Answers to Additional Questions - Education Scotland

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Chemistry


Answers to Additional Questions






[ADVANCED HIGHER]








THERMOCHEMISTRY

ECONOMI CS: MI CROECON
OMI CS ( AH)

i i i

































Acknowledgements

This document is produced by Learning and Teaching Scotland as part of the Nationa
l
Qualifications support programme for Chemistry. The work of John Briggs and Douglas
Buchanan in particular is acknowledged with thanks.


First published 2003

Electronic version 2003


© Learning and Teaching Scotland 2003


This publication may be reproduc
ed in whole or in part for educational purposes by
educational establishments in Scotland provided that no profit accrues at any stage.


THE THEO
RY OF PERFECT COMPET
ITION



CONTENTS



Unit 1
: Electronic Structure and the Periodic Table

1


Unit 2
: Principles of Chemical Reactions

27


Unit 3
:

Organic Chemistry

45


Extra Questions

63






vi

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003



ELECTRONIC STRUCTURE

AND THE PERIODIC TAB
LE

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


1


© Learni ng and Teachi ng Scot l and 2003

UNIT 1



1.

Br

Br bond enthalpy = 194 kJ mol

1

(1)


E

=

Lh


for one mole of bonds (See units above.)

(1)




=

E/Lh

= 194000/6.63


10

34


6.02


10
23

J/Js




(Showing these units helps to get the units ri
ght for




the next line.)

(1)



=

4.86


10
14

s

1

(or Hz)

(1)





(4)


2.

(a)

An excited electron returns to ground state, emitting



energy difference as visible light of a specific wavelength.


1



(b)

The energy gaps between energy levels decrease
with



increasing energy, i.e. the higher energy levels get



closer and closer together.


1



(c)

(i)

E

=

h







=

6.63


10

34



1.26


10
15
J





=

8.35


10

19

J

(1)



But IE

=

L




E





=

6.02


10
23



8.35


10

19

J mol

1





=

50.27


10
4

J mol

1





=

502.7 kJ mol

1

(1)





2




(ii)

The first ionisation energy of the element.





(Check page 10 of the Data Booklet to confirm




that this is a likely answer.)


1





(5)


3.

(a)

Line A has a longer wavelength than all the others



shown
.

(1)



(Since it is the Balmer series, ground state is
n

= 2.)



This represents the smallest energy jump,



i.e. from
n

= 3 to
n

= 2.

(1)





2



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2

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003


(b)

E

=

Lhc
/


(1)



(Note: The question tells you to include ‘
L
’ in this



equation.)




=

(6.02


10
23



6.63


10

34



3


10
8

)/656


10

9

(1)




=

0.1825


10
6

kJ mol

1




=

182.5 kJ mol

1

(1)







3




(5)


4.

(a)

The ultra
-
violet region


1



(Se
e page 14 of the Data Booklet: ultra
-
violet includes





= 310 nm; visible runs from ~700 nm to ~400 nm.)



(b)

E

=

Lhc
/


(1)




=

6.02


10
23



6.63


10

34



3


10
8
/284


10

9


(1)




=

0.422


10
(23


34 + 8 + 9)

J mol

1





=

422 kJ mol

1

(1)







3



(c)

Energy from the spark excites some electrons to a higher



energy level.

(1)



When these electrons return to ground state a specific



amount of energy is released and this shows up as a



line of measurable wavelength (or frequency) in the



spectrum.

(1)





2



(d)

To gain a wider range of properties, e.g. harder,



resistant to corrosion, etc.


1




(7)


5.

(a)

The Balmer series is, for
n
1
= 2:



1/


=

R
h

(1/2
2



1/4
2
)




=

1.097


10
7



(1/4


1/16)




=

1.097


10
7



3/16




=

2.06


10
6


(1)





=

1/(2.06


10
6
) = 0.485


10

6




=

485 nm

(1)




This line will be blue
-
green (see page 14 of the Data



Booklet).

(1)





3



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3


© Learni ng and Teachi ng Scot l and 2003


(b)

n
1

= 1

(1)



All
the jumps must be of shorter wavelength,



i.e. of higher energy than those for
n
1
= 2.

(1)





2




(5)


6.

(a)

An electron


1



(b)

Each letter represents an orbital orientated along the
x
-
,



y
-

or
z
-
axis.


1



(c)

(i)

s orbitals are spherical and s
ymmetrical around




the nucleus.

(1)




p orbitals are dumb
-
bell shaped and are




symmetrical around each axis.

(1)






2




(ii)

The p orbitals are arranged mutually at right angles.


1




(d)

Electrons are placed singly in degenerate orbitals befor
e pairing
occurs in one orbital.


1



(e)

Error 1:

The ‘4p’ orbitals should be labelled ‘3p’.

(1)



Error 2:

Hund’s rule states that electrons will occupy




degenerate orbitals singly before any one is




doubly filled.

(1)






The correct configurat
ion is:





1s

2s 2p
x

2p
y

2p
z


3s 3p
x

3p
y

3p
z





2




(8)


7.

(a)

(i)

1

1



(ii)

5

1



(b)

(i)

6

1



(ii)

2

1



(iii)

18

1



(c)

(i)

1s

1



(ii)

4s 4p 4d 4f

1





(7)

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ANSWERS TO ADDI TIONA
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© Learni ng and Teachi ng Scot l and 2003

8.

(a)

(ii) is wrong because the 2s electrons have identical ‘addresses
’,
i.e. all four quantum numbers the same



for both.



or
Electrons cannot have


parallel spin in the same



orbital.

(1)



This violates Pauli’s exclusion principle.

(1)






2


(b)

The two 2p electrons should occupy two degenerate 2p orbitals
(with pa
rallel spin).

