Chapter 7 Structure of atoms and chemical bonding
1. Use the following data to calculate the lattice energy of MgF2(s) (kJ mol-1)
Heat of sublimation of Mg:
148 kJ mol-1
1st ionization potential of Mg:
738 kJ mol-1
2nd ionization potential of Mg:
1450 kJ mol-1
Bond Dissociation of F2:
159 kJ mol-1
Electron affinity of F:
-328 kJ mol-1
Heat of formation of MgF2(s) from the elements: -1123 kJ mol-1
Mg(s)  Mg(g)
+148 kJ mol-1
Mg(g)  Mg+(g) + e¯
+738
+
+2
Mg (g)  Mg (g) + e¯
+1450
F2(g)  2 F(g)
+159
2 F(g) + 2 e¯  2 F¯(g)
2(-328)
+2
Mg (g) + 2 F¯(g)  MgF2(s)
-U
Mg(s) + F2(g)  MgF2(s)
-1123
Thus, U = -(-1123-(148+738+1450+159+2(-328)) = +2962 kJ mol-1
2.
What kind of hybrid orbitals does the S atom use to bond with the H atoms in a molecule
of H2S?
2(1) + 6 = 8 valence electrons. Arranging the H atom around the S atom with single bonds
leaves 4 electrons, which are placed as two lone pairs on the S atom. The S atom therefore
has four electon clouds around it, and must therefore be using sp3 hybrids to bond with
the H atoms.
3.
Why do atomic light sources have only discreet frequencies?
The light from an atomic source is produced when electrons fall from a higher energy
shell to a lower energy one. The energy, hence the wavelength of the photon, corresponds
to the difference in energies of the upper and lower shell. Since these energies are
quantized, so is the energy difference, and therefore so is the energy or wavelength of the
photon.
4.
A p-orbital is shown below. An electron in this orbital occupies both lobes. How can the
electron get from one lobe to the other if the node is a place where the electron can never
exist?
Node
5.
6.
Electrons have wavelike properties. Since waves can propagate through a node, so can
electrons.
The ground electronic configuration for copper (Cu) is predicted by AUFBAU to be
[Ar] 4s23d9. However, Cu actually has a different configuration. State what you think the
actual configuration is, and state why it is different from what AUFBAU predicts.
It is likely [Ar]4s13d10. In this configuration, there is one half-filled subshell, and a filled
subshell, which are more energetically stable configuration that a filled 4s subshell and a
9/10ths filled 3d subshell.
Explain how the ionization energy changes going from left to right across a period, and
briefly explain why this happens.
The ionization energy (or potential) increases going left to right across a period. This is
because electrons are being added to the same principal shell, and so have approximately
the same energies. However, protons are being added to the nucleus at the same time. The
effective nuclear charge therefore increases going left to right, creating a larger energy of
attraction for the electrons, thus requiring a larger energy to remove one.
7.
SF6 is an octahedral molecule. What type of hybrid orbitals does the S atom use to bond
with the F atoms?
Since there are six charge clouds around the central S atom, it must be using sp3d2 hybrid
orbitals.
8.
(a) Use VSEPR theory to predict the shape of the SF3+ ion.
Valence electrons = 6 + (3 x 7) – 1 = 26.
Arranging the F atoms around the central S atom and completing the octets on the F
atoms uses 24 electrons. The other two are a lone pair on the S atom. The molecule is
therefore of the form AX3E, and is trigonal pyramidal.
(b) Use VSEPR theory to predict the shape of AsCl5.
Valence electrons = 5 + (5 x 7) = 40.
Arranging the Cl atoms around the central As atom and completing the octets on the Cl
atoms uses all 40 electrons. The molecule is therefore of the form AX5, and is trigonal
bipyramidal.
(c) A molecular orbital (MO) diagram for oxygen (O2) is shown below. What will happen
to the bond length and bond energy if we ionize O2 to O2+ and why?
The electron that is lost during ionization will come from the highest energy MO, i.e.
the *2p. Since this is an antibonding orbital, the bond order will increase upon ionization.
Thus, the bond will be shorter and stronger in the ion.
(d) How can you deduce from the MO diagram that O2 has a double bond?
The bond order is BO = ((8 – 4)/2) = 2, thus the molecule has a double bond.
2p
2p
2p
2s
2s
(b) What is the energy of the photons having a wavelength of 97.20 nm? Express in kJ
mol-1.


