A-LEVEL CHEMISTRY TOPIC FIFTEEN,15, TRANSITION METALS.

Question 1

D-BLOCK ELEMENTS, LOCATION.

Answer 1

THE D-BLOCK ELEMENTS SIT IN THE MIDDLE OF THE PERIODIC TABLE.

SOME OF THESE ELEMENTS ARE TRANSITION METALS.

Question 2

WHAT IS A TRANSITION ELEMENT?

Answer 2

A TRANSITION ELEMENT IS A D-BLOCK ELEMENT THAT CAN FORM AT LEAST ONE,1, STABLE ION WITH A PARTIALLY FILLED, INCOMPLETE, D-SUBSHELL.

Question 3

ELECTRON CONFIGURATION ORDER.

Answer 3

1S2, 2S2, 2P6, 3S2, 3P6, 4S2, 3D10.

Question 4

PERIODIC TABLE BLOCK.

Answer 4

SPDF.

Question 5

D-SUBSHELL.

Answer 5

THE D-SUBSHELL CAN HOLD UP TO TEN, 10, ELECTRONS.

Question 6

D-BLOCK ELEMENTS THAT ARE NOT TRANITION ELEMENTS.

Answer 6

FOR PERIOD 4 D-BLOCK ELEMENTS ONLY EIGHT,8, OF THESE ARE TRANSITION ELEMENTS.

SCANDIUM AND ZINC ARE NOT TRANSITION ELEMENTS.

THIS IS BECAUSE THEY DO NOT FORM A STABLE ION WITH A PARTIALLY FILLED, INCOMPLETE D-SUBSHELL.

Question 7

CHROMIUN AND COPPER.

Answer 7

AN ELECTRON FROM THE 4S ORBITAL MOVES INTO THE 3D SUBSHELL TO CREATE A MORE STABLE HALF FULL OR FULL 3D SUB-SHELL.

LOOSE FROM THE 3D SUB-SHELL FIRST.

Question 8

ELECTRON LOSS D-BLOCK ELEMENTS.

Answer 8

YOU ALWAYS LOOSE FROM THE 4S SUB-SHELL FIRST.

Question 9

TRANSITION METALS PROPERTIES.

Answer 9

TRANSITION METALS HAVE SPECIFIC PROPERTIES WHICH INCLUDE VARIABLE OXIDATION STATES, FORM COLOURED IONS IN SOLUTION AND ARE GOOD CATALYSTS.

Question 10

TRANSITION METALS VARIABLE OXIDATION STATES.

Answer 10

TRANSITION METALS HAVE VARIABLE OXIDATION STATES.

THIS IS BECAUSE THE ELECTRONS SIT IN 4S AND 3D ENERGY LEVELS WHICH ARE VERY CLOSE.

THE ELECTRONS ARE GAINED AND LOST USING A SIMILAR AMOUNT OF ENERGY.

Question 11

D-BLOCK ELEMENTS, THE IONS FORMED.

Answer 11

VANADIUM, V
2+
3+

VO,
2+
3+

CHROMIM, Cr,
3+

CrO,

Cr2O7 2-

MANGANESE, Mn

2+

MnO,

MnO4 2-

MnO4 -

IRON, Fe

2+
3+

COLBOLT, Co
2+

NICKLE, Ni
2+

COPPER, Cu
2+

TITANIUM, Ti
2+
3+

Question 12

TRANSITION METALS IN SOLUTION.

Answer 12

THEY FORM COLOURED IONS IN SOLUTION.

COLOURED IONS IN AQEOUS SOLUTION.

Question 13

VANADIUM COLOURS.

Answer 13

V2+
VIOLET.

V3+,
GREEN.

VO2+,
BLUE.

VO2 +,
YELLOW.

Question 14

CHROMIUM COLOURS.

Answer 14

Cr3+,
GREEN/VIOLET.

VIOLET WHEN SURROUNDED BY 6H2O.
THEY ARE NORMALLY SUBSTITUTED SO USUALLY LOOK GREEN.

Cr2O7 2-,
ORANGE.

Question 15

IRON COLOURS.

Answer 15

FE2+,
GREEN.

FE3+, YELLOW.

Question 16

COLBOLT COLOURS.

Answer 16

Co2+,
PINK.

Question 17

COPPER COLOURS.

Answer 17

Cu2+,
BLUE.

Question 18

TRANSITION METALS, COMPLEX IONS.

Answer 18

TRANSITION METALS CAN FORM COMPLEX IONS.

