transition metals Flashcards
transition element
a d block element that forms one or more stable ions with an incomplete d sub shell
why do transition elements have variable oxidation states
variable oxidation= because the energy of the 3d and 4s orbitals are similar. There isn’t a huge increase in energy to remove a thrid electron than second or first
smaller difference in IE
ligand
species with one or more lone pairs of electrons that form dative covalent bonds to a central transition element atom or ion
complex ion
central transition metal ion surrounded by ligands, bonded to the central ion by dative/coordiate covalent bonds
why from complex ions
ions have empty d orbitals available for lone pairs of electron
“energetically accessible and can form dative bonds with ligands”
why behave as catalysts
have more than one stable oxidation state. can gain or lose electrons from going to one oxidation state to another. (oxidising or reducing agents)
“energetically accessible and can form dative bonds with ligands”
why form coloured ions
have unfilled d orbitals, cause electrons to absorb energy and jump to higher energy levels. this light not absorbed is reflected and is what we perceive as colour.
Copper sulfate + sodium hydroxide
[Cu(H2O)6]2+ + 2OH- –> Cu(OH-)2(H2O)4(s) + 2H2O
NH3 to copper blue ppt
Cu(OH-)2(H2O)4(s) + 4NH3 –> Cu(NH3)4(H2O)2(aq)2+ + 2H2O + 2OH-
deep blue solution from pale blue ppt
copper sulfate?? + hydrochloric acid
[Cu(H2O)6]2+ + 4Cl- –> Cu(Cl-)4(aq)2- + 6H2O
blue + yellow = green solution
cobalt sulfate + sodium hyrdoxide
[Co(H2O)6]2+ + 2OH- –> Co(OH-)2(H2O)4(s) + 2H2O
pink solution to blue ppt
turns red when warmed if the alkali is in excess
cobalt sulfate + conc NH3
[Co(H2O)6]2+ + 6NH3 –> Co(NH3)6(aq)2+ + 6H2O
pink solution to brown solution
cobalt sulfate + conc hydrochloric acid
[Co(H2O)6]2+ + 4Cl- –> Co(Cl-)4(aq)2- + 6H2O
pink to blue solution
Cobalt sulfate with cl-
tetrahedral
Cobalt sulfate with scn-
tetrahedral
Cobalt sulfate with oh-
tetrahedral
degenerative orbitals
there are 5 d orbitals in an isolated transition element atom or ion in the same energy level
non degenerative orbitals
the 5 d orbitals split after ligands bond and are in different energy levels
octahedral complexes, two higher and three lower d orbita
3dx2 - y2 and 3dz2 have lobes on the axis where the ligands attached or approaching
therefore the electrons in these orbitals are closer to bonding electrons so more repulsion and higher energy level
tetrahedral complexes, three higher and two lower d orbitals
3dx2 - y2 and 3dz2 have lobes between the axis where the ligands attached or approaching
therefore the electrons in these orbitals are further from bonding electrons so less repulsion and lower energy level
explain why transition elements form coloured compounds in terms of the frequency of light absorbed as an electron is promoted between two non-degenerate d orbitals
have incomplete d shells = have space for electrons to be promoted and then absorb light.
when an electrons is promoted to the higher energy level, the difference between the two non-degenerate d orbitals corresponds to a particular frequency of light absorbed and reflects the light complementary.
ligand copperexcahnge and colur chanrg
The ligand exchange of [Cu(H2O)6]2+ and [Co(H2O)6]2+ by NH3 ligands causes a change in the colour of the solutions
[Cu(H2O)6]2+ is light blue in colour whereas [Cu(NH3)4(H2O)2)]2+ is deep blue in colour
[Co(H2O)6]2+ is a pink solution whereas [Co(NH3)6]2+ is a brown solution
the ligand causes d orbitals to split by a different amount of energy and so different ΔE and light frequency absorbed
NH3 to copper sulfate
[Cu(H2O)6]2+ + 4NH3 –> Co(NH3)4(H2O)2(aq)+ + 4H2O
hydrochloric acid to copper sulfate
[Cu(H2O)6]2+ + 4Cl- –> Cu(cl-)4(aq)+ + 6H2o
Kstab
equilibrium constant for the formation of the
complex ion in a solvent (from its constituent ions or molecules)
Kstab expression
same as Kc,
water not included = large in excess its concentration is considered constant
high stability
higher value of Kc, higher its stability
feasibility
the two values, positive to right, negative to left added to gether gives a positive value = feasible
half equation of C2O4 2-
C2O4 2- –> 2CO2 + 2e-
