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Chemistry Module 5 > 5.3.1 Transition elements > Flashcards

5.3.1 Transition elements Flashcards

(54 cards)

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1
Q

why do transition metal elements form certain characteristics

A

due to incomplete d sub-level in ions

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2
Q

characteristics of transition metals

A

formation of coloured ions
variable oxidation states
catalytic activity
complex ion formation

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3
Q

which elements aren’t classed as transition elements

A

zinc

scandium

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4
Q

why is zinc not classed as a transition metal

A

Zn can only form a 2+ ion
in this ion Zn2+ has a complete d orbital and so doesn’t meet the criteria of having an incomplete d orbital in one of its ions

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5
Q

why is scandium not classed as a transition metal

A

Sc can only form 3+ ion, in this ion Sc3+ has an empty d orbital so doesn’t meet the criteria of having an incomplete d orbital in one of its ions

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6
Q

3 typical properties of transition elements

A

existence of more than one oxidation state in its compounds
form coloured ions
behave as catalysts

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7
Q

existence of more than one oxidation state in its compounds

A

e.g. iron normally forms +2 and +3

chromium can form +2 +3 +6

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8
Q

form coloured ions

A

the colours can vary depending on the oxidation state

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9
Q

behaviour as catalysts

A

iron in haber process to make ammonia

manganese dioxide to decompose hydrogen peroxide

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10
Q

complex ion definition

A

central metal ion surrounded by ligands

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11
Q

ligand definition

A

an atom, ion or molecule which donate an electron pair

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12
Q

co-ordinate bonding

A

when shared pair of electrons in the covalent comes from only one of the bonding atoms

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13
Q

co-ordination number

A

number of co-ordinate bonds formed to a central metal ion

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14
Q

what can ligands be

A

unidentate
bidentate
multidentate

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15
Q

unidentate

A

form one co-ordinate bond per ligand

H2O, NH3 and Cl-

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16
Q

bidentate

A

two atoms with lone pairs and can form two co-ordinate bonds per ligand
NH2CH2CH2NH2 and ethanedioate ion C2O4 2-

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17
Q

multidentate

A

can form multiple bonds per ligand

EDTA4- can form 6

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18
Q

EDTA4-

A

6 donor sites (4O and 2N)