E = h = hc/

= 6.63 x 10-34 J s (3.00 x 108 m s-1) / (97.2 x 10-9 m)
= 2.05 x 10-18 J (per photon)
x 6.02 x 1023 mol-1 = 1231870 J (mol photons)-1
= 1232 kJ mol-1
9. Born Haber cycles were used to make the first reliable prediction of an electron affinity.
Calculate the electron affinity of fluorine (F) given the following data:
Heat of sublimation of potassium:
+90 kJ mol-1
Ionization potential of potassium:
+419 kJ mol-1
Bond dissociation energy of fluorine (F2):
+159 kJ mol-1
Lattice energy of KF(s):
+821 kJ mol-1
Net energy of formation of KF(s) from K(s) and F2(g):
-569 kJ mol-1
K(s)  K(g)
K(g)  K+(g) + e½ F2(g)  F(h)
F(g) + e-  F-(g)
K+(g) + F-(g)  KF(s)
+90 kJ mol-1 (vaporization of K(s))
+419 kj mol-1 (ionization of K(g))
+159/2 = +79.5 kJ mol-1 (dissociation of F2(g), per mole of F)
?? (electron affinity of F)
-821 (opposite of lattice energy of KF(s))
Adding the above equations results in the equation for the formation of KF(s) from the elements:
K(s) + ½ F2(g)  KF(s)
-569 kJ mol-1
Thus, the electron affinity of fluorine is –569 – (90 + 419 + 79.5 – 821) = -335.6 kJ mol-1
10.
State the Pauli Exclusion Principle.
No two electrons in a single atom may have the same four quantum numbers.
11.
Which of LiF(s) and LiCl(s) has the larger lattice energy and why? Which of KF(s) and
SrF2(s) has the larger lattice energy and why?
ULiF > ULiCl since F¯ is a smaller ion than Cl¯, resulting in a smaller bond length, hence a
larger electrostatic energy
USrF2 > UKF since Sr+2 is doubly charged, whereas K+ is only singly charged, resulting in a
larger electrostatic energy.
12.
Write the four oxygen-only reactions in the Chapman cycle.
O2 + hυ  O + O
O + O 2  O3
O3 + hυ  O2 + O
13.
O + O 3  2 O2
Which is larger: the van der Waals constant ‘a’ for Ne(g) or that of Kr(g)? Explain any
differences.
Do the same for the ‘b’ constant for these two gases.
aKr > aNe, since Kr is a larger, more polarizable atom
bKr > bNe, since Kr is a larger atom
In a molecule of NO, the highest energy electron is alone in a *2p molecular orbital. Is
the bond energy of NO greater or less than that of NO? How do you know?
Since the electron will be removed from this antibonding MO during ionization, the bond
energy of NO+ will be greater than that of NO.
15. (a) Use VSEPR theory to predict the shape of SeO3-2.
14.
Valence electrons = 6 + (3 x 6) + 2 = 26
Making single bonds to the O atoms from the Se atom and completing the octets around the O
atoms uses 24 electrons. The last two are placed as a lone pair on the Se atom, resulting in an
AX3E configuration. This is trigonal pyramidal.
(b) Which of CH2Cl2, CHCl3 and CCl4 are polar?
If you draw the Lewis structures of all three, it is apparent that the CH2Cl2 is bent, the CHCl3
is a trigonal pyramid and the CCl4 is tetrahedral. Only the CCl4 is symmetrical and only it will be
non-polar. The other two are polar.
(c) Calculate the formal charge on the B atom in BF3 and use this result to explain why
the B atom does not follow the octet rule.
BF3 is a trigonal planar molecule having 3 + (3 x 7) = 24 electrons. All 24 will be used by
forming single B-F bonds and completing the octets on the F atoms. The B atom is
assigned half of the bonding electrons (3). B is in group 3 and has 3 valence electrons in
the free atom. Its formal charge in BF3 is thus 3-3 = 0, even though it does not have an
octet.
16. State Hund’s rule.
Hund’s rule says that electrons are placed in degenerate orbitals one orbital at a time, until each
orbital has one. Only then are electrons paired up.
17. (a) Use VSEPR to predict the shapes of the following two species. Wrong name = zero
marks.
ClF3 valence electrons = 7 + (3 x 7) = 28
Making single bonds and completing the octets around the F atoms uses 24 electrons.
