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TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 1
A2 Level
REACTIONS OF PERIOD 3 ELEMENTS WITH WATER
a) (i) Sodium melts into a ball / heat is given out (1)Effervescence / gas evolved (1)Sodium skates across the surface (1)Ignition may occur (1)Maximum 3 marks
Quality of language: two or more sentences with correct spelling, punctuation and grammar in which themeaning is clear (1).
(ii) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H
2(g) (1)
(iii) 13-14 (1)
(iv) Redox / oxidation-reduction (1)
b) (i) Less reactive / less powerful reducing agents / higher ionisation energies / lower in electrochemical series /less –ve Eê values (1)
(ii) Pass steam (1)over the heated metal (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 2
A2 Level
REACTIONS OF PERIOD 3 ELEMENTS WITH OXYGEN
a)
Reaction conditions
Formula of oxide(or lower oxide)
Na Mg Al Si P S Cl
(6)
Ignite in air or O2
Heat in O2
Ignite in air or O2
(1) (1) (1)
Na2O MgO Al
2O
3SiO
2P
4O
6 or
P2O
3(½) (½) (½) (½)
SO2
(½)(½)
b) (i) 4Na + O2 → 2Na
2O (1)
(ii) 2Mg + O2 → 2MgO (1)
(iii) 2Al + 3O2 → 2Al
2O
3 (1)
(iv) Si + O2 → SiO
2 (1)
(v) 2P + 3O2 → 2P
2O
3 (1)
(vi) S + O2 → SO
2 (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 3
A2 Level
STRUCTURE AND BONDING OF PERIOD 3 OXIDES
a)
b) From Na2O → MgO → Al
2O
3 there is an increase in m.p. (1)
Reason: ionic bonding becomes stronger (1)(Or lattice enthalpy increases in magnitude (1))because cations decrease in size (½)and increase in charge (½)(Or because surface charge density / charge:radius ratio increases (1))
SiO2 has the highest m.p. (1)
because all / many strong covalent bonds must be broken for it to melt (1)
Remainder have low m.p. (1)because the only forces to be overcome on melting are relatively weak (1)intermolecuar forces / van der Waals’ forces (1)
c) Melting point of MgO is higher than that of CaO (1)
Reason: the ionic bonding in MgO is stronger than that in CaO (1)(Or the lattice enthalpy of MgO is greater than that of CaO (1))because Mg2+ is smaller than Ca2+ / Mg2+ has a higher charge density (1)
Structure
Bonding
Na2O MgO Al
2O
3SiO
2P
4O
10SO
2Cl
2O
Covalent(1)
Ionic(1)
Ionic lattice(1)
Macro-molecular
(1)
Simple molecular(1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 4
A2 Level
PERIOD 3 OXIDES WITH WATER
a) Oxides on LHS react / dissolve in water to give alkaline solutions / pH >7.0 (1)e.g. Na
2O / Li
2O (1)
Na2O(s) + H
2O(l) → 2NaOH(aq) (1)
Oxides further along a period are insoluble in water (1)e.g. BeO / MgO / Al
2O
3 (1)
Oxides on RHS give acidic solutions / pH < 7.0 (1)e.g. SO
2 / SO
3 / other acidic oxide (1)
SO2(g) + H
2O(l) → H
2SO
3(aq) (1)
b) Acidity increases with oxygen content / oxidation number (1)CO: no reaction with water (1)pH = 7 (1)
CO2(g) + H
2O(l) → H
2CO
3(aq) (1)
pH = 5-6 (1)
P4O
6(s) + 6H
2O(l) → 4H
3PO
3(aq) (1)
pH = 2-4 (1)
P4O
10(s) + 6H
2O(l) → 4H
3PO
4(aq) (1)
pH = 0-2 (1)
Note Examiners may accept equations without state symbols.
