Transition metals Flashcards

(57 cards)

1
Q

What are D block elements

A

Elements found in the middle of the periodic table

Most are known as transition elements

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2
Q

Electron configuration for chroniuM?

A

Has one electron in each orbital of the 3d subshell and one is 4s

Adds stability

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3
Q

Copper electron configuration

A

FULL 3d subshell, and 1 electron in 4s

Adds stability

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4
Q

How to form ions- TM

A

S electrons are removed first

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5
Q

What is the definition of a TM

A

D block elements that have an ion with an incomplete d-sub shell

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6
Q

What are the 2 exceptions of d block elements that are not transition metals

A

Scandium and zinc

DON’T form stable ions with incomplete d sub shells

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7
Q

Why is scandium not a transition metal?

A

Only forms one ion- Sc3+ , which has an empty d subshell

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8
Q

why is Zinc not a transition metal?

A

forms one ion - Zn 2+ , whcih has a full d subshell. When is becomes 2+, it just looses 4s so D continues to be full

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9
Q

3 key properties of TM metals and their compounds

A

Form compounds where the TE has a different oxidation sate

Form colored compounds

Elements and compounds can act as catalysts

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10
Q

Different oxidation states -why?

A

Compounds with more than one oxidation state e.g. iron forms 2 chlorines - iron 2 and iron 3 chloride

The elements show variable oxidation states because the energy levels of 4s and 3d subshells are very close to one another, so different numbers of electrons can be gained or lost using similar amounts of energy

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11
Q

The trend in number of oxidation states ACROSS the Tm series

A

Increases to manganese, then decreases

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12
Q

Formation of coloured compounds?

A

When mix an aqeuous solution of TM ions with aqueous sodium hydroxide or aqueous ammonia you get a coloured hydroxide precipitate

In aqueous solutions, TM metals take on the form [M(h20)]6 n +

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13
Q

TM as good catalysts

A
  • TM and their compounds make good catalysts because they can change oxidation states by gaining or losing electrons within their d orbitals, and then bind with reactants to form intermediates as part of a reaction pathway that has a lower activation energy
  • Can therefore transfer electrons to speed up reactions

Good at absorbing substances onto their surfaces to lower the AE and providing a surface for the reaction to take plpace - use d orbitals to do this

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14
Q

Catlayst examples - haber and contact process

A

HABER- Iron used, NH3+ 3H2– 2NH3

CONTACT - Vanadium oxide – SO2+1/2 o2 – So3

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15
Q

Complex ion

A

A metal ion surrounded by coordinately bonded ligands

One or more molecules or negatively charged ions bond to central metal ion

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16
Q

Ligand

A

A molecules or ion that donates a pair of electrons to central metal ion to a coordinate covalent bond

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17
Q

Dative/coordinate covalent bond

A

Special kind of covalent bond, formed when one of the bonded atoms provides both of the electrons fro the shared pair

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18
Q

Coordination number

A

Indicates number of coordinate bonds attached to the central metal ion

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19
Q

How do you write formula for complex ion

A

Square brackets

Metal then ligand with number of ligands

Overall charge outside

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20
Q

Monodentate ligand

A

Have one pair able to donate to central metal ion - availiable for bonding

e.g. water

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21
Q

Cyanide and ammonia charge

A

-1 and 0

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22
Q

Bidentate ligands

A

Some ligands can donate 2 lone pairs of electrons to the central metal ion, forming 2 coordinate bonds

E.g. 1,2 diaminoethane- ‘en’
In en

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23
Q

EN

A

1,2 diaminoethane

Each N atom donates a pair of electrons to teh central metal ion forming a coordinate bond

:Nh2h2ch2nh2:

24
Q

Copper with en

A

[Co(Nh2CH2CH2NH2)3]3+

Oxidation number of co is 3+

Coordination number is 6 because there are 3 En’s and each ligand forms 2 coordinate bonds