A COMPLEX ION IS WHERE A CENTRAL TRANSITION METAL ION IS SURROUNDED BY LIGANDS BONDED BY DIATIVE, COORDINATED, BONDS.

THE LIGANDS HAVE AT LEAST ONE,1, LONE PAIR OF ELECTRONS WHERE THEY ARE USED TO FORM A DIATIVE COVALENT, COORDINATE, WITH A METAL.

A SQUARE BRACKET SHOWS THE FULL COMPLEX AND OVERALL CHARGE OF THE COMPLEX SITS OUTSIDE OF THIS.

COMPLEX IONS CAN COME IN DIFFERENT SHAPES.

Question 19

LIGAND.

Answer 19

A LIGAND IS AN ION, ATOM OR MOLECULE THAT HAS AT LEAST ONE,1, LONE PAIR OF ELECTRONS.

THEY CAN BE MONODENTATE, BIDENTATE OR POLYDENTATE.

Question 20

MONODENTATE LIGAND.

Answer 20

LIGANDS WHICH ONLY HAVE ONE,1, LONE PAIR OF ELECTRONS ARE CALLED MONODENTATE, UNIDENTATE, LIGANDS.

H2O:

:NH3

:Cl-

:OH-

Question 21

BIDENTATE LIGAND.

Answer 21

LIGANDS WHICH HAVE TWO,2, LONE PAIRS OF ELECTRONS ARE CALLED BIDENTATE LIGANDS.

ETHANEDIOATE.

ETHANE-1,2-DIAMINE.

Question 22

MULTIDENTATE LIGANDS.

Answer 22

LIGANDS WHICH HAVE MORE THAN ONE,1, COORDINATE BOND ARE CALLED MULTIDENATE LIGANDS.

EDTA 4- IS AN EXAMPLE OF A MULTIDENATE LIGAND.

IT CAN FOR SIX,6, COORDINATE BONDS WITH THE CENTRAL METAL ION.

Question 23

COMPLEX SHAPES.

Answer 23

THE SHAPE IS DEPENDANT ON THE SIZE OF THE LIGANDS AND THE COORDINATION NUMBER.

THE COORDINATION NUMBER IS THE NUMBER OF COORDINATE BONDS IN A COMPLEX.

OR THE NUMBER OF LIGANDS.

SOME LIGANDS ARE SMALL AND YOU CAN FIT SIX,6, OF THEM AROUND A CENTRAL METAL ION.

WATER, AMMONIA AND HYDROXIDE IONS.

SOME LIGANDS ARE LARGER AND YOU CAN ONLY FIT FOUR,4, OF THEM AROUND THE CENTRAL METAL ION,

:Cl-

ETHANEDIOATE AND ETHANE-1,2-DIAMINE ARE LARGER STILL.

NORMALLY YOU HAVE THREE,3, OF THEESE AROUND A CENTRAL METAL ION

Question 24

OCTAHEDRAL.

Answer 24

COMPLEXES WITH A COORDINATION NUMBER OF SIX,6, FORM OCTAHEDRAL SHAPES.

ALL BOND ANGLES IN AN OCTAHEDRAL COMPLEX ARE NINTY,90, DEGREES.

Question 25

TETRAHEDRAL AND SQUARE PLANAR SHAPES.

Answer 25

COMPLEXES WITH A COORDINATION NUMBER OF FOUR,4, FORM TETRAHEDRAL AND SQUARE PLANAR SHAPES.

WILL USUALLY HAVE LARGER LIGANDS AROUND THEM, THIS IS WHY IT HAS LESS COORDINATIONS.

BOND ANGLES IN A TETRAHEDRAL COMPLEX ARE 109.5 DEGREES.

SQUARE PLANAR COMPLEXES,

PLATNIUM, Pt,
WITH TWO,2, CHLORINE LIGANDS AND TWO,2, AMMONIA LIGANDS.

THIS IS AN ANTI CANCER DRUG CIS-PLATIN.

BOND ANGLES IN A SQUARE PLANAR COMPLEX ARE ALWAYS NINTY,90, DEGREES.

Question 26

COMPLEX SHAPES, OVERALL CHARGE.

Answer 26

COMPLEXES HAVE AN OVERALL CHARGE WHICH IS THE SAME AS IT’S TOTAL OXIDATION STATE.

TOTAL OXIDATION STATE OF METAL.= TOTAL OXIDATION STATE - TOTAL OXIDATION STATE IF LIGANDS.