(1)



Hund’s rule

(1)






2



(c)

(vi) has violated the Aufbau principle since the s



sublevel

(1)



is not yet full but the 2p sublevel is filling.

(1)






2





(6)



9.

(a)

















2



(b)

A region wh
ere one or (at most) two electrons are likely



to be found.


1



(c)

It signifies the second energy level.


1



(d)

Of equal energy


1


z

z

z

x

y

z

2s

x

y

2p
x

y

y

x

x

2p
z

2p
y

ELECTRONIC STRUCTURE

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5


© Learni ng and Teachi ng Scot l and 2003

(e)













(

1 for any error)


2




(7)


10.








only one area correctly labelled

(1)






(2)


11.

(a)

Manganese


1


(b)

1s
2
2s
2
2p
6

3s
2

3p
6

4s
2

3d
5

or

[Ar] 4s
2
3d
5



1


(c)

[Ar] 4s
0

3d
4

(the 4s term may be omitted)


1




(3)


12.

(a)

380 kJ mol

1

(

20 kJ mol

1

)


1



(b)

There is a huge energy requirement to break the noble



gases into a stable octet.



or

It is very difficult to remove an electron from a full



energy level.


1



(c)

(i)

The Group 1 metal has the largest radius in that period




and

has the smallest nuclear charge in that period.




Both facts lead to a lesser attraction for the




outermost electron.


1




(ii)

Each new energy level means a larger radius (less attraction
for the outermost electron) and provides




a grea
ter shielding effect (again reduced attraction




by the nucleus).


1

ELECTRONIC STRUCTURE

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6

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003


(d)

Two factors apply: the steady increase in nuclear charge



and
the slight decrease in atomic radius from Li to Ne



makes the attraction of the nucleus for outer electrons greate
r.


1



(e)

(i)

Be 1s
2

2s
2




B 1s
2

2s
2

2p
1
, i.e. B has started a new subshell so




its outermost electron is relatively easier to remove than
that of Be, where a complete subshell has to be broken




into.


1




(ii)

Half
-
full shells are relative
ly stable so N (with a half
-




filled p subshell) has a higher IE than O, which has




one electron more.


1





(7)


13.

(a)

Electrons are excited by electric discharge to a higher



level.

(1)



These electrons emit energy as they return to a lower




energy level.

(1)



The quantity of energy emitted depends on the energy



values of the two energy levels involved. (Many lines



may be produced and each will represent a specific



electronic jump.)

(1)





3




(b)

Pass the light through a pri
sm and examine the



spectrum produced on a screen.

(1)



Spectral lines characteristic of sodium and of neon



would be seen.

(1)





2




(5)


14.

(a)

Since
E

=
hc
/


or E


1
/

,

shorter wavelengths



correspond to higher energy.

(1)



or

It can be assumed that these lines represent a part



of a series of lines which converge (at a continuum, not



shown) at the higher energy end, i.e. to the left on the



diagra
m shown in the question. Both arguments lead to



the conclusion than 393 nm represents the highest



energy value shown.

(1)





2


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7


© Learni ng and Teachi ng Scot l and 2003


(b)

Instead of coloured lines on a black background, there



would be black lines on a coloured background



(of the

visible spectrum).


1



(c)



=

1/

= 1/620


10

7

(1)




=

16129 cm

1

(Remember the units.)

(1)







2



(d)

The orange
-
red of the 620 lines would probably swamp



the less intense blue
-
green lines.


1




(6)


15.

(a)

By spraying as a solution into a Bunsen flame



or

by elect
ric discharge through a gaseous sample



or
by electric sparking between graphite electrodes


1



(b)

Valence electrons are excited and promoted to higher



energy levels.


1



(c)

(i)

Electrons return to a lower energy level, including




the ground s
tate, emitting energy equal to the




energy difference between the two levels involved




as light.


1




(ii)

A spectral line on the spectrum for each ‘jump’




or
a series of characteristic spectral lines





or

the intensity of light of one spectra
l line.


1



(d)

By using an appropriate filter.


1



(e)

The intensity of the light emitted.


1



(f)

Make up standard solutions of Ca
2+
(aq).

(1)



Use the solutions to make a graph of intensity of



radiation vs concentration of solution.

(1)





2





(g)

Measure the intensity of radiation of the water sample



(being tested).

(1)



The concentration is read from the graph.

(1)





2

ELECTRONIC STRUCTURE

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(h)

(i)

Each lamp gives out radiation characteristic of a




specific metal. Particles of the metal in the sample
absorb




a measurable amount of this light in proportion to their
concentration as electrons are promoted.


1




(ii)

The quantity of energy absorbed, i.e. the difference in




intensity, between the incident light and the transmitted
light.


1




(12)


16.

(a)








1



(b)

(i)









2




(ii)





1




(iii)

In (ii) two of the bonding electrons are delocalised




over the two oxygen atoms and the carbonyl carbon:










so the negative charge is considered to be shared equally
between all t
hree atoms and
not

to be residing on either
oxygen.





(In (i) the negative charge is attached completely,




i.e. localised, on only one of the oxygens.)


1





(5)

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17.

(a)









1



(b)

(i)









1




(ii)









1




(3)


18.

(a)

Trigonal
planar (trigonal is acceptable).

1



(b)

In CF
4
all four outer electrons in carbon are used in



bonding pairs with fluorines. All four bonds are



identical. The bond angle will be the tetrahedral angle



of 109.5°.

(1)




In NF
3

two of the electrons

in nitrogen form a lone



pair. The three others each form bonding pairs.



The lone pair exerts a greater repulsion ‘downwards’



so the bond angle is slightly less than 109.5°.