25
shapes of complex ions- 4 coordinate shapes
Tetrahedral or square planar
26
Tetrahedral shape
bond angle of 109.5 e.g. Cocl4 2- or CuCl4 2- REMEMBER THE BRACKETS and charge
27
Square planar
90 degrees occurs in complex ions of transition metals with 8-d electrons in the highest energy d subshell e.g. platnium and gold [Pt(NH3)4]2+
28
6 coordinate complex
octahedral e.g. [Cu(h2o)6]2+ or [Fe(h20)6]3+
29
what is stereoisomers
Same structural formula but different arrangement of the atoms CIS TRANS OPTICAL
30
When is stereoisomerism shown
CIS TRANS - 4 coordinate and 6 coordinate containing 2 different monodentate ligands 6 coordinate complexes containg monodentate and bidentate can show BOTH optical and cis trans Also 6 coordinator that can 3 bidentate show opitcal
31
Cis trans isomerism
CIS on same | Trans on opposite side
32
e.g. of a sqaure planar and octrahedral that shows cis trans?
Pt(Nh3)2cl2 [Cu(Nh3)4(H20)2]2+
33
CAN tm metals be toxic
yes
34
optical isomerism ? and example
Only includes octahedral complexes contain 2 or more bidentate ligands Entantiomers are non superimposable mirror images of each other [Ni(NH2CH2CH2NH2)3]2+
35
Cis platin
A complex of platinum(II) witt 2 chloride ions and 2 ammonia SQUARE PLANAR Anti-cancer drug chloride ligands easy to displace, so platin looses and bonds to 2 nitrogen atoms on DNA inside cancerous cell Block on its DNA prevents cell from reproducing,s o dies BUT also prevents normal cells from reproducing - including the blood which can suppress immune system
36
What is ligand subsitution
When one ligand in a complex ion is replaced by another
37
AQUEOUS copper 2 ions-
When copper 2 sulfate is dissolved in h2o Pale blue [Cu(h2o)6]2+
38
Aqueous copper 2 ions with ammonia
pale blue preciptate of Cu(OH2) forms then to darker blue when the precipitate dissolves in excess ammonia [Cu(h2o)6] 2+ + 4Nh3 --- [Cu(nh3)4(h20)2] 2+ + 4h2o
39
Aqeous copper 2 ions with chloride ions
Yellow solution forms 6 waters replaced with 4 chlorides ( chloride ligands larger) Tetrahedral shape , reversible Cu(H2O)6]2+ (aq) + 4Cl (aq)== [CuCl4]2- (aq) + 6H20 (l)
40
Forming chromium(III) ions - 2 ways
chromium(III) potassium sulfate/chrome alum KCr(So4)212(h2o) is dissolved in water , [Cr(h2o)6]3+ is formed, a PALE PURPLE violet solution When chronimium(iii) sulfate is dissolved in water, a green solution containing chronium(III) is formed. {Cr(H20)5SO4]+
41
Chromium(III) in ammonia
[Cr(h2o)6]3+ , INITIALLY the drop wise ammonia formed Cr(OH)3, then it dissolves in the excess ammonia to form [Cr(NH3)6]3+, purple Cr(h2o)6]3+ (aq) + 6NH3 ---- [Cr(NH3)6]3+ + 6H20
42
Ligand substitution- haemoglobin
4 protein chains held together by weak inter molecular forces, each chain has a haem molecule within its structure ( fe2+) Oxygen can bind but SO CAN CO- more strongly than oxygen, forming carboxyhaemoglobin Irreversible
43
What is a precipitation reaction?
Occurs when 2 aqueous solutions containing ions react to form an insoluble ionic solid ( precipitate)
44
Cu2+ in NaOH
Cu(OH)2 precipitate BLue Insoluble in excess
45
Fe2+ in NaOH
Fe(OH)2 precipitate Pale green--- green Insoluble in excess Turns bronw on surface as oxidised to iron(III)
46
Fe3+ in NaOH
Fe(OH)3 precipitate Pale yellow--- orange brown INSOLUBLE in excess
47
Mn2+ in NaOH
Pale pink --- brown Mn(OH) 2 Insoluble in excess
48
Cr3+ in NaOH
Violet --- grey green Cr(OH)3 Precipitate soluble in excess -- forming DARK GREEN = Cr(OH3) + 3OH- -- [Cr(OH)6]3-
49
which precipitates react with an excess of ammonia and what are their colours
Cu(OH) 2 is a blue precipitation dissolving to form a deep blue solution of [Cu(NH3)4(h2o)2]2+ Cr(OH)3 is a green precipitate which dissolves to form [cr(NH3)6]3+ which is a purple solution
50
How do Fe2+, 3+ and Mn2+ react with ammonia
SAME way as sodium hydroxide | NOT further reaction with aqueous ammonia
51
Oxidation of iron 2 to iron 3
MnO4- reduced to Mn2+ Iron 2 oxidised to iron 3 purple mn04 goes to colourised
52
Reduction of iron 3 to iron 2
Fe3+ with iodide ions, orange brown fe3+ reduced to pale green fe2+ Iodide form iodine = brown
53
Reduction of Cr2o72- to cr3+
Aqeuous cr2o7 2- orange chromium 111 are green Cr3+ - orange to green Excess of zinc- chromium reduced to chromium 11 , pale blue
54
Cr3+ oxdised to cr04-
using h2o2
55
Reduction of Cu2+ to Cu+
When aqueous Cu2+ react with excess iodide ions, I- is oxidised to BROWN I2 and Cu2+ goes to I+ Pale blue --- white precipitate of CuI and brown
56
DISPROPORTIATION | with cu2o
When Cu20 ( solid copper 1 oxide) reacts with hot dilute sulfuric acid, a brown precipitate of copper is formed and a blue solution of copper 11 sulfate CU+ is oxidized and reduced Cu2o+ + h2so4---- Cu + CuSO4 + h20
57
testing for ammonium ions
NH4+ -- When heated with hydroxide ions, reacts to produce ammonia gas ( NH3) To test for the ammonium ion, NaOH is heated with the solution, if ammonia is evolved, damp red pH indicator paper will turn blue