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 5
A2 Level
ACID - BASE CHARACTER OF PERIOD 3 OXIDES
a)
b) MgO(s) + H2SO
4(aq) → MgSO
4(aq) + H
2O(l) (1)
Al2O
3(s) + 6HCl(aq) → 2AlCl
3(aq) + 3H
2O(l) (1)
Al2O
3(s) + 2NaOH(aq) + 7H
2O(l) → 2Na[Al(OH)
4(H
2O)
2](aq)(1)
Accept Na[Al(OH)4]
SO3(s) + 2NaOH(aq) → Na
2SO
4(aq) + H
2O(l) (1)
Or SO3(s) + NaOH(aq) → NaHSO
4(aq) (1)
Accept similar equations for SO2(g)
(Throughout, accept similar equations for other acids and bases. Examiners may accept equations without state symbols)
c) MgO O2- + 2H+ → H2O(1)
O2- accepts protons (1)and is therefore a base on the BrØnsted-Lowry theory (1)
Or donates lone pairs of electrons to H+ (1)and is therefore a base on the G.N. Lewis theory (1)
SO3 SO3 + 2OH- → SO
42- + H
2O (1)
Or SO3 + OH- → HSO
4- (1)
SO3 accepts a lone pair of electrons from OH- (1)
and is therefore an acid on the G.N. Lewis theory (1)
(Accept similar explanation for SO3 + H
2O, or for SO
2 with either OH- or H
2O)
Acid - base character
Oxide Na2O MgO Al
2O
3SiO
2P
4O
6
P4O
10
SO2
SO3
Cl2O
ClO2
Cl2O
7
Basic (½) Basic (½) Amphoteric(1)
Acidic (½) Acidic (½) Acidic (½) Acidic (½)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 6
A2 Level
STABILITY OF GROUP 4 OXIDATION STATES
a) (i) +2 (1)+4 (1)Signs must be given.
(ii) More stable state of tin is +4 (1)More stable state of lead is +2 (1)
b) (i) SnCl2
Reducing agent (1)PbCl
2Neither (1)
SnO2
Neither (1)PbO
2Oxidising agent (1)
(ii) Accept SnCl2 with KMnO
4, K
2Cr
2O
7, FeCl
3, HgCl
2, C
6H
5NO
2, etc.
Correct reactant (1) Correct product (1)
Accept PbO2 with conc HCl(aq), Na
3[Cr(OH)
6] etc.
Correct reactant (1) Correct product (1)
c) +2 state increases in stability (1)because this is essentially an ionic state / M2+ ions are concerned (1)Ionisation energies decrease / ions are formed more easily on descending the group (1)
+4 state decreases in stability (1)because energy is needed to promote an ns electron to a vacant np orbital (1)and, for larger atoms, less energy is released when (weak) covalent bonds are formed (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 7
A2 Level
HYDROLYSIS OF GROUP 4 CHLORIDES
(ii) H2O cannot coordinate to the C atom / nucleophilic attack by H
2O cannot occur (1)
because C does not have a vacant orbital in its outer shell / 2d orbital to accept a lone pair (1)
b) (i) Observation A white suspension / cloudy solution (1)
Equation SnCl2(aq) + H
2O(l) SnCl(OH)(s) + HCl(aq) (1)
Prevention By adding hydrochloric acid (1)which disturbs equilibrium to the LHS (1)
(ii) PbCl2 has an ionic structure (1)
but SnCl2 is predominantly covalent (1)
With H2O acting as a nucleophile, substitution occurs (1)
Examiners may accept equations without state symbols.
