Download Topic 9.4: Transition Elements Chemistry

Transition Elements Chemistry - Suggested Solutions
Topic 9.4: Transition Elements Chemistry
AJC 2009/P3/Q3c,d
1
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Transition Elements Chemistry - Suggested Solutions
CJC 2009/P2/Q3a-c
2 (a) Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 or [Ar] 3d5 4s1
Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 or [Ar] 3d3
(b)
(i)
(ii)
Cr
S
Mols 0.298 1.19
Ratio 1
4
x+y+1=7
x:
4
N
2.09
7
y:
2
+3
(iii)
-
NH
.. 3
SCN:
Cr
(III)
:NCS
:NCS
SCN:
..
NH3
(c)
(i) K2Cr2O7 + 2 HBr
(ii)
→ 2 KCrO3Br + H2O
No, oxidation no. of Cr doesn’t change, remains at +6
(iii) Eocell = +1.33 – (+1.07) = +0.26 V >0
Products should be Cr3+(aq) and Br2(l)
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Transition Elements Chemistry - Suggested Solutions
(iv) Reaction didn’t take place at standard condition.
OR The reaction in step I took place under cooled conditions.
(v)
AgBr
(vi)
Cr is reduced from +6 (orange solution of Cr2O72-) to +3 Cr3+ (deep green solution)
DHS 2009/P3/Q4a-c
(a) Explain why transition metals are denser than s-block elements.
[3]
They have relatively smaller atomic radius and higher relative atomic
mass. Hence, they have a close-packed structure.
(b) Describe the bonding in the following substances and explain why it
contributes to the specified property in glass:
substance
(i) silicon (IV) oxide
(ii) lead (IV) oxide
property
hardness
Electrical conductivity (if any)
[4]
SiO2 has a giant molecular structure with extensive covalent bonding in a
giant three-dimensional structure.
PbO2 has a giant ionic structure. In the solid state, the ions can only vibrate
about fixed positions.
(c)
The window frames enclosing the stained glass is made of aluminium due to
its low density and high melting point.
Account for the high melting point of aluminium.
[2]
High amount of energy is required to overcome the strong electrostatic
forces of attraction between the cations and sea of delocalised
electrons.
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Transition Elements Chemistry - Suggested Solutions
HCI 2009/P2/Q6a,b
4
A: [Cr(H2O)6]3+
B: [Cr(H2O)6]2+
C: Cr(OH)3 or Cr(H2O)3(OH)3
D: CrO42– or Na2CrO4
E:
3+
H2
N
H2N
NH2
Cr
N
H2
NH2
H2N
Cr3+ has a high charge density. Hence [Cr(H2O)6]3+ can undergo hydrolysis in
water to produce H+ ions, forming CO2 with carbonate ions.
Ligand exchange
Cr3+ has d3 electronic configuration.
In an octahedral ligand field, the 6 NH3 ligands will split the five degenerate 3d
orbitals into 2 groups of different energy levels.
The difference in the two energy levels, ∆E, falls within the visible region of the
electromagnetic spectrum. An electron in a lower d orbital energy level can
absorb radiation in the visible spectrum and be promoted into the higher d orbital
energy level.
This d→d electron transition gives rise to the colour as the complement of the
absorbed colour.
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Transition Elements Chemistry - Suggested Solutions
HCI 2009/P3/Q4a,b,e
5
(a) Any two properties:
Fe has a higher melting/boiling point than Al.
Fe has stronger metallic bonds as both the 3d and 4s electrons can be used in
metallic bonding due to their proximity in energies, hence more energy is required to
overcome the stronger bonds.
Fe has a greater density than Al.
Fe has a greater atomic mass but its atomic radius is smaller. Hence atomic volume
is smaller. Since density = mass/volume, density of Fe is greater than Al.
Fe has better conductivity than Al as both the 3d and 4s electrons are available in the
mobile sea of electrons due to the proximity in energy of 3d and 4s orbitals.
