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Chemistry > 24. Transition Elements > Flashcards

24. Transition Elements Flashcards

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1
Q

Define ‘transition element’.

A

D-block element forming one or more stable ions (variable oxidation states) with an incomplete d subshell.

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2
Q

What are the two elements which are an exception to the ‘transition’ naming?

A

Sc and Zn.

  • Sc only forms Sc3+ with an empty d-subshell.
  • Zn only forms Zn2+ with a full d-subshell.
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3
Q

Which transition elements have unusual electronic configurations?

A

Cr - [Ar] 3d5 4s1
Cu - [Ar] 3d10 4s1
This is because half and full d-subshells are more stable.

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4
Q

How are transition element ions formed?

A

Electrons are lost from the 4s subshell first, then 3d. The maximum oxidation state for each element at the start of Period 4 involves all the 4s and unpaired 3d electrons.

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5
Q

What is the most common oxidation state, and why?

A

+2, formed when two 4s electrons are lost. This state dominates from Fe onwards as the 3d electrons become harder to remove due to increasing nuclear charge.

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6
Q

In which species is it more likely for higher oxidation states to be found?

A

Complex ions or compounds with O or F due to their higher electronegativities.

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7
Q

What are the physical properties of transition elements?

A
  • High melting point
  • High density
  • Hard and rigid
  • Good conductors
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8
Q

What is unusual about the chemical properties of transition elements?

A

Their first IE, atomic radii and ionic radii do not vary much across Period 4.

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9
Q

What are the differences in MP and density between transition elements and S-block?

A
  • MP higher than Ca due to more 3d electrons - stronger metallic bonding.
  • Density higher than Ca, because radius decreases as mass increases.
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10
Q

What are the differences in radius, first IE and conductivity between transition elements and S-block?

A
  • Atomic and ionic radii lower than Ca.
  • First IE higher than Ca.
  • Electrical conductivity lower than Ca (except for Cu).
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11
Q

How can transition elements have multiple oxidation states?

A

The difference in successive ionisation energies is not very large, but it results in a larger lattice enthalpy / enthalpy of hydration. Therefore, the overall enthalpy changes for different oxidation states are not very different, so it is not difficult to convert between different states.

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12
Q

Define ‘ligand’.

A

A species containing a lone pair of electrons, that forms a coordinate bond with a central metal atom/ion by donating its lone pair.

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13
Q

Define ‘coordination number’.

A

The number of dative bonds a ligand forms with the central metal atom/ion.

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14
Q

Define ‘monodentate’, ‘bidentate’ and ‘polydentate’ ligands.

A

Mono - forms one coordinate bond.
Bi - forms two coordinate bonds.
Poly - forms multiple coordinate bonds.

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15
Q

Define ‘complex ion’.

A

An ion formed by a central metal, surrounded by ligand/s.

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16
Q

How do you find the charge on a complex ion?

A

Sum of the charges on the central metal atom/ion and each ligand. The ion is written as [M(ligand)] ^ charge.

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17
Q

What are octahedral complexes? Give some examples.

A

Coordination number of 6.

  • [Fe(H2O)6]2+ (hexaaquairon (II) )
  • [Co(en)3]2+ (contains bidentate NH2CH2CH2NH2)
  • [Mn(ox)3]3- (contains bidentate -OOCCOO-).
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18
Q

What are tetrahedral complexes? Give some examples.

A

Coordination number of 4. Ligands too big to fit 6 around central atom.

  • [CuCl4]2-
  • [CoCl4]2-
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19
Q

What are square planar complexes? Give some examples.

A

Coordination number of 4.

  • [Ni(CN)4]2-
  • cis-platin.
20
Q

What are linear complexes? Give some examples.

A

Coordination number of 2. Occurs mainly with transition ions.
- [Ag(NH3)2]+ (diamminesilver (I) ion).

21
Q

What is EDTA 4-?

A

Hexadentate ligand.

- NCH2CH2N with two -CH2COO- on each N.

22
Q

How do you name complex ions?

A

Ligand:
- NH3 = ammine, OH = hydroxo, CN = cyano
- Prefix = number of ligands (eg. hexaaqua)
- Multiple - alphabetical order of ligand.
Metal (cationic complex):
- Name of metal + ON after the ligands.
Metal (anionic complex):
- Name of metal with special suffix (eg. Cu = cuprate, V = vanadate, Fe = ferrate) and ON, after the ligands.

23
Q

What is ‘ligand exchange’?

A

The substitution of ligands in a complex for others - can be partial or total. The new complex formed will be more stable than the original.

24
Q

Describe the reaction between [Cu(H2O)6]2+ and NaOH (aq), then conc. NH3.

A

NaOH acts as a base, removing H+ from two H2O ligands.
- [Cu(H2O)6]2+ + 2OH- -> Cu(H2O)4(OH)2 + 2H2O
- blue solution -> light blue precipitate.
Ammonia replaces OH and two H2O.
- Cu(OH)2(H2O)4 + 4NH3 -> [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-
- ppt dissolves, deep blue solution formed.

25
Q

Describe the reaction between [Cu(H2O)6]2+ and conc. NH3.

A
  • Ammonia acts as a base, removing H+.
    [Cu(H2O)6]2+ + 2NH3 -> [Cu(H2O)4(OH)2] + 2NH4+
  • Then, ammonia acts as a ligand.
    [Cu(H2O)4(OH)2] + 4NH3 -> [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-
  • blue solution -> light blue ppt -> dark blue solution in excess ammonia.
26
Q

Describe the reaction between [Cu(H2O)6]2+ and conc. HCl.