(1)





2




(3)


ELECTRONIC STRUCTURE

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10

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


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19.

(a)












(1)

(1)



(Do not rely on you
r skill as an artist


label the shapes also.)


2



(b)

Nitrogen has an extra pair of electrons (a lone pair).



These exert a strong repulsive force downwards on the bonding
pairs hence they are pushed down ‘below’ the tetrahedral angle,
creating a pyr
amid:

(1)















Boron has only three outermost electrons, so BCl
3

has



only three pairs of bonding electrons. They spread themselves
symmetrically (or as far from each other as possible), i.e.



pointing to the corners of an equilateral triang
le.

(1)






2




(4)


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11


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20.

(a)

There are five pairs of electrons around the central



chlorine atom (since an atom of Cl has seven outermost electrons
and each F atom contributes one electron to



the total

10 electrons


5 pairs).


1



(b)

Five pairs lead to a trigonal bipyramid.


1



(c)

















(Any
two

of these three shapes are acceptable.)



2




(4)


ELECTRONIC STRUCTURE

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12

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L QUESTI ONS ( AH CHEM
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© Learni ng and Teachi ng Scot l and 2003

21.

(a)

Sodium chloride has a face
-
centered cubic arrangement.



or

Each ion is surrounde
d by six near neighbours of the
oppositely charged ions.

(1)





(Any
one

of these equivalent sketches is acceptable.)

(1)





2



(b)

The different sizes (radius ratio) of the two ions involved cause
CsCl to have a co
-
ordination of 8:8 (and not 6:6



a
s in NaCl).


1



(c)

The closer match of iron(II) oxide to the radius ratio of



NaCl suggests a sodium chloride structure.




Na
+
/Cl


=

95/181

=

0.52



Cs
+
/Cl


=

174/181

=

0.96



Fe
2+
/O
2


=

61/136

=

0.45

2




(5)

ELECTRONIC STRUCTURE

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13


© Learni ng and Teachi ng Scot l and 2003

22.

(a)

Each Na
+


ion

has six Cl


ions as near neighbours and each



Cl


ion has six Na
+
ions


similarly arranged. Each ion



therefore has six near neighbours


6:6 co
-
ordination.


1



(b)

Since Na and Cl have a valency of 1 they cannot have



six bonds.



or

In the sketch each Na
+

has 6 Cl


near neighbours and



none of these can be considered to be the partner for



the Na
+
any more than any other therefore the lines



represent the direction along which the forces of ionic bonding
apply (and do not represent shared pairs of

electrons as in
covalent bonds of covalent molecules.)


1



(c)

Molecules of sodium chloride do not exist. Only



covalent substances can be correctly labeled as molecular,



but no single Na
+


has a single identifiable special



neighbour among the s
ix Cl


ions.


1




(3)


23.

(a)

H
2

+ SiHCl
3



Si + 3HCl



2



(

1 for each error)




(b)

(i)

Boron


1



(ii)

Phosphorus


1



(c)

(i)

Boron


p
-
type semiconductor.



(ii)

Phosphorus


n
-
type semic
onductor.


1




(both correct for 1)






(5)


24.

(a)

C, E

(both correct for 1)


1




(b)

F, G

(both correct for 1)


1




(c)

B



1



(d)

A, D

(both correct for 1)


1




(Note: while graphite is undoubtedly a non
-
metal it



conducts electricity by a
mechanism that we associate



with metals, called metallic conductivity.)






(4)


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25.

(a)

Doping


1



(b)











1



(c)

(i)

An n
-
type semiconductor.

(1)




P has five outer electrons (one more than Si) and




these extra electrons are the c
harge carriers.

(1)






2




(ii)

Boron


1



(d)

n
-
type semiconductors have a surplus of electrons (from



the dopant) in some areas and these fill the lowest



unfilled conduction band, making the semiconductor a



better conductor.

(1)





p
-
type se
miconductors have positive holes (introduced



by the dopant) and these are the charge carriers.

(1)






2



(e)

(i)

A layer of an n
-
type semiconductor, e.g. Si




containing P impurities, is attached to a layer of




a p
-
type semiconductor, e.g. Si co
ntaining B




impurities.


1




(ii)

When a p

n junction is irradiated with light,




electron hole pairs are formed and electrons




migrate.

(1)





The electrons migrate


n
-
type semiconductor




and the holes


p
-
type. The n
-
type is therefore




now negative with respect to the p
-
type, i.e. a




potential difference has been created.

(1)






2





(10)


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26.

(a)

T
c

is the temperature at which the electrical resistance falls



to zero.


1



(b)

Superconductors are materials that present no re
sistance



at all to the flow of an electric current.


1



(c)

Liquid helium has the lowest boiling point and this was



needed to achieve the very low temperatures that were needed
initially.


1



(d)

T
c

for some new materials was above the boiling po
int of



nitrogen


a much cheaper coolant.


1



(e)

(i)

This would bring superconductivity within reach at




ambient temperatures, with no need for expensive
coolants.


1




(ii)

Distribution of electricity, electric trains/trams,




electric moto
rs (big and small), etc.



1




(6)


27.

(a)

B


1


(b)

A


1


(c)

C


1


(d)

A


1


(e)

D



1




(5)


28.

(a)

Graph 1


1


(b)

Graph 3


1


(c)

Graph 2


1




(3)


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29.

(a)

C (diamond or

graphite) has a covalent network



structure so covalent bonds
have to be broken on



melting, i.e. the melting point is very high.

(1)




N
2

has a covalent molecular structure so only van der



Waals’ forces need to be broken to melt solid nitrogen,



i.e. the melting point is very low.

