5.3.1 Transition Elements Flashcards
what are d-block elements
- located between group 2 and 13
- where the 3d subshell has the highest energy, and electrons are added to this orbital
explain characteristics of d-block elements, since they are metallic
- high melting/boiling point
- shiny in appearance
- conduct electricity/heat
what is electron configuration
shows the arrangement of electrons in shells and sub-shells
- electrons occupy orbitals in order of increasing energy
what is the electron configuration of Zn
1s2 2s2 2p6 3s2 3p6 4s2 3d10 (could also just write 3d10 and then 4s2)
what’s important to remember when transition metals form ions
- they lose their 4s electrons before 3d
- (when making the atoms, reverse, so 4s fills before 3d)
what’s special about the electron configurations of chromium and copper
- they are 4s1 3d5 and 4s1 3d10
- as having a half filled/fully filled 3d subshell gives them extra stability
what are transition elements
d-block elements that form at least 1 ion with a partially filled d-subshell
give examples of 2 d-block elements that are not transition metals
zinc and scandium
why is zinc and scandium not transition metals
- Zinc only forms Zn2+ ions, so lose both 4s electrons, leaving behind 3d10 (FULLY FILLED D ORBITAL)
- Scandium only forms Sc3+ ions, so loses its 4s electrons, as well as its only 3d1 electron (FULLY EMPTY D ORBITAL)
what are the 3 properties of transition metals
1) form compounds in which the transition element has different oxidation state
2) form coloured compounds
3) the elements and their compounds can act as catalysts
explain transition metals having variable oxidation states
- e.g. iron (II) chloride and iron (III) chloride
- all transition metals form 2+ ions
- the number of oxidation states a transition metal has increases across the series to manganese (+5 to +7), then decreases to +3
- each oxidation state has a distinct colour
- species with transition metals in their highest oxidation states are really strong oxidising agents
explain transition metals forming coloured compounds
- the colour of a solution is linked to the partially filled d-orbital of the ion, and varies with different oxidation states
- e.g. Fe(II) - pale yellow - 3p6 3d6
- Fe(III) - yellow - 3p6 3d5
- e.g. Cr (III) - green
- Cr (VI) - yellow/orange
what is a catalyst
increases the rate of chemical reaction via providing an alternative reaction pathway, but remains unchanged itself
explain transition metals as catalysts
- can be HETEROGENEOUS catalysts , where in a different state to the reactants
- e.g. in Haber process: N2 + 3H ⇌ 2NH3 - uses IRON CATALYST
- e.g. catalytic decomposition of hydrogen peroxide: 2H2O2 → 2H2O + O2 - uses MnO2 CATALYST
- e.g. contact process of producing sulfur trioxide via oxidation: 2SO2 + O2 ⇌ 2SO3 - uses VANADIUM OXIDE ION
- e.g. hydrogenation of vegetable fats in making margarine - uses NICKEL CATALYST
- can be HOMOZYGOUS catalysts, where they are in the same state as the reactants
- e.g. reaction of zinc with acids: Zn + H2SO4 → ZnSO4 + H2 - needs Cu2+ CATALYST
- e.g. S2O8 2- + I- → 2SO4 2- + I2 - uses Fe2+ catalyst, which reacts in the first step of the reaction to Fe3+, and then is regenerated in the second step
what type of special ions can d-block elements form
- complex ions, though not restricted to d-block
how are complex ions formed
when one or more molecules or negatively charged ions bond to a central metal ion
what are ligands
a molecule or ion that donates a pair of electrons to a central metal ion to form a coordinate bond (dative covalent bond)
what is a coordinate bond
a special kind of covalent bond formed when one of the bonded atoms provides both of the electrons in a shared pair
what is coordination number
the number of coordinate bonds attached to the central metal ion
how should you show complex ions in a formula
- complex ion is enclosed inside square brackets
- the overall charge is shown outside the bracket
- the ligand is found inside round brackets, with the number outside
- e.g. [Cr(H2O)6]3+ ( where Cr3+ is the central metal ion, and each H2O molecule donates a lone pair of electrons from the O atom to Cr3+)
what would the overall charge of a complex ion be
the sum of the charges of the central metal ion and any ligand present
what are the two charges a ligand can have
neutral or negative
what is a monodentate ligand
a ligand that is able to donate one pair of electrons to a central metal ion
what are examples of monodentate ligands
- H2O:
- :NH3
- :Cl-
- :CN-
- :OH-
what are bidentate ligands
ligands that can donate 2 lone pairs of electrons to the central metal ion, forming 2 coordinate bonds each
what are examples of bidentate ligands
- 1,2-diaminoethane, shortened to en (NH2CH2CH2NH2)
- ethanedioate ion, also known as oxolate (C2O2) 2-
how does cn behave as a ligand