FIND THE OVERAL CHARGE, AND MINUS THE CHARGE OF EACH THE LIGANDS TIMES BY HOW MANY THERE ARE.

Question 27

HAEMOGLOBIN.

Answer 27

HAEM IS A MULTIDENTATE LIGAND THAT IS FOUND IN THE MOLECULE HAEMOGLOBIN.

HAEMOGLOBIN IS A PROTEIN USED TO TRANSPORT OXYGEN AROUND THE BODY IN BLOOD.

THE STRUCTURE IS OCTAHEDRAL.

FOUR,4, OF THE NITROGEN COMES FROM ONE MULTIDENTATE LIGAND CALLED HAEM.

ONE OF THE COORDINATE BONDS COMES FROM A LARGE PROTEIN CALLED GLOBIN.

THE FINAL COORDINATE BONDS COMES FROM EITHER AN OXYGEN OR WATER MOLECULE.

Question 28

HAEMOGLOBIN, OXYGEN TRANSPORT.

Answer 28

OXYGEN SUBSTITUTES THE WATER LIGAND IN THE LUNGS WHERE OXYGEN CONCENTRATION IS HIGH TO FORM OXYHAEMOGLOBIN.

THIS IS TRANSPORTED AROUND THE BODY.

OXYHAEMOGLOBIN GIVES UP OXYGEN TO A PLACE WHERE IT IS NEEDED.

WATER TAKES THE PLACE AND HAEMOGLOBIN RETURNS BACK TO THE LUNGS TO START THE PROCESS AGAIN.

Question 29

HAEMOGLOBIN AND CARBON MONOXIDE.

Answer 29

CARBON MONOXIDE IS A POISONOUS GAS THAT CAUSES HEADACHES, UNCONSCIOUSNESS AND EVEN DEATH.

IT IS KNOWN AS THE SILENT KILLER.

IF CARBON MONOXIDE, CO, IS INHALED THE WATER LIGAND IS REPLACED WITH A CARBON MONOXIDE, CO, LIGAND.

UNFORTUNATELY CARBON MONOXIDE BONDS STRONGY SO IT IS NOT READILY REPLACED BY OXYGEN OR WATER.

THIS MEANS THAT OXYGEN CAN NOT BE TRANSPORTED AND LEADS TO OXYGEN STARVATION IN ORGANS.

HENCE WHY CARBON MONOXIDE IS POISONOUS.

Question 30

OPTICAL ISOMERISM.

Answer 30

COMPLEX IONS SHOW OPTICAL ISOMERISM, A TYPE OF STEREOISOMERISM.

COMPLEXES ARE OPTICAL ISOMERES WHEN THEY ARE NON SUPERIMPOSABLE MIRROR IMAGES.

REFLECTED IN THE MIRROR LINE, SO THEY ARE MIRROR IMAGES.

OCTAHEDRAL COMPLEXES WITH THREE,3, BIODENTATE LIGANDS SHOW OPTICAL ISOMERS.

Question 31

CIS-TRANS ISOMERISM.

Answer 31

COMPLEX IONS ALSO SHOW CIS-TRANS ISOMERISM, ANOTHER TYPE OF STERERISOMERISM, WHICH IS A BRANCH OF E./.=Z ISOMERISM.

OCTAHEDRAL COMPLEXES WITH FOUR,4, LIGANDS OF THE SAME TYPE TWO,2, LIGANDS OF A DIFFERENT TYPE DISPLAY CIS-TRANS ISOMERISM.

IF THE TWO,2, DIFFERENT LIGANDS ARE OPPOSITE EACH OTHER YOU HAVE A TRANS ISOMER.

IF THE TWO,2, DIFFERENT LIGANDS ARE ADJACENT TO EACH OTHER YOU HAVE A CIS ISOMER.

Question 32

SQUARE PLANAR COMPLEX IONS, CIS-TRANS ISOMERISM.

Answer 32

SQUARE PLANAR COMPLEX IONS ALSO HAVE CIS-TRANS ISOMERISM.

SQUARE PLANAR COMPLEXES WITH TWO,2, LIGANDS OF THE SAME TYPE AND TWO,2, LIGANDS OF A DIFFERENT TYPE DISPLAY CIS-TRANS ISOMERISM.

IF THE TWO,2, DIFFERENT LIGANDS ARE OPPOSITE EACH OTHER YOU HAVE A TRANS ISOMER.

