CHAPTER 14
Practice Exercises
14.1
White phosphorus consists of individual P4 molecules. Red phosphorus consists of linear chains
of P4 molecules. Black phosphorus consists of cross-linked chains of P4 groups.
Black phosphorus has a higher melting point than the white or red allotropes.
The order of increasing density is : white P < red P < black P.
Review Questions
14.1
14.3
Oxidation states are shown beneath the atoms:
Sulfur is oxidised and oxygen is reduced.
The sulfur in H2S is oxidised (loses electrons), and the sulfur in SO2 is reduced (gains electrons).
14.5
The correct answer is D
14.7
The correct answer is D
Review Problems
14.9
The oxidation reactions and their standard potentials are as follows:
Even though Cl2 is thermodynamically harder to oxidize than H2O, the reaction kinetically is
much faster, so Cl2is preferentially oxidized in aqueous solution containing Cl2. Oxidation of F2 is
thermodynamically is very difficult such that water is preferentially oxidized in aqueous solutions
containing F2.
14.11
14.11 Each phosphate ion has 5 + 4(6) + 3 = 32 valence electrons and a total of 96 electrons.
When three phosphate ions condense to form a ring, the ring has six atoms, three each of P and O,
with two additional outer O atoms bonded to each P atom. Add in lone pairs to each oxygen atom
(2 pairs for the inner atoms, and 3 pairs for the outer atoms).
FCP = 5-4 = 1. Make a double bond to each P atom using a lone pair from one of its
oxygen atoms. This will make FCP = 0 and complete the Lewis structure:
14.13
outer
The conversion of calcium phosphate into phosphoric acid proceeds through several
steps. First, calcium phosphate is reacted with coke:
This is a redox reaction in which phosphorus is reduced from the +5 oxidation state to
an
oxidation state of zero. Next, phosphorus is burned in another redox reaction, which
returns
phosphorus to the +5 oxidation state: P4 (g) + 5 O2 (g) → P4O10 (s). There are
no Brønsted
acid-base reactions, but when water is added to generate phosphoric acid, a
Lewis acid-base
reaction occurs:
P4O10 (s) + 6 H2O (l) → 4 H3PO4 (aq)
14.15
A water molecule reacts with the phosphate group at the end of ATP, breaking a P – O
bond to form ADP and phosphoric acid:
14.17
(a)
S
S
S
S
S
S
S
S
Electron pair geometry around S will be tetrahedral (four sets of electron pairs), modified by
BP-LP, LP-LP and BP-BP repulsions. Molecular shape is cyclic, bond angles around S will be
less than 109.5o. No permanent dipole moment.
(b)
2-
O
S
O
O
This is one of 3 resonance structures. Electron pair geometry around S is tetrahedral (four sets of
electron pairs), modified by BP-LP repulsions. Molecular shape is trigonal pyramidal, bond
angles around S will be less than 109.5o. No permanent dipole moment.
(c)
2-
O
O
S
O
O
This is one of 2 resonance structures. Electron pair geometry around S will be exactly tetrahedral
(four sets of electron pairs). Molecular shape is tetrahedral, bond angles around S will be 109.5o.
No permanent dipole moment.
(d)
CH 3
S
CH3
Electron pair geometry around S will be tetrahedral (four sets of electron pairs), modified by
BP-LP, LP-LP and BP-BP repulsions. Molecular shape is bent, bond angles
around S are
less than 109.5o. Dimethylsulfide has a permanent dipole moment.
14.19
(a)
4.47  10-7 =
3.98  10-8  [HCO 3- ]
[H 2CO3 ]
[HCO 3 ]
therefore
=11.2
[H 2CO3 ]
and
[H 2CO3 ]
=0.0893
[HCO3  ]
7.08  10-8  [HPO4 2- ]
6.31  10 =
[H 2 PO 4 - ]
-8
(b)
therefore
and
[HPO 4 2- ]
=0.891
[H 2 PO 4- ]
[H 2 PO 4 - ]
= 1.12
[HPO 4 2- ]
(c) Because the pKa lies far from the desired pH. Buffers work best when the pKa of the weak acid
is approximately equal to the desired pH.
Additional Exercises
14.21
8 H2S + 8 Cl2 → S8 + 16 HCl
To determine the amount of Cl2 required, first calculate the H2S content in the water:
 103 mL   1.00 g   25 g H 2S 
2
mass H 2s = 6.0 x 10 L 

 6
 = 1.5 x 10 g

 1 L   1 mL   10 g water 
3
Next, do appropriate mole-mass conversions:
 1 mol   1 mol C12   70.91 g 
mC12 = 1.5 x 102 g H 2S 
= 3.1 x 102 g



 35.08 g   1 mol H 2S   1 mol 
14.23
Silver is a soft Lewis acid, which reacts much more readily with soft Lewis bases such
as hydrogen sulfide with hard Lewis bases such as oxygen.
14.25
Each of these phosphates can undergo a Brønsted acid-base reaction in which each ammonium
cation transfers a proton to the phosphate anion, generating neutral phosphoric acid and ammonia:
(NH4)3PO4 → 3 NH3 + H3PO4
(NH4)2HPO4 → 2 NH3 + H3PO4
(NH4)H2PO4 → NH3 + H3PO4
The first reaction generates more ammonia than either of the other two, accounting for
the higher vapour pressure of ammonia shown by (NH4)3PO4.
14.27
Determine the amounts of Al and Na using standard stoichiometric techniques.
Al3+ + 3 eAl
For every 1 mole of charge, 1/3 mole of aluminium will be refined:
 1 mol A1   26.98g 
mA1 = 1 mol e- 
 = 8.99g
- 
 3 mol e   1 mol A1 
Na+ + eNa
For every 1 mole of charge, 1 mole of sodium will be refined: 23.0 g
14.29
The reactions can be balanced by inspection:
2 Al2O3 + 6 Cl2 → 4 AlCl3 + 3 O2
AlCl3 + 3 Na → Al + 3 NaCl
14.31
The reaction consumes 12 H atoms for 4 P atoms, or 3 H atoms for every P atom. Thus,
the product is H3PO4: P4O10 + 6 H2O → 4 H3PO4