Identify the following substances and write balanced equations for any reactions mentioned. a. Two anions that give carbon dioxide when their salts are treated with dilute acids.
There are three possible answers here, carbonate, hydrogen
carbonate and cyanate:
CO3-2(aq) + 2H+(aq) ---> H2O(l) + CO2(g)
HCO3-(aq) + H+(aq) ---> H2O(l) + CO2(g)
CNO-(aq) + 2H+(aq) + H2O(l) ---> NH4+(aq) + CO2(g)
(The last reaction is slower than the others and rather
obscure, I wouldn't expect Chem 2210 students to know it.)
b. An oxide of a group 14 element that will oxidise Mn+2 to MnO4- in acidic solution.
Lead dioxide, PbO2, will do this: 5PbO2(s) + 2Mn+2(aq) + 2H+(aq) ---> 5Pb+2(aq) + 2MnO4-(aq) + H2O(l)
c. A simple hydride that spontaneously inflames or explodes in air.
Most hydrides of silicon (or boron, but that is group 13): SiH4(g) + 2O2(g) ---> SiO2(s) + 2H2O(l)
d. A non-metal chloride that does NOT hydrolyse with water to give hydrochloric acid and another acid.
Carbon tetrachloride: CCl4(l) + 3H2O(l) XXX> H2CO3(aq) + 4HCl(aq) (or anything else!)
e. Two elements which have one allotrope which is a metallic type conductor and one which isn't.
Carbon (diamond - insulator & graphite - metallic type conductor) and tin (white - metallic and grey - diamond structure probably semi-conductor, but I'm not sure!)
f. An oxide of a group 14 element in which its formal oxidation state is +1.3333. (It's only stable below 0ºC)
Carbon sub-oxide, C3O2, is the only one I can think of.
g. A compound of these elements that can be found in toothpaste.
Calcium carbonate is often used as a filler and some fluoridated
toothpastes contain tin(II) fluoride as the source of fluoride
ion.
h. Two substances containing two group 14 elements (and no
others) that might be found in a workshop. What properties make
them useful?
Carborundum, SiC, is a very hard compound with the diamond structure. It is often used in grinding stones and sand papers.
Solder contains a mixture of tin and lead and is used because it is low melting and binds readily to many other metals.
(Note my careful use of the word "substance" in the question, this covers elements, compounds and mixtures.)
N.B. These are not necessarily model answers, more a list of
points that should be included. Refer to class notes and text for
details.
a. Carbon dioxide is a gas but silicon dioxide is a solid.
Carbon dioxide contains pp
-pp
double bonding between C=O atoms while the p
p-pp overlap between Si
and O is much weaker due to the large size of the 3p orbitals of
Si. However, the Si-O single bond can be strengthened by dp-pp bonding and the
thermodynamics favour the formation of two double bonds for CO2
and four single bonds for each Si in SiO2. (For a full answer you
should explain pp-pp
and dp-pp bonding, with
diagrams.) The crucial point, however, is that in the case of CO2
all the bonding requirements are satisfied in a small molecule -
the oxygen does not have to bond to another atom and so it is a
gas. In the case of SiO2 each oxygen is bound to two silicons,
building up a macromolecule, which has to be a solid. (There is
nothing inherently gaseous about double bonds, it is the
combination of oxygen being divalent and forming strong double
bonds that does it.)
b. As you decend the group oxidation state II becomes less
reducing and oxidation state IV changes from being stable to
highly oxidising.
You have to analyse the formation of compounds in terms of
thermodynamic cycles. Basically for compounds to form the energy
released when the compound's bonds (covalent or ionic) are formed
has to be greater than the energy required to atomise or ionise
the elements. As you descend a group the atoms of the elements
get larger and while factors like sublimation enthalpy and
ionisation enthalpy get smaller the strengths of bonds with other
elements decreases faster. There isn't enough energy released in
bond formation to compensate for raising the element to its
highest oxidation state. (This is sometimes referred to as "the
inert pair effect" but this isn't really an explanation, just
another name for the phenomenon!)
c. Lead has oxides in which its oxidation state is 2.67 and 3.