IF THE TWO,2, DIFFERENT LIGANDS ARE ADJACENT TO EACH OTHER YOU HAVE A CIS ISOMER.

Question 33

D-ORBITAL SPLITTING.

Answer 33

THE D-SUBSHELL IS SPLIT INTO TWO,2, WHEN LIGANDS BOND WITH THE CENTRAL METAL ION.

FOR EXAMPLE, YOU ATTACH LIGANDS AND SOME ORBITALS GAIN ENERGY. AN ENERGY GAP IS CREATED.

Question 34

ENERGY GAP REPRESENTATION.

Answer 34

THE ENERGY GAP IS REPRESENTED BY DELTA E.

Question 35

D-ORBITAL SPLITTING, ENERGY.

Answer 35

WHEN ELECTRONS ABSORB LIGHT ENERGY SOME MOVE FROM THE LOWEST ENERGY LEVEL, GROUND STATE, TO HIGHER ENERGY LEVEL ORBITALS, EXCITED STATE.

IN ORDER FOR THIS TO HAPPEN THE ENERGY FROM THE LIGHT MUST EQUAL THE ENRGY GAP, DELTA E.

Question 36

DELTA E, SIZE DEPEND.

Answer 36

THE SIZE OF DELTA E IS DEPENDANT ON:

THE CENTRAL METAL ION AND ITS OXIDATION SATE.

THE TYPE OF LIGAND.

COORDINATION NUMBER.

Question 37

COLOURED COMPLEXES.

Answer 37

SOME FREQUENCIES OF VISIBLE LIGHT ARE ABSORBED BY TRANSITION METALS COMPLEXES.

THE FREQUENCIES ABSORBED DEPENDS ON THE SIZE OF THE ENERGY GAP, DELTA E.

THE LARGER THE ENERGY GAP, DELTA E, THE HIGHER THE FREQUENCY OF LIGHT ABSORBED.

ANY FREQUENCIES WHICH ARE NOT ABSORBED ARE REFLECTED OR TRANSMITTED.

THE COMBINATIONS OF ALL THESE FREQUENCIES CREATE A COMPLIMENTARY COLOUR THAT WE OBSERVE WITH SOME COMPLEXES.

WE OBSERVE THE REFLECTED OR TRANSMITTED FREQUENCY THAT WAS NOT ABSORBED.

FOR THOSE COMPLEXES WHERE WE HAVE A FULL OR EMPTY 3D SUBSHELL NO ELECTRONS CAN MIGRATE TO THE HIGHER ENERGY LEVEL.

THIS MEANS WE SEE THESE COMPLEXES AS COLOURLESS OR WHITE.

Question 38

COMPLEMENTARY COLOURS.

Answer 38

THE COLOURS WE SEE ARE COMPLIMENTS OF THE COLOURS ABSORBED BY THE SOLUTION.

WHITE LIGHT IS MADE OF THE COLOURS WE SEE IN THE LIGHT SPECTRUM. WHEN THIS HITS A TRANSITION METAL SOLUTION ONLY ONE FREQUENCY IS ABSORBED.

ANY FREQUENCIES WHICH ARE NOT ABOSRBED ARE REFELECTED OR TRANSMITTED.

FOR EXAMPLE, A GREEN COLOUR OBSERVED WOULD MEAN A MAGENTA COLOUR IS BEING ABSORED.

Question 39

TRANSITION METALS AND REDOX POTENTIALS.

Answer 39

REDOX POTENTIALS TELL US HOW EASILY AN ION IS REDUCED WHICH IS THE SAME AS ELECTRODE POTENTIALS.

THE LEAST STABLE IONS HAVE THE LARGEST REDOX POTENTIAL AND ARE MORE LIKELY TO BE REDUCED.

MORE POSITIVE, LESS STABLE.

THE MOST POSITIVE, THE ONE THAT IS REDUCED.

REMEMBER E THETA VALUES HAVE TO HAVE STANDARD CONDITIONS:

TEMPERATURE AT 298 K, KELVINS.
PRESSUE AT 100KPa, KILLER PASCALS.
CONCENTRATIONS OF IONS AT ONE,1, moldm^-3.

Question 40

VANDIUM CHEMISTRY.

Answer 40

VANADIUM IS A TRASNITION METAL THAT HAS A WIDE RANGE OF DIFFERENT OXIDATION STATES AND COLOURS.

V2+,
OXIDATION STATE, +2.
VIOLET.