Fe is harder than Al due to stronger metallic bonding in Fe as Fe can use both its 3d
and 4s electrons for metallic bonding due to the proximity in energy of 3d and 4s
electrons.
(b)
26Fe
26Fe
3+
1s2 2s2 2p6 3s2 3p6 3d6 4s2
1s2 2s2 2p6 3s2 3p6 3d5
13Al
3+
13Al
1s2 2s2 2p6 3s2 3p1
1s2 2s2 2p6
Although Fe has one more quantum shell than Al and higher nuclear charge, the
increase in shielding effect due to the d electrons is proportionately less than the
increase in nuclear charge, hence Fe experiences higher effective nuclear charge
and is smaller.
Fe3+ has one more quantum shell than Al3+ and the additional shielding is due to s
and p electrons which are more effective in shielding compared to d electrons.
(i)
This is due to the presence of partially-filled 3d subshells that can accept electrons from lone
pairs or form temporary bonds with reactant molecules during adsorption, thus weakening the
bonds in the reactant molecules and lowering the activation energy of the reaction.
(ii)
Energy / kJ
Ea
N2 +
Ea (catalysed)
∆H
NH3
Reaction
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Transition Elements Chemistry - Suggested Solutions
IJC 2009/P2/Q2c
6
IJC 2009/P3/Q1e
7
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Transition Elements Chemistry - Suggested Solutions
IJC 2009/P3/Q2d
8
IJC 2009/P3/Q3b
9
JJC 2009/P2/Q2a,b
10
(a)
Fe(H2O)63+ = Fe(H2O)5(OH)2+ + H+
(bi) 1s22s22p63s23p63d10
(ii)
Colourless because Cu+ does not have unfilled/ partially filled d-orbital
JJC 2009/P2/Q4a
(a)
(i)
Mn
O
Mass
63.8
36.2
Ar
54.9
16.0
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Transition Elements Chemistry - Suggested Solutions
Mole
1.16
Ratio
1
Empirical Formula: MnO2
(ii)
Identity of Z: MnO4−
2.26
2
[1]
[1]
Type of reaction Disproportionation [1]
Equation 3MnO42−− + 2H2O → 2MnO4− + MnO2 + 4OH− [1]
MI 2009/P3/Q1b
12
b (i)
[Al(H2O)6]3+ + H2O ƒ [Al(H2O)5(OH)]2+ + H3O+
b (ii)
Size of anion: U3+(aq) < Al3+(aq)
[½]
U3+(aq) has a lower charge density
[½]
less able to distort the electron cloud of H2O
[½]
less weakening of O-H bond
[½]
+
less able to produce H
MJC 2009/P2/Q3a
13
Coordination number of Co = 6
In Co2+ ions, the d orbitals are split into two groups due to the ability of the ligands to
split them into the energy levels. d The d electrons undergoes d-d transition and is
promoted to the higher d orbital. During the transition, the d electron absorbs a certain
wavelength of light from the visible region of the electromagnetic spectrum and emits
the remaining wavelength which appears as the colour of the complex observed.
NYJC 2009/P2/Q2b
14
(bii)
2 Cu+ Cu + Cu2+
Cu+ + e
Cu2+ + e
Cu
Cu+
+ 0.52 V
+ 0.15 V
Eθ = + 0.52 – (+ 0.15) = + 0.37 V > 0 (feasible)
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Transition Elements Chemistry - Suggested Solutions
PJC 2009/P3/Q5a,b
15
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Transition Elements Chemistry - Suggested Solutions
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Transition Elements Chemistry - Suggested Solutions
RI 2009/P3/Q3a,b
16
(a)
The electronic configuration of V3+ is [Ar]3d2.
Hence its 3d−subshell is only partially filled.
V3+ exists as aqua complex, [V(H2O6)]3+ in aqueous solution.
H2O ligand splits the 3d−subshell into two sets of different energy.
Electron in the lower set absorbs visible light energy corresponding to the small
energy gap between the 2 sets of energy and is promoted to the upper set.
