topic 15: transition metals Flashcards
transition metals are:
d-block elements that form
ion(s) with incompletely-filled d-orbitals
Why is Zn not a transition metal?
- forms Zn 2+
- complete d subshell
Why is Sc not a transition metal?
- forms Sc 3+
- empty d orbital
ligand
a species which can donate a lone electron pair to form a dative covalent bond
monodentate
one covalent bond per ligand
bidentate
2 covalent bonds per ligand
complex ion
a central metal ion surrounded by ligands
how does colour arise? (4)
- LIGANDS cause d orbital to split into different energy levels
- TM has incomplete d orbital
- electrons gain light energy, excited, promoted to higher energy d orbital levels
- the light NOT ABSORBED is transmitted and seen as colours
if coordination number is 6 you are…
octahedral
- usually with smaller ligands eg oh- , h2o, nh3
if coord number is 4 you are
tetrahedral
usually with larger ligands eg cl-
cis platin
- SQUARE PLANAR
Pt(CL-)2(NH3)2
WHY IS ONLY THE CIS platin used
- in cancer treatment
- ?????
coord number of EDTA 4-
6 per ligand !!!
heamoglobin ligand
- multidentate ligand
- fe 2+ complex
what happens with haemoglobin and co
- ligand exchange!!!!! co replaces o2
- dative covalent bond formed w CO stronger than with O2 therefore prevents O2 binding to haemoglobin
- reduces amount of o2 that can be carreid to cells
vanadium +5 usual compound and colour
V(O2)+
YELLOW
vanadium +4 usual compound and colour
(VO2)+
Blue
vanadium +3 usual compound and colour
- V 3+
green
vanadium +2 usual compound and colour
- V 2+
violet
vanadium from +5 to +2 colours
yellow
blue
green
violet
what do u usually use to reduce the vanadium
zinc
how to go from Cr2O7 2- to Cr2+ or 3+
- zinc
acidic conditions
why anaeroboic conditions when reducing cr2o7 2- to cr2+
to prevent atmospheric oxygen from re oxidising the cr2+ to cr 3+
how to stabilise the cr 2+ so its doesnt re oxidise
add a ligand eg ethanedioate as bidentate
cr2o7 2- colour
orange
cr3+ cloour
green
cr 2+ colour
blue
how to make it easier to oxidise a transition metal in a low oxidation state?
- give it alkali conditions
- eg Cr(h2o)6 3+ vs cr(oh)6 3-
- easier to remove electrons from a negative complex
what t use to oxidise cr(oh)6 3-
- hydrogen peroxide
forms CrO4 2-
CrO4 2- colour
yellow
equilibrium of the chromate vs dichromate
2(CrO4)2- + 2H+ -> Cr2O7 2- + h2o
how to produce Cr2O7 2- ??
- alkali conditions
- oxidise cr(oh)6 3- using H2O2
- forms Cro4 2- (yellow). then considering the eqm, add acid to get Cr2O7 2-
Fe 2+ aqua complex AND COLOUR
Fe(H2O) 2+
green
Co 2+ aqua complex AND COLOUR
Co(H2O) 2+
PINK
Cu 2+ aqua complex AND COLOUR
Cu[H2O]6 2+
blue
Fe 3+ aqua complex AND COLOUR
[Fe(H2O)6]3+
brown
Cr 2+ aqua complex AND COLOUR
[Cr(H2O)6]3+
green
equilkibria of the aqueous metal complexes in water
[Fe(H2O)6]2+ + H2O = H3O+ + [Fe(H2O)5(OH)]+
explain how the equilibrium of the aqeous ions shows acidity and which of the 3+ or 2+ complexes are more acidic?