V3+,
OXIDATION STATE, +3.
GREEN.

VO2+,
OXIDATION STATE, +4.
BLUE.

VO2 +,
OXIDATION STATE,+5.
YELLOW.

Question 41

REDUCTION POTENTIALS.

Answer 41

THE REDUCTION PORTENTIALS CAN BE USED TO WORK OUT IS A REACTION INVOLVING TRANSITION METAL IS LIKELY TO HAPPEN.

IN ORDER TO OCCUR THE E THETA VALUE, MUST BE POSITIVE.

E THETA.= REDUCTION- OXIDATION.

Question 42

CHROMIUM CHEMISTRY.

Answer 42

Cr2O7 2-,
OXIDATION STATE, +6.
ORANGE.
ALSO KNWON AS DICHROMATE (VI) IONS, THEY ARE GOOD OXIDISING AGENTS.

CrO3 2-.
OXIDISIND STATE, +G.
YELLOW.
ALSO KNOWN AS CHROMATE (VI), THEY ARE ALSO GOOD OXIDISING AGENTS.

Cr3+,
OXIDATION STATE, +3.
GREEN.
THIS IS THE MOST STABLE IONS, WHEN SURROUNDED BY SIX,6, WATER LIGANDS A VIOLET SOLUTION IS FORMED.

HOWEVER Cl- IMPURITIES REPLACE THE WATER LIGANDS TO MAKE IT LOOK GREEN.

Cr2+,
OXIDATION STATE,+2.
BLUE.

Question 43

AMPHOTERIC CHROMIUM HYDROXIDES.

Answer 43

CHROMIUM HYDROXIDES CAN ACT AS AN ACID OR A BASE WHICH MEANS THEY ARE AMPHOTERIC.

WE CAN HYDROLYSE METAL AQUA IONS TO FORM INSOLUBLE CHROMIUM HYDROXIDES BY ADDING A BASE, OH- IONS OR NH3.

CHROMIUM ADDED TO WATER, WHICH IS THEN ADDED TO HYDROXIDE,

THREE,3, WATER LIGANDS
AND THREE,3, HYRDOXIDE LIGANDS.
THEY FORM A SOLID COMPLEX BECAUSE IT NOW HAS NO CHARGE.
THIS IS WHY WE SEE A PRECIPITATE.

WE ALSO GET WATER AS A PRODUCT, 3H2O.

CHROMOIUM ADDED TO WATER WHICH IS ADDD TO AMMONIA,

THREE,3, WATER LIGANDS
AND THREE,3, HYDROSIDE LIGANDS.
FORMS A SOLID COMPLES.
NOW HAS NO CHARGE.
THIS IS WHY WE SEE A PRECIPITATE.
WE ALSO GET AMMONIUM AS A PRODUCT, 3NH4+.

CHROMIUM HYDRTOXIDES ARE AMPHOTERIC:

WHEN WE ADD A BASE IT ACTS AS AN ACID AND DONATES H+ IONS TO REACT WOTH THE OH- AND IT DISSOLVES.

WHEN WE ADD AN ACID IT ACTS AS A BASE, BY ACCEPTING H+ IONS AND IT DISSOLVES.

THE REACTIONS FORM A GREY-GREEN SOLID OF CHROMIUM AND WATER.

Question 44

ADDING AQEOUS AMMONIA TO THE CHROMIUM.

Answer 44

IF WE ADD EXCESS AMMONIA, NH3, TO SOLID Cr(H2O)3(OH)3 WE DO GET A LIGAND EXCHANGE REACTION.

PURPLE Cr(NH3)6 3+ SOLUTION IS FORMED,
AND WATER AND HYDROXIDE IONS.
THREE,3, OF EACH.

Question 45

MAKING CHROMIUN COMPLEXES.

Answer 45

MOST COMPLEXES CAN BE MADE IN A BEAKER, HOWEVER SOME COMPLEXES SUCH AS CHROMIUM (II) ETHANOATE Cr2(CH3COO)4 (H2O)2 REUIRE A MORE COMPLEX METHOD.

FIRST ACIDIFIED SODIUM DICHROMATE (VI), ORNAGE, IS REDUCED USING ZINC.

IT FORMS:

A GREEN SOLUTION CONTAINING Cr 3+ IONS.

A BLUE SOLUTION CONTAINING Cr 2+ IONS.

SECONDLY, SODIUM ETHANOATE REACTS WITH Cr 2+ TO FORM A RED PRECIPITAE OF CHROMIUM (II) ETHANOATE.