SiCl
ClCl
Cl (1) OH
H
SiCl
ClCl
OH
(1)+ HCl
Repeated × 3
Si(OH)4 (1)SiO
2
-2H2O (1)
↔
a) (i) SiCl4(l) + 2H
2O(l) → SiO
2(aq)(s) + 4HCl(aq) (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 8
A2 Level
DEFINITIONS AND ELECTRONIC CONFIGURATIONS
a) A d-block element is one with its highest energy electron in a d-orbital (1)A transition element is one which can form one or more stable ions with a partially occupied d-subshell (1)
b) (i) Yes (1)Configuration is [Ar] 3d1 4s2 (1)
(ii) No (1)Its only cation has the configuration [Ar] 3d0 4s0 (1)
c) (i)
(ii) Unusual featureTheir atoms have only one 4s electron (1)while atoms of other transition elements have two 4s electrons (1)
Reason for copperConfiguration as shown rather than [Ar] 3d9 4s2 (1)because of exceptional stability associated with a fully occupied 3d subshell (1)
Reason for chromiumConfiguration as shown rather than [Ar] 3d4 4s2 (1)because there is no mutual electrical repulsion in any orbital and the electron spin is parallel/ [Ar] 3d5 4s1 is a more stable electronic configuration (1)
d) (i) Iron(III) ion (1)Manganese(II) ion (1)Accept Fe3+ & Mn2+
(ii) Half-filled 3d subshell (1)
[Ar]
4s3d 4p
[Ar]
[Ar]
[Ar]
Cu
Cu2+
Cr
Cr3+
(1)
(1)
(1)
(1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 9
A2 Level
BONDING IN COMPLEX IONS
a) (i) A molecule or anion (1)which can donate one or more lone pairs of electrons to a cation (1)
(ii) Complex cation Species with a positive charge (1)formed from ligands bonded to a central cation (1)
Example [Ag(NH3)
2]+ or any other (1)
Complex anion Species with a negative charge (1)formed from ligands bonded to a central cation (1)
Example [Fe(CN)6]4- or any other (1)
b) (i)
III
H2O
H2O OH
2
Cr
OH2
H2O
H2O
3+
(2) Deduct 1 mark if charge is missing
Octahedral shape (1)
(ii) Coordinate / dative covalent (1)
(iii) Lone pair of electrons on the O atom of a H2O molecule (1)
Cr3+ ion has vacant orbitals which can accept 6 pairs of electrons (1)
[Ar]
4s3d 4p
(1) for unpaired electrons(1) for electron pairs(1) for labels
Electronspossessed byCr3+ ion.
Electrons originating from the ligands
(iv)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 10
A2 Level
c) (i)
Ti4+ [Ar]
4s3d 4p
Zn2+[Ar]
Ti2+ [Ar]
Ti4+ [Ar]
Cu+ [Ar]
4s3d 4p
(1)
(1)
(1)
(1)
(ii) Only Ti2+ (1)
(iii) Colour results from the absorption of one or more frequencies of white light (1)in promoting d-electrons from a low energy orbital to a higher energy orbital (1)Not possible if the 3d subshell is empty / rules out Ti4+ (1)Not possible if the 3d subshell is full / rules out Cu+ & Zn2+ (1)Only Ti2+ has a partially filled d-subshell (1)Maximum 4 marks
COLOUR OF COMPLEX IONS
a)ION Cr3+(aq) Mn2+(aq) Fe2+(aq) Fe3+(aq) Cu2+(aq)
COLOUR green (pale) pink (pale) green yellow blue or purple or brown
b) [Cu(H2O)
6]2+ → [CuCl
4]2- or other example
Blue Yellow or green (1) for formulae (1) for colours
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 11
A2 Level
a) Test 1Reagent BaCl
2(aq) / Ba(NO
3)
2(aq) (1)