(1)





2




(b)

(i)

FCl



1




(ii)

LiCl



1




(iii)

BeCl
2



1




(iv)

Polarity falls from Li to N since the difference in
electronegativity decreases to zero.

(1)





Polarity rises from N to F since the difference in
electronegativity increases again (with the polarity
revers
ed).


(1)







2






(7)


30.

The H
+

ion would consist of only a nucleus and as such


cannot have a separate existence.


(In fact it exists in aqueous solution attached to a water


molecule, i.e. H
3
O
+
.)

(1)


31.

(a)

AlCl
3
= 27 + (3


35.5) = 133.5
but the relative molecular



mass = 267, i.e. twice the value for the empirical formula.



The structure must therefore be a dimer of AlCl
3
,




i.e. Al
2
Cl
6
.


1



(b)

Covalent molecular structure with dipole

dipole



interactions.


1



(c)

Aluminium hyd
roxide.

(1)




The chloride is hydrolysed by the water and HCl(g) is



evolved.

(1)





2




(4)

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© Learni ng and Teachi ng Scot l and 2003

32.

(a)

Li
+
H



or

Mg
2+
(H

)
2


(1)



Each reacts vigorously (at least) with water, releasing



H
2

gas, leaving the metal hydroxide in solution.

(1)





2



(b)

Al
2
O
3

+

6HCl



2AlCl
3

+

3H
2
O

(1)



base

+

acid



salt

+

water

(½)



(Note: there is no need to show the dimeric nature of



the salt since it will be in solution as separate ions.)





Al
2
O
3

+

3H
2
O

+

2NaOH



2NaAl(OH)
4

(1)

acid



+

b
ase



salt

(½)




sodium aluminate




or

sodium tetrahydroxoaluminate(III)


3



(c)

SiCl
4
, PCl
3
, SCl
2
are all hydrolysed by water.

(1)



Fumes of HCl(aq) are seen and an oily liquid falls to



the bottom of the tube.

(1)



(BCl
3
and CCl
4

are not hydrol
ysed; the chlorides of As,



Se, Br, Sb and Te are all covalent chlorides and simply



dissolve in non
-
polar solvents without reaction, but are
hydrolysed by water, since it is polar, giving fumes of



HCl(aq), as well as an oxyacid or an oxychloride


both



seen as an oily droplet.)


2



(d)

The hydrogen bond is the force of attraction between



an H atom of one molecule (which is directly attached



to an element of high electronegative value)

(1)



and an atom of high electronegative value on an
other



molecule.

(1)



For water (or HF or HCl)

(1)



the boiling points are higher than predictions based



only on molecular size.

(1)



(This is also true of melting point, of viscosity, of



surface tension and of specific heat capacity.



Hydro
gen bonding also causes water to have its



maximum value of density at 4
o
C, (ice floats), by holding
neighbouring molecules a little further apart than liquid
molecules.)


4



ELECTRONIC STRUCTURE

AND
THE PERIODIC TABLE

18

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

(e)


Na
2
O

MgO

Al
2
O
3

SiO
2

P
2
O
3

SO
2

Cl
2
O



strong

weak

amphoteric

acidic

a
cidic

acidic

acidic




base

base


or

strong or

alkali


amphoteric




strong



base







acid




1



(12)


33.

(a)






(1)







(1)





2



(b)

Fe
3+

is the more stable because it has a d sub
-
shell that



is exactly half
-
filled. This is a more st
able arrangement



than that with one more electron.


1




(3)


34.

(a)

+1


1



(b)

10


1



(c)













When the ion is irradiated with white light, energy is absorbed
and promotes one electron from a lower 3d



energy level to the higher 4s ene
rgy level.



1




(3)


ELECTRONIC STRUCTURE

AND THE PERIODIC TAB
LE

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


19


© Learni ng and Teachi ng Scot l and 2003

35.

(a)

Water molecules and hydroxide ions.

(1)



They use a lone pair of electrons to form a dative



covalent bond with the central metal ion, Fe
2+
.

(1)





2


(b)






(1)

octahedral

(1)






2




(c)

When a water molecu
le acts as a ligand its O

H bonds



will become more polarised than before so H
+
ions



will tend to form.

(1)






However, the hydroxide ion donates a complete negative



charge to the central metal ion, effectively reducing the



positive charge on
it and so reducing the polarising



effect on the O

H bonds. H
+

ions are therefore less



likely to form, i.e. the solution is less acidic.


(1)





2




(6)


36.

(a)

The 3d orbitals are each half
-
filled before one is doubly



filled, i.e. maximum unp
airing before pairing, etc.


1



(b)









(Note: the 4s box may precede the 3d or it may be



omitted altogether since it is empty.)


1



(c)

Only one


1




(3)


ELECTRONIC STRUCTURE

AND
THE PERIODIC TABLE

20

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

37.


(a)

1.

[Cu(NH
3
)
4
]
2+


(1)



2.

[Cu(NH
3
)
3
(Cl)]
+


(1)



3.

[Cu(NH
3
)
2
(Cl)
2
]
0

(You ma
y omit the zero in





this formula.)

(1)




Ligands are held to a central metal ion by dative covalent



bonds, not by ionic bonds. The ionic chloride ions are



free to form a precipitate with Ag
+

ions whereas the



covalently bonded chlorides are n
ot.

(1)





4



(b)

The degree of d

d splitting is controlled by the ligands



involved.

(1)



NH
3

on its own will therefore be different from NH
3




with varying proportions of Cl.

(1)





2




(6)


38.

(a)

CrO
4
2





Cr

=

? oxidation state




=

? + (4




2) =

2



?

