Download Periodic Chemical Properties

Chemistry 1050
Periodic Chemical Properties
Winter 2014
Text: Tro, Fridgen, Shaw Canadian Edition
Additional Problems
1.
Write ground state electron configurations in spdf notation for the following: Use shorthand for
elements found in period # $ 4
cations:
Be2+, Na+, Zn2+, Ti3+, Cr2+, Cu+, Ag+
anions:
O2-, P3-, IG
2.
Write equations to represent the four successive ionization energies of beryllium.
3.
Write balanced equations to represent the following processes:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
The first ionization energy of F, I1 = 1680 kJ molG1
The third ionization energy of K, I3 = 4410 kJ molG1
The fifth ionization energy of P, I5 = 6240 kJ molG1
The first ionization energy of Na+, I1 = 4560 kJ molG1
The first ionization energy of ClG, I1 = 348 kJ molG1
The electron affinity of C, EA = !123 kJ molG1
The electron affinity of OG, EA = 744 kJ molG1
The electron affinity of Al3+, EA = !2750 kJ molG1
4.
Arrange the following in order of increasing atomic radius as well as increasing first ionization
energy: He, As, F, P, and N. Use general period and group trends.
5.
Explain why the effective nuclear charge felt by valence electrons is fairly constant for elements in
the same group but tends to increase across a period.
6.
(a)
(b)
(c)
(d)
On what 2 factors does the electron affinity of an atom depend?
Explain why the noble gases and the alkaline earth metals have no tendency to gain an
electron.
The electron affinity of anions is very endothermic while the electron affinity of cations is
very exothermic. Explain.
The electron affinity of nitrogen is slightly endothermic. Suggest a reason for this.
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7.
Give brief explanations for the following observations: Hint: First determine the electron
configuration of any element or ion named. It is always an important clue to any explanation. Also
look to see if a pair of elements are in the same period or family or if a pair of species are
isoelectronic.
(a)
The first ionization energy of oxygen (1314 kJ molG1) is greater than the first ionization
energy of sulfur (999.5 kJ molG1).
(b)
The first ionization energy of beryllium (899 kJ molG1) is greater than the first ionization
energy of lithium (520 kJ molG1).
(c)
The first ionization energy of helium (2373 kJ molG1) is much greater than the first
ionization energy of lithium (520 kJ molG1).
(d)
The fourth ionization energy of boron (25,000 kJ molG1) is much greater than the fourth
ionization energy of carbon (6220 kJ molG1).
(e)
The atomic radius of argon (98 pm) is larger than the atomic radius of neon (70 pm).
(f)
The atomic radius of aluminum (143 pm) is larger than the atomic radius of silicon (132
pm).
(g)
The first ionization energy of aluminum (577.9 kJ molG1) is less than the first ionization
energy of magnesium (738.1 kJ molG1).
(h)
The first ionization energy of sulfur (999.5 kJ molG1) is less than the first ionization energy
of phosphorus (1012 kJ molG1).
(i)
The ionic radius of the calcium ion, Ca2+, (99 pm) is smaller than the atomic radius of the
calcium atom (197 pm).
(j)
The ionic radius of the sulfide ion, S2G, (184 pm) is larger than the atomic radius of the
sulfur atom (127 pm).
(k)
Determine which of the following species are isoelectronic with neon: Mg2+, ClG, Kr,
N3G, K+.
(l)
The ionic radius of the oxide ion, O2G, (140 pm) is larger than the ionic radius of the
aluminum ion, Al3+, (50 pm).
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8.
(m)
The first ionization energy of neon (2080 kJ molG1) is greater than the first ionization energy
of the fluoride ion, FG (333 kJ molG1) but less than the first ionization energy of the sodium
ion, Na+, (4560 kJ molG1).
(n)
Sodium reacts vigorously withnon-metals to formionic compounds containing the Na+ ion.
No ionic compound containing the Na2+ ion exists.
(o)
Potassium is more reactive than calcium but less reactive than cesium in all respects. (Hint:
these metals react with non-metals to form cations).
(p)
The electron affinities of the halogens are very exothermic.
Element M is a third period element with the following successive ionization energies:
I1, = 786 kJ molG1
I2 = 1577 kJ molG1
I3 = 3232 kJ molG1
I4 = 4355 kJ molG1
I5 = 16,091 kJ molG1
I6 = 19,784 kJ molG1
I7 = 23,766 kJ molG1
I8 = 29,251 kJ molG1
(a)
(b)
Identify element M. Justify your choice.
Determine the energy required in kJ molG1 for the following process:
M(g)+ 6 M5+(g) + 4 eG
(c)
Explain why the second ionization energy of element M is greater than it's first ionization
energy.
9.
(a)
(b)
Which groups of the periodic table contain diamagnetic atoms?
Classify the following as paramagnetic or diamagnetic: F, Zn, Ti2+, S2G,Na+, Ba2+.
