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Chemistry May > Transition Metals > Flashcards

Transition Metals Flashcards

1
Q

lewis acid vs b-l acid

A 
L = accepts e pairs
B-L = donates hydrogens
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2
Q

ions

A

lose electrons form 4S first

normally 4s 3d but becomes 3d

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3
Q

colour

A

3d orbitals split
light energy absorbed = energy gap
e’s jump from ground to excited state
reflected light is seen - energy in visible range

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4
Q

coordinate bond

A

shared pair of electrons from one atom in a covalent bond

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5
Q

transition metal definition

A

1 or more ions formed with incomplete D sub shells

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6
Q

ligands bond

A

COVALENT not ionic

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7
Q

tm complex

A

central metal ion surrounded by coordinately bonded ligands

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8
Q

ligand

A

atom/molecule that donates electron pair

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9
Q

Sc
Cr
Cu
Zn

A

not a TM
4s1 3d5
4s1 3d10
not a TM

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10
Q

chemical properties

A

variable oxidation states
coloured compounds
catalysts
complex ions

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11
Q

catalysts

A

iron - haber process
vanadium oxide - contact process
rhodium - catalytic converters

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12
Q

unidentate
bidentate
multidentate

A

ligand that forms 1 CO-B
forms 2
more than one CO-B

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13
Q

unidentate ligand

A

h20 - lone pairs v close together
:Cl- - chloride ion
NH3
:CN- - cyanide

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14
Q

bidentate ligand

A

en = ethane-1,2,diamine :NH2

ethanedioate (2-) - C2O4 (one O double bond on C) :O-

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15
Q

multidentate ligand

A

EDTA 4- = 6 CO-B’s

incr entropy so incr stability

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16
Q

TM dissolved in water

A

aqua ion
6 H2O ligands
octahedral shape

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17
Q

TM ion shape w/ 2 ligands

A

linear

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18
Q

TM ion shape w/ 4 ligands

A

tetrahedal
but platin is square planar -> cisplatin = anti-cancer drug that stop replication of cancerous cells but suppresses immune system

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19
Q

TM ion shape w/ 6 ligands

A

octahedral

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20
Q

stereoisomerism

A

cis/trans
optical isomerism in octahedral w/ 2 or more bidentate ligands…… using CIS bc unsymmetrical + non superimposable mirror images

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21
Q

:Cl- vs NH3 and H2O

A

4 coordinate bonds vs 6

since Cl is large

22
Q

factors of TM ion colour

A

coordination number
ligands
oxidation states

23
Q

colour energy

A

energy needed to jump = energy gap ΔE
ΔE = h v ( plancks constant x freq of light absorbed)
ΔE = h c / λ ( pc x speed of light / wavelength)

24
Q

if no 3d electrons or 3d sub level is full…

A

no electron jump
no energy absorbed
compound = white or colourless cuz all light reflected

25
Q

colorimetry/spectroscopy - why

A

calculate conc of complex by measuring how much light absorbed
colour intensity ∝ conc

26
Q

colorimetry/spectroscopy - steps

A

white light shone through filter
filter only lets colour of light absorbed pass
passes through sample to colorimeter
add on ligand to intensify colour if only TM sol

27
Q

variable oxidation states

A

have partially filled d-orbitals, so can lose 4s and 3d electrons

28
Q

vanadium and ox states

A

VO2 + - 5+ - yellow
VO 2+ - 4+ - blue
V 3+ - 3+ - green
V 2+ - 2+ - violet

29
Q

vanadium oxide + zinc

A 
half eq
VO2 (+) + 2H(+) + e- ----> VO (2+) + H2O
VO (2+) + 2H(+) + e- ----> V(3+) + H2O
V3(+) + e- ----> V(2+)
3e-'s all together

Zn —-> ZN (2+) + 2e-

3:2 ratio so… V’s x2 and Zn x3

overall:
8H(+) + 3Zn + 2VO2 (+) —-> 3Zn (2+) + 4H2O + 2V(2+)

30
Q

tollen’s reagent

A

[Ag(NH3)2]+

Ag = (1+)
Ag(+) + e- —> Ag
when added to aldehyde - Ag+ is reduced
aldehyde oxidises to carboxylic acid = silver mirror

31
Q

reducing/oxidising agents

A

reducing agent SO is oxidised SO loses electrons
Fe(2+) –> Fe(3+) + e-

oxidising agent SO is reduced SO gains electrons
Cl2 + 2e- –> 2Cl(-)

32
Q

redox

A

balance O’s with H2O
balance H’s with H+
balance charges with e-

ratio of e-‘s

33
Q

MnO4(-) –> Mn(2+)

A

deep purple to pink

5e- + 8H(+) + MnO4(-) –> Mn(2+) + 4H2O

34
Q

Cr2O7(2-) –> Cr(3+)

A

orange to green

6e- + 14H(+) + Cr2O7(2-) –> 2Cr(3+) + 7H2O

35
Q

aqua metal ions in acidic/alkaline conditions

A

acidic - reduced

alkaline - oxidised

36
Q

oxylate ions

A

C2O4(2-) –> 2CO2 + 2e-

37
Q

iron ions and manganate ions

A

Fe(2+) –> Fe(3+) + e-
5e- + 8H(+) + MnO4(-) –> Mn(2+) + 4H2O
e’s are 1:5

5Fe(2+) + MnO4(-) + 8H(+) –> 5Fe(3+) + Mn(2+) + 4H2O

38
Q

equilibrium displaced

A

no. reactant molecules smaller and no. product molecules bigger
disorder increase
entropy increase

39
Q

iron importance

A

needed for haemoglobin

40
Q

define homogenous vs heterogenous

A

same state/phase vs different phase to the reactants

41
Q

without catalyst, activation energy is high

A

negative ions which repel

42
Q

catalyst properties

A

has variable oxidation states
ea lowers due to oppositely charged ions attracting
catalyst can be reduced and oxidised back

43
Q

heterogenous catalyst

A

reactants adsorb to its active site
bonds weaken
desorb

44
Q

contact process

A

V2O5 catalyst

SO2 + 1/2 O2 —> SO3

STEP 1: SO2 + V2O5 → SO3 + V2O4
STEP 2: V2O4 + ½O2 → V2O5

45
Q

iodine clock

A

S2O8(2-) + 2 I(-) → 2SO4 (2-) + I2
v high ae since -ve repel
SO iron ii catalyst

S2O8 (2-) + 2Fe (2+) → 2Fe(3+) + 2SO4 (2-)
2Fe(3+) + 2I(-) → 2Fe(2+) + I2

46
Q

auto catalysis

A

Mn2+ = catalyst

5e- + 8H(+) + MnO4(-) –> Mn(2+) + 4H2O
C2O4(2-) –> 2CO2 + 2e-

2:5 ratio SO
5C2O4(2-) + 16H(+) + 2MnO4(-) –> 2Mn(2+) + 8H2O + 10CO2

Mn(2+) —> Mn(3+) + e-

STEP 1: MnO4(-) + 4Mn(2+) + 8H(+) → 4H2O + 5Mn(3+)
STEP 2: 2Mn3(+) + C2O4(2-) → 2CO2 + 2Mn(2+)

47
Q

autocatalysis

A

product of a reaction is also a catalyst for that reaction

48
Q

impurities in heterogenous reaction

A

block active sites/deactivates catalyst

49
Q

graph

A

reaction is slow/shallow grad as conc of reactant used

faster/gradient incr as catalyst forms

50
Q

fe cr2o7

A

1:6

Chemistry May (3 decks)

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