Observation with A White (½) precipitate (½)Observation with B No precipitate / solution remains clear (1)
Test 2Reagent AgNO
3(aq) (1)
Observation with A No precipitate / solution remains clear (1)Observation with B Cream / off-white / pale yellow (½) precipitate (½)
b) (i) A [Pt(NH3)
6]4+ B [PtCl(NH
3)
5]3+
C [PtCl2(NH
3)
4]2+ D [PtCl
3(NH
3)
3]+
Accept cis or trans isomer Accept fac or mer isomer
(ii) Dissolve equal masses / amounts of each in water (1)Either Titrate a suitable volume of each with AgNO
3(aq) (1)
Titres for A, B, C and D are in the ratio 4 : 3 : 2 : 1 (1)
Or Add excess AgNO3(aq) to each, then filter, wash, dry and weigh precipitates (1)
Masses for A, B, C and D are in the ratio 4 : 3 : 2 : 1 (1)
IVPt
NH3
4+
(1)
NH3
H3N
H3N
NH3
NH3
IVPt
NH3
3+
(1)
NH3
H3N
Cl
NH3
NH3
IVPt
Cl
2+
(1)
NH3
H3N
Cl
NH3
NH3
IVPt
Cl
+
(1)NH3
Cl
NH3
NH3
Cl
c) (i)
(ii)
IINi
Cl (1)
NH3
H3N
Cl
trans (½) cis (½)
cis (½)
IINi
Cl (1)
NH3
Cl
NH3
Cl
IIICo
Cl
+
(1)
NH3
NH3
H3N
H3N Cl
IIICo
+
(1)
NH3
NH3
H3N Cl
IIICo
+
(1)
NH3
NH3
H3N
ClNH
3
NH3
trans (½)
ISOMERISM
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 12
A2 Level
LIGAND EXCHANGE REACTIONS
a) Definition A reaction in which one or more coordinated water molecules (1)is replaced / substituted by other ligands (1)
Reason 1 Relatively strong ligands tend to displace weaker ones / water is a weak ligand (1)
Reason 2 Ligand exchange is reversible / if a ligand is added in high concentration, equilibrium will be disturbed toRHS (1)
Quality of language:two or more sentences with correct spelling, punctuation and grammar in which the meaning is clear (1).
b) (i) Ionic equation [Cu(H2O)
6]2+(aq) + 6NH
3(aq) → [Cu(NH
3)
6]2+(aq) + 6H
2O(l) (1)
Colour change From blue (½) to royal blue/purple (½)
Shape of resulting complex ion Octahedral (1)
(ii) Ionic equation [Cu(H2O)
6]2+(aq) + 4Cl-(aq) → [CoCl
4]2-(aq) + 6H
2O(l) (1)
Colour change From blue (½) to green/yellow (½)
Shape of resulting complex ion Tetrahedral (1)
c) If ligands are relatively large (1)fewer of them can be accommodated around a cation (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 13
A2 Level
DEPROTONATION THEORY
a) (i) pH decreases from 7.0 (1)
(ii) Reasons A proton (1)is transferred from [Fe(H
2O)
6]3+ / H
2O ligand (1)
to a molecule of solvent water (1)Hence [H
3O+] > [OH-] (1)
Maximum 3 marks
Equation [Fe(H2O)
6]3+(aq) + H
2O(l) ¾ [Fe(OH)(H
2O)
5]2+(aq) + H
3O+(aq) (1)
(Examiners may accept equation without state symbols)
(iii) Type of reaction Deprotonation / acid-base (1)Function of hydrated metal ions Acid (1)Function of water molecules Base (1)
b) (i) [Fe(OH)(H2O)
5]2+(aq) + H
2O(l) ¾ [Fe(OH)
2(H
2O)
4]+(aq) + H
3O+(aq) (1)
[Fe(OH)2(H
2O)
4]+(aq) + H
2O(l) ¾ [Fe(OH)
3(H
2O)
3](s) + H
3O+(aq) (1)
(Examiners may accept equations without state symbols)
(ii) OH- is a stronger base than H2O (1)
c) (i) Solution is acidic (1)[Cu(H
2O)
6]2+ behaves in a similar manner to [Fe(H
2O)
6]3+ (1)
(ii) FeCl3 has the lower pH (1)
Fe3+ has a higher charge than Cu2+ (1) and is smaller (1)Or Fe3+ has a higher charge density than Cu2+ (2)
