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Edexcel Chemistry A LEVEL - RM > TOPIC 15 - TRANSITION METALS > Flashcards

TOPIC 15 - TRANSITION METALS Flashcards

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1
Q

Transition metal

A

is a d-block element that has an incomplete d-subshell as a stale ion

found in btw Gp2 and 3
- lose e- from 4s first

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2
Q

properties of transition metals

A
  • have different oxidative states
  • form complexes
  • coloured compounds
  • acts as catalysts - good catalyst as they exist in variable oxid stats so provide alternative pathways so Gp1 etc not good catalysts
  • as well as other metal porperties
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3
Q

why is Sc not a transition metal element?

A

Sc3+ as a stable ion it’s d block subshell is empty and it only forms Sc3+ ion

Sc - 4s2,3d1
Sc 3+ - 3p6

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4
Q

why is Zn not a transition metal element?

A

Zn only forms Zn2+ and as it’s stable ion it’s d-subshell is completely filled
- 3d10
zn - [Ar] 4s0,3d10

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5
Q

why do transition metals have variable oxidation states?

A

4s and 3d orbitals are really close in energy levels - is possible for electrons to be lost from both orbitals easily so remaining electrons can form stable configurations

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6
Q

exceptional e- config

A

Cr - 24 as a ion 4s1 3d5
Cu - 29 as a ion 4s1 3d10

5 subshells in d = total 3d10

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7
Q

what is a complex ion

A

a transition metal ion bonded to 1/ more ligands by co-ordinate bonds/dative covalent bonds

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8
Q

ligand def

A

molecule or an ion that can donate a pair of electrons to the transition metal ion to form a co-ordinate bond

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9
Q

co-ordinate number

A

the total number of c-ordinate bonds formed between a central metal ion and its ligands

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10
Q

solid wedge

A

bonds coming out of the plane of the paper

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11
Q

hatched wedge

A

bonds going behind the plane of the paper

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12
Q

ligands and their charge

A 
H2O = 0   aqua
-OH = -1     hydroxo
NH3 = 0    ammine
Cl - = -1       chloro
Cl- ligand is much larger and change in coord no
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13
Q

hexaaquairon(II)

A

[Fe(h2o)6] 2+

coordinate no = 6
6 dative covalent bonds

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14
Q

tetrachloroferrate (III)

A

[Fecl4]-

depends on size of ligand that determine the coordinate no

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15
Q

[Cu(NH3)4(H20)2] 2+

A

tetraammineaquacopper (II)

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16
Q

why is partially filled d-orbital in transition element is responsible for the colour. (e- occupied)

A

when white light passing through sol. containing transition metal ions some wavlengths of visible light from EM spec are absorbed
colour observed is a mix of wavelength of light that have not been absorbed

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17
Q

complementary colours

A

absorbs red - look blue/ green

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18
Q

increase in frequency absorbed violet/indigo

A

the lower the wavelength absorbed 400nm

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19
Q

sol. CuSO4 appears blue

A

because sol absorbs red/organge region of EM and refelcts/transits blue colour

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20
Q

Sc(III) 4s23d0

A

is colourless in sol. as there is no partially filled d-orbital for the colour

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21
Q

uses of iron

A

vehicle bodies to reinforce concrete

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22
Q

uses titanium

A

jet engine parts

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23
Q

uses of copper

A

water pipes

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24
Q

example of catalysts in transition metals

A

iron - harber process
vanadium oxide - contact process
MnO2 - decompo H2O2

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25
Q

how colour emitted

A

movement of lower energy e- to higher energy level = promotion
- the amount of energy it absorbs depends on the difference in energy btw the two levels - bigger diff, more energy ab

energy gained by electron directly prop to freq of absorbed light and inversely prop of the wavelength

26
Q

shape of complex

octahedral

A 
6 ligands
e- donated 12
bond angle 90
[Co{NH3)6]2+
Mn[OH2] 2+
Al[OH] 3-
Fe[oh2(h20)4]
27
Q

tetrahedral

A

4 ligands
8 electrons donated
109.5
[CuCl4]2-

28
Q

linear

A

2 ligands
4 e- donated
180
[Ag(NH3)2]+

29
Q

ions and complexes are colourless

A

no available electrons to excite and move around cannot absorb light = colourless

30
Q

square planar complexes

A

Pt and Ni bond angle 90 only 4 coordinate bonds
- cisplatin isomer = Ptcl2nh32 - side effcts so given small amounts and cure cancer therapy not given as a single isomer and not in a mixture with trans form

31
Q

monodentate ligand

A

can form 1 coordiante bond per ligand

32
Q

bidentate

A

have 2 atoms with lobe pairs and can form 2 coordinate bonds per ligand
e.g. Cr(C2O4)3]3- ethanedioate
Cr(NH2CH2CH2NH2)3]3+

33
Q

multidentate

A

EDTA4- can form 6 coordinate bonds per ligand
4O and 2N to donate
- haemoglobin - o2 transport
CO can form strong coordinate bond with Hb and replace O2

