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CHAPTER 5
Practice exercises
5.1
F
F
S
F
F
5.3
O
O
O
O
O
O
5.5
Ethane is symmetrical, so does not have a dipole moment. However, ethanol has a polar
—O—H group at one end and so has a dipole moment.
5.7
Oxygen has the valence electron configuration 2s22p4, so it uses four sp3 hybridised
orbitals. Two of these overlap with 1s orbitals from the hydrogen atoms to form σ bonds
and the other two sp3 hybrid orbitals contain lone pairs.
5.9
H2 has two electrons which occupy a bonding orbital, but H2– has a third electron which
must occupy an antibonding orbital, hence H2 has the stronger bond.
Review questions
5.1
(a)
(b)
(c)
(d)
O: 1s22s22p4; its six n = 2 electrons are involved in bond formation.
P: 1s22s22p63s23p3; its five n = 3 electrons are involved in bond formation.
B: 1s22s22p1; its three n = 2 electrons are involved in bond formation.
Br: 1s22s22p63s23p63d104s24p5; its seven n = 4 electrons are involved in bond
formation.
5.3
(a)
(b)
Al, group 13, 3 valence electrons
As, group 15, 5 valence electrons
(c)
(d)
F, group 17, 7 valence electrons
Sn, group 14, 4 valence electrons
5.5
(a)
(b)
(c)
(d)
N (3.0) attracts electrons more than C (2.5)
S (2.5) attracts electrons more than H (2.1)
I (2.5) attracts electrons more than Zn (1.6)
S (2.5) attracts electrons more than As (2.0)
5.7
The order of bond polarity is: PH3 < H2S < NH3 < H2O.
5.9
Ba, Be, and Bi, all metals, are found in ionic compounds as cations.
Br is found in ionic compounds as an anion.
B is not found as an atomic ion.
5.11
(a)
(b)
(c)
(d)
5.13
5.15
32 valence electrons
90 valence electrons
24 valence electrons
32 valence electrons
5.17
5.19
5.21
(a)
(b)
(c)
(d)
T-shaped
Seesaw-shaped
Trigonal planar
Square pyramid
5.23
(a)
(b)
(c)
(d)
5.25
Three sets of electron pairs around the central atom, giving trigonal planar
electron group geometry. One lone pair gives a bent shape, with ideal angle of
120˚.
There are five sets of electron pairs around the central atom, giving trigonal
bipyramidal shape and ideal angles of 90˚ and 120˚
A total of five sets of electron pairs, giving trigonal bipyramidal electron group
geometry. One lone pair gives a seesaw shape, with ideal angles of 90˚ and 120˚.
There are six sets of electron pairs around the central atom giving octahedral
electron group geometry. Two lone pairs result in a square planar shape, with
ideal bond angles of 90˚.
The Lewis structure of CO2 shows no lone pairs on the C atom, thus there are only two
sets of electron pairs around C resulting in a linear shape. The two C=O bonds point
opposite each other, so bond polarities cancel. The Lewis structure of SO2 below shows a
lone pair on the S atom, resulting in a bent molecule whose polar bonds do not cancel
each other.
S
O C O
5.27
O
O
The order of increasing bond strength is:
Bond
C—C <
H—N <
C=C <
C=O
Energy (kJ mol–1)
345
390
750
615
<
N≡N
945
Strong bonding and bond polarity of the n = 1 orbital makes H—N > C—C, multiple
bonding makes C=C > H—N, bond polarity makes C=O > C=C, and multiple bonding
makes N≡N > C=O.
5.29
Hydrogen always uses its 1s orbital to form bonds. Chlorine, the halogen in the third row,
has seven electrons in its n = 3 valence orbitals. Just as in Cl2, chlorine uses the 3p orbital
pointing along the bond axis, which overlaps with the 1s orbital of the hydrogen atom:
5.31
(a)
(b)
(c)
This atom must have four sets of electron pairs, requiring sp3 hybridisation.
Four sets of electron pairs, requiring sp3 hybridisation.
Three sets of electron pairs, requiring sp3 hybridisation.
5.33
(a)
Four sets of electron pairs around N, sp3 hybrids.
H
H C
H
(b)
N
H C H
H
Three sets of electron pairs around S, sp2 hybrids.
S
O
(c)
H
O
SN = 2, sp hybrids.
S C S
5.35
A description of bonding begins with the Lewis structure of the molecule, from which
steric numbers of inner atoms translate into hybridisation. Outer atoms use atomic
orbitals for bond formation. The N atoms in hydrazine have SN = 4, use sp3 hybrids and
have tetrahedral geometry: 4 sp3 (N) – 1s (H) σ bonds, 1 sp3 (N) – sp3 (N) σ bond, 2 lone
pairs in sp3 hybrids. The Lewis structure of hydrazine (showing the approximate
geometry) and orbital overlap diagrams are:
5.37
The steric number of each carbon atom is determined from the line structures of the
compounds. These are shown on the line structures below:
C atoms designated ‘4’ have only single bonds and thus four sets of electron pairs, sp3
hybridisation and tetrahedral geometry. Those with double bonds, designated ‘3’, have
three sets of electron pairs, sp2 hybridisation and trigonal planar geometry. Two C atoms,
designated ‘2’, have two double bonds, sp hybridisation and linear geometry. The σ
bonds are described by overlap of hybrid orbitals, with each C—H bond described as a
hybrid overlapping with a hydrogen 1s orbital. The π bonds are described by side-by-side
overlap of 2p orbitals. In pent-1-yne, the two C atoms designated ‘2’ have three bonds: a
σ bond formed from sp hybrids and two π bonds, at right angles to each other, as in
acetylene. The orbital overlap diagrams for acetylene are:
5.39
From weakest to strongest bond, the order is H22– < H2– < H2.
