Transition metals

What are transition metals?

D block elements that are between group 2 and 13. They ‘re electron with the highest energy is in the 3d subshell.They can form at least 1 ion with partially filled d.

Which elements are exceptions within the d block that aren’t metals?

Sc and Zn. When they form ions that don’t ge partially filled d shells. They still loose the 2 from the 4s shell leaving them as 3p6 and 3d10 ions.

How do we write electron configuration of transition metals?

Opposite spins, singularly fill each shell first , fill from lower energy orbital first (4s<3d) Cr and Cu prioritise getting half filled and fully filled orbital so they are exceptions: written 3d5 4s1 and 3d10 4s1 respectively

Characteristics of transition metals

Metallic, high melting point , shiny in appearance, conduct electricity and heat. Form compoundswhere elements have different oxidation states ( higher ON, better oxidising agent) , forming coloured compounds, act as catalysts

Uses of transition metals

Cu Ag Ni Zn in making coins

Fe in construction and toolsCu in electrical cables heat and water pipesd

Ti in space and medical apparatus

How are transition elements and ions used as catalysts?

Haber process ; Fe is used to make ammonia N2+3H2=2NH3

Contact process; uses vanadium(v) oxide V2O5 within the process of making sulfuric acid. 2SO2+O2=2SO3

Hydrogenation; uses Ni for turningalkenes into alkanes

Decomposing H2O2; using MnO2 to make oxygen 2H2)2= 2H2O +O2

Transition metals ions colours. Ti3+ / V2+, V3+, V4+, V5+ /Cr2+, Cr3+, Cr6+ /Mn2+, Mn4+, Mn6+, Mn7+ /Fe2+, Fe3+ /Co2+, Co3+ / Ni2+ / Cu2+

Ti3+ LILAC/ V2+ LILAC PURPLE, V3+ GREEN, V4+ BLUE, V5+ YELLOW /Cr2+ BLUE, Cr3+ GREÈN, Cr6+ ORANGE /Mn2+ LIGHT PINK, Mn4+ MAROON, Mn6+ GREEN, Mn7+ LILAC /Fe2+LIGHT GREEN , Fe3+ PALE YELLOW /Co2+BRIGHT PINK, Co3+ GREEN / Ni2+ GREEN / Cu2+ BLUE

Solution to precipitate for: Cu(||) Fe(||)Fe(|||)Mn(||)Cr(|||)

Cu(||) from pale blue to pale blue

Fe(||) from pale green to dark green

Fe(|||)from yellow to orange brown

Mn(||) from light pink to pale brown

Cr(|||) from greenbto dark green.

What is the difference between a ligand and a dative bond?

L= a molecule or ion that donates a pair of electrons to central metal ion, and forms a dative bond

D= covalent bond of which both electrons came from one of the atoms

Monodente means they have donated just 1 pair, bicentennial means they have donated 2 and multidente means they’ve donated various.

What is the difference between optical and sterio- isomers?

S=Molecules or complexes with same structure but different spacial arrangements. Cis-same side 90° between them, Trans-opposite side 180° between them. They have different colours an properties

O= molecules that can have nonsuperimposible mirror images of each other (enantionmers) either 3 bidente ligand, 2 bidente and 2 monodente, or 1hexadente.only is molecules can have opticalisomers since the trans is symmetrical.Have different chemical and biological and physical properties

What is cis-plantin?

The Cisisomer of [Pt(Cl)2(NH3)2] and is a drug used to aid cancer treatment often in chemo

Give examples of mono- bi- and hexa- dente molecules.

Mono-

Bi- ethanedioate ion -:OOCCOO:- 2- and the ethane-1,2-diamine EN :NH2CH2CH2NH2:

Hexa- ethylenediaminetetraacetic acid EDTA forms EDTA 4-which donates 6 lone pair electrons

Reacting Cu (||) (aq) and HCl(aq)

LIGAND SUBSTITUTION; 6 H2O get replaced by 4 Cl- (Cl- is bigger has more repulsion bc of more electrons).

[Cu(H2O)6]2+ +4Cl- = [CuCl4]2- +6H20 from pale blue octrahedral to yellow tetrahedral

Reacting Cu (||) (aq) and a little NH3 solution

PRECIPITATION; ammonia is just a base here and accepts 2 protons.

[Cu(H2O)6]2+ +2NH3 = [Cu(OH)2(H2O)4] + 2NH4+ from pale blue to a dark blue precipitate ( hydrated copper hydroxide is octrahedral)

When we bubble NH3 in water we get OH- (proton acceptor) it can hydrate it again and make water

Hydrated copper (||) and excess NH3

Four NH3 replace 2 of the OH- ligand and 2 of the H2O ligand which is fully dissolved . This looks like..

[Cu(OH)2(H2O)4] + 4NH3 = [Cu(NH3)4(H2O)2]2+ +2H2O +2OH- from a blue ppt to a neutral dark blue solution with octraheral ligands

If we add even more ammonia solution we get LIGAND SUBSTITUTION replacing 4 NH3 with 4H2O

[Cu(H2O)6]2+ +4NH3 = [Cu(NH3)4(H2O)2]2+ +4H2O from pale blue precipitate to deep blue solution

Reacting Cu2+ (aq) with NaOH

Cu2+ +2OH- = Cu(OH)2

This is insoluble in excess OH-

==)

Reactions of Cr3+ and small bit of NH3