ch14 Even
ch14 Even
ch14 Even
Exercises
14.2 (a) Be2C(s) + 4 H2O(l) 2 Be(OH)2(s) + CH4(g)
(b) CO(g) + Cl2(g) COCl2(g)
(c) 2 Mg(s) + CO2(g) 2 MgO(s) + C(s)
(d) Na2CO3(s) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l)
(e) BaCO3(s) BaO(s) + CO2(g)
(f) CS2(g) + 3 Cl2(g) CCl4(g) + S2Cl2(l)
(g) SnO(s) + 2 HCl(aq) SnCl2(aq) + H2O(l)
14.6 (a) Diamond has a high thermal conductivity because its structure consists
of carbon atoms held in an infinite network by single covalent bonds.
Thus any thermal motion at one side of the crystal lattice is quickly
transferred through to the other side, making it an excellent thermal
conductor.
(b) According to Le Chatelier’s principle, high pressure would favor
formation of diamond from graphite because diamond has the higher
density. High temperature would be favored, for the phase change has an
extremely high activation energy because rearrangement of covalent bonds
must occur.
69
The Group 14 Elements 70
70 Chapter 14
14.10 The unit cell of calcium carbide contains four pairs of ions (as NaCl
structure). Thus: 1
4 64.1 g mol = 4.26 10 22 g
Mass = 1
23
6.02 10 mol
4.26 10 22 g = 1.92 10 22 cm3 = 1.92 108 pm3
Volume = 2.22 g cm 3
3 8 3
Length of side = (1.92 10 pm ) = 577 pm
Sum of ionic radii = (577 pm)/2 = 288 pm
2
Thus r(C2 ) = 288 pm 114 pm = 174 pm
14.12 The formal charge representation is as follows, showing the carbon with a
formal negative charge.
14.14 Both gases are colorless and odorless, but carbon monoxide is poisonous
as a result of its strong reaction with hemoglobin; carbon dioxide is
nontoxic. Carbon monoxide is chemically reactive; for example, it burns
in air:
2 CO(g) + O2(g) → 2 CO2(g)
Carbon dioxide is chemically unreactive; for example, it does not react
with dioxygen.
14.20 The ―a‖ and ―b‖ designation distinguishes structural isomers. The only
possible structural isomer of CF3 CH2F would be CHF2 CHF2, so this
must be the formula of HFC-134b.
14.24 The bond angle will be 180º because there are only the two bonding pairs
(no lone pairs) around the central carbon atom.
+ 3+ 4+ 2 2
14.26 x × (Na ) + 6 × (Al ) + 6×(Si ) + 24 × (O ) + 1 × (S2 ) = 0
x(+1) + 6(+3) + 6(+4) + 24( 2) + 1( 2) = 0
x=8
14.32 Tin forms two oxides, SnO and SnO2, while lead forms three, PbO,
Pb3O4, and PbO2. For tin, the tin(IV) oxide is the thermodynamically
stable oxide; lead(II) oxide is the most stable for lead, the lead(IV) oxide
being a good oxidizing agent. The oxide Pb3O4 contains alternating Pb2+
and Pb4+ ions.
14.38 Be2C and Al4C3. They are diagonally related (see Chapter 9).
14.44 Lead is found in former industrial sites, in old lead-based paint, and in
cigarettes. In many parts of the world, lead-containing gasolines are still
used; thus the motor vehicles in those countries produce high levels of tiny
lead particles.
14.48 The structure of methyl isocyanate is what one would expect, but the
linear arrangement of silyl isocyanate suggests some multiple bonding
involving the d orbitals of the silicon.
14.52 Because carbonate ion is the conjugate base of the very weak acid,
hydrogen carbonate, the following equilibrium lies to the right, and it is
the hydroxide ion from this equilibrium that precipitates the aluminum
ion. The hydroxide of aluminum must be more insoluble than the
carbonate.
2
CO3 (aq) + H2O(l) HCO3 (aq) + OH (aq)
3+
Al (aq) + 3 OH (aq) Al(OH)3(s)
14.54 We can find the moles of gas X using the Ideal Gas Equation, PV = nRT:
V (100 kPa )(0.244 L) 9.85 10 3 mol
n P
1 1
RT (8.31 kPa L mol K )(298 K)
1
Thus the molar mass of gas = 62.9 g·mol
Next we calculate the moles of hydrogen gas:
2
n PV (100 kPa )(0.730 L) 2.95 10 mol
1 1
H2
RT (8.31 kPa L mol K )(298 K)
and the moles of silicon:
1 mol SiO 1 mol Si
1.200 g 2.00 10 2 mol
n Si 2
1 mol SiO
60.1 g
2
14.56 Aluminum has one electron less than silicon, and phosphorus has one
electron more than silicon, so one can use the ―combo‖ element analogy
similar to the resemblance of boron-nitrogen compounds to those of
carbon (Chapter 9). Thus it is not unreasonable for this compound to have
the quartz structure with the framework of alternating aluminum and
phosphorus atoms substituting for silicon atoms.
14.58 The covalent radius of silicon is much greater than that of carbon. Thus
the greater mass is more than compensated for by the larger bond lengths.
14.60 With each terminal oxygen atom having an oxidation state of –2 and the
bridging peroxo- oxygen atoms having a –1 oxidation state, then that of
each carbon has to have an oxidation state of +3 (see below).