A

[Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O

  • colour change from blue to yellow
  • reversed by adding excess water
  • tetrahedral structure more stable
27
Q

What is produced when you mix ammonium chloride and copper sulfate solutions?

A

(NH4)2CuCl4 - ligand exchange producing yellow-green crystals.

28
Q

Describe the reaction between [Co(H2O)6]2+ and NaOH (aq).

A

[Co(H2O)6]2+ + 2OH- -> [Co(H2O)4(OH)2] + 2H2O

- pink solution -> blue ppt, turns red when warmed w/ excess.

29
Q

Describe the reaction between [Co(H2O)6]2+ and conc. NH3.

A

[Co(H2O)6]2+ + 2NH3 -> [Co(H2O)4(OH)2] + 2NH4+
- in small amounts, ammonia acts as a base, removing H+
[Co(H2O)6]2+ + 6NH3 -> [Co(NH3)6]2+ + 6H2O
- excess ammonia
- green ppt, then brown solution

30
Q

How are Cl- and H2O ligands exchanged in [Co(H2O)6]2+ ?

A 
Concentrated HCl. 
[Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O
- colour change from pink to blue
- reversed by adding excess water
- Co usually forms tetrahedral complexes with anionic ligands.
31
Q

In which ways can water ligands be replaced in [Cr(H2O)6]3+ ?

A

Excess concentrated ammonia.
- [Cr(H2O)6]3+ + 6NH3 ⇌ [Cr(NH3)6]3+ + 6H2O
Warming chromium(III) sulfate solution.
- [Cr(H2O)6]3+ + SO4 2- -> [Cr(H2O)5SO4]+ + H2O
Warming chromium(III) chloride solution.
- [Cr(H2O)6]3+ + 2Cl- -> [Cr(H2O)4Cl2]+ + 2H2O

32
Q

How do you test for iron(III) ions using ligand exchange?

A

[Fe(H2O)6]3+ + SCN- -> [Fe(SCN)(H2O)5]2+ + H2O

- colour change from yellow to blood red

33
Q

How is geometric isomerism shown in transition metal complexes?

A

Cis- and trans-platin.

  • Pt(NH3)2Cl2, square planar shape.
  • cis - ammonias on same side, chlorides on same side
  • trans - opposite sides
34
Q

How is optical isomerism shown in transition metal complexes?

A

Octahedral complexes with bidentate ligands. Forms entaniomers which are mirror images of each other and cannot be superimposed.
- eg. [Ni(NH2CH2CH2NH2)3]2+ or [Cr(C2O4)3]3-.

35
Q

How is cisplatin used as an anti-cancer drug?

A

Much smaller than cells so can fit through CSM.

  • Cl- concentration much lower inside cell, ligand exchange occurs with water.
  • cisplatin binds with A and G in cancer DNA, water molecules fall off (more loosely bound).
  • coordinate bonds form between Pt and N
  • helix shape changed, DNA cannot replicate, cell cannot divide.
  • major side effects as cisplatin doesn’t only enter cancer cells.
36
Q

Define ‘stability constant’.

A

EQM constant for the formation of complex ions in a solvent, from its constituent ions/molecules.

  • EQM lies in direction of more stable complex.
  • Water is not included in this expression due to its large excess.
  • Higher Kstab = higher stability.
37
Q

Compare Kstab values for monodentate and bidentate ligands.

A

Bidentate higher Kstab.

38
Q

How do you find overall Kstab from individual values?

A

Ligands are replaced one at a time in a stepwise process. With competing equilibria, the one with the higher Kstab value wins.
To find overall Kstab, multiply all the individual values together, and intermediates will cancel out.

39
Q

Why do we see compounds as being coloured?

A

White light passes through a sample, the sample absorbs a range of frequencies within the visible spectrum. The light that passes through and isn’t absorbed forms the colour we see. This is usually the complementary colour.
- yellow + blue, red + cyan, magenta + green.

40
Q

How do transition metal complexes form colours? (degenerate d orbitals)

A

The five d-orbitals in an isolated transition ion/atom are degenerate (all at the same energy level), but in the presence of ligands, they split into two non-degenerate sets of orbitals.
This is caused by repulsion between the ligand’s ions and the d-orbital ions. Two have a higher energy level than the other three.

41
Q

How do transition metal complexes form colours? (electron promotion)

A

There is an E gap between the two orbital levels. When light is applied, an electron absorbs some of this E and jumps into a higher energy level.
The colour of the light is matched to the E difference due to different frequencies in the spectrum.

42
Q

What are the 5 shapes of the d-orbitals?

A

Dxy, Dxz, Dyz.
- four lobes lie in the specified plane.
- the lobes point BETWEEN axes rather than along them.
Dx2-y2, Dz2
- lobes point along the axes
- z looks like p-orbital with ring around middle.
- when ligands are added, these feel more repulsion as they are directly along the axes. So, they are promoted to the higher energy level.

43
Q

How does colour work in tetrahedral complex ions?

A

Same way as octahedral, but three orbitals are in the higher level and two are in the lower level.

44
Q

How does ligand identity affect complex colour?

A

Smallest splitting -> largest (increasing field strength):

- Cl-, F-, OH-, H2O, NH3, CN-

45
Q

How does metal ON affect complex colour?

A

Higher ON = larger splitting.

46
Q

How does coordination number affect complex colour?

A

Octahedral causes more splitting than tetrahedral.

Chemistry (12 decks)

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