Since visible light is absorbed, aqueous solutions of vanadium ions are coloured,
with the colour observed being the complementary colour of the light absorbed.
(b)(i) SO2 (g) + 2VO2+ (aq) → SO42− (aq) + 2VO2+ (aq)
SO2 reduces yellow VO2+ to blue VO2+.
As the reaction proceeds, the presence of both yellow VO2+ and blue VO2+ causes
the mixture to appear green until all VO2+ is used up and the solution appears blue
due to VO2+.
(ii) SO2 can also react with KMnO4. It must be boiled off completely to ensure that none
of it remains so that the KMnO4 used in titration reacts only with vanadium ions and
hence the amount of vanadium ions can then be determined accurately.
(iii)
Amount of vanadium ions = 5 x amount of MnO4− used
33.00
x 0.0200 = 3.300 x 10−3 mol
=5x
1000
Mass of vanadium in sample = 3.300 x 10−3 x Ar of V = 3.300 x 10−3 x 50.9 =
0.1680g
0.1680
Percentage by mass of V =
x 100 = 56.0%
0.300
SAJC 2009/P3/Q2b
17
From data booklet,
Co3+ + e
I2 + 2e
S2O82- + 2e
Co2+
E0 = +1.82 V
2I-
E0 = + 0.54 V
2SO42-
E0 = + 2.01V
2Co3+ + 2I-
+
2Co2+
I2
0
E = + 1.82 – 0.54 = +1.28 V > 0
2Co2+ + S2O82-
2Co3+ + 2SO42-
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Transition Elements Chemistry - Suggested Solutions
E0 = + 2.01 – 1.82 = +0.19 V >0
(ii)
[Cu(H2O)6]2++ 4NH3
[Cu(NH3)4(H2O)2]2+ + 4 H2O
[Cu(NH3)4(H2O)2]2+ + edta4-
[Cuedta]2- + 2 H2O + 4 NH3
Strength of ligands : H2O < NH3 < edta4-
SRJC 2009/P3/Q1d
dative / coordinate/covalent bond
18
Rh has a giant metallic structure with strong electrostatic forces of attraction between
cations and sea of delocalised electrons.
Both NH3 and H2O have simple molecular structures with intermolecular hydrogen bonds.
For the identification of hydrogen bonds, either mentioned here or in part (iv).
Number / Extent of hydrogen bonds: NH3 < H2O
Energy required: NH3 < H2O
Boiling point: NH3 < H2O
NH3 is soluble in water
as favourable hydrogen bonds between NH3 and water / solute-solvent interaction can be
formed.
SRJC 2009/P3/Q4b
19 (b) Transition metals possess variable oxidation states due to the small energy difference
between the 3d and 4s electrons. Thus different number of 3d and 4s electrons may be
lost to form stable ions or compounds of different oxidation states.
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Cu2+(aq) + 2OH-(aq)
Cu(OH)2 (s) ----- (1)
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Transition Elements Chemistry - Suggested Solutions
When NH3(aq) is added gradually, [OH-] will increase
Ionic product of Cu(OH)2 > Ksp of Cu(OH)2
Pale blue ppt, Cu(OH)2 is formed
[Cu(H2O)6]2+(aq) + 4NH3(aq)
[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) -----(2)
deep blue
When excess NH3 is added,
NH3 ligands replaces the H2O ligands, forming a more stable deep blue
[Cu(NH3)4(H2O)2]2+ complex with Cu2+(aq).
[Cu2+] decreases as it is being used to form the complex, equilibrium position in (1)
shifts left to increase [Cu2+]
Pale blue ppt dissolves.
The d orbitals of Cu2+ (aq) are split into two different energy level due to presence of
H2O ligands.
The d electron undergoes d-d transition and is promoted to a higher energy d orbital.
During the process, red wavelength of light energy from the visible region of the
electromagnetic spectrum is absorbed and blue wavelength is transmitted which
appears as the colour observed.
