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A-LEVEL CHEMISTRY > Chemistry of transition elements > Flashcards

Chemistry of transition elements Flashcards

1
Q

Transition elements

A

d-block elements which form one or more stable ions with an incomplete d subshell

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2
Q

The d-block elements which aren’t classed as transition metals

A

Scandium and Zinc

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3
Q

Scandium is not classed as a transition element because

A

It only forms one ion, Sc3+, that has no electrons in its 3d subshell; it has the electronic configuration of [Ar]

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4
Q

Zinc is not classed as a transition element because

A

It forms only one ion, Zn2+, that has a complete 3d subshell; it has the electronic configuration [Ar]3d10

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5
Q

The five orbitals in a d subshell

A

3dyz
3dxz
3dxy
3dx2 - y2
3dz2

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6
Q

Special properties of transitional elements

A

~Variable oxidation states
~Behave as catalysts
~Form complex ions
~Form coloured compounds

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7
Q

The most common oxidation states of Titanium/TI

A

+3 and +4

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8
Q

The most common oxidation states of Vanadium/V

A

+2, +3, +4 and +5

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9
Q

The most common oxidation states of Chromium/Cr

A

+3 and +6

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10
Q

The most common oxidation states of Manganese/Mn

A

+2, +4, +6 and +7

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11
Q

The most common oxidation states of Iron/Fe

A

+2 and +3

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12
Q

The most common oxidation state of Nickel/Ni

A

+2

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13
Q

The most common oxidation states of Copper/Cu

A

+1 and +2

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14
Q

Why do transitional metals make excellent catalysts

A

~because of their variable oxidation states: during catalysis, the transition element can change to various oxidation states by gaining electrons or donating electrons from reagents within the reaction
~substances can also be adsorbed onto their surface and activated in the process

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15
Q

Why transitional elements can form ions with variable oxidation states

A

variable oxidation states can be formed as the 3d and 4s atomic orbitals are similar in energy - this means that a similar amount of energy is needed to remove a different number of electrons

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16
Q

When the transition elements form ions, the electrons of

A

the 4s subshell are lost first, followed by the 3d electrons

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17
Q

The most common oxidation state is

A

+2, which is usually formed when the two 4s electrons are lost

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18
Q

Transition elements can easily form complex ions because

A

they have empty d orbitals that are energetically accessible- the empty d orbitals are therefore not too high in energy and can accommodate a lone pair of electrons

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19
Q

ligand

A

a molecule or ion (species) that has one or more lone pairs of electrons. These lone pairs of electrons are donated by the ligand, to form dative covalent bonds to a central metal atom or ion

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20
Q

Monodentate ligands

A

can form only one dative bond to the central metal ion

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21
Q

Bidentate ligands

A

can each form two dative bonds to the central metal ion- this is because each ligand contains two atoms with lone pairs of electrons

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22
Q

complex

A

a molecule or ion formed by a central metal atom or ion surrounded by one or more ligands

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23
Q

complex ion

A

if a complex has an overall charge it is called a complex ion

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24
Q

coordination number

A

the number of dative bonds formed between the central metal ion and the ligands

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25
Q

Shapes of complexes with a coordination number of 6

A

octahedral

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26
Q

Shapes of complexes with a coordination number of 4

A

tetrahedral or square planar

27
Q

bond angles in linear complexes are

A

180 degrees

28
Q

bond angles in tetrahedral complexes are

A

109.5 degrees

29
Q

bond angles in square planar complexes are

A

90 degrees

30
Q

bond angles in octahedral complexes are

A

90 degrees

31
Q

isolated transition element

A

an atom or ion that is not bonded to anything else

32
Q

degenerate orbitals

A

orbitals are all at the same energy level (they are equal in energy)

