Recitation Activity 7 (Chem 121)
Chapters 7-8
Periodic Trends
1. Consider the successive ionization energies of four elements from the 4th period (row) of the
periodic table, The s-block metal, Calcium (Ca), the transition metal, Titanium (Ti), the
metalloid, Germanium (Ge), and the nonmetal, Selenium (Se).
Element
1st IE
(kJ/mol)
2nd IE
(kJ/mol)
3rd IE
(kJ/mol)
4th IE
(kJ/mol)
5th IE
(kJ/mol)
6th IE
(kJ/mol)
7th IE
(kJ/mol)
Ca
590
1145
4912
6491
8153
---
---
Ti
659
1309
2652
4174
9581
---
---
Ge
762
1537
3302
4411
9020
---
---
Se
941
2045
2973
4144
6590
7880
14990
a. Write the electron configurations for Ca, Ti, Ge and Se.
Ca: [Ar] 4s2
Ge: [Ar] 4s23d104p2
Ti: [Ar] 4s23d2
Se: [Ar] 4s23d104p4
b. Write equations for the first and fourth ionization energies of titanium.
1st: Ti → Ti+ + e∆H = 659 kJ/mol
th
3+
4+
4 : Ti → Ti
+ e
∆H = 4174 kJ/mol
c. Explain the fact that the 1st, 2nd and 5th ionization energies for calcium are smaller than
observed for Ti and Ge, while the 3rd and 4th ionization energies for calcium are higher than
Ti and Ge.
Ca has two valence electrons in the 4s subshell. The energy required to remove
these electrons is smaller than for Ti and Ge, because Ca is the leftmost element (more
metallic). The 3rd and the 4th IE for Ca are very large because these processes require
removal of core electrons.
For Ti we are removing valence electrons until we get to the 5th IE, which is very
high because this requires removal of a core electron. Since we are removing core
electrons for both Ca and Ti at this point once again the IE for Ti is higher, because it
is further to the right in the periodic table.
Germanium contains four valence electrons (4s2 and 4p2), so the first four IE are
not abnormally high, but higher than for Ti because Ge is further to the right in the
periodic table and more electronegative. Upon reaching the 5th IE an electron must be
removed from a completely filled 3d orbital. Once filled the d-orbitals act as core
orbitals and this is reflected in the large value of the 5th IE for Ge, which is higher
than the 5ht IE for Ca and almost as high as the 5th IE for Ti.
d. How many valence electrons do germanium and selenium have? What are the electron
configuration for Ge4+ and Se6+? Are your answers consistent with the observation that the
ionization energy for Ge jumps dramatically between the 4th IE and the 5th IE, and for Se
between the 6th and 7th IE?
Ge has 4 valence eSe has 6 valence eGe4+: [Ar] 3d10
Ge6+: [Ar] 3d10
The answers are consistent with the dramatic jump in IE because the 5th electron for
Ge and the 7th electron from Se must be removed from the filled 3d subshell. As
described in the previous problem, the filled d-subshell acts as core electrons.
2. Consider the 1st ionization energies of 3 unidentified elements from the 3rd period.
Element
X
Y
Z
1st IE (kJ/mol)
578
1521
1012
Based on ionization energies which of these elements would you expect to have the smallest
atomic radius? Which one do you expect to have the largest radius? Can you explain your
choices in terms of effective nuclear charge?
The effective nuclear charge is a measure of how strongly the outermost electron is
attracted to the nucleus. As the attraction between the outermost electron and the
nucleus increases the atom becomes smaller (atomic radius decreases) and it requires
more energy to remove that electron from the atom (the first ionization energy
increases). Therefore, as the effective nuclear charge increases the radius decreases
and the 1st IE increases.
Atomic Radius:
Largest X > Z > Y Smallest
Ionic Bonding
3. Consider the ionic compounds calcium oxide and titanium (II) oxide, both of which adopt the
rock salt crystal structure like sodium chloride.
a. What is electron configuration of the cation in each compound?
Ca2+: [Ar]
Ti2+: [Ar] 3d2
b. Classify each compound as either diamagnetic (no unpaired electrons, effectively nonmagnetic) or paramagnetic (contains unpaired electrons and interacts with a magnetic
field).
Ca2+ is diamagnetic (all of its electrons are paired)
Ti2+ is paramagnetic (it has 2 unpaired d electrons)
c. In which compound will the cation-anion distances be the shortest?
Atomic Radius: Ca2+ > Ti2+
M-O Distance: Ca2+ > Ti2+
d. Which compound will have the largest lattice energy?
The lattice energy is a measure of how much energy is required to dissociate a crystal
into its component ions, for example what is the energy of the following reaction:
CaO(s) → Ca2+(g) + O2-(g)
As the strength of the ionic bonding interaction increases the lattice energy increases.
The most important factor is the charges of the ions. In this case both TiO and CaO
contain +2 cations and -2 anions, so they will be comparable on that count. The second
most important factor is the distance between the ions. As the distance between ions
decreases, the ionic bonding becomes stronger and the lattice energy goes up. In part
c we noted that Ca2+ is larger than Ti2+, and since we expect O2- to be the same size
in each case the Ti-O distance is shorter than the Ca-O distance, and the lattice
energy of TiO will be larger.
Covalent Bonding and Lewis Structures
4. Draw Lewis dot structures for an O2 molecule and for hydrogen peroxide, H2O2. In which of
these molecules will the oxygen-oxygen distance be the shortest?
O O
H
O
O
H
O2 has a bond order of two (a double bond), while H2O2 has a bond order of one (a single
bond). Therefore the H2O2 O-O bond distance is longer than it is in O2.
5. Consider the molecule S2Cl2. Construct Lewis structures and use the concept of formal
charge to determine the most stable arrangement of atoms: S-Cl-S-Cl or S-Cl-Cl-S or Cl-SS-Cl.
-1
+1
0
0
S Cl S Cl
-1
+1
+1
-1
S Cl Cl S
0
0
0
0
Cl S S Cl
The Cl-S-S-Cl arrangement is the best because it minimizes the formal charges (shown
above each Lewis structure)
6. Using known bond enthalpies estimate the enthalpy associated with the transformation of
methyl isocyanide (H3CNC) to acetonitrile (H3CCN). Based on this result which of these two
molecules is more stable? Based on formal charges, which molecule would you expect to be
more stable?
H
H
C
H
N
C
H
Methyl isocyanide
H
C
C
N
H
Acetonitrile
Reaction enthalpy = Bonds broken – Bonds formed
Reaction enthalpy = {3(C-H single bonds) + (C-N single bond) + (C-N triple bond)}
– {3(C-H single bond) + (C-C single bond) + (C-N triple bond)}
Reaction enthalpy = (C-N single bond) – (C-C single bond) = 293 kJ/mol – 348 kJ/mol
Reaction enthalpy = -55 kJ/mol
The negative reaction enthalpy means that the reaction is exothermic, and to a good
approximation we can take the products (acetonitrile) to be more stable than the reactants
(methyl isocyanide).
The formal charges for methyl isocyanide put a +1 charge on the nitrogen and a –1 charge
on the terminal carbon. The formal charges for acetonitrile are zero for all atoms
involved. Thus formal charges predict that acetonitrile will be more stable, in agreement
with the bond enthalpy predictions.