Hence Fe3+ increases polarisation of O⎯H bonds (1)which weakens these bonds / facilitates loss of a proton (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 14
A2 Level
REACTIONS OF CATIONS WITH OH- AND NH3
a) (i) Both OH- and NH3 are bases (1)
(ii) Green precipitate (½) Cr(OH)3 (½)
White precipitate (½) Mn(OH)2 (½)
Green precipitate (½) Fe(OH)2 (½)
Brown precipitate (½) Fe(OH)3 (½)
Blue precipitate (½) Co(OH)2 (½)
Green precipitate (½) Ni(OH)2 (½)
Blue precipitate (½) Cu(OH)2 (½)
Any 3Accept full formulae
c) (i) Aluminium hydroxide (1)Lead(II) hydroxide (1)Tin(II) hydroxide(1)Tin(IV) hydroxide (1)Beryllium hydroxide (1)Any 2
(ii) [Al(OH)4(H
2O)
2]- (1)
[Pb(OH)4]2- (1)
[Sn(OH)4]2- (1)
[Sn(OH)6]2- (1)
[Be(OH)4]2- (1)
Any 2
b)
Co(OH)2 / Ni(OH)
2 / Cu(OH)
2
(Accept Cr(OH)3) (1)
NH3 is a ligand /
ammine formation (1)
Ligand exchange / substitution (1)
CoII – pale brown solution /NiII – blue solution /CuII – dark blue solution (1)
[Co(NH3)
6]2+/
[Ni(NH3)
6]2+ /
Cu(NH3)
4(H
2O)
2]2+ (1)
Cr(OH)3 (1)
Cr(OH)3 is amphoteric (1)
Deprotonation / acid-base (1)
Green solution (1)
[Cr(OH)4(H
2O)
2]- /
[Cr(OH)5(H
2O)]2- /
[Cr(OH)6]3- (1)
Example of a compound which will dissolve
Reason for dissolving
Type of reaction occurring
Observation
Formula of the ion produced
Excess NH3 (aq) Excess dilute NaOH(aq)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 15
A2 Level
a) (i) +6
(ii) +6
(iii) +3
(iv) +5
(v) +5
(vi) +3
(vii) +4
(viii) +3
(ix) +6
(x) +2
(½) each x 10 = 5 marks
b) Energy levels of 4s and 3d electrons are very similar (1)hence 3d as well as 4s electrons can be used in bonding (1)Although more energy is required to remove more electrons (1)this is often compensated by the release of latttice enthalpy (1)or hydration enthalpy (1)Maximum 3 marks
OXIDATION STATES
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 16
A2 Level
COLORIMETRY
a) (i) Volume of 0.1 M CuSO4(aq) 20 cm3 (½)
Volume of 0.1 M NH3(aq) 80 cm3 (½)
(ii) n (Cu2+) = 0.1 x 20/1000 = 0.002 mol (1)n (NH
3) = 0.1 x 80/1000 = 0.008 mol (1)
∴ mole ratio Cu2+ : NH3 = 1.4 (1)
(iii) [Cu(NH3)
4(H
2O)
2]2+ (1)
Or [Cu(NH3)
4]2+ (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 17
A2 Level
LIGAND EXCHANGE AND STABILITY CONSTANTS
(a) (i) Yellow colour is a complex with ammonia (1) because the ammonia is a more reactive ligand than water (notsulfate)/ the ammonia complex is more stable the aqua complex (1).
Blue colour is a cyanide complex (1) because the cyanide is a more reactive ligand than ammonia / the cyanidecomplex is more stable the ammonia complex (1).
No further reaction (1) because ammonia is a less reactive ligand than the cyanide (1).
Allow 1 mark for well constructed answer and use of three terms like: ligand, reactive, powerful,displacement, complex, stable, aqua.
(ii) Mauve colour is an ethane-1,2-diamine complex (1) because the ethane-1,2-diamine is a more reactive ligandthan ammonia/ the ethane-1,2-diamine complex is more stable the ammonia complex (1).