33
Q

The five d orbitals in an isolated transition element are

A

degenerate orbitals

34
Q

The dative bonding from the ligands causes

A

the five d orbitals to split into two sets of non-degenerate orbitals

35
Q

non-degenerate orbitals

A

orbitals that are not equal in energy

36
Q

Splitting of orbitals in octahedral complexes

A

1)The lone pairs of electrons of the six ligands repel the electrons in the x2-y2 and z2 orbitals of the metal ion more than they repel the electrons in the 3dyz, 3dxz, and 3dxy orbitals
2) This is because the ligands are attached to or approaching the central metal ion along the x, y and z axes, and the 3dx2-y2 and 3dz2 orbitals have lobes along these axes, thus the 3dx2-y2 and 3dz2 orbitals line up with the dative bonds in the complex’s octahedral shape
3)The electrons in these two orbitals are closer to the bonding electrons, so there is more repulsion
4)This means that when the d orbitals split, the 3dx2-y2 and 3dz2 orbitals are at a slightly higher energy level than the other three

37
Q

ΔE

A

The difference in energy between the non-degenerate d orbitals

38
Q

Splitting of orbitals in tetrahedral complexes

A

1)The bonding pair of electrons from the four ligands now line up with the 3dyz, 3dxz, and 3dxy orbitals of the central metal ion
2)Now, the 3dx2-y2 and 3dz2 orbitals lie between the metal-ligand bonds
3)Therefore, there is less repulsion with the 3dx2-y2 and 3dz2 orbitals
4)When the d orbitals split this time, the 3dx2-y2 and 3dz2 orbitals are at lower and more stable energy level than the other three

39
Q

Why are transition element complex solutions coloured

A

they absorb part of the electromagnetic spectrum in the visible light region

40
Q

The observed colour in coloured solutions

A

is the complementary colour which is made up of light with frequencies that are not absorbed

41
Q

copper(II) ions absorb light from the red end of the spectrum so the complementary colour observed is therefore

A

pale blue (cyan)

42
Q

When light shines on a solution containing a transition element complex

A

an electron will absorb the frequency of light which corresponds to the exact amount of energy (ΔE) between their non-degenerate d orbitals

43
Q

amount of energy absorbed by an electron in a complex can be worked out by the equation

A

ΔE = h x v

44
Q

h =

A

Planck’s constant (6.626 x 10-34 m2 kg s-1)

45
Q

v =

A

frequency (Hertz, Hz or s-1)

46
Q

The electron uses the energy from the light to

A

jump into a higher, non-degenerate energy level- electro promotion

47
Q

The other frequencies of light which are not absorbed combine to make the

A

complimentary colour

48
Q

ΔE is affected by

A

the different ligands which surround the transition element ion

49
Q

What causes the size of ΔE and thus the frequency of light absorbed by the electrons to be slightly different?

A

Different ligands split the d orbital by a different amount of energy depending on the repulsion that the d orbital experiences from these ligands

50
Q

depending on the size of ΔE

A

a different colour of light is absorbed by the complex solution and a different complementary colour is observed

51
Q

complexes with similar transition elements ions, but different ligands, can have

A

different colours

52
Q

colour of [Cu(H2O)6]2+ solution

A

light blue

53
Q

colour of [Cu(NH3)4(H2O)2)]2+ solution

A

deep blue

54
Q

colour of [Co(H2O)6]2+ solution

A

pink

55
Q

colour of [Co(NH3)6]2+ solution

A

brown

56
Q

Ligand exchange

A

when one ligand in a complex is replaced by another

57
Q

Ligand exchange forms

A

a new complex that is more stable than the original one

58
Q

During ligand exchange, if the ligands are of a similar size

A

there are no changes in the coordination number or the geometry of the complex

59
Q

During ligand exchange, if the ligands are of a different size

A

then a change in coordination number and the geometry of the complex will occur

60
Q

When a transition element ion is in solution, it can be assumed that it exists as a

A

hexaaqua complex ion (i.e. it has six water ligands attached to it)

61
Q

[Cu(H2O)6]2+ (aq) complex ion colour

A

blue

62
Q

What happens upon dropwise addition of sodium hydroxide (NaOH) solution to [Cu(H2O)6]2+ (aq)

A

[Cu(H2O)6]2+(aq) + 2OH-(aq) → Cu(OH)2(H2O)4(s) + 2H2O(l)
-a light blue precipitate is formed
-partial ligand substitution of two water ligands by two hydroxide ligands has occurred

63
Q

What happens upon the addition of excess concentrated ammonia (NH3) solution to Cu(OH)2(H2O)4(s)

A

Cu(OH)2(H2O)4(s) +

A-LEVEL CHEMISTRY (14 decks)

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