ANSWERS TO EXERCISES AND REVISION QUESTIONS(iii) (iv) 8 (v)[BeCl 4 ]^2 –Beryllium has a more stable electronic configuration in
[BeCl 4 ]^2 – , because the atom is surrounded by eight
electrons.
5C
(i) (a) (b)(c) (d)(e) (f)(ii)Nitrogen (1s^2 2s^2 2p^3 ) cannot expand its octet as
there are no available d orbitals to do so.
(iii)Chlorine can expand its octet, but it is too small an
atom to accommodate seven atoms of fluorine around it.
Iodine is a much bigger atom.
5D
(i)trigonal planar (ii)tetrahedral
(iii)roughly tetrahedral (not all the bond lengths are the
same)
(iv)tetrahedral (v)octahedral.
5E
(i)Distorted tetrahedral with respect to the electron
pairs; ‘bent’ with respect to H–S bonds (similar to water).
(ii)Distorted tetrahedral with respect to the electron
pairs; trigonal pyramidal with respect to its bonds
(similar to ammonia).
(iii)Distorted octahedral with respect to the electron
pairs.
(iv)Distorted tetrahedral with respect to the electron
pairs; ‘bent’ with respect to its bonds.
(v)As for (iv).
5F
(i)trigonal planar (remember that all the bond lengths
are the same because of resonance)
(ii)trigonal planar (iii)linear (iv)tetrahedral
(v)trigonal planar.
5G- (i)no – linear (ii)yes (iii)no – tetrahedral
(iv)yes (v)yes. - (i)yes (ii)H+–F–.
5H
(i)two
(ii)As Fig. 5.6, with ‘2’ ions and two free electrons for
each ion in the structure.
(iii)three
(iv)Na, Mg, Al (the more valence electrons that are free,
the stronger the bonding).
(v)Al (it has the strongest metallic bonding, therefore it
requires most energy to break down the structure).
5I
····CC–CC–CC–C····
5J
(i)Kr, Ar, Ne, He (the smaller the atom, the fewer the
electrons and the weaker the forces that attract the
atoms together).
(ii)Again, I 2 is the largest molecule and has the greatest
number of electrons, so the forces that attract I 2
molecules together are the strongest and it has the
highest melting and boiling points. The other halogens
follow this trend.
Revision questions
5.1
(i) (ii)(iii) (iv)(v)5.2The shapes with respect to the electron pairs around
the central atom are roughly the following:
(i)trigonal bipyramidal (ii)linear
(iii)trigonal planar
(iv)trigonal bipyramidal (v)tetrahedral.
5.3linear
5.4 (i)no (ii)no (iii)no (iv)yes (v)yes.
5.5Substances which exist as atoms or non-polar
molecules (such as the noble gases or iodine), liquefy or
solidify at appropriate temperatures.
5.6Ca because it has two valence electrons and
stronger metallic bonding (more electron density to
‘glue’ the positive ions together).
5.7Each carbon atom in graphite has one free, mobile
electron. There are no free electrons in diamond, they
are all involved in covalent bonding.
5.8A CO 2 molecule is linear and non-polar. Only London
forces attract the molecules together in the solid state.
5.9The oxygen atom of water has two lone pairs of
electrons and can therefore form two hydrogen bonds
per molecule. The nitrogen atom in ammonia has only
one lone pair and ammonia molecules can only form one
hydrogen bond per molecule. In addition, oxygen is more
electronegative than nitrogen.
5.10No, there is not enough electronegativity difference
between carbon and hydrogen in the C–H bond to
produce a hydrogen that is sufficiently positively charged
to hydrogen bond to the oxygen that is bonded to carbon
in a neighbouring molecule.
5.11If you make a model you will find there are 8
hexagons and 12 pentagons.Unit 6
Exercises
6A
(i)
Ca^2 +,2NO H^2 O
3 – (s) —Ca^2 +(aq)2NO 3 – (aq)
(ii)
2K+,SO H^2 O
42 – (s) —2K+(aq)SO 42 – (aq)
(iii)
2Na+,CO H^2 O
32 – ·10H 2 O(s) —2Na+(aq)
CO 32 – (aq)10H 2 O(l)6B
The nitrate ion is the spectator ion. Overall ionic reaction
is:
Cu^2 +(aq)2NO 3 – (aq)Zn(s)Zn^2 +(aq)
2NO 3 – (aq)Cu(s)
or
Cu^2 +(aq)Zn(s)Zn^2 +(aq)Cu(s)
This is the same ionic equation as for the reaction
between copper sulfate solution and zinc solid.
6C
(i)carbonate, sulfide and hydroxide.
(ii)We need to mix a solution of a soluble barium salt
(such as barium ethanoate, (CH 3 COO–) 2 Ba^2 +) with a
solution of a soluble sulfate (such as magnesium
sulfate, Mg^2 +,SO 42 – ).
(iii)We need to mix a solution of a soluble iron salt (such
as iron(II) sulfate, Fe^2 +,SO 42 – ) with a solution of a soluble
hydroxide (such as sodium hydroxide, Na+,OH–).
6D
Coordinate bond.
6E
HBr(g)H 2 O(l)H 3 O+(aq)Br–(aq)
or
HBr(g) —H^2 OH+(aq)Br–(aq)6F
(i)Calcium chloride will form a white precipitate,
whereas magnesium nitrate will not. Potassium sulfate
may also give a cloudiness of (slightly soluble) silver
sulfate depending upon the concentrations of reactants
used.
(ii)The precipitates are silver bromide (AgBr) and silver
iodide (AgI):
Ag+(aq)Br–(aq)AgBr(s)
Ag+(aq)I–(aq)AgI(s)
These precipitates are useful in confirming the presence
of bromides and iodides in solution.
6G
(i)The rest do not contain a sulfate ion.
6H
(ii)Ammonium carbonate:
NH 4 +(aq)OH–(aq)NH 3 (g)H 2 O(l)6I- (i)two (ii)one.
- (i)ethanoates (acetates) (ii)nitrates.
6J
K+,OH–(s) —H^2 OK+(aq)OH–(aq)
Ca^2 +,2OH–(s) —H^2 OCa^2 +(aq)2OH–(aq)6K
(i)Fe(s)H 2 SO 4 (aq)FeSO 4 (aq)H 2 (g)
(ii)Fe(s)2H+(aq)Fe^2 +(aq)H 2 (g)
(SO 42 – (aq) is the spectator ion).
6L
(i)D is Zn^2 +(solid ZnO turns yellow upon heating). The
white precipitate suggests that C is Cl–.
(ii)Dilute sulfuric acid. Confirm this by adding BaCl 2 /HCl
(white precipitate observed).
(iii)Mix the solutions together one by one. The copper
solution is coloured blue. Both sodium carbonate and
sodium hydroxide will form a blue precipitate with copper
ions since both copper(II) hydroxide and copper(II)
carbonate are insoluble (see Table 6.2). However, only
sodium carbonate will ‘fizz’ with the remaining solution,
dilute hydrochloric acid.4350230_000118_2 9 _Ans.qxd 3/2/06 2:22 pm Page 435