- each nitrogen atom has a lone pair of e-, and donates these to the central metal ion
- so forms a like square shape
how does the oxolate ion behave as a ligand
- each negatively charged oxygen atom donates a lone pair of electrons
- forms a square with the Cs, and then each C has a =O going off it
what does the shapes of complex ions depend on
the coordination number
what shape does a 6 coordination complex ion form
- octahedral
- 90 degree bond angles
what 2 shapes do 4 coordination complex ions form
tetrahedral (more common)
square planar
what does a tetrahedral complex ion look like
bond angles of 109.5 degrees
what does a square planar complex ion look like
- each atom around the metal ion is on the corners of square
- only present when the transition metal in the centre has 8-d electrons in the highest d-sub shell
- e.g. platinum (II), palladium (II) and gold (II)
- bond angles of 90 degrees
what are stereoisomers
have the same structural formula but a different arrangement of atoms in space
what 2 types of stereoisomers are there
- cis-trans
- optical
- the type of isomerism present depends on the number of ligands attached to the central metal ion
what ions show what types of isomerism
- both 4 and 6 show cis-trans if different monodentate ligands present
- 6 show cis-trans and optical with both monodentate and bidentate
how are cis-trans isomers in complex ions different to organic molecules
- organic: require a C=C double bond to prevent rotation of groups around each C atom
- in complex ions, no double bond needed
how do square planar complex ions show cis-trans isomers
CIS = 2 identical groups are adjacent to each other (their coordinate bonds are 90 degrees apart)
TRANS = 2 identical groups are opposite each other (180 degrees apart)
how do octahedral complex ions show cis-trans isomers
- if all monodentate, you have 4 types of one ligand, and 2 types of the other
- CIS = next to each other (90 degrees)
- TRANS = opposite each other (180 degrees)
- if 2 are monodentate, and 2 are bidentate, follows same pattern
what are optical isomers
non-superimposable mirror images of each other
- also called enantiomers
where are optical isomers present in complex ions
- octahedral with 2 or more bidentate ligands
- if 2 ligands, must be CIS isomers, as if trans, reflecting would give same pattern
name the drug for cis-trans isomers used in medicine
cis-platin
- Pt central ion
- NH3 on one side
- Cl on the other
- square planar
what is cis-platin and how does it work
- an anti-cancer drug
- attacks tumours and shrinks their size
- via forming a platinum complex inside the cell which binds to DNA and prevents it from replicating, stopping cell division, eventually leading to apoptosis
what are the drawbacks of cis-platin
- has unpleasant side effects
- can lead to kidney damage
- HOWEVER, still used, but research underway to find another platinum based drug to inhibit growth but show no side effects
what is a ligand substitution reaction
where one ligand in a complex ion is replaced by another ligand
why is it important to carry out any qualitative analysis drop by drop
so you can record ALL observations (see all stages of the reaction)
how can you form a Copper (II) complex ion
- dissolve copper (II) sulfate in water, forming a pale blue complex ion
- forms [Cu(H2O)6]2+
how does [Cu(H2O)6]2+ react with excess aqueous ammonia
[Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O
- pale blue solution to dark blue solution
- ammonia replaces 4 of the H2O ligands
what are the observations you should see when [Cu(H2O)6]2+ react with excess aqueous ammonia
1) first, a pale blue precipitate should form, as Cu(OH)2 is being formed
2_ then the Cu(OH)2 dissolves in excess, giving a dark blue solution
how does [Cu(H2O)6]2+ react with excess of aqueous Cl- ions
[Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O
- pale blue solution to yellow solution
- can use conc. excess HCl as a source of Cl- ions
what observations should you see when [Cu(H2O)6]2+ react with excess of aqueous Cl- ions
1) an intermediate green solution forms when the yellow and blue solutions mix
2) yellow solution
3) if you add more water, the solution should go back to pale blue, though lighter than before
explain how the addition of Cl- ions causes a change in shape and coordination number of [Cu(H2O)6]2+
- chloride ligands are larger in size than water ligands
- so fewer Cl- ions can fit around the central Cu2+ ion
- so changes coordination number from 6 to 4
- and changes shape from octahedral to tetrahedral
how can you get a chromium(III) complex ion form
1) dissolve chromium (III) potassium sulfate KCr(SO4)2 ‘ H2O, also known as chrome alum, in water to form [Cr(H2O)6]3+, giving pale purple solution
2) or dissolve chromium (III) sulfate in water, forming [Cr(H2O)5(SO4)]+, which is a green solution
how does [Cr(H2O)6]3+ react with aqueous excess ammonia