TJC 2009/P3/Q5a,b
20
(a) (i)
• Mn is a transition element which forms Mn2+ in which the d subshell is
partially filled with 5 electrons.
• Water molecule is a ligand and has a lone pair of electrons on O atom that
can form a dative bond with the central metal ion Mn2+.
(ii)
•
When the ligands approach the central metal ion, splitting of the d-orbitals
occur. The energy gap, ∆E
between the non-degenerate orbitals
corresponds to the wavelength of light in the visible region of the
electromagnetic spectrum.
When d-d transition of electrons takes place, radiation in the visible region
of the electromagnetic spectrum corresponding to ∆E is absorbed.
The light energy not absorbed will be seen as the colour of the complex.
•
Empirical formula of A = MnO
•
•
(b)
(i)
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Transition Elements Chemistry - Suggested Solutions
(ii)
•
•
•
B = MnO4−
It is a disproportionation reaction
3 MnO42−− + 2H2O → 2 MnO4− + MnO2 + 4OH−
TPJC 2009/P2/Q3b
21
(b) (i)
1s2 2s2 2p6 3s2 3p6 3d3
(ii)
Cr2O72- + 14H+ + 3Zn 2Cr3+ + 7H2O + 3Zn3+
2HCl + Zn ZnCl2 + H2
(iii)
To let out the hydrogen gas produced.
Or to prevent a build up of pressure due to evolution of hydrogen gas.
(iv)
Cr3+ + e- Cr2+ Eθ = -0.41V
Zn2+ + 2e- Zn
Eθ = -0.76V
θ
E cell = -0.41 – (-0.76) = + 0.35V .
Therefore, reduction of Cr3+ to Cr2+ is still feasible.
(v)
The hydrogen produced exerts greater pressure in the flask.
(vi)
Ability of chromium compound to vary oxidation state or to display colors.
TPJC 2009/P3/Q1a-c
22
(a)
• X is a transition metal with a partially filled d-subshell.
• In the presence of ligands, the d orbitals (of the transition metal ion) split into 2 energy
levels with an energy gap that falls within the visible light spectrum.
• Electrons are able to be promoted from a lower energy d-orbital to the higher energy one by
absorbing energy from the visible spectrum.
OR
When an electron from the d-orbital of lower energy is promoted to one of higher energy (dd electronic transition), an amount of energy, ∆E, in the visible region of the
electromagnetic spectrum is absorbed.
• The light energy not absorbed will be seen as the colour of the complex.
OR
We observe the complementary colour / the wavelengths that are transmitted.
(b)
Cr: [Ar]3d54s1
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Transition Elements Chemistry - Suggested Solutions
Al: [Ne]3s23p1
Atomic radius of chromium is smaller than that of aluminium as it has higher nuclear
charge and 3d orbitals provide poor shielding
effective nuclear charge higher hence radius smaller.
Cr3+: [Ar]3d3
Al3+: [Ne]3s0
Cr3+ has one more quantum shell compared to Al3+, hence the bigger ionic radius.
(c)
Step 1
Relevant data:
3+
–
Mn + e = Mn
+
–
2+
–
E = +1.49 V
2+
E = +1.52 V
MnO + 8H + 5e = Mn + H O
4
2
–
Overall equation for step 1:
E
cell
Relevant data:
3+
4
2
2–
–
2CO + 2e = C O
2
3+
–
Mn + e = Mn
Overall equation for step 2:
cell
2+
= +1.52 – 1.49 = +0.03 V > 0 AND reaction is feasible.
Step 2
E
+
MnO + 8H + 4Mn → 5Mn + 4H O
E = –0.49 V
2 4
2+
E = +1.49 V
2–
3+
2Mn + C O
2
4
2+
→ 2Mn + 2CO
2
= +1.49 – (–0.49) = +1.98 V > 0 AND reaction is feasible.
YJC 2009/P2/Q1a
23
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Transition Elements Chemistry - Suggested Solutions
YJC 2009/P3/Q1a,c,d
24
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183