Question 46

MAKING CHROMIUM COMPLEXES, CONDITIONS.

Answer 46

Cr 2+ IONS ARE EASILY OXIDISED BACK TO Cr 3+ SO THE WHOLE EXPERIMENT MUST BE DONE IN INERT CONDITIONS.

NORMALLY A NITROGEN ATMOSPHERE IS USED.

Question 47

WHEN CAN A COLOUR CHANGE EXIST?

Answer 47

A COLOUR CHANGE CAN EXIST WHEN LIGANDS IN A COMPLEX EXCHANGE./.= SUBSTITUTE.

THESE SUBSTITUTION REACTIONS SHOW LIGANDS OF A SIMILAR SIZE BEING EXCHANGED.

YOU CAN IN FACT GET A PARTIAL SUBSTITUTION WHEN NOT ALL OF THE LIGANDS HAVE BEEN SUBSTITUTED.

Question 48

LIGAND STRENGTH.

Answer 48

DIFFERENT LIGANDS CAN FORM DIFFERENT STRENGTH BONDS TO THE METAL ION.

THIS MEANS THAT REACTIONS ARE NOT EASILY REVERSED, AS THE NEW COMPLEX FORMED IS MORE STABLE.

MULTIDENTATE LIGANDS FORM COMPLEXES THAT ARE MORE STABLE THAN MONODENTATE LIGANDS.

Question 49

ENTROPY IMPACT ON COMPLEX.

Answer 49

ENTROPY INCREASING FORMS A MORE STABLE COMPLEX.

IN A LIGAND EXCHANGE REACTION BONDS ARE BROKEN IN THE ORIGINAL COMPLEX AND NEW ONES ARE FORMED TO MAKE THE NEW COMPLEX.

QUITE OFTEN THE ENERGY NEEDED TO BREAK THE BONDS IS SIMILAR AS THE ENRGY REALESED WHEN NEW ONES ARE FORMED.

THE ENTHALPY CHANGE, DELTA H, IS SMALL.

Question 50

REVERSIBLE REACTIONS.

Answer 50

EVEN IF A REACTION HAS A SMALL ENTHALPY CHANGE, IF THE PRODUCT FORM IS MORE STABLE THE REACTION WILL NOT BE REVERSIBLE.

Question 51

THE CHELATE EFFECT.

Answer 51

THE INCREASE IN STABILITY IS KNOWN AS THE CHELATE EFFECT.

WHEN WE SUBSTITUTE MONODENATE LIGANDS WITH BIDENATE AND MULTIDENATE LIGANDS WE CREATE A SOLUTION WITH MORE PARTCILES IN IT.

THIS MEANS THAT WE HAVE INCREASED THE ENTROPY.

REACTIONS THAT HAVE AN INCREASE IN ENTROPY ARE MORE LIKELY TO HAPPEN.

IF THE NUMBER OF PARTICLES INCREASES, AS DOES THE ENTROPY.

THIS IS REALLY DIFFIFULT TO REVERSE, AS THIS WOULD MEAN IN DECREASE IN ENTROPY.

AN ENTROPICALLY UNFAVOURABLE REACTION, WOULD BE IF YOU WERE TO REVERSE THIS REACTION.

Question 52

WHAT DOES THE COLOUR OF COMPLEX IONS DEPEND ON?

Answer 52

COLOUR OF COMPLEX IONS DEPENDS ON THE SIZE OF DELTA E, ENERGY DIFFERENCE WHICH IS AFFECTED BY CHANGE OF OXIDATION STATE, CO-ORDINATION NUMBER AND CHANGE OF LIGAND.

Question 53

WHAT USUALLY CHANGES WHEN YOU CHANGE THE COORDINATION NUMBER?

Answer 53

WHEN YOU CHANGE THE COORDINATION NUMBER, THE SHAPE OF THE COMPLEX ALSO USUALLY CHANGES.

WE USUALLY SEE A CHANGE IN COORDINATION NUMBER BEACUSE THE SIZE OF THE LIGAND CHANGES.

Question 54

COPPER, COLOUR CHANGE COMPLEX EQUATIONS.

Answer 54

COPPER,
Cu2+.

IN AQEOUS SOLUTION,
BLUE.
IN AQEOUS SOLUTION MEANS DISSOLVES IN WATER OR IN WATER.
[Cu(H2O)6] 2+
OR Cu2+.
THEY ARE BOTH AQEOUS, SO HAVE AQEOUS STATE SYMBOLE, (aq).