=


2 + 8




=

+ 6

(1)




i.e. Cr atom has lost six electrons.



1s
2

2s
2
2p
6

3s
2
3p
6
3d
0

4s
0

(You may omit the 3d
0
4s
0
.)

(1)





2



(b)

Cr in CrO
4
2



is in oxidation state +6 (see above).



In Cr
2
O
7
2


, Cr = ?



(2


?) + (7



2)

=

2



? = +12/2 = +6



So Cr in Cr
2
O
7
2


is also in oxidation state +6.

(1)



This is therefore not a redox reaction.

(1)





2



(c)

d

d splitting can only apply when there is at least one electron



in a d orbital (this theory similarly does not

apply to a



completely filled d sub
-
shell).


1





(5)


ELECTRONIC STRUCTURE

AND THE PERIODIC TAB
LE

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


21


© Learni ng and Teachi ng Scot l and 2003

39.

(a)

Colour intensity is at a maximum when the concentration



of the complex is at a maximum.

(1)



Ratio of [Ni
2+
]/[ligand] = 25/75 = 1/3.

(1)



i.e.
x

= 1,
y

= 3.

(1)





3



(b)

The N atom
s in diaminoethane each have one lone pair.

(1)



Each lone pair forms a dative covalent bond with the



Ni
2+

ion.

(1)





2



(c)

The complex is purple, i.e. it absorbs in the green



region.

(1)



The filter transmits green to get the best absorption



from the complex.

(1)

2




(7)


40.

The H
2
O ligands

(1)


split the degenerate d orbitals.

(1)


Energy from the red end of the visible spectrum is absorbed


as electrons are promoted across the small energy gap,

,


now existing in the d orbitals. Hence the green colour is


seen.

(1)




3

ELECTRONIC STRUCTURE

AND
THE PERIODIC TABLE

22

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

41.

(a)

Green


1



(b)



















Labels/units on axes

(½ + ½)



Subtracting 3 from each colorimeter reading

(1)



Concentration of unknown = 0.0068


0.0001 g l

1


(1)



(Remember to subtract 3 from the colorimeter reading



for the unknown.)


3



(c)

There is some iron in the water.



or

Impurities with similar absorption patterns are present.


1



(d)

Duplicates should be prepared




or

Large volumes should be
used (to reduce error).




or
A blank should be carried out on the buffer and solvent



without any sample present.


1



(e)

d

d splitting
or

d

d transitions


1




(7)


42.

(a)

Five degenerate orbitals are split into two orbitals of



higher energy
and three of lower energy. Electrons can



be promoted across this gap by absorbing energy from



the visible spectrum.

(1)




The peak around 410 nm represents the wavelengths absorbed
(and equals the energy value of the d

d split).

(1)





2


ELECTRONIC STRUCTURE

AND THE PERIODIC TAB
LE

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


23


© Learni ng and Teachi ng Scot l and 2003


(b)

S
ince the chloro complex leads to a lower value of



d

d splitting

(1)



less energy is needed to make the jump

(1)



so absorption moves to the lower or red end of the



spectrum.

(1)





3




(5)


43.

(a)

[VO
2
]
+

V + (2



2)

=

+5





V

=

+5 (or V)

(1)



[V(H
2
O)
6
]
2+

V

=

+2 (or II)

(1)





2



(c)

Hexaaquavanadium(III)

(1)















(Make sure you show that the O atom

(1)



is donating the lone pair,
not

the H atom.)


2



(c)

(i)

1s
2

2s
2
2p
6

3s
2
3p
6

4s
0

3d
0

(
or

o
mit the last two terms)


1



(ii)

It has no d electrons to be promoted.


1



(d)

Absorption should cover the red/yellow and possibly the green
regions.















1




(7)

ELECTRONIC STRUCTURE

AND
THE PERIODIC TABLE

24

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

44.

(a)

Ni
2+

[Ar] 4s
2

3d
8
or

1s
2
2s
2

2p
6

3s
2

3p
6

4s
2

3d
8
.


1



(b)

C
l


splits the d orbitals and energy (a selection of frequencies)
from the visible spectrum promotes



electrons across the gap.

(1)



Remaining frequencies are transmitted and these



produce the colour.

(1)






2



(c)

The size of the d

d split is al
tered by the presence of different
ligands, so frequencies absorbed are different



and the colour transmitted is altered.


1




(4)


45.

(a)

[Cu(Cl )
4
]
2


((1) for formula and (1) for charge)


2



(b)

(i)

Hexaamminecopper(II)


1




(ii)















1




(4)


46.

(a)

E

=

Lhc
/



(1)





=

Lhc
/
E

= 6.02


10
23



6.63


10

34


3


10
8
/239


10
3

(1)




= 119.74

10

6
/239




= 5.01


10

7
m




= 501 nm

(Note: this is in the visible spectrum






so it is a sensible answer.)

(1)








3



(b)

Green is absorbed

(1)



so purple o
r magenta seen.

(1)





2




(5)

ELECTRONIC STRUCTURE

AND THE PERIODIC TAB
LE

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


25


© Learni ng and Teachi ng Scot l and 2003

47.

(a)

Green (since red is absorbed).


1



(b)

E

=

Lhc
/



(1)




=

6.02


10
23



6.63


10

34



3


10
8
/540


10

9
J mol

1

(1)




=

0.222


10
6

J mol

1




=

222 kJ mol

1

(This is a likely order of magnitude








see Data Booklet.)

(1)








3







(4)


48.

(a)

It has a full d sub
-
shell.


1



(b)

It contributes no colour to compounds. It has only one valency.
Neither the element nor its compounds show significant



catalytic properties.



(any two)


2



(c)

Fe

Haber process



Pt

Ostwald process/oxidation of ammonia.



