Transition Metals: Colours & Reactions

Cr(OH)₃

grey-green precipitate

Cr(OH)₆^3-

dark green solution

CrO₄^2-

yellow solution

Cr₂O₇^2-

orange solution

Fe(H₂O)₆³⁺

yellow solution

Fe(OH)₂

green precipitate

Fe²⁺

pale green solution

Fe(SCN)(H₂O)₅²⁺

blood red solution

Fe(OH)₃

reddish-brown precipitate

Mn²⁺

colourless if dilute

faint pink

MnO₂

dark brown precipitate

MnO₄^2-

dark green solution

MnO₄^-

purple solution

Cu(H₂O)₆²⁺

pale blue

Cu(NH₃)₄(H₂O)₂²⁺

dark blue solution

[CuCl₄]^2-

yellow solution

Cu(OH)₂

pale blue precipitate

V³⁺

green solution

V²⁺

violet solution

Na₂CO₃ + Cr(NO₃)

Grey-green precipitate & CO₂ formed

Cr(H₂O)₆³⁺ ⇌ Cu(H₂O)₅(OH)²⁺ + H⁺

Due to the high charge density of Cr³⁺, under acidic hydrlysis in water to form H⁺

H⁺ reacts with CO₃^2- to form CO₂

2H⁺ + CO₃^2- ⇌ H₂O + CO₂

[Cr(H₂O)₆]³⁺ + 3OH^- (dropwise)

Grey-green precipitate formed, Cr(OH)₃

[Cr(H₂O)₆]³⁺ + 3OH^- (in excess)

dark green Cr(OH)₆^3- solution formed

Cu

pink solid

[CuCl₂]^-

colourless solution (Cu(I))

CuO

black solid

Cu₂O

red / yellow solid

CuI

white

acid + CrO₄^2-

orange Cr₂O₇^2- solution formed, oxidation state stays the same

excess ammonia added to CuSO₄ (aq), followed by H₂SO₄

pale blue precipitate formed first: Cu²⁺ + 2OH^- →Cu(OH)₂

dark blue solution formed via ligand exchange reaction: Cu²⁺ + 4NH₃ ⇌ Cu(NH₃)₄(H₂O)₂²⁺ + 4H₂O

Since Cu²⁺ ions used to form complex ion, [Cu²⁺] decreases, so POE of equation 1 shifts to the left, causing precipitate to dissolve

When H₂SO₄ is added,

H⁺ reacts with NH₃ and OH^-- in solution, [NH₃] decreases, POE of equation 2 shifts to the left

[Cu²⁺] increases, causing POE of equation 1 to shift right, causing reformation of pale blue ppt

When H₂SO₄ added in excess, Cu(OH)₂ neutralised to form pale blue solution CuSO₄

HCl + Cu²⁺

ligand exchange reaction takes place to form yellow [CuCl₄]^2- solution

pale-blue —> green —> yellow

green is formed due to a mixture of preceding Cu(H₂O)₆ complex, and yellow [CuCl₄]^2- product