ADD SOME OH- (aq) OR NH3 (aq) TO AQEOUS SOLUTION,
PALE BLUE PRECIPITATE.
Cu(OH)2(H2O)4 (s)

ADD EXCESS OH- (aq) TO THE PRECIPITATE,
INSOLUBLE IN EXCESS NaOH,
NO CHANGE.

ADD EXCESS NH3 (aq) TO THE PRECIPITATE,
DARK BLUE SOLUTION,
PARTIAL LIGAND SOLUTION.
[Cu(NH3)4 (H2O)2] 2+ (aq).

Question 55

IRON, COLOUR CHANGE COMPLEX EQUATIONS.

Answer 55

IRON,
FE 2+.

IN AQEOUS SOLUTION,
PALE GREEN.
[Fe(H2O)6] 2+ (aq)
OR Fe 2+ (aq).

ADD SOME OH- (aq) OR NH3 (aq) TO AQEOUS SOLUTION.
DIRTY GREEN PRECIPITATE.
Fe(OH)2(H2O)4 (s).

ADD EXCESS OH- (aq) TO THE PRECIPITATE,
INSOLUBLE TO NaOH.
NO CHANGE.

ADD EXCESS NH3 (aq) TO THE PRECIPIATE,
INSOLUBLE TO NH3,
NO CHANGE.

IRON,
Fe 3+.

FIRST EQUATION IS THE SAME JUST TO A 3+ CHARGE,
YELLOW.

SECOND ONE,
FE(OH)3 (H2O)3 (s)
ORANGE PRECIPIATE.

SAME RESULTS FOR THE OTHER TO,2, EQUATIONS.

Question 56

COLBOLT, COLOUR CHANGE COMPLEX EQUATIONS.

Answer 56

COLBOLT,
Co 2+.

IN AQEOUS SOLUTION,
[Co(H2O)6] 3+ (aq)
OR Co 2+ (aq).

ADSOME OH- (aq) OR NH3 (aq) TO AQEOUS SOLUTION.

Co(OH)2(H2O)4 (s)
BLUE PRECIPIATE.
TURNS BRON AFTER A WHILE.

NO CHANGE.
TO EXCESS OH- (aq).

ADD EXCESS NH3 (aq) TO THE PRECIPITATE,

Co(NH3)6 2+ (aq),
TURN BROWN AFTER A WHILE.

Question 57

WHAT CAN CATALYSTS BE?

Answer 57

CATALYSTS CAN BE HETEROGENEOUS OR HOMOGENEOUS.

HETEROGENEOUS IS A CATLAYST THAT IS IN A DIFFERENT PHASE FROM THE REACTANTS.

HOMOGENEOUS IS A CATLAYST THAT IS IN THE SAME PHASE AS THE REACTANTS.
THEY FORM AN INTERMEDIATE SPECIES BY REACTANTS COMBINING WITH THE CATALYST WHICH REACT TO FORM PRODUCTS.

THE CATALYST IS REFOMED AGAIN.

Question 58

INCREASING THE SURFACE AREA OF A HETERGENEOUS CATALYST.

Answer 58

INCREASING THE SURFACE ARE OF THE HETEREOGENEOUS CATALYST WILL INCREASE THE RATE OF REACTION.

MORE PARTICLES CAN REACT WITH THE CATALYST AT THE SAME TIME.

Question 59

TRANSITION METALS AS CATALYSTS.

Answer 59

AS TRANSITION METALS HAVE VERIABLE OXIDATION STATES THEY ARE GOOD CATALYSTS BY RECIEVING AND LOOSING SOME ELECTRONS IN THE D-ORBITALS TO SPEED UP REACTIONS.

Question 60

THE CONTACT PROCESS.

Answer 60

THE CONTACT PROCESS USES VANADIUM (V) TO MAKE SULFURIC ACID.

IN THE MAUFACTURE OF SULFURIC ACID WE USE V2O5, VANADIUM (V) AS A CATALYST.

THIS IS A HETEROGENOUS CATALYST.

IT USES TO CATALYSE SO2 TO SO3.

FIRST V205 OXIDISES SO2 TO SO3 AND IS ITSELF TO REDUCE V2O4.

V2O4 IS OXIDISED BY OXYGEN TO REFORM V2O5.

REMEBER CATALYSTS MUST ALWAYS BE REGENERATED.