34
Q

colour change arises in changes of

A

oxidation state
co-ordinate no
ligand

35
Q

changing ligand or co-ordinate no will

A

change the energy split btw d-orbitals and change the freq of light absorbed

36
Q

chelate effect in terms of entropy

A

multidentate ligands are favoured as they have greater no. of coordinate binds per ligand

EDTA displaces ligands that form fewer coordinate bonds per molecule so significant increase in entropy - gibbs free energy <0 so reaction is feasible so more stable complex ion is formed

37
Q

Zn in acidic conditions will reduce VO2+ all the way to V2+

A

alkaline conditions are required for ions to be oxidised so thats the role of pH

  • Zn has more negative electrode pot than V half eq Zn2+ + 2e- = Zn E=-0.76v
38
Q

colour of VO2 +

A

yellow

39
Q

colour of V02+

A

blue

40
Q

colour of V3+

A

green. tin metal Sn reduces VO2+ to V3+

41
Q

colour of V2+

A

violet

42
Q

Ecell value shows that the reduction of V

A

becomes less favourable as the oxid state of V decreases to the point where V2+ to V3+ oxidation is more favourable than the other way around as it is more +ve compared to -ve

43
Q

[Ag(NH3)2)]+

linear

A

tollens reagent to test for aldehyde/ketones

44
Q

Cr2O7 2- (orange) reduced with zinc in acid conditions as Fe is less strong reducing agent and only reduce to Cr3+

A

to Cr3+(green) and Cr2+ (blue)

45
Q

Cr 2+ can form a ligand with

A

sodium ethanoate - red precip

46
Q

Cr3+ can be oxidised by hydrogen peroxide in alkaline conditions into

A

Cr2O7 2-

47
Q

2CrO4 2- by acidification can turn into

A

Cr2O7 2-
equilibrium add 2H+ = h20

Acidification cause equil shift to the right so Cr2O7 2- increase

48
Q

amphoteric metal hydroxides

Cr(OH)3(H2O)3

A

dissolve in acid - accept protons

+ 3H+ -> [Cr(H2O)6] 3+

49
Q

Cr(OH)3(H2O)3 with base

A

+ 3OH- –> [Cr(OH)6]3- + 3H2O

50
Q

enthalpy change for ligand subst

A

very small - close to 0 as bonds formed are similar to bonds broken

51
Q

advantages of catalysts for reactions

A
  • proceed at lower temp and press

- saves energy and valuable resources

52
Q

heterogenous catalysts

haber process

A

e- are ransferred to produce reactive intermediate and speed up rate of reaction
e.g. contact process V2O5 (reduced from V5+ to V4+ ) convert SO2 -> SO3

53
Q

catalytic converters

A

adsorption use 3d 4s e- to make bonds/ correct orientation/ proximity/ weaken/ desorb from active site
Pt/ Rd/Pd
CO + NO —-> N2 + CO2

54
Q

advantage of using heterogeneous catalyst

A

no need for seperatin of products from catalyst

55
Q

good properties of catalysts

A
  • cant adsorb too strongly

- cant adsorb too weakly not held on for long enogh

56
Q

increase efficiency of heterogenous catalysts

A
  • increase SA = inc no of active site

- spread onto inert medium

57
Q

catalyst poisoning

A

poisoned by impurities (S) which do not desorb and cause block of active site
- increase chemical costs

58
Q

homogenous catalyst

A

S2O8 2- + I- where Fe2+ used as catalyst and these negative ions would repel naturally and never react so high Ea which overcome

  • form intermediates - low Ea
  • homogenous cat its e- pot must lie btw the e- pot of the 2 reactants so reduce reactant with more positive e- pot

form 2SO4 2- + I2
STAGE1 => 2SO42- + 2Fe3+
STAGE2 => 2Fe2+ + I2

59
Q

autocatalysis

A

when product of a reaction is also a catalyst for that reaction

60
Q

graph for autocatalysed reaction

A

intiallyat concen y-axis the is slow and uncatalysed not as much had been formed. rate inc as catalyst is made ; catalysed react faster and slows down as reactants are used up (MnO4 -) drops
so opposite sigmail curve
- high Ea due to 2 -ve ions

61
Q

autocatalysis by Mn2+ in titrations of C2O4 2- with Mno4 -

A

2 MnO4- + 5 C2O42- + 16H+ –> 2Mn2+ + 10 CO2 + 8 H2O
Catalysed alternative route
Step 1 4Mn2+ + MnO4- + 8 H+ –> 5Mn3+ + 4 H2O
Step 2 2Mn3+ + C2O4 2- —> 2Mn2+ + 2 CO2

62
Q

monitor concen of MnO4 -

A

using colorimeter measure the intensity of purple

- quicker determination of concen as it does not disrupt the reaction mixture

Edexcel Chemistry A LEVEL - RM (12 decks)

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