No further reaction with the yellow cyanide complex (1) because ethane-1,2-diamine is a less reactive ligand thanthe cyanide (1).
(b) (i) No (1) the complex with ethane-1,2-diamine has a larger stability constant that the EDTA complex (1).
(ii) No (1) the complex with cyanide has a larger stability constant that the ammonia complex (1).
(iii) [Co(CN)6]3- (1) The cyanide is the more stable complex and is formed in preference to ammonia complex (1)
All the [Co(H2O)
6]3+ has reacted with the cyanide (1)
(iv) [Co(NH3)]3+ (1) The [Co(NH
3)]3+ is the more stable complex and is formed in preference to [Co(NH
3)]2+
complex (1)All the ammonia has reacted with the [Co(H
2O)
6]3+ (1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 18
A2 Level
d-ORBITAL SPLITTING AND LIGHT ABSORPTION
a) (i)
Adding ligands Absorbing light
dx2y2 dx2 dx2y2 dx2
dxy dxz dyz dxy dxz dyzGround state Excited state
EnergyCu2+ uncomplexed
(ii)
Adding ligands Absorbing light
dx2y2 dx2 dx2y2 dx2
dxy dxz dyz dxy dxz dyzGround state Excited state
EnergyFe3+ uncomplexed
(iii)
Adding ligands Absorbing light
dx2y2 dx2 dx2y2 dx2
dxy dxz dyz dxy dxz dyzGround state
EnergyV3+ uncomplexed
Excited state
-1 mark for every mistake. Maximum 6 marks.
b) (i) No electrons in the 3d orbitals (1) No electrons available for promotion (1)
(ii) All 3 d orbitals are full (1) No empty higher level orbital for an electron to be promoted to(1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 19
A2 Level
VANADIUM CHEMISTRY I
a)
b) Oxidation Number Formula Colour Name+2 [V(H
2O)
6]2+ (1) (Violet) (Vanadium(II))
+3 ([V(H2O)
6]3+) Green (1) (Vanadium(III))
+4 VO2+(aq) (1) Blue (1) Oxovanadium(IV) (1)+5 VO
2+(aq) (1) Orange / yellow (1) Dioxovanadium(V) (1)
c) Orange / yellow (1) → green (1) → blue (1) → green (1) → violet / lavender (1)
V [Ar]
V3+ [Ar]
4s3d 4p
(1)
(1)
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 20
A2 Level
VANADIUM CHEMISTRY II
a) Shape Octahedral (1)
Types of bonding Covalent (within the ligand) (1)and dative covalent / coordinate (between central metal and ligand) (1)
b) (i) Prediction Acidic (1)
Equation [V(H2O)
6]3+ + H
2O ¾ [V(OH)(H
2O)
5]2+ + H
3O+ (1)
(ii) Identity Vanadium(III) hydroxide / V(OH)3 / [V(OH)
3(H
2O)
3] (1)
Equation [V(H2O)
6]3+ + 3OH- → [V(OH)
3(H
2O)
3] + 3H
2O (1)
c) (i) VO2+(aq) + 2H+(aq) + e- → VO2+(aq) + H
2O(l) (2)
(ii) SO32-(aq) + H
2O(l) → SO
42-(aq) + 2H+(aq) + 2e- (2)
(iii) 2VO2
+(aq) + SO3
2-(aq) + 2H+(aq) → 2VO2+(aq) + SO4
2-(aq) + H2O(l) (2)
(iv) KMnO4 / K
2Cr
2O
7 / any other oxidising agent with Eê > + 1.0 V (1)
In the above equations, award (1) for formulae and (1) for balance.Examiners may accept equations without state symbols.