Pt

catalytic converters in car exhausts.



Ni

hardening of vegetable oils

fats.



(accept others)


3



(d)

Vacant d orbitals are readily available for (reversible)



bonding, i.e. show variable oxidation states.


1




(7)


49.

(a)

They are compounds of the transition metals.


1



(b)

One reactant may be adsorbed (or held b
y covalent



bonds, often with d orbitals of the catalyst)

(1)



in a

suitable orientation for a more probably successful




collision.




or

in such a way that bonds within the reactant are



weakened (and there is a new arrangement of bonds,



i.e.
a reaction occurs).

(1)





2




(3)


26

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

PRINCIPLES OF CHEMIC
AL REACTIONS

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


27


© Learni ng and Teachi ng Scot l and 2003

UNIT 2



1.

(a)

Iodide ions


1



(b)

Starch indicator


1



(c)


n

=
v



c

= 24.8


10

3



0.1 = 2.48


10

3



1



(d)

Moles of Cu
2+

=

2.48


250/25

=

2.48


10

2

(1)



Mass of Cu

=

2.48


10

2



63.5

=

1.575 g

(1)



Percentage of Cu

=

1.575/2.63


100

=

59.89%

(1)






3





(6)


2.

Fe(NH
4
)
2
(SO
4
)
2
.6H
2
O = 392


Moles of
Fe
2+

=

1.8/392

=

4.6


10

3

(1)


From redox equation Fe
2+
:

MnO
4



=

5
:
1

(1)


Moles of MnO
4



=

4.6


10

3
/5

=

9.2


10

4

(1)


Concentration

=

9.2


10

4
/0.04

=

0.023 mol l

1

(1)





(4)


3.

(a)

(i)

AgCl =143.4




Mass of Ag in precipitate

=

10
7.9/143.4


0.6





=

0.451 g

(1)




Mass of Ag in coin = 1000/100


0.451

=

4.51 g

(1)




Percentage of Ag in coin = 4.51/10


100

=

45.1%

(1)






3




(ii)

Add some AgNO
3

to filtrate. There should be no




more precipitate.


1



(b)

CuCNS = 121.6



Mass of Cu = 63.5/121.6


0.31

=

0.1618

(1)



Mass of Cu in coin = 1000/100


0.1618

=

1.62 g

(1)



Percentage of Cu in coin

=

16.2%

(1)






3





(7)


PRINCIPLES OF CHEMIC
AL REACTIONS

28

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

4.

(a)

Add more AgNO
3



no further precipitate.


1



(b)

2AgNO
3

+ MgCl
2



2AgCl + Mg(NO
3
)
2




Moles of AgCl

=

2.01/143.4

=

0.014

(1)



Moles of MgCl
2

=

0.014


1/2


500/100

=

0.035

(1)



Mass of MgCl
2

=

0.035


95.3

=

3.336 g

(1)



%MgCl
2

=

3.336/4.5


100

=

74.1%

(1)






4



(c)

Moles AgNO
3

= 0.014/4 = 0.0035


(1)



Volume = moles/conc. = 0.0
035/0.1 =

0.035 litres





(35 cm
3
)

(1)






2






(7)


5.

(a)

Moles of HCl

=

15


10

3


1.0

=

1.5


10

2

(1)



Moles of Na
2
CO
3

=

1/2


1.5


10

2

=

7.5


10

3


(1)



Mass of Na
2
CO
3

=

7.95


10

3



106


250/25

(1)





= 7.95 g



(1)










4



(b)

Ma
ss of water

=

16


7.95

=

8.05 g



Moles of water

=

8.05/18

=

0.447



Moles of Na
2
CO
3

=

7.95/106

=

0.075

(1)



Ratio of moles

=

0.447/0.075

=

5.96 = 6 (i.e. 6H
2
O)

(1)










2









(6)


6.

(a)

Moles of NaOH

=

18.2/1000


0.1

=

1.82


10

3

(1)



Moles

of CH
2
(COOH)
2

=

9.1


10

4




(1)










2



(b)

Mass of acid

=

9.1


10

4



250/25


104

=

0.946 g

(1)



Mass of water

=

1.28


0.946

=

0.334 g

(1)










2



(c)

Moles of water

=

0.334/18 = 0.0186



Moles of acid

=

0.946/104 = 0.0091



(1)





n

=

0.0186/0.0091 = 2



(1)






2






(6)



PRINCIPLES OF CHEMIC
AL REACTIONS

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


29


© Learni ng and Teachi ng Scot l and 2003

7.

(a)

Mass of BaSO
3

=

1.09 g



(1)



Moles of BaSO
3

=

1.09/217 = 0.005



(1)










2



(b)

Moles of Na
2
SO
3

=

0.005


250/50

=

0.025

(1)



Mass of Na
2
SO
3

=

0.025


126

=

3.15 g

(1)



%Na
2
SO
3

=

3.15/5.02


100

=

62.75%

(1)






3


(c)

It is a good oxidising agent.



or

It can oxidise SO
3
2


but not SO
4
2

.

(1)



It acts as its own indicator.

(1)






2





(7)


8.

(a)

(i) Yield = 50%

1



(ii) side reactions
or

impurities

1



(b)

The suggestion is wrong.

(1)



A catalyst only brings the reaction to the
same




equilibrium more quickly.

(1)






2



(c)

K

is a constant at a fixed temperature and altering the



alcohol concentration will not change the value of
K
.

(1)



It will, however, increase the yi
eld of the ester as the



forward reaction will be increased.

(1)






2





(6)


9.

