chapter 24

Question 1

special case of Chromium and Copper

Answer 1

• do not follow expected principal
• stability
• half filled d5 sub shell and a fully filled d10 subshell give additional stability

Question 2

transition elements

Answer 2

d-block elements that can form an ion with a partially filled d-orbital.
Sc and Zn are d-block elements but not transition metals

Question 3

common oxidation states- Ti

Answer 3

+2, +3 (violet), +4, +5

Question 4

common oxidation states V

Answer 4

+2 (violet), +3 (green), +4 blue, +5 (yellow)

Question 5

common oxidation states Cr

Answer 5

+2 (blue), +3 (green), +4, +5, +6 (red/orange)

Question 6

common oxidation states Mn

Answer 6

+2 (pink), +3, +4 (dark pink), +5, +6 (green), +7 (violet)

Question 7

common oxidation states Fe

Answer 7

+2 (cream), +3 (pale yellow), +4, +5, +6

Question 8

common oxidation states Co

Answer 8

+2 (bright pink), +3 (green), +4, +5

Question 9

common oxidation states Ni

Answer 9

+2 (green), +3, +4

Question 10

Common oxidation states Cu

Answer 10

+1, +2 (blue), +3

Question 11

Chromium

Answer 11

2 oxidation states in its compounds, +3 & +6
3+- green
6+- orange

Question 12

transition metals as catalysts examples

Answer 12

• iron in the Haber process
• vanadium (V) oxide in the contact process
• Ni in hydrogenation
• manganese (IV) oxide in decomposition of hydrogen peroxide

Question 13

ligand

Answer 13

a molecule or ion that donates a pair of electrons to a central metal ion to form a coordinate bond (dative covalent)

Question 14

monodentate ligand

Answer 14

a ligand that is able to donate one pair of electrons to a central metal ion eg water, ammonia, chloride, cyanide & hydroxide

Question 15

bidentate ligand

Answer 15

a ligand that can donate two lone pairs of electrons to the central metal ion, forming 2 coordinate bonds eg 1,2-diaminoethane or ethanedioate ion (oxalate)

Question 16

square planar complexes

Answer 16

occur in complex ions of transition metals w/ 8 d-electrons in the highest energy d-subshell- Platinum (II), palladium (II) and gold (III)

Question 17

cis-trans isomerism

Answer 17

eg square planar- in cis- isomer the 2 identical groups are adjacent to each other- 90 degrees apart, trans- isomer 2 identical groups are opposite each other- 180 degrees apart

Question 18

optical isomerism in octahedral complexes

Answer 18

only occurs in complexes containing 2 or more bidentate ligands
optical isomers are non-superimposable mirror images of each other
trans isomers cannot form optical isomers

Question 19

ligand substitution reaction

Answer 19

one in which one ligand in a complex ion is replaced by another ligand

Question 20

aq copper (II) ions with ammonia, colours and stages

Answer 20

pale blue complex ion [Cu(H2O)6}2+– changes to dark blue [Cu(NH3)4(H20)2]2+, 4NH3 ligands are substituted

2 diff reactions:

1) pale blue precipitate of Cu(OH)2 formed
2) Cu(OH)2 precipitate dissolves in excess ammonia to from a dark blue solution

Question 21

aq copper (II) ions w/ cl- ions

Answer 21

HCl used
pale blue--> yellow
change in shape and coordination number
[Cu(H20)6]2+(aq) + 4Cl-(aq) ---> [CuCl4]2-(aq) + 6H20
octahedral--> tetrahedral

Question 22

aq Chromium (III) ions w/ ammonia

Answer 22

[Cr(H2O)6]3+ + 6NH3 –> [Cr(NH3)6]3+ + 6H2O
violet–> purple

1) initially grey-green precipitate Cr(OH)3 formed
2) Cr(OH)3 dissolves in excess ammonia to form [Cr(NH3)6]3+

Question 23

precipitation reaction

Answer 23

occurs when 2 aq solutions containing ions react together to form an insoluble ionic solid- precipitate

Question 24

precipitation reaction w/ NaOH- Cu2+

Answer 24

blue solution–> blue precipitate
insoluble in excess NaOH
Cu2+ + 2OH- –> Cu(OH)2(s)

Question 25

precipitation reaction w/ NaOH- Fe2+

Answer 25

``` pale green sol--> dark green precipitate turns brown in air as Fe2+ is oxidised to Fe3+ Fe2+ + 2OH- --> Fe(OH)2 in air; Fe(OH)2 --> Fe(OH)3 ```

Question 26

precipitation reaction w/ NaOH- Fe3+

Answer 26

pale yellow sol--> orange-brown | Fe3+ + 3OH- --> Fe(OH)3

Question 27

precipitation reaction w/ NaOH- Mn2+

Answer 27

pale pink sol--> light brown precipitate | Mn2+ +2OH- --> Mn(OH)2

Question 28

precipitation reaction w/ NaOH- Cr3+

Answer 28

violet sol--> green precipitate Cr3+ + 3OH- --> Cr(OH)3 green precipitate soluble in excess NaOH to form dark green sol Cr(OH)3 + 3OH- --> [Cr(OH)6]3-(aq) dark green sol

Question 29

precipitation w/ ammonia- Cu

Answer 29

Cu2+ + 2OH- --> Cu(OH)2 blue precipitate which dissolves in excess ammonia to form a deep blue solution [Cu(NH3)4(H2O)2]2+

Question 30

precipitation w/ ammonia- Cr

Answer 30

Cr3+ + 3OH- --> Cr(OH)3 green precipitate dissolves in excess ammonia to form [Cr(NH3)6]3+ purple solution

Question 31

precipitation w/ ammonia- Fe2+, Fe3+ & Mn2+

Answer 31

react w/ excess ammonia in the same way they react w/ aq sodium hydroxide, precipitates of Fe(OH)2, Fe(OH)3 and Mn(OH)2 no further reaction w/ aq ammonia, precipitates do not dissolve

Question 32

reduction of Cr2O7 2- to Chromium (III) Cr3+

Answer 32

orange--> green acidified Cr2O7 2- can be reduced to 3+ by the addition of zinc Cr2O7 2-(aq) + 14H+(aq) + 3Zn(s)--> 2Cr3+(aq) + 7H2O + 3Zn2+(aq) excess of Zn Cr3+ can be reduced further to Cr2+ Zn(s) + 2Cr3+(aq) --> Zn2+(aq) + 2Cr2+(aq) green--> pale blue Cr2O7 has more positive E value so is more likely to gain electrons, then Cr3+, then Zn

Question 33

oxidation of Cr3+ to CrO42-

Answer 33

hydrogen peroxide is a powerful oxidising agent 3H2O2 + 2Cr3+ + 10OH- --> 2CrO4 2- + 8H2O Cr- +3-> +6 oxidised O- -1-> -2 reduced

Question 34

Reduction of Cu2+ to Cu+

Answer 34

``` react w/ excess iodide ions - I- is oxidised to brown I2 - Cu2+ is reduced to Cu+ + Cu+ forms a white precipitate of copper (I) iodide 2Cu+ + 4I- --> 2CuI + I2 ```