-[M(H2O)6]3+ more acidic than [M(H2O)6]2+
-3+ metal ions higher charge and smaller size
- greater polarising power
- stronger esa to water molecule (in the equilibria)
- strains O-H bond more so it breaks more easily releasing H+ ions
Cu 2+
Fe 2+
Co 2+
from aqeous complex to with limited OH- (include colours)
[M(H2O)4(OH)2] (s)
Cu:blue ppt
Co:blue ppt
Fe: green ppt
Fe 3+
Cr 3+
from aqeous complex to with limited OH- (include colours)
[M(OH)3(H2O)3]
Fe: bronwn ppt
Cr: green ppt
generic equation M2+ from aquoeous complex to with ammonia
[M(H2O)6]2+ + 2NH3 = [M(OH)2(H2O)4] + 2NH4+
of all of the ions, the only one with a change w excess OH- is
Cr3+
forms a green solution
complexes which do ligand exchange with exces NH3:
- Cr3+ (6x nh3)
- Co2+ (6x NH3)
Cu2+ (4x nH3)
Cr3+ with EXCESS NH3
[Cr(OH)3(H2O)3] + 6NH3 = [Cr(NH3)6]3+ + 3H2O+ 3OH-
PURPLE SOLUTION
Co2+ with EXCESS NH3
[Co(OH)2(H2O)4] + 6NH3 = [Co(NH3)6]2+ + 4H2O + 2OH-
PALE YELLOW SOLUTION
Cu2+ with EXCESS NH3
[Cu(OH)2(H2O)4]2+ + 4NH3 = [Cu(NH3)4(H2O)2]2+ + 2H2O + 2OH-
deep blue solution
adding Cl- ligand changes IMPORTANCE
- Cu2+ and Co2+
- coordinate number from 6 to 4
[(Cu)(Cl)4]2- - BECAUSE Cl- much larger than H2O so more steric hindrance
copper complex: yellow/green
cobalt: blue
TETRAHEDRAL
HOW DOES A CATALYSTIC ONCVERTER WORK?
- decrease CO and NO emissions by
-adsorption of CO and NO molecules onto the surface of the catalyst
-weakening of bonds and chemical reaction
-desorption of CO2 and N2 product molecules from the surface of the catalyst
2 CO + 2 NO = 2 CO2 + N2
reaction between Iodide and persulfate ions
- equation
- catalysis
(S2O8)2- + 2I- = 2(SO4)2- + I2
- rate is v slow due to repulsion between the negatively charged ions
CATALYSED PROPOSITION:
- 1. (S2O8)2- + 2Fe2+ = 2(SO4)2- + 2Fe3+
-2. 2I- + 2Fe3+ = 2Fe2+ + I2
possible becase Fe has variabke oxidation states
essential point to note about electrode potential of homogeneous transition metal catalysts
- e cell must lie between the oxidised and reduce
- so can first reduce the more positive one, then oxidise the more negative one
autocatalysis exampke Mn 2+ , mno4 - and ethanedioate
2 MnO4- + 5 (C2O4)2- + 16 H+ = 2Mn2+ + 10 CO2 + 8 H2O
- REDUCES THEN OXIDISES
coordinate number defintion
number of dative covalent bonds to the central metal atom
ethanedioate is
bidentate o-c=o - o-c=o
only d be not transition metals
Sc and Zn
Why are Zn2+ and Sc3+ colourless?
- complete/empty 3d subshell
- electrons can not be promoted to higher d orbitals
why do diff ligands cuase diff colours?
causes diff splitting in d orbitals so diff energy diff between them
difference in colour from flame test vs tm
SIMILARITIES:
- electrons promoted to higher E level
- difference in E levels determines colour
DIFF
- TM: light energy absorbed, splitting of d orbital by ligand, colour not absorbed is transmitted and seen
- FT: heat energy absorbed, electrons move back down to ground state
how to deterine how many cl - are acting as ligand
- add AgNo3
- those NOT acting as ligands will react
vanadium colours, 5+ down to 2+
Yellow
Blue
Green
Violet
if the acid you’re adding is dilute, then its
an acid base
NOTTT with Cl- ligands
QUALITIES CHARACTERISTIC of transition metals
- form coloured complexes
cis v trans isomers
cis = ligand on same side
what metals are used in a catalytic converter
platinum
palladium
rubidium
copper 2+ in xs oh-
no effect
zn 2+ in xs oh-
dissolves to form colurless soln
isomerism in complexes
ONLY IN OCTAHEDRAL
if next to each other, then cis
if opposite (top v bottom) then trans