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 21
A2 Level
CHROMIUM CHEMISTRY I
a) Equation [Cr(H2O)
6]3+ + H
2O ¾ [Cr(OH)(H
2O)
5]2+ + H
3O+ (1)
Explanation The solution contains [Cr(H2O)
6]3+ ions which are purple / violet / blue (1)
and [Cr(OH)(H2O)
5]2+ ions which are green (1)
b) Observation Green / grey-green (½) precipitate (½)which dissolves / is soluble in excess NaOH(aq) (1)to give a green (½) solution (½)
Equations [Cr(H2O)
6]3+
+ 3OH- → [Cr(OH)
3(H
2O)
3] + 3H
2O (1)
[Cr(OH)3(H
2O)
3] + OH- → [Cr(OH)
4(H
2O)
2]- + H
2O (1)
Accept equations for the formation of [Cr(OH)5(H
2O)]2- or [Cr(OH)
6]3-
Examiners may require state symbols of complexes, e.g. [Cr(OH)4(H
2O)
2]-(aq)
c) Observation From green / blue green (1) to blue (1)
Half-equation [Cr(H2O)
6]3+ + e- → [Cr(H
2O)
6]2+
Or Cr3+(aq) + e- → Cr2+(aq) (1)
d) (i) Chromate(VI) ion (1)
(ii) Name Dichromate(VI) ion (1)
Equation 2 CrO4
2-(aq) + 2H+(aq) → Cr2O
72-(aq) + H
2O(l) (2)
Award (1) for formulae and (1) for balance.Examiners may accept an equation without state symbols.
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 22
A2 Level
CHROMIUM CHEMISTRY II
a) Cr Cl H2Om/100g 19.5 40.1 40.4n 19.5/52 40.1/35.5 40.4/18 (1)
= 0.375 1.13 2.24 (1)ratio 1 3 6∴ empirical formula = CrCl3(H2O)6 (1)
b) X = [Cr(H2O)
6]3+ (1)
Hexaaquachromium(III) ion (1)
Y = [CrCl2(H
2O)
4]+ (1)
Tetraaquadichlorochromium(III) ion (1)
c) The cations have different ligands attached to them (1)causing different degrees of splitting of energy levels of d orbitals (1)
d) In Y, the chloride ions / chlorine atoms can be close to each other (cis isomer) (1)or across the structure from each other (trans isomer) (1)
e) (i) [CrCl(H2O)
5]2+ (Cl-)
2 (1) Ionic charges are not essential
(ii) Two (1)
H2O
H2O Cl
Cr
OH2
Cl
H2O
III
H2O
H2O OH
2
Cr
ClCl
H2O
III
+ +
(1) (1)
cis trans
TOPIC 21 ANSWERS & MARK SCHEMES
QUESTIONSHEET 23
A2 Level
COBALT CHEMISTRY I
a) Transition element An element which forms one or more stable ions (1)with incompletely filled d-subshells (1)
Example Co2+ [Ar] 3d7 / Co3+ [Ar] 3d6 (1)
Cationic complex Species with an overall +ve charge (1)formed by coordination of ligands to a central cation (1)
Example [Co(H2O)
6]2+ / [Co(NH
3)
6]2+ / [Co(NH
3)
6]3+ (1)
Anionic complex Species with an overall –ve charge (1)formed by coordination of ligands to a central cation (1)
Example [CoCl4]2- / [Co(OH)
4]2- (1)
b)
c) Type of bonding Coordinate / dative covalent (1)
Feature Lone pair of electrons (1)
d) (i) Blue (½) precipitate (½)which dissolves in / is soluble in excess conc. NH
3(aq) (1)
to give a pale brown / red-brown (½) solution (½)
(ii) Replacement / substitution (1)of one type of ligand by another (1)[Co(H
2O)
6]2+ + 6NH
3 → [Co(NH
3)
6]2+ + 6H
2O (1)
e) Observation Solution becomes dark brown (1)
Explanation Oxidation of CoII to CoIII / formation of [Co(NH3)
6]3+ (1)
H2O
H2O OH
2
Co
OH2
H2O
H2O
II
2+
(1)
Co
Cl
Cl Cl
○
○
○Cl
12121212
II
2-
(1)