5.41
5.43
(a)
The molecular geometry is tetrahedral.
Cl
Cl Si Cl
Cl
(b)
The molecular geometry is seesaw.
F
F
Se
F
F
(c)
The molecular geometry is T-shaped.
Cl
Cl
Cl
Cl
5.45
There are two isomers:
5.47
Both ions have the same number of valence electrons, 5 + 4 + 6 + 1 = 16. Two pairs are
required to complete the bonding framework and the remaining six pairs are placed on
the two outer atoms:
Shift electron pairs to make two additional bonds to the central atom, completing its
octet. The cyanate ion has two near-equivalent structures, each of which has one atom
with formal charge –1:
The isocyanate ion also has two near-equivalent structures, but the formal charge on the
nitrogen atom cannot be reduced to zero:
5.49
Consult table 5.3 for the correspondence between molecular shapes and lone pairs.
(a)
(b)
(c)
This is the trigonal pyramid, derived from the tetrahedron when there is one lone
pair. The ideal bond angles are 109.5˚ and an example is NH3.
This is the trigonal planar shape associated with SN = 3 and no lone pairs. The
ideal bond angles are 120˚ and an example is the nitrate anion, NO3–.
This is the T-shape, the shape associated with with five sets of electrons pairs
around the central atom and two lone pairs. The ideal bond angles are 90˚ and
120˚ and an example is BrF3.
5.51
VSEPR theory predicts tetrahedral bond angles (109.5˚) for both these molecules. A bond
angle of 104.5˚ is smaller than this, because of the larger repulsion of lone pairs. An
angle of 92.2˚ indicates that the bonding is described using p orbitals rather than spacing
the electron pairs as far apart as possible. Space-filling models of the two molecules show
that the smaller oxygen atom cannot accommodate two H atoms at right angles, but the
larger S atom can.
5.53
Describe bonding and geometry starting with a Lewis structure and a count of bonds and
non-bonding pairs around inner atoms. The provisional structure of each compound has a
positive formal charge on the row 3 sulfur atom, which is reduced to zero by making
double bonds to each oxygen atom:
The sulfur atom in each compound has three sets of electron pairs around the central
atom, so sp2 hybrids overlapping with oxygen 2p orbitals describe the σ bonds: two in
SO2 and three in SO3. SO2 is bent, like O3 and SO3 is trigonal planar, like NO3–. The
Lewis structures indicate the presence of two π bonds in SO2 and three π bonds in SO3.
All the π orbitals extend over the entire molecule. Because sulfur is a third-row element,
its 3d orbitals contribute to the extended π bonding orbitals, which form from side-byside overlap of oxygen 2p orbitals with sulfur 3p and 3d orbitals.
5.55
5.57
(a)
diamagnetic
(b)
paramagnetic
Determine the Lewis structures using the standard procedure. Both molecules have
2(5) + 6 = 16 valence electrons. Two pairs are required for the bonding framework and
each outer atom has three pairs:
Because these molecules contain only row two atoms, the octets must be completed by
shifting two electron pairs to make double bonds:
Molecules have dipole moments only if they are unsymmetric. A linear N—O—N
structure does not have a dipole moment, because the N—O dipole exactly cancels the
O—N dipole. Thus N2O must have the structure N—N—O to have a dipole moment.
5.59
The Lewis structures of molecules with formula XF3 show octets around the inner atom
and FCX = 0, making them stable. Compounds with formula XF5 also have
FCX = 0 but have five electron pairs associated with the inner atom. This is possible for
phosphorus, a third row element that has d orbitals available for bonding. It is not
possible for nitrogen, a second row element that lacks valence d orbitals:
5.61
To generate the configuration of an excited state, move an electron from an occupied
orbital to an unoccupied orbital. The most stable excited state results from moving an
electron from the least stable occupied orbital to the most stable unoccupied orbital. The
excited state has one more antibonding electron and one less bonding electron than the
ground state, so the excited state has a weaker N—N bond than the ground state.
5.63
A description of bonding and geometry starts with determination of the Lewis structure.
ClO2 has 7 + 2(6) = 19 valence electrons. The bond framework requires two pairs, six
additional electrons are placed on each O atom, leaving three lone electrons
on the Cl atom. FCCl = 7 – 2 – 3 = +2, so shift two pairs to make double bonds:
The Cl atom has four sets of electron pairs (remember that a lone electron requires an
orbital, just as an electron pair does), so its geometry is tetrahedral, the σ bonds can be
described using sp3 hybrids from Cl and the molecule has a bond angle near 109o. There
is an extended π system formed from d orbitals on Cl overlapping side-by-side with 2p
orbitals from the two O atoms. This molecule is considered unusual because it has an odd
number of electrons.
5.65
Visualise what happens when protons add to oxalate anions using Lewis structures:
All the C—O bonds in the oxalate anion, with extended π orbitals, have equal lengths and
strengths, with a C—O bond order of 1.5. Adding two protons removes two oxygen
atoms from the extended π system and converts the C—O bonds into single bonds. The
other two C—O bonds become double bonds. Thus, the C—OH bonds in oxalic acid are
longer than the C—O bonds in oxalate anions, whereas the C=O bonds in oxalic acid are
shorter.