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

TOPIC 15 - TRANSITION METALS Flashcards

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
how colour emitted
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
shape of complex | octahedral
``` 6 ligands e- donated 12 bond angle 90 [Co{NH3)6]2+ Mn[OH2] 2+ Al[OH] 3- Fe[oh2(h20)4] ```
27
tetrahedral
4 ligands 8 electrons donated 109.5 [CuCl4]2-
28
linear
2 ligands 4 e- donated 180 [Ag(NH3)2]+
29
ions and complexes are colourless
no available electrons to excite and move around cannot absorb light = colourless
30
square planar complexes
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
monodentate ligand
can form 1 coordiante bond per ligand
32
bidentate
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
multidentate
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
colour change arises in changes of
oxidation state co-ordinate no ligand
35
changing ligand or co-ordinate no will
change the energy split btw d-orbitals and change the freq of light absorbed
36
chelate effect in terms of entropy
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
Zn in acidic conditions will reduce VO2+ all the way to V2+
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
colour of VO2 +
yellow
39
colour of V02+
blue
40
colour of V3+
green. tin metal Sn reduces VO2+ to V3+
41
colour of V2+
violet
42
Ecell value shows that the reduction of V
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
[Ag(NH3)2)]+ | linear
tollens reagent to test for aldehyde/ketones
44
Cr2O7 2- (orange) reduced with zinc in acid conditions as Fe is less strong reducing agent and only reduce to Cr3+
to Cr3+(green) and Cr2+ (blue)
45
Cr 2+ can form a ligand with
sodium ethanoate - red precip
46
Cr3+ can be oxidised by hydrogen peroxide in alkaline conditions into
Cr2O7 2-
47
2CrO4 2- by acidification can turn into
Cr2O7 2- equilibrium add 2H+ = h20 Acidification cause equil shift to the right so Cr2O7 2- increase
48
amphoteric metal hydroxides | Cr(OH)3(H2O)3
dissolve in acid - accept protons | + 3H+ -> [Cr(H2O)6] 3+
49
Cr(OH)3(H2O)3 with base
+ 3OH- --> [Cr(OH)6]3- + 3H2O
50
enthalpy change for ligand subst
very small - close to 0 as bonds formed are similar to bonds broken
51
advantages of catalysts for reactions
- proceed at lower temp and press | - saves energy and valuable resources
52
heterogenous catalysts | haber process
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
catalytic converters
adsorption use 3d 4s e- to make bonds/ correct orientation/ proximity/ weaken/ desorb from active site Pt/ Rd/Pd CO + NO ----> N2 + CO2
54
advantage of using heterogeneous catalyst
no need for seperatin of products from catalyst
55
good properties of catalysts
- cant adsorb too strongly | - cant adsorb too weakly not held on for long enogh
56
increase efficiency of heterogenous catalysts
- increase SA = inc no of active site | - spread onto inert medium
57
catalyst poisoning
poisoned by impurities (S) which do not desorb and cause block of active site - increase chemical costs
58
homogenous catalyst
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
autocatalysis
when product of a reaction is also a catalyst for that reaction
60
graph for autocatalysed reaction
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
autocatalysis by Mn2+ in titrations of C2O4 2- with Mno4 -
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
monitor concen of MnO4 -
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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