(a)

(i)

K

= [SO
3
]/[SO
2
]
2
[O
2
]


1



(ii)

No units


1



(b)

(i)

Le Chatelier’s principle states that if a system is subjected
to any change, the system readjusts itself

to counteract the
applied change.


2




(ii)

Increased temperature favours

H

+ve. Thus the




backward reaction is favoured and the equilibrium




position moves to the left.



1




(iii)

K

will decrease.


1






(6)


PRINCIPLES OF CHEMIC
AL REACTIONS

30

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

10.

(a)

A 1; B 3; C 1; D 2; E 2



Reactions A and C are going from a larger volume to a smaller
volume theref
ore an increase in pressure favours the products.




Reaction B goes from a small volume (no gas) to a large volume
therefore an increase in pressure favours the reactants.




Reaction D has two volumes of gas on both sides therefore
pressure does not af
fect the result.




Reaction E has no gas therefore pressure change has no effect.




(5


(1) for correct answer with correct explanation)


5



(b)

It will have the same general shape but it will be less



steep.


1





(6)


11.

(a)

K

= [NO
2
]
2
/[NO]
2

[O
2
]


2



(b)

K

> 1 therefore equilibrium favours the products.


1



(c)

Increasing temp
erature favours

H

+ve



therefore the products are favoured

(1)



therefore
K

decreases.

(1)






2


(d)

15

=

[NO
2
]
2
/[0.1]
2



[0.1]





[NO
2
] =

0.015

(1)






= 0.12 mol l

1

(1)






2





(7)


12.

(a)

[Ba
2+
] =

1


10

10

= 1


10

5

mol l

1



1



(b)

BaSO
4

= 233.4 g mol

1

(1)



Mass dissolving = 233.4


1


10

5

(1)





= 2.334


10

3
g

(1)






3


PRINCIPLES OF CHEMIC
AL REACTIONS

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


31


© Learni ng and Teachi ng Scot l and 2003


(c)

(i)

K
sp

is a constant therefore stays the same.


1




(ii)

The equilibrium moves to the left therefore [Ba
2+
]



decreases.


1






(6)


13.

(a)

(i)

Ether


1



(ii)

Ether is less dense than water.


1



(b)

K

=

[X
et her
]/[X
wat er
]

=

12 (let
V

grams be extracted)



12

=

(
V
/100)
/
(1.1


V
)/100

=

V
/(1.1


V
)

(1)



V

=

1.0154 g



(1)






2



(c)

12

=

V
/50
/
(1.1


V
)/100



V

=

0.943 g



Mass
now dissolved in the water = 1.1


0.943 = 0.157 g

(1)



12

=

V
/50
/
(0.157


V
)/100



V

=

0.106 g



Total extracted = 0.943 + 0.106 = 1.049 g

(1)






2


(d)

Many organic compounds are non

polar and dissolve



in non
-
polar solvents such as ether (water is

polar).


1





(7)


14.

(a)

Moles of acid originally = 2.36/118 = 2


10

2



1




(b)

(CH
2
COOH)
2

+ 2NaOH


(CH
2
COONa)
2

+ 2H
2
O



Moles of NaOH = 34.8/1000


1 = 3.48


10

2


(1)



Moles of acid = 1.74


10

2

(1)



Moles in ether = (2


10

2
)


(1.74


10

2
) = 2.6


10

3

(1)






3



(c)

K

= 2.6


10

3
/0.1
/
1.74


10

2
/0.1 = 0.149


1



(d)

Use two 50 cm
3

portions of ether (or 4


25 cm
3
),




i.e. do two or four extractions from the aqueous layer.


1





(6)


PRINCIPLES OF CHEMIC
AL REACTIONS

32

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

15.

(a)

(i)

The solvent mixture


1



(ii)

The chromatography paper


1



(iii)

3


1



(iv)

Glycine


if the solvent were allowed to move




further up the paper then the component may




not travel the same distance as glycine.


2




(v)

Solvent front


1









(vi)

Base line








1



(b)

(i)

To prevent the solvent evaporating away
or

to ensure the
atmosphere in the tank is saturated with the




solvent.


1






(ii)

R
f

values are used to identify substances. The same
substance has the same
R
f

value for the same




conditions.


1






(iii)

The ink may dissolve in the solvent.


1




(iv)

The component is held very strongly by the paper on the
baseline. (Do not answer in terms of solubility.)


1






(v)

A mixture of solvents gives better separation; more
components are likely to be s
oluble.


1




(vi)

Yes (mobile and stationary phases)


1




(vii)

The components are colourless and so must be




reacted with a developing agent to make them




visible.


1





(14)


16.

(a)

(i)

Nitrogen
or

argon


1



(ii)

It will not react with th
e alkanes.


1

PRINCIPLES OF CHEMIC
AL REACTIONS

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


33


© Learni ng and Teachi ng Scot l and 2003


(b)

The time it takes for the compound to move through the



column from the point it was injected at to the point where



it is detected.


1



(c)

(i)

A


methane, E


2,2
-
dimethylpropane,




F


2
-
methylbutane


3




(ii)

Peak F is at

9.3, which is the value for 2
-
methylbutane.




Peak E is at about 8.7 and therefore the alkane must




be an another isomer of pentane.




Peak A is at about 1.7 and therefore the molecule




must be smaller than ethane.


3





(9)


17.

(a)

H
2
O



1


(b)

OH





1


(c)

NH
3




1





(3)


18.

(a)

(i)

The HSO
3

(aq) provides an H
+
(aq) ion.


1



(ii)

The SO
3
2

(aq) is the conjugate base.


1



(b)

HSO
3

(aq) + H
+
(aq) H
2
SO
3
(aq)

1






(3)


19.