- NH3 replaces all of the H2O ligands
- [Cr(H2O)6]3+ + 6NH3 → [Cr(NH3)6]3+ = 6H2O
- violet to purple
what observations do you see when [Cr(H2O)6]3+ reacts with aqueous excess ammonia
1) initially forms a grey-green precipitate of Cr(OH)3
2) the precipitate then dissolves in the excess ammonia, to form the purple solution of complex ion
what is the structure of haemoglobin
- blood carries oxygen around the body due to haemoglobin, which is an iron containing protein in red blood cells
- made of 4 protein chains held by weak intermolecular forces
- each chain has a haem group, containing central metal ion Fe2+, which binds to O2 gas
how does haemoglobin carry O2 around the body
- when blood passes through the lungs, haemoglobin bonds to O2 due to the increases oxygen pressure
- forms oxyhaemoglobin
- can also bond to CO2 and carry out of lungs to be exhaled
what can haemoglobin also bond to
carbon monoxide CO
- forms carboxyhemoglobin
why is CO a problem
- when breathed in
- a ligand substitution reaction takes place, and the O2 is replaced with CO2
- CO binds more strongly than O2 as well, so only a little amount of CO breathed in prevents large amounts of haemoglobin from carrying O2
- bonds are so strong, they are irreversible, so if too high conc., causes death
what is a precipitation reaction
when 2 aqueous solutions containing ions react together to form insoluble ionic solid (precipitate)
what do transition metals react with to form precipitates
NaOH (aq)
NH3 (aq)
- some precipitates formed will go on to dissolve in excess
how does Cu2+ react with NaOH
Cu2+ + 2OH- → Cu(OH)2 (s)
- blue solution to
- blue precipitate
- INSOLUBLE IN EXCESS
how does Fe2+ react with NaOH
Fe2+ + 2OH- → Fe(OH)2 (s)
- pale green solution to
- green precipitate
- INSOLUBLE IN EXCESS
- if left standing in the air, will be oxidised further
- Fe(OH)2 →Fe(OH)3
- green precipitate to
- orange brown precipitate
- Iron (II) is oxidised to Iron (III)
how does Fe3+ react with NaOH
Fe3+ + OH- → Fe(OH)3
- pale yellow solution
- to orange brown precipitate
- INSOLUBLE IN EXCESS
how does Mn2+ react with NaOH
Mn2+ +OH- → Mn(OH)2
- pale pink solution to light brown precipitate, darkens standing in air
- INSOLUBLE IN EXCESS
how does Cr3+ react with NaOH
Cr3+ + 3OH- → Cr(OH)3
- violet solution to
- green precipitate
- IS SOLUBLE IN EXCESS:
- Cr(OH)3 + 3OH- → [Cr(OH)6]3-
- green precipitate to
- green solution
how does Cu2+ react with NH3
- via the pathway of [Cu(H2O)6]2+ reaction with excess aqueous ammonia
1) precipitation: Cu2+ + 2OH- → Cu(OH)2
- blue solution to blue precipitate
2) Cu(OH)2 will dissolve in excess ammonia to produce [Cu(NH3)4(H2O)2]2+
- blue precipitate to deep blue solution
how does chromium react with excess aqueous ammonia
- via same pathway as the complex ion
1) Cr3+ + 3OH- → Cr(OH)3
- violet solution to green precipitate
2) Cr(OH)3 dissolves in excess ammonia, to make [Cr(H2O)6]3+
- green precipitate to purple solution
how do Fe2+, Fe3+, Mn2+ react with excess ammonia
same way they do with NaOH
explain how Fe2+ can be oxidised into Fe3+
- METHOD USED IN REDOX TITRATIONS (under acidic conditions) TO SHOW COLOUR CHANGE
- MnO4- + 8H+ + Fe3+ →Mn2+ + 5Fe3+ + 4H2O
- the MnO4- is being reduced
- Fe2+ is being oxidised
- PURPLE TO COLOURLESS (due to MnO4- turning into Mn2+
how can Fe3+ be reduced into Fe2+
- via adding iodide ions, which later turn into iodine
- 2Fe3+ + 2I- → 2Fe2+ + I2
- the irons go from orange brown to pale green, but colour change is masked via the production of brown iodine
explain using E° how Fe2+ and Fe3+ can be both oxidised and reduced
- the E° value of Mno4-/Mn2+ is more positive than the irons, so Fe2+ is oxidised, and MnO4- is reduced
- the E° value of the irons is more positive than I-/I2, so I- is oxidised snd Fe3+ is reduced
explain how dichromate ions can be reduced into chromium ions
- via the addition of Zn
- Cr2O7 2- + 14H+ + 3Zn → 2Cr3+ + 7H2O + 3Zn2+
- colour change of orange to green of the chromate ion
- dichromate is reduced
- Zn is oxidised
how can Cr3+ ions be further reduced
- with an excess of Zn
Zn(s) + 2Cr3+ → Zn2+ _ 2Cr2+ - green to pale blue colour change of Cr
explain with E° values why Cr2O7- can be reduced so well with Zn
- Zn is a very strong reducing agent
- has the lowest E° value of both equations
- so can reduce all the way down to Cr2+
how can Cr3+ be oxidated to form CrO4 2-
- uses hot alkaline H2O2
- 3H2O2 + 2Cr3+ +10OH- → 2CrO4 2- + 8H2O
- H2O2 is reduced, and Cr3+ is oxidised
how can Cu2+ be reduced to Cu+
- using iodide ions
2Cu2+ + 4I- → 2CuI + I2 - copper goes from pale blue solution to white precipitate
- formation of iodine produces brown colour
explain the disproportionation reaction of copper oxide with sulfuric acid
Cu2O + H2SO4 → Cu + CuSO4 + H2O
- forms a brown solid and a blue solution
- is reduced into Cu
- is oxidised into CuSO4