Question 61

CATALYSTS USE.

Answer 61

CATALYSTS ARE USED TO MAKE PRODUCT FATER AND CAN BE USED TO LOWER THE TERMPERATURE REQUIRED FOR A REACTION.

THIS SAVES ENERGY AND MONEY.

BETTER FOR THE ENVIRONMENT.

Question 62

HETEROGENEOUS CATALYSTS.

Answer 62

HETEROGENEOUS CATALYSTS CAN BE POISONED BY IMPURITIES.

IMPURITIES CAN BIND TO THE SURFACE OF A CATALYST AND BLOCK ACTIVE SITES FOR REACTANTS TO ADSORB.

WHEN AN IMPURITY BLOCKS A SITE WE CALL THIS POISONING.

CATALYTIC POISONING REDUCES THE SURFACE AREA OF THE CATALYST FOR THE REACTANTS TO ADD.

THIS SLOWS DOWN THE REACTION.

Question 63

THE HARBOUR PROCESS.

Answer 63

THE HARBOUR PROCESS,

N2 (g) + 3H2 (g) -> 2NH3 (g).

THEY HYFROGEN IS MADE FROM METHANE.

METHANE CONAINS SULFUR IMPURITIES.

Question 64

CATALYST USED IN THE HARBOUR PROCESS.

Answer 64

A SOLID IRON CATALYST.

Question 65

POSIONED CATALYST EFFECT.

Answer 65

A POISONED CATALYST MEANS LESS PRODUCT IS MADE,
THE CATALYST NEED TO BE PREPLACED ORCLEANED MORE OFTEN,
INCREASED COST OF THE CHEMICAL PROCESS.

AN IRON CATALYST, THE HARBOUR PROCESS,

ANY SULFUR THAT IS NOT REMOVES WILL ADSORB TO THE SURFACE FORMING IRON SULFIDE.

THE CATALYST IS LESS EFFICENT.

Question 66

HETEROGENEOUS CATALYST, FUNCTION.

Answer 66

SUBSTANCES ADSORB ONTO THE SURFACE OF SOLID HETEROGENEOUS CATALYSTS.

REACTIONS OCUR ON SOLID HETEROGENEOUS CATALYSTS.

THEY BOND WITH THE SURFACE OF THE CATALYST.
THIS IS CALLED ADSORPTION.

THE BONDS IN THE REACTANTS WEAKEN AND BREAK TO FORM RADICALS.

THE RADICALS WILL REACT WITHH EACH OTHER TO MAKE NEW SUBSTANCES.

THE NEW MOLECULES ARE THEN RELEASED FROM THE SURFACE OF THE CATALYST IN A PROCESS CALLED DESORPTION.

Question 67

LEAD POISONING.

Answer 67

LEAD POISONS PLATNIUM CATALYSTS IN A CATALYTIC CONVERTER IN A CAR.

LEAD WAS FOUND IN PETROL HOWEVER THIS IS NO LONGER WIDELY USED.
IT REDUCED THE EFFECTIVNESS OF THE CATALYST.

THE POISON ADSORBS TO THE SURFACE OF THE CATALYST BETTER THSN THE REACTANT CAN.

Question 68

HOMOGENEOUS CATALYSTS.

Answer 68

HOMOGENEOUS CATALYSTS HAVE A UNIQUES ENERGY PROFILE DIAGRAM.

WE HAVE ENTHALPY ON THE Y AXIS.
REACTION PROGRESS ON THE X AXIS.

REACTANTS AND PRODUCTS LABELLED.

LABEL THE ACTIVATION ENERGY WITH AND WITHOUT A CATALYST.

WE FORM AN INTERMEDIATE SPECIES WHEN WE USE HETEROGENEOUS CATALYSTS.

THEY FORM BY REACTANTS COMBINING WITH THE CATALYST TO FORM PRODUCTS.

THE CATALYST IS REFORMED AGAIN.

Question 69

CATALYST, ACTIVATION ENERGY.

Answer 69

THE CATALYST LOWER THE ACTIVATION ENERGY SO MORE PARTICLES NOW HAVE ENOUGH ENERGY TO REACT.

Question 70

AMMONIUM ION, LIGAND.

Answer 70

THE AMMONIUM ION CAN NOT ACT AS A LIGAND BECAUSE IT IS A POSITIVE ION, AND SO DOES NOT HAVE A LONE PAIR OF ELECTRONS TO FORM A COORDINATE BOND.