(a)

pH

=

½p
K
a



½ log
c

(1)




= ½


4.77


½ log0.01

(1)




= 2.38
5


(

1)




=

3.4

(1)



or K
a

=

[H
+
]
2
/[CH
3
COOH]

(1)



Therefore [H
+
]

=


(1.75


10

5



0.01)

(1)






=


1.7


10

7







=

4.12


10

4




pH

=



log H
+

=


log 4.12


10

4

= 3.4

(1)






3


(b)

CH
3
COOH(aq) CH
3
COO

(aq) + H
+
(aq)

acid



CH
3
COO

Na
+
(
s)


CH
3
COO

(aq) + Na
+
(aq)

salt




(i)

Add HCl


the H
+
(aq) ions added react with




the excess CH
3
COO

(aq) from the salt.

(1 + 1)

2





(ii)

Add NaOH


the OH

(aq) added reacts with




the H
+
(aq) from the acid and more acid ionises





to replac
e the H
+
(aq) ions removed.

(1 + 1)

2

PRINCIPLES OF CHEMIC
AL REACTIONS

34

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003


(c)

pH = p
K
a



log[acid]/[salt]

=

4.77


log 0.25/0.15

(1)





=

4.55

(1)






2





(9)


20.

(a)

The product of [H
+
(aq)]


[OH

(aq)]


1



(b)

[H
+
]
2

= 51.3


10

14
, therefore [H
+
] = 7.16


10

7

and



pH = 6.15


(1+1+1)






3



(c)

An increase in temperature favours

H +ve. As the temperature
increases so does the value of
K
w

and therefore more ions form.
Thus the ionisation is endothermic.


1





(5)


21.

(a)

NaH
2
PO
4
; Na
2
HPO
4
;

Na
3
PO
4


(any two, 1+1)






2



(b)

K
a

= [H
+
][ H
2
PO
4

]/[H
3
PO
4
]

(1)



7.08


10

3
=

[H
+
]
2
/0.1



[H
+
]
2


=

7.08


10

4




[H
+
]


=

2.66


10

2


(1)



pH


=

1.6

(1)








3



(You could use pH = ½p
K
a



½log[C].)


(5)


22.

K
w

at 288K


=

0.45


10

14

(1)


[H
+
]



=

6.71


10

8


(1)


pH



=


log 6.71 x 10

8

= 7.17

(1)







(3)


23.

(a)

[Ca(OH)
2
]

=

0.126/74 moles per 100 cm
3






=

0.017 mol l

1


(1)



[OH

] = 2


0.017 = 0.034 mol l

1

(1)






2



(b)

[H
+
]

=

10

14
/0.034 = 2.94


10

13

(1)




pH

=



log 2.94


10

13

= 12
.53

(1)






2





(4)


PRINCIPLES OF CHEMIC
AL REACTIONS

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


35


© Learni ng and Teachi ng Scot l and 2003

24.

(a)

Methyl yellow pH = 3.3; bromothymol blue pH = 7;



thymol blue pH = 8.9 (3 correct = 2; 1 or 2 correct = 1)


2



(b)

Thymol blue.

(1)



The titration involves a weak acid and strong alkali



therefore the salt will
have a pH above 7.

(1)






2





(4)


25.

(a)

25


0.1/20 = 0.125 mol l

1



1



(b)

Weak

(1)



Acid neutralised at pH 9

(1)






2



(c)

Indicator must change colour at a pH corresponding



to the neutralisation point.


1



(d)





















2




(e)

No suitable indicator


1





(7)


Small point of inflection about pH7.

PRINCIPLES OF CHEMIC
AL REACTIONS

36

ANSWERS TO ADDI TIONA
L QUESTI ONS ( AH CHEM
ISTRY)


© Learni ng and Teachi ng Scot l and 2003

26.

(a)

1.6/64.1

=

0.025 mol in 250 cm
3


(1)




=

0.1 mol l

1


(1)






2



(b)

pH

=

½ p
K
a



½log
c


(1)




=




1.8)


(½ log 0.1)

(1)




=

1.4

(1)


or

K
a

=

[H
+
][HSO
3

]/[H
2
SO
4
]

(1)



H
+

=


K
a



[H
2
SO
4
] =

(1.5


10

2
)


(0.1)

(1)




=

3.87


10

2




pH

=



log [H
+
] = 1.4

(1)



(If the wrong equation is used then zero marks.)


3



(c)

It is volatile, i.e. SO
2

is given off from solution.
or

It is



unstable.


1



(d)

Accept range within 7.5


10.5 but must have at least 1.5



of a differen
ce between values.


1





(7)


27.

K
a

=

[H
+
][A

]/[HA] = [H
+
]
2
/[HA]

(1)


[H
+
]

=


K
a



[HA] =

1.3


10

5



0.1

(1)



=

1.14


10

3


pH

=

2.9

(1)

or

pH

=

½p
K
a



½log
c


(1)



=

½


4.9


½log 0.1

(1)



=

2.95

(1)


(If the wrong equation is used then zero marks. )


(3)


28.

(a)

Find


MgCl
2

(s)



Mg
2+
(aq)

+

2Cl

(aq)



Gi
ven (1)

Mg
2+
(g)



Mg
2+
(aq)




H


1923




(2)

Cl

(g)



Cl

(aq)




H


338




(3)

Mg
2+
(g)

+

2Cl

(g)



Mg
2+
(Cl

)
2

(s)


H


2326




H

=




H
3

+

H
1

+ 2

H
2




(1)




=


(

2326) + (

1923) + 2(

338)



(1)




=



273 kJ mol

1






(1)











3



(b)

MgCl
2
(s)
+ 6H
2
O(l)


MgCl
2