10.
Define or explain van der Waals radius, metallic radius, covalent radius, ionic radius, effective
nuclear charge, electron shielding, electron penetration, first ionization energy, electron affinity,
diamagnetic species and paramagnetic species.
11.
Explain why the atomic radii of transition elements tend to be the same across a period. A few
unusual exceptions occur. Explanations for the exceptions are beyond Chem 1050.
12.
Are Cr and Fe2+ isoelectronic? Briefly explain your answer.
13.
Arrange the following in order of increasing size.
(a)
(b)
(c)
P3G, N3G, As3G
Ca2+, Ca, Mg2+
He, Ne, Li+, O2G, O
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14.
(a)
(b)
Using Slater’s rules, determine the screening constant S and Zeff for the electrons of B, Mg,
Ca, Zn.
The successive ionization energies of Be are 900, 1757, 14,849 and 21,007 kJ molG1.
Rationalize these values. Slater’s rules may help.
Answers
1.
cations
Be2+ (Z = 4, 2eG) 1s2
Na+ (Z= 11, 10eG) 1s22s22p6
Zn2+ (Z = 30, 28eG) [Ar] 3d10
Te3+ (Z = 22, 19eG) [Ar] 3d1
Cu+ (Z = 29, 28 eG) [Ar] 3d10
Ag+ (Z = 47, 46 eG) [Kr] 4d10
anions
O2G (Z = 8, 10 eG) 1s22s22p6
P3G (Z = 15, 18 eG) 1s22s22p63s23p6
IG (Z = 53, 54 eG) [Kr] 4d105s25p6
2.
3.
4.
5.
6.
Be(g) → Be+(g) + eG
Be+(g) → Be2+(g) + eG
Be2+(g) → Be3+(g) + eG
Be3+(g) → Be4+(g) + eG
(a)
F(g) 6 F+(g) + eG
(b)
K2+(g) 6 K3+(g) + eG
(c)
P4+(g) 6 P5+(g) + eG
(d)
Na+(g) 6 Na2+(g) + eG
(e)
ClG(g) 6 Cl(g) + eG
(f)
C(g) + e! 6 CG(g)
(g)
OG(g) + eG 6 O2G(g)
(h)
Al3+(g) + eG 6 Al2+(g)
lowest I1 As < P < N < F < He
highest I1
smallest r He < F < N < P < As
largest r
As one goes down a group, the increase in nuclear charge is nearly balanced by a corresponding
increase in the number of core electrons. As one goes across a period, additional electrons go into
the same shell where they provide insufficient additional shielding against the increasing nuclear
charge.
(a)
Factors:
1.
nuclear attraction for added electron
2.
repulsions of atom’s electrons, especially the valence electrons for
added electron.
(b)
The added electron for the Noble gases must enter the ns orbital of the next higher electron
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7.
8.
9.
shell. The added electron for Group 2A goes into the valence np subshell where it is well
shielded by the core and valence ns electrons from the nucleus.
(c)
For anions, added electron is repelled by anion charge. For cations, added electron is
strongly attracted by cation charge.
(d)
Added electron must be paired up with a N electron in a half!filled 2p orbital. The
increased repulsions decreases the attraction of a nitrogen atom for an extra electron.
(a)
same family, Zeff of valence electron removed about the same. I1 decreases with n-value
as an electron in shell further from nucleus is easier to remove.
(b)
same period, n(constant). Zeff Be > Zeff Li (period trend)
(c)
ls electrons of He are poorly screened from nucleus. They are very close to nucleus and
strongly held. The 2s electron of Lithium is well screened by its core electrons, is in a shell
further from the nucleus and it is weakly held.
(d)
A strongly held core electron is removed from B vs well screened valence electron
removed from C.
(e)
same family, argon has one more electron shell.
(f)
same period, n(constant). Zeff Si > Zeff Al (period trend)
(g)
same period, 3p electron of Al does not penetrate inner shells as well as 3s electron of
Mg. It feels a lower Zeff and a weaker nuclear attraction.
(h)
same period, paired electron removed from oxygen experiences greater repulsions due
to close proximity of electrons in same orbital vs the unpaired electron removed from
nitrogen. This lowers its Zeff and the nuclear attraction for it.
(i)
Ca2+ has only 3 electron shells occupied vs 4 for Ca.
(j)
S2G has extra 2 electrons. They increase repulsions and decrease Zeff of the valence
electrons. The electrons are not held as tightly and they spread out.
(k)
Mg2+, N3G
(l)
isoelectronic species: Al3+ has more protons (13) vs O2G which has only 8. A larger #
protons means a greater nuclear attraction pulling the electrons in closer.
(m)
isoelectronic species: Na+ has the most protons (11) and the greatest nuclear attraction
to overcome. FG has the least protons (9) and the weakest nuclear attraction to
overcome.