forms a 1:1 complex with metal (11) ions

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19
Q

what shape complexes do transition metals form with small ligands

A

octrahedral

with H2O and NH3

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20
Q

what shape complexes do transition metals form with larger ligands

A

tetrahedral

with Cl-

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21
Q

what shape is cisplatin

A

square planar

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22
Q

what shape does Ag+ normally form

23
Q

what types of stereoisomerism can complexes show

Study These Flashcards
A

cis-trans and optical

24
Q

cis isomerism

Study These Flashcards
A

normally known as Z isomers

on zee-zame-zide

25
trans isomerism
normally known as E isomers
26
complexes with 3 bidentate ligands
can form 2 optical isomers | non-superimposable mirror images
27
use of cisplatin
anticancer drug
28
how does cisplatin work
binds to the DNA of cancer cells and stops cancer cells dividing causes cell apoptosis
29
reactions with excess NH3
ligand substitution occurs with several transition aqueous ions [Cu(H2O)6]2+ + 4NH3 -> [Cu(NH3)4(H2O)2]2+ +4H2O blue to deep blue
30
why is ligand substitution with Cu incomplete?
as not all the water molecules are substituted
31
ligand exchange Co and Cr with excess NH3
NH3 and H2O are similar in size and are uncharged | so ligand exchange occurs without change of co-ordination number
32
ligand reactions with chloride ions
addition of a high conc of chloride ions (from a conc chlorine containing acid) to an aqueous ion leads to a ligand substitution reaction in Cu and Co changes coordination number from 6 to 4
33
why in reactions with chloride ions can co-ordinate number change
because the Cl- ligand is larger and charged compared to H2O and NH3
34
copper reaction with chloride ions
[Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O blue to yellow/green
35
cobalt reaction with chloride ions
[Co(H2O)6]2+ + 4Cl- -> [CoCl4]2- + 6H2O pink to blue
36
what is formed if solid metal chloride is dissolved in water
aqueous [METAL(H2O)6]2+ complex rather than the | chloride [METALCl4]2-
37
biological complexes
iron 2 ion in haemoglobin haem is an iron 2 complex with a multidentate ligand O2 bonds to Fe2+ ions in the haemoglobin CO is toxic to humans as it forms a strong coordinate bond with haemoglobin, stronger than with oxygen so prevents the binding
38
what are the ions involved in the precipitation reactions with sodium hydroxide and ammonia
``` Cu2+ Mn2+ Fe2+ Fe3+ Cr3+ ```
39
basic form of the full equation for ppt reactions with NaOH
[metal (H2O)6] ^metal charge + 2OH- -> metal (H2O)6- charge of metal(OH)charge of metal + charge of metal H2O
40
copper ppt reaction with NaOH
Cu2+ (aq) + 2OH- -> Cu(OH)2 (s) blue solution to blue ppt ppt is insoluble in excess sodium hydroxide
41
manganese ppt reaction with NaOH
Mn2+ (aq) + 2OH- (aq) ->Mn(OH)2 (s) very pale pink solution to pale brown ppt, darkens when in contact with air ppt is insoluble in excess sodium hydroxide
42
iron(II) ppt reaction with NaOH
Fe2+(aq) +2OH- (aq) ->Fe(OH)2 (s) pale green solution to green ppt ppt insoluble in excess but in contact with air will turn brown as Fe(II) is oxidised to Fe(III), orange-brown Fe(OH)2 (s) -> Fe(OH)3 (s)
43
iron (III) ppt reaction with NaOH
Fe3+ (aq) + 3OH- (aq) -> Fe(OH)3 (s) pale yellow to orange-brown ppt is insoluble in excess sodium hydroxide
44
chromium (III) reaction with NaOH
Cr3+ (aq) +3OH- (aq) -> Cr(OH)3 (s) violet to grey-green ppt is soluble in excess and will form dark green solution Cr(OH)3 (s) + 3OH- (aq) -> [Cr(OH)6]3+ (aq)
45
What happens to copper and chromium in excess ammonia
Their hydroxide ppts will dissolve
46
Chromium in excess ammonia
Cr(OH)3(H2O)3 (s) + 6NH3 (aq) -> [Cr(NH3)6]3+ (aq) +3H2O (l) + 3OH- (aq) Cr becomes [Cr(NH3)6]3+ purple solution
47
Copper in excess ammonia
Cu(OH)2(H2O)4 (s)+ 4NH3 (aq) -> [Cu(NH3)4(H2O)2 (aq) +2H2O (l) + 2OH- (aq)` Becomes deep blue
48
Which electrons do transition elements lose first
Their 4s before the 3D
49
Iron (II) oxidation
Easily oxidized to Fe3+ green -> brown, normally use potassium manganate (VII) but oxygen in air also works MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+ (aq) +4H2O (l) + 5Fe3+ (aq) Purple to colourless
50
Iron (III) reduction
Brown to green using potassium iodide Brown colour of iodine can mask the colour change 2FE3+ (aq) + 2I- (aq) -> I2 (aq) + 2Fe2+ (aq)
51
Reducing chromium
Cr3+ (green) and Cr2+ (blue) commonly formed by reduction of Cr2O7 2- (orange) by strong reducing agent zinc in HCl acid solution. Fe2+ is a less strong reducing agent so will only reducing dichromate to Cr3+ Cr2O7 2- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+ Orange to green
52
Oxidizing chromium
``` Alkaline chromium (III) can be oxidized using hydrogen peroxide, easier to oxidize a negative ion Chromium after being in excess NaOH ``` Cr(OH)6 3- (aq) —-(H2O2)—> CrO4 2- (aq) Green to yellow Reduction in this reaction: H2O2 + 2e- -> 2OH-
53
Reduction of Cu2+ to Cu+
Potassium iodide Copper (II) is blue 2Cu2+ (aq) + 4I- (aq) -> I2 (aq) + 2CuI (s) Brown solution and white ppt
54
Disproportionation of copper(I) ions
When reacting with sulfuric acid will disproportionate to Cu2+ and Cu metal 2Cu+ -> Cu + Cu2+ Full equation: Cu2O + H2SO4 -> CuSO4 + Cu + H2O

Chemistry Module 5 (7 decks)

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