Chapter 28 Transition Elements Flashcards
Define transition element.
- d-block element
- which forms one or more stable ions
- with incomplete d-orbitals (therefore, 3d^0 & 3d^10 are excluded)
What are the 5 d-orbitals?
- x^2 - y^2
- xy
- yz
- xz
- z^2
Sketch the shape of a 3dxy orbital & 3dz^2 orbital.
- refer to slides
State the properties of transition elements.
- variable oxidation states
- behave as catalysts
- form complex ions
- form coloured compounds
Why do transition elements have variable oxidation states?
- Easier to remove the third electron compared with the 1st & 2nd.
- 4s orbital & the 3d orbitals have a small energy difference.
Why are Scandium & Zinc not transition metals?
- Sc3+ = 3d^0
- Zn2+ = 3d^10
- Both ions have no incomplete d-orbitals.
How do we predict oxidation states?
- Remove all electrons from 4s orbital. Therefore, all transition metals have a minimum of +2 oxidation state (4s^2).
- Remove unpaired electrons from 3d orbital.
State the electronic configuration & oxidation states for Titanium (Ti)
(Ar = 22).
- Ti: [Ar] 4s2 3d2
- Oxidation states: +2, +3, +4
State the electronic configuration & oxidation states for Chromium (Cr)
(Ar = 24).
- Cr: [Ar] 4s1 3d5
- Oxidation states: +2, +3, +4, +5, +6
State the exception for Chromium (Cr), which cannot have a +1 oxidation state.
- Not stable.
State the electronic configuration & oxidation states for Manganese (Mn)
(Ar = 25).
- Mn: [Ar] 4s2 3d5
- Oxidation states: +2 to +7
Why can’t we have Mn7+?
- Mn7+ cannot exist as a monoatomic ion. It has to form a polyatomic ion. Hence, MnO4-
- Because across the period, shielding effect remains constant, nuclear charge increases, & nuclear attraction increases.
- Not stable.
State the electronic configuration & oxidation states for Iron (Fe)
(Ar =26).
- Fe: [Ar] 4s2 3d6
- Oxidation states: +2 to +6
State the electronic configuration & oxidation states for Cobalt (Co)
(Ar = 27).
- Co: [Ar] 4s2 3d7
- Oxidation states: +2 to +5
State the electronic configuration & oxidation states for Copper (Cu)
(Ar = 29).
- Cu: [Ar] 4s2 3d7
- Oxidation states: +1, +2
- Copper is an exception to the rules because it can exist as +1.
Why are transition metals good catalysts?
- They have variable oxidation states.
- Can form complexes - easily change to various oxidation states by gaining or donating electrons from reagents within the reaction.
- Good catalytic site - substances can be adsorbed onto their surface & activated.
Define complex.
- central metal ion
- surrounded by ligands
- by forming dative bond(s)
Define ligand.
- species with lone pair
- forms dative bond
- with central metal ion
Draw the [Fe(H2O)6]2+ complex ion.
- 6 H2O molecules
- 2 lone pairs of electrons on each oxygen atom in each H2O molecule
- 6 dative bonds
- charge = 2+
- refer to class notes
Name some common ligands.
- H2O
- Cl-
- NH3
- CN-
- OH-
- CH3OH
- CH3NH2
Relationship between relative formula mass (Mr) & number of ligands formed.
- Greater Mr, fewer ligands.
- Mr H2O = 18, 2 to 6 ligands
- Mr NH3 = 17, 2 to 6 ligands
- Mr OH- = 26, 4 to 6 ligands
- Mr CN- = 26, 4 to 6 ligands
- Mr Cl- = 35.5, up to 4 ligands
How do complexes form? Use Fe2+ & H2O as an example.
- Electronic configuration of Fe2+: [Ar] 3d6
- Fe2+ has 4 unpaired electrons in the 3d orbital.
- When H2O approaches, the 6 lowest orbitals from subshells 4s, 4p & 4d hybridise.
- The lone pairs of electrons from 6 H2O molecules go into the 6 hybridised orbitals (that are of the same energy level).
Define monodentate ligand.
- 1 ligand uses
- 1 lone pair to form
- 1 dative bond with central metal ion.
- E.g. NH3 (has 1 lone pair)
Define bidentate ligand.
- 1 ligand uses
- 2 lone pairs to form
- 2 dative bonds with central metal ion
- Lone pairs must have a gap of more than 1 atom between each other.
Examples of bidentate ligands.
- -OOCCOO- (diethanoate ion)
- HOCH2CH2OH (ethylene glycol)
- H2NCH2CH2NH2 (ethylene diamine)
Draw the [Fe(HOCH2CH2OH)3]2+ ion.
- refer to class discussion
Define coordination number.
- Number of coordinate bonds surrounding the metal.
[Fe(CN-)6]3-:
- oxidation state of metal?
- no. of ligands?
- type of ligand?
- coordination number?
- shape + angle?
- is it coloured? why?
- Oxi. state = +3
- No. of ligands = 6
- Type of ligand = monodentate
- Coordination number = 6
- Shape + angle = octahedral & 90°
- Coloured? = Yes. Fe3+ has incomplete d-orbitals.
Which Group & which metals forms complex ions with a square planar shape?
- Group 10
- Nickel, palladium, platinum
- Because 4 bond pairs & 2 lone pairs.
[Cu(H2O)6]3+:
- oxidation number of metal
- no. of ligands
- type of ligands
- coordination number
- shape + angle
- is it a coloured complex? explain
- draw structure
- oxi no. - +3
- no. of ligands - 6
- type of ligand - monodentate
- coordination number - 6
- shape + angle: octahedral, 90°
- Yes. Because Cu3+ has an incomplete d-orbital.
- Refer to WS 31.1
[Ni(CN)4]2-:
- oxidation number of metal
- no. of ligands
- type of ligands
- coordination number
- shape + angle
- is it a coloured complex? explain
- draw structure
- oxi no. - +2
- no. of ligands - 4
- type of ligand - monodentate
- coordination number - 4
- shape + angle: square planar, 90°
- Yes. Because Ni2+ has an incomplete d-orbital.
- Refer to WS 31.1
[Co(NH2CH2CH2NH2)3]2+:
- oxidation number of metal
- no. of ligands
- type of ligands
- coordination number
- shape + angle
- is it a coloured complex? explain
- draw structure
- oxi no. - +2
- no. of ligands - 3
- type of ligand - bidentate
- coordination number - 6
- shape + angle: octahedral, 90°
- Yes. Because Co2+ has an incomplete d-orbital (3d7).
- Refer to WS 31.1
Explain why transition elements behave as catalysts. (2)
- more than 1 (stable) oxidation state
- empty d-orbitals can form dative bonds with ligands