This refers to the compounds Pb2O3 and
Pb3O4. They are distinct compounds, although
Pb3O4 is better known than
Pb2O3. Although they are often represented
as 2PbO.PbO2 and PbO.PbO2 they are not
just mixtures, but have crystal structures of their own with
Pb+2 and Pb+4 ions in distinct
locations.
d. The bond dissociation enthalpy for carbon monoxide is the
largest known for any diatomic molecule.
This is because CO has a 10 electron triple bond like N2. A full
answer should show how the pattern of molecular orbitals is
arrived at and how 10 electrons give the maximum bond strength.
We won't deal with this until we get to nitrogen.
e. The electrical conductivity of silicon increases with
temperature while that of tin decreases.
This is a question about band theory. Si is a semiconductor with a smallish band gap and Sn a metal with a Fermi sea of electrons. This difference in behaviour distinguishes semi-conductors and metallic conductors. A full answer would explain all these terms with diagrams.
Identify the lettered substances in the following and write balanced equations for the reactions. (The "rules" for inorganic roadmaps are that all reaction are assumed to go to completion giving easily separable, pure products.)
a. A is a white powder only moderately soluble in water to give a neutral solution. A gives a bright yellow flame in a flame test. When dilute sulfuric acid is added to A a gas B is evolved which turns blue litmus red and turns lime water milky. When A is heated gently more B is produced and a white solid C remains. C is soluble in water giving a basic solution. When C is treated with dilute sulfuric acid B is again produced. When aqueous solutions of A & C are each treated with calcium chloride solution C gives a precipitate D, but A does not.
The most concise way to answer these is to give a list of equations (preferably balanced), identifying each lettered compound at its first appearance. Make additional comments where appropriate.
NaHCO3(s) ---> Na+(aq) + HCO3-(aq): Na gives a yellow flame.
A
NaHCO3(s) + H+(aq) ---> Na+(aq) + H2O(l) + CO2(g)
B
/\
2NaHCO3(s) ---> Na2CO3(s) + H2O(g) + CO2(g)
C
CO3-2(aq) + H2O(l) <===> HCO3-(aq) + OH-(aq): (hydrolysis)
Na2CO3(s) + 2H+(aq) ---> 2Na+(aq) + H2O(l) + CO2(g)
CO3-2(aq) + Ca+2 ---> CaCO3(s)
D
Ca(HCO3)2 is not known in the solid state,
but can exist in solution.
b. P is a naturally occuring substance which has been mined in
Newfoundland. When P is treated with dilute hydrochloric acid a
foul smelling gas, Q, is produced and a white crystalline solid,
R. R is soluble in hot water and hot solutions give a white
precipitate, S when aqueous sodium carbonate is added. S
dissolves in dilute nitric acid giving B and a colourless
solution, T, which gives yellow precipitates, X & Y, when
potassium chromate or potassium iodide are added.
When P is heated in air a sharp smelling, acidic gas, U,
is given off and a yellowish solid, V, remains. When the right
proportions of P & V are mixed together and heated more
U is formed and a greyish, metal, W, remains which is
soft enough to leave marks when it is rubbed on paper.
PbS(s) + 2HCl(aq) ---> H2S(g) + PbCl2(s)
P Q R
Pb+2(aq) + CO3-2(aq) ---> PbCO3(s)
S
PbCO3(s) + 2H+(aq) ---> Pb+2(aq) + CO2(g) + H2O(l)
T (as nitrate)
Pb+2(aq) + CrO4-2(aq) ---> PbCrO4(s)
X
Pb+2(aq) + 2I-(aq) ---> PbI2(s)
Y
2PbS(s) + 3O2(g) ---> 2PbO(s) + 2SO2(g)
V U
PbS(s) + 2 PbO(s) ---> 3Pb(s) + SO2(g)
W
H.J.Clase 1999.03.03