(n)
weak nuclear attraction for well screened valence 3s electron. Very strong nuclear
attraction for core 2p electrons
(o)
metals react to form K+, Ca2+, Cs+ ions. I1(Cs) < I1 (K) < (I1 + I2) (Ca). Reactivity
would increase as the energy required in the reaction to form the product cation decreases
making Cs most reactive and Ca least reactive.
(p)
The effective nuclear charge on the valence np electrons of a halogen is quite large so the
nuclear attraction for the added electron is relatively strong. Also, a filled np subshell and
a noble gas electron configuration results which is very stable.
Silicon I5 >> I4
(b)
E = 25,255 kJAmolG1 (Sum of I2, I3, I4 and I5) Use Hess’s Law.
(c)
The second electrons must be removed from a positively charged species (M+). This is
more difficult to do and requires more energy. An alternate explanation is that the remaining
2 valence electrons are now less well shielded from the nuclear charge and more strongly
held.
(a)
Groups 2, 12, 18
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(b)
10.
paramagnetic: F, Ti2+ diamagnetic: Zn, S2G, Na+, Ba2+
van der Waals radius:
metallic radius:
covalent radius:
One half of the internuclear distance between neighbouring atoms
in a solid noble gas.
One half the internuclear distance between identical atoms in a
metallic crystal.
One half the internuclear distance between identical atoms joined
by a single covalent bond.
ionic radius:
radius of an ion as determined by properly apportioning the distance
between the nuclei of a cation and anion in an ionic bond. By making an
initial guess of the radius of a single ion, e.g. O2G, one can estimate the size
of other ions bonded to it. These estimated values can be used to
determine the size of additional ions. The initial guess can then be revised
to give consistent results.
effective nuclear charge:
The number of protons an electron actually feels is pulling it towards the
nucleus.
electron shielding: The screening or shielding of the nuclear charge by electrons between a
given electron and the nucleus.
electron penetration: Ability of an electron to get close to the nucleus. This ability decreases as the
electron’s l-value increases.
The energy required to remove an electron froman isolated, gaseous atom
first ionization energy:
(or ion).
electron affinity:
The energy change that results when an isolated, gaseous atom (or ion)
gains an electron.
diamagnetic species: An atom, ion (or molecule) in which all electrons are paired. It is weakly
repelled by a magnetic field.
An atom, ion (or molecule) which has one or more unpaired electrons. It
paramagnetic species:
is attracted by an external magnetic field. The more unpaired electrons
present, the greater the attraction.
Additional electrons go into an inner shell, i.e. they are added to the (n-1) d subshell where they
11.
effectively screen the ns valence electrons from the increase in nuclear charge. The ns valence
electrons (usually two, sometimes one) feel a roughly comparable force of attraction to the nucleus.
Cr: (Z = 24, 24eG) [Ar] 3d54s1
12.
Fe2+: (Z = 26, 24eG) [Ar] 3d6
They both have 24eG but are not isoelectronic as they do not have the same electron configuration.
Smallest to largest: (a) N3G, P3G, As3G (b) Mg2+, Ca2+, Ca (c) Li+, He, Ne, O, O2G
13.
14. (a)
B (2s, 2p) S = 2.40
Zeff = 2.60 B(1s) S = 0.30 Zeff = 4.70
Mg (3s) S = 9.15 Zeff = 2.85 Mg (2s, 2p) S = 4.15 (ave) Zeff = 7.85 (ave)
Mg (1s) S = 0.30 Zeff = 11.70
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Ca (4s) S = 17.15 Zeff = 2.85 Ca (3s, 3p) S = 11.25 (ave) Zeff = 8.75 (ave)
Ca (2s, 2p) S = 4.15 (ave) Zeff = 15.85 (ave) Ca (1s) S = 0.30 Zeff = 19.70
Zn (4s) = S = 25.65 Zeff = 4.35 Zn (3d) S = 21.25 Zeff = 8.75 Zn (3s, 3p) S = 11.25 (ave)
Zeff = 18.75 (ave) Zn (2s, 2p) S = 4.15 (ave) Zeff = 25.85 (ave) Zn (1s) S = 0.30 Zeff = 29.70
14. (b)
I1 represents loss of first 2s electron of Be (1s22s2)
S (2s electron of Be) = 2.05 Zeff = 1.95
I2 represents loss of 2s electron of Be + (1s22s1)
S ( 2s electron of Be +) = 1.70 Zeff = 2.30
I3 represents loss of a 1s electron of Be 2+ (ls2)
S (1s electron of Be 2+) = 0.30 Zeff = 3.70
I4 represents loss of only 1s electron of Be 3+ (1s1)
S (1s electron of Be 3+) = 0 Zeff = Z = 4.00
I3 and I4 are large vs I1 and I2 due to much larger Zeff values of electrons removed and the fact that they
belong to the inner core shell of Be. The ls electrons are much closer to the nucleus than the 2s electrons.
Developed by: Dr. Chris Flinn
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