There are therefore two lone pairs on the Cl atom. This is thus of the form AX3E2, which
is T-shaped.
XeF4 valence electrons = 8 + (4 x 7) = 36
Making single bonds and completing the octets around the F atoms uses 32 electrons.
There are therefore two lone pairs on the Xe atom. This is thus of the form AX4E2, which
is square planar.
(b) What hybridization would you expect for the central atom in each of the following
two species?
H2C=O Since the C is surrounded by three bonding groups and no lone pairs, it is using sp2
hybrids.
BH4¯
hybrids.
Since the B is surrounded by four bonding group and no lone pairs, it is using sp3
(c) Referring to the MO diagram for N2 at the below,
2p
2p
2p
2s
2s
(i) Calculate the bond order of N2.
BO = (8 – 2)/2 = 3
(ii) Would N2+ have a longer or shorter bond than N2? Why?
Creating N2+ requires removal of a 2p (bonding) electron. This will reduce
The bond order to (7 – 2)/2 = 2.5. This will therefore have a longer bond
than N2+.
(iii) Is N2+ paramagnetic? Why?
Yes, because of the unpaired electron in the 2p orbital.
18.
Ammonia, NH3, has a tetrahedral arrangement of charge clouds around the nitrogen
atom. Why is the
H-N-H bond angle (107.3°) less than the ideal tetrahedral angle (109.5°)?
In NH3, there is a lone pair of electrons on the N atom. This pair repels the N-H bonding
pairs more strongly than they repel one another, resulting in the H-atoms being pushed
together slightly, reducing the bond angle.
19.
What two concepts are brought together in the Planck equation, E = h?
This equation brings together the ideas that light is wavelike, hence has a wavelength, ,
and has particle-like properties, i.e. is composed of photons having energy E.
20.
21.
Name three physical properties of molecules that can be explained using molecular
orbital theory
MOs correctly predict the magnetic properties of molecules (i.e. para- or diamagnetism),
as well as the variation of bond length and bond energies when the molecules are ionized.
Also, by calculating the bond order, we can preedict whether the molecule has a single,
double or triple bond (any 3 of these 4).
Explain how the electron affinity changes going from left to right across a period, and
why.
Electron affinity increases (i.e. becomes more negative) going from left to right across a
period. This is because the effective nuclear charge increases in this direction due to an
increasing number of protons in the nucleus, i.e. less shielding of the nucleus by the
valence electrons going L to R. The energy released upon addition of an electron is
therefore greater.
22. (a) Predict the formulas of compounds formed from:
(i)
Mg and S
MgS
(ii) Na and Se
Na2Se
(iii) V with S
V2S5
(iv) Y with O
Y2O3
(v) Al with N
AlN
(b) Why is the ionization potential of Zn greater than those of its two neighbours, Cu and Ga?
Zinc’s electronic structure is [Ar] 4s2 3d10. Since this is two filled sub-shells, the electrons
are energetically stable, i.e. at lower energies than those of Cu ([Ar] 4d13d10) or Ga
([Ar]4d23d104p1), and are therefore more difficult to remove, i.e. require more energy to
remove.
(c) Which of As or Se has the larger atomic radius? Why?
As is a larger atom since Se has more protons, which draw the electrons inwards.
(d) Why is the second ionization potential of each element greater than the first?
The first electron is being removed from a singly charged positive ion. The second is
being removed from a doubly charged ion, which requires more energy.
(e) Which of LiF or KBr would have the higher lattice energy? Why?
23.
24.
25.
Li and F are smaller than K and Br respectively. Hence, the Li is closer to the F than the
K is to the Br, resulting in a larger lattice energy in LiF.
Why is it not possible for an electron in an atom in the third row of the periodic table to
have the quantum numbers: n = 3, l = 1, ml = -2, ms = ½ ?
If l = 1, ml = -1, 0 or +1 only.
AUFBAU predicts that Gold (Au) has the electronic configuration: [Xe] 6s24f145d9, but it
is actually
[Xe] 6s14f145d10. Why?
6s24f145d9 is two filled and one 9/10 ths filled subshell. 6s14f145d10 is two filled and one
half-filled subshell, which is more energetically stable.
Zn, Cd and Hg have anomalously high ionization energies. Why?
All three of the elements have s2d10 electronic configurations. The two filled subshells are
energetically stable, i.e. at a lower energy. Hence, more energy is required to remove an
electron from them.