Cl, Co
Ca (2)
C (4)
Cl (7)
Co (9)
List all of the atoms/ions from question 2 that are paramagnetic.
Sb
[Kr]5s24d105p3
Se
[Ar]4s23d104p4
Fr
[Rn]7s1
Fe3+ [Ar]3d5
Fe2+
C
[He]2s22p2
Cl
[Ne]3s23p5
Co
List all of the atoms/ions from question 2 that are diamagnetic.
Sn2+ [Kr]5s24d10
S2[Ne]3s23p6
F[He]2s22p6
Ca
[Ar]4s2
[Ar]3d6
[Ar]4s23d7
Consider H and He+ in the ground state.
Which would you expect to have a higher ionization energy?
He2+ (Both are one-electron atoms/ions, but He2+ has two protons in the nucleus so the
electron should be more strongly attracted to the nucleus therefore more difficult
to remove.)
Calculate the ionization energy for H.
First ionization energy is the energy required to remove an electron from an atom (far
enough away that it no longer feels any attraction to the nucleus. Mathematically, we can
approximate this as the energy required to excite an electron from n = 1 to n = ∞ (where
∞ stands for infinity). Recall that 1/0 = ∞ therefore 1/∞ = 0.
For a hydrogen atom, Z = 1 (1 proton in the nucleus), n1 = 1 and n2 = ∞
∆E
= En2 - En1
∆E
= Ry × Z2 × [(1/n1)2 - (1/n2)2]
= (2.179 × 10-18 J) × (1)2 × [(1/1)2 - (1/∞)2]
∆E
(c)
= 2.179 × 10-18 J
Calculate the ionization energy for He+.
For a helium cation, Z = 2 (2 protons in the nucleus), n1 = 1 and n2 = ∞
∆E
= En2 - En1
∆E
= Ry × Z2 × [(1/n1)2 - (1/n2)2]
= (2.179 × 10-18 J) × (2)2 × [(1/1)2 - (1/∞)2]
∆E
(d)
= 8.716 × 10-18 J
Did your calculations in parts (b) and (c) support your answer to part (a)?
Yes. The ionization energy for He+ was four times higher than that for H.
5.
(a)
(b)
6.
(a)
(b)
(c)
(d)
(e)
List and name the four quantum numbers.
n
principal quantum number
l
angular momentum quantum number
ml
magnetic quantum number
ms
electron spin magnetic quantum number
Explain briefly what each quantum number tells you about an electron.
n
indicates the approximate energy of an electron (therefore its approximate average
distance from the nucleus); as such, it can be considered to indicate the size of the
orbital in which the electron resides
l
indicates the shape of the orbital in which the electron resides (s, p, d, f, etc.)
ml
indicates the orientation of the orbital in which the electron resides (e.g. px vs. py)
ms
indicates the spin of the electron (+½ or –½)
Give the name and symbol for the neutral element matching each of the following
descriptions.
“I have 8 neutrons and 6 valence electrons.”
oxygen (O)
6 valence electrons = group 6 or 16; the only element in either
group likely to have only 8 neutrons (i.e. mass of about 2×8 = 16
g/mol) is oxygen
“I have a full octet and belong to the same period as nickel.”
krypton (Kr)
full octet = noble gas (group 18); nickel is in period 4
“Half of my electrons are valence electrons.”
beryllium (Be)
If half the electrons are valence, then half are core. There are only
a few possible values for the number of core electrons. Any
element in period 1 has 0 core electrons; any element in period 2
has 2 core electrons (1s2); any element in period 3 has 10 core
electrons (1s22s22p6); any element in period 4 has 18 core electrons
(1s22s22p63s23p6); etc. Beryllium has 2 core electrons plus 2
valence electrons. Note that there is no other element that meets
these criteria (calcium has 20 electrons, but 18 are core – not 10).
“I have twice as many valence electrons as core electrons.” see logic for 6(c)
carbon (C)
2 core electrons and 4 valence electrons.
“I have twice as many core electrons as valence electrons.” (bonus: name all three
elements fitting this description) see logic for 6(c)
lithium (Li)
2 core electrons + 1 valence electron
phosphorus (P)
10 core electrons + 5 valence electrons
cobalt (Co)
18 core electrons + 9 valence electrons
***note that xenon (Xe) has 8 valence electrons – not 18!***
7.
(a)
(b)
(c)
For a neutral ground state nickel atom,
Write the complete electron configuration.
1s2 2s2 2p6 3s2 3p6 4s2 3d8
Write the electron configuration using noble gas notation.
[Ar] 4s2 3d8
Draw an orbital box diagram showing the valence electrons. Label each box.
4s
(d)
3dxy 3dxz 3dyz 3dz2 3dx2-y2
Use the table below to list a set of quantum numbers describing the valence electrons.
Use as many spaces as necessary.
Electron
n
l
ml
ms
1 (4s)
4
0
0
+½
2 (4s)
4
0
0
–½
3 (3d)
3
2
+2
+½
4 (3d)
3
2
+1
+½
5 (3d)
3
2
0
+½
6 (3d)
3
2
-1
+½
7 (3d)
3
2
-2
+½
8 (3d)
3
2
+2
–½
9 (3d)
3
2
+1
–½
10 (3d)
3
2
0
–½
11
12