Transition Metals Pt2

Question 1

What happens when transition metals form ions?

Answer 1

They lose the 4s electrons before the 3d

This process influences their oxidation states and reactivity.

Question 2

Why can transition metals easily oxidize and reduce?

Answer 2

Ions contain partially filled sub-shells of d electrons that can easily lose or gain electrons

This characteristic allows for a variety of oxidation states.

Question 3

What is the trend in the relative stability of oxidation states across the period?

Answer 3

Relative stability of +2 state with respect to +3 state increases across the period

This indicates that the +2 oxidation state becomes more favorable as you move across the transition metals.

Question 4

What do compounds with high oxidation states tend to be?

Answer 4

Oxidising agents

MnO is a compound with a high oxidation state.
Examples include V2+ and Fe2+ for reducing agents

Question 5

How many main oxidation states does vanadium have?

Answer 5

Four main oxidation states

These include +5, +4, +3, and +2.

Question 6

What is the color of the VO3- ion (oxidation state +5)?

Answer 6

Yellow solution

This ion is often found as ammonium vanadate (V).

Question 7

What is the color of the VO2+ ion (oxidation state +4)?

Answer 7

Blue solution

This is one of the oxidation states of vanadium.

Question 8

What color corresponds to the V3+ ion (oxidation state +3)?

Answer 8

Green solution

This indicates the presence of vanadium in the +3 state.

Question 9

What is the color of the V2+ ion (oxidation state +2)?

Answer 9

Violet solution

This is the lowest oxidation state for vanadium.

Question 10

What happens when acid is added to NH4VO3?

Answer 10

Turns into a yellow solution containing the VO2+ ion

This demonstrates the behavior of vanadium in acidic conditions.

Question 11

What occurs when zinc is added to vanadium (V) in acidic solution?

Answer 11

Reduces the vanadium down through each successive oxidation state

The color changes from yellow to blue to green to violet.

Question 12

What does the E value being more negative tell us?

Answer 12

Better reducing agent

Other one is better oxidising agent

Question 13

Reaction when zinc in HCl added to a dichromate (VI) ion - Cr2O7 2-? What happens if Fe2+ is used instead of zinc?

Answer 13

Reduced Cr2O7 2- (orange) into Cr3+ (green) and Cr2+ (blue)

Fe2+ weaker reducing agent so only reduce dichromate to Cr3+

Question 14

Conditions needed when zinc and dichromate react?

Answer 14

Keep zinc/dichromate under hydrogen atmosphere so can be reduced into Cr2+
As O2 in air will oxidise Cr2+ to Cr3+
- ACIDIC CONDITIONS

Cr2O7 2- + 14H+ + 3Zn —> 2Cr3+ + 7H2O + 3Zn2+
Cr2O7 2- + 14H+ + 4Zn —> 2Cr2+ + 7H2O + 4Zn2+

Question 15

How can Cr2+ be stabilised?

Answer 15

Form a stable complex ion with ligand, ethanoate ion
- bubble Cr2+ ions through SODIUM ETHANOATE —> stable red precipitate of chromium(II) ethanoate forms

Question 16

What is the reaction between Fe2+ and Cr2O7 2- an example of?

Answer 16

Quantitative redox titration - doesn’t need indicator

Cr2O7 2- + 14H+ + 6Fe2+ —> 2Cr3+ + 7H2O + 6Fe3+
ORANGE GREEN

Question 17

In terms of electrode potentials, why does zinc reduce dichromate down to Cr2+ and iron only to Cr3+?

Answer 17

Electrode potential of iron(II) is between the 2 chromium half equations - so reduce down to Cr3+

Zinc has more negative electrode potential than all chromium half equations so zinc will reduce chromium down to Cr2+

Question 18

How is chromium oxidised?

Answer 18

Acidified (not easy to oxidise) so add excess NaOH to form alkaline (easier to oxidise)
[Cr(H2O)6]3+ —-> [Cr(OH)6]3-

Oxidised by using ox agent - hydrogen peroxide
2 [Cr(OH)6]3- + 3H2O2 —> 2CrO4 2- + 2OH- + 8H2O
GREEN SOLUTION. YELLOW SOLUTION

Question 19

How can chromate (CrO4 2-) and dichromate ions (Cr2O7 2-) be converted into each other ?

Answer 19

Equilibrium reaction
2CrO4 2- + 2H+ —> Cr2O7 2- + H2O
Yellow solution. Orange solution

NOT REDOX as both have ox number +6 so ACID BASE reaction
- adding acid/alkali can shift position of equilibrium left and right respectively

Question 20

Iron (II) Metal aqua ions with limited OH- and NH3

Answer 20

[Fe(H2O)6]2+ (aq) + 2OH- (aq) →** [Fe(H2O)4(OH)2]**(s) + 2H2O (l)
GREEN solution—> GREEN PRECIPITATE darkens on standing

[Fe(H2O)6]2+ (aq) + 2NH3 (aq) → [Fe(H2O)4(OH)2](s) + 2NH4+ (aq)
GREEN SOLUTION —> GREEN PRECIPITATE

Question 21

Fe3+ Metal aqua ions with OH- /NH3?

Answer 21

[Fe(H2O)6]3+ (aq) + 3OH- (aq) → [Fe(H2O)3(OH)3] (s) + 3H2O (l)
YELLOW SOLUTION —> BROWN PRECIPITATE

[Fe(H2O)6]3+ (aq) + 3NH3 (aq) → [Fe(H2O)3(OH)3] (s) +3NH4+ (aq)
YELLOW solution —> BROWN PRECIPITATE

Question 22

Copper (II) metal aqua ions with OH- and NH3?

Answer 22

[Cu(H2O)6]2+ (aq) + 2OH- (aq) →** [Cu(H2O)4(OH)2]**(s) + 2H2O (l)

[Cu(H2O)6]2+ (aq) + 2NH3 (aq)—> [Cu(OH)2(H2O)4](s) + 2NH4 + (aq)
BLUE SOLUTION —> pale BLUE PRECIPITATE

Excess AMMONIA, copper hydroxide:
Cu(OH)2(H2O)4 + 4NH3 (aq) —> [Co(NH3)4(H2O)2]2+ (aq) + 2OH- (aq) + 4H2O (l)
** BLUE PRECIPITATE —> DEEP BLUE SOLUTION**
Ligand exchange

Question 23

Cobalt (II) with OH- and NH3?

Answer 23

[Co(H2O)6]2+ (aq) + 2OH- (aq) →** [Co(H2O)4(OH)2]**(s) + 2H2O (l)

[Co(H2O)6]2+ (aq) + 2NH3 (aq) →** [Co(H2O)4(OH)2]**(s) + 2NH4 + (l)
Pale pink solution —> blue precipitate (turns brown on standing)

Excess NH3 with cobalt hydroxide :
Co(H2O)4(OH)2 + 6NH3(aq) —> [Co(NH3)6]2+ + 2OH- (aq) + 4H2O (l)
Ligand exchange reaction : BLUE PRECIPITATE dissolved —> YELLOW SOLUTION

Question 24

Chromium (III) with OH- and NH3?

Answer 24

[Cr(H2O)6]3+ (aq) + 3OH- (aq) —> Cr(OH)3(H2O)3 + 3H2O (l)

[Cr(H2O)6]3+ (aq) + 3NH3 (aq) —> Cr(OH)3(H2O)3 + 3NH4 + (aq)
Green solution —> grey/green precipitate