Topic 9.4: Transition Elements...

Transition Elements Chemistry - Suggested Solutions

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Topic 9.4: Transition Elements Chemistry

AJC 2009/P3/Q3c,d 1


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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) Cr S N

Mols 0.298 1.19 2.09 Ratio 1 4 7 x + y + 1 = 7

x: 4 y: 2

(ii) +3

(iii)

Cr(III)

NH3

NH3

NCS

NCSSCN

SCN

-

(c)

(i) K2Cr2O7 + 2 HBr →→→→ 2 KCrO3Br + H2O

(ii) 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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(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 Cr2O7

2-) 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 property (i) silicon (IV) oxide hardness (ii) lead (IV) oxide 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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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: CrO4

2– or Na2CrO4 E:

Cr

NH2

H2N NH2

NH2

H2N

H2N

3+

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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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 1s2 2s2 2p6 3s2 3p6 3d6 4s2 13Al 1s2 2s2 2p6 3s2 3p1

26Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5 13Al3+ 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)

N2 +

NH3

∆H

Ea

Ea (catalysed)

En

erg

y /

kJ

Reaction


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IJC 2009/P2/Q2c 6

IJC 2009/P3/Q1e 7


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IJC 2009/P3/Q2d 8

IJC 2009/P3/Q3b 9

JJC 2009/P2/Q2a,b

10 (a)

(bi)

(ii)

Fe(H2O)63+ ==== Fe(H2O)5(OH)2+ + H+

1s22s22p63s23p63d10

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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Mole 1.16 2.26

Ratio 1 2

Empirical Formula: MnO2 [1]

(ii) Identity of Z: MnO4−−−− [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 � [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 � Cu + 0.52 V

Cu2+ + e � Cu+ + 0.15 V

Eθ = + 0.52 – (+ 0.15) = + 0.37 V > 0 (feasible)


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PJC 2009/P3/Q5a,b 15


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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) → SO4

2− (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

= 5 x 33.00

1000x 0.0200 = 3.300 x 10−3 mol

Mass of vanadium in sample = 3.300 x 10−3 x Ar of V = 3.300 x 10−3 x 50.9 = 0.1680g

Percentage by mass of V = 0.1680

0.300x 100 = 56.0%

SAJC 2009/P3/Q2b

17

2Co3+

+ 2I- 2Co

2+

+ I2

E0

= + 1.82 – 0.54 = +1.28 V > 0

2Co2+

+ S2O8

2- 2Co

3+ + 2SO4

2-

From data booklet,

Co3+

+ e Co2+

E0

= +1.82 V

I2 + 2e 2I- E

0 = + 0.54 V

S2O82-

+ 2e 2SO42-

E0

= + 2.01V


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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 18 dative / coordinate/covalent bond

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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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.

(b) (i) • Empirical formula of A = MnO


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(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) Cr2O7

2- + 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

Eθ

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 (d-d 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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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: Mn3+

+ e–

= Mn2+

E� = +1.49 V

MnO4

–

+ 8H+

+ 5e–

= Mn2+

+ H2O E�

= +1.52 V

Overall equation for step 1: MnO4

–

+ 8H+

+ 4Mn2+

→ 5Mn3+

+ 4H2O

E�cell

= +1.52 – 1.49 = +0.03 V > 0 AND reaction is feasible.

Step 2

Relevant data: 2CO2

+ 2e–

= C2O

4

2–

E� = –0.49 V

Mn3+

+ e–

= Mn2+

E� = +1.49 V

Overall equation for step 2: 2Mn3+

+ C2O

4

2–

→ 2Mn2+

+ 2CO2

E�cell

= +1.49 – (–0.49) = +1.98 V > 0 AND reaction is feasible.

YJC 2009/P2/Q1a 23


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YJC 2009/P3/Q1a,c,d 24


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Topic 9.4: Transition Elements...

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