Transition Metal Chemistry Flashcards
(40 cards)
What are transition metals? What are d block elements?
Transition metals: metals which can form ions with a partially filled d or f subshell
D block elements: elements whose highest energy electron is in the d subshell
What happens to the energy of the 3d vs 4s orbital across the period?
En = -C (Zeff/n)^2
The 4s orbital is more penetrating than the 3d, meaning the 4s electrons can penetrate the core region more greatly
This mean the 4s is at a lower energy, as it has a greater Zeff
Across the period, the nuclear charge increases, resulting in orbital contraction, and electron shielding of the 4s increases by the 3d
As on average the 3d orbital is closer to the nucleus, it is more greatly stabilised than the 4s, and when the Zeff is large enough, by Sc, the energy of the 3d falls below the 4s
i.e e=-(zeff/n)^2, the increase in zeff for 3d is larger than 4s due to be on average closer, and so eventually becomes lower in energy
How does electron repulsion vary between 4s/3d orbitals?
S/S<S/D<D/D
This is because the 4s orbital is more diffuse, so less coulombic repulsion
How is exchange energy calculated? And for the d subshell?
E exchange: sum of ((n up ( n up -1)) /2)K
And added to electrons down
Across the D subshell, the exchange energy increases, except 5–6 where it stays the same
What is the electron configuration of Mn / Cr ?
4s1 3d5/ 3d10
To maximise exchange energy across the d subshell
What happens to I1 + I2 across the period? Why?
Generally a removal of the 4s
There is a general increase across the period, with I2 dominating the trend
This is because nuclear charge increases, resulting in Zeff increasing, and so greater nuclear attraction to overcome so larger ionisation energies
As Cu will remove an electron from the 3d, as will Cr, as so there is a very large exchange energy to overcome, therefore the I1+ I2 value is larger than the others
What happens to I3 across the period and why?
General increase across the period as nuclear charge increases, resulting in Zeff increasing, and so greater nuclear attraction to overcome so larger ionisation energies
Larger increase for Mn and Zn, as they have maximum exchange energy to overcome
And for Fe, removing a d electron will not alter exchange energy, emphasising the change
What happens to hydration enthalpies across the group and why?
Formation of M(H2O)6 complexes, i.e an octahedral crystal field
The hydration enthalpies generally become more exothermic across the period
This is because of an increase in Zeff, so smaller ionic radii, more exothermic ion-dipole interactions, and so a linear increase
However, this linear increase is only present at d0, d5, and d10
There is additional stabilisation of these complexes via LFSE, seen in the table below. However, the most stable complex is expected to be d8 and d3 rather than d9 and d4
This arises due to Jahn teller distortion, with the axial bonds extending, ending the degeneracy of unequally filled orbitals, stabilising the complex
What happens to MCl2 lattice enthalpies across the group and why?
Via crystal structure, MCl2 is an MX6 octahedral in geometry, and so experiences an octahedral crystal field
The lattice enthalpies generally become more exothermic across the period
This is because of an increase in Zeff, so smaller ionic radii, more exothermic ion-ion interactions, and so a linear increase
However, this linear increase is only present at d0, d5, and d10
There is additional stabilisation of these complexes via LFSE, seen in the table below. However, the most stable complex is expected to be d8 and d3 rather than d9 and d4
This arises due to Jahn teller distortion, with the axial bonds extending, ending the degeneracy of unequally filled orbitals, stabilising the complex
What happens to the M2+/M3+ electrode potential across the group and why?
Using the cycle, the enthalpy change accompanying the reduction is related to the hydration enthalpies and - I3
Across the period, generally I3 increases, resulting in more negative reaction enthalpies for the reduction, and so the reduction is favoured, resulting in more positive electrode potentials for this reduction
However, there are anomalies such as increases in LFSE in (II) for d4 and d8
What is the variation in stability of MX3 compounds across the period and why?
Looking at the Hess cycle, a balance of a higher I3, and whether the exothermic lattice enthalpy can compensation for this
Up until Mn, most complexes with halides can compensation
At Mn, much higher I3 arising from much higher exchange energy (II), and so only F has an exothermic enough formation/lattice enthalpy to compensate
As there is no reduction in exchange energy for Fe, more halides can
After Fe, the I3 is too high for any halide to compensate via lattice enthalpy so complexes do not form
Why does splitting of the d orbitals occur?
When ligands coordinate, they are treated as point charges, with the d electrons destabilised by the repulsion
In a hypothetical uniform field, the d orbitals would be stabilised by the same amount, acting on the centre
However, as the orbitals lay in different planes and different geometries of the ligands, the orbitals will be destabilised by different amounts, resulting in splitting
What happens to the d orbitals in an octahedral crystal field?
The dz2 and dx2-y2 align with the ligands so raised in energy =
eg by 3/5 ΔO
And the others lowered = t2g
by -2/5 ΔO
What happens to the d orbitals in a tetrahedral crystal field?
The dz2 and dx2-y2 align away so ligands so lowered in energy =
eg by -3/5 Δt
And the others raised, as align more so, but not as much as when octahedral = t2g
by -/5 Δt
What happens to the d orbitals in a square planar crystal field?
Start with octahedral, and z plane removed, dz2 and both z terms fall
xy increases, top x2y2
What happens to the d orbitals in 5 coordinate complexes crystal field?
Square-based pyramid : effectively a less stabilized square planar
Trigonal-bypyramidal: most in z2, and then xy planes
What do the d orbitals look like?
Align in the planes named, apart from x2y2 which are along the axis named
Easier to draw when two axis on, draw orbital, then add the 3rd axis
What happens to atomisation across the period and why?
Metal d orbitals overlap forming bonding and antibonding orbitals
There is an expected rise and fall, from bond strength rising and falling as bonding and subsequently antibonding orbitals are filled
So an expected peak at d5, 4s1, Cr and minimum at d10 4s2, Zn
Actually peak at V, and dip at Mn, due to randomisation of spin leading to increased electron-electron repulsion
Mn actually maximises exchange energy, and so fill antibonding, destabilising the bond
For larger n , non-Aufbau filling is rare, with reduced electron-electron repulsion
What is ligand field stabilisation energy? How does it arise?
Additional stabilisation of a complex contributing to lattice enthalpies and hydration enthalpies
Due to splitting d subshell, and so filling of the lower energy orbitals stabilises the complex
What factors affect the splitting parameter?
Charge of the cation: larger charge, ligands drawn closer, and so greater repulsion, greater splitting
Larger n: more diffuse orbitals result in smaller ligand - orbital distances, so greater repulsion and splitting
More ligands, and type of ligand: if backbonding with a pi acceptor this will increase splitting
How do Δtd and Δo compare?
Δt= 4/9Δo
How can the ligand trends be explained in terms of splitting parameter?
Halogens<water<ammonia< en<pph3< CN-<CO
pi donors, anything with lone pairs, raise the energy of the homo and so reduce the splitting parameter
pi acceptors lower the energy of the homo increase the splitting parameter
What is the difference between high spin and low spin complexes? What are the two energy factors involved?
High spin: fills both t2g and eg
Low spin: pairs the electrons first before filling both orbitals
A balance of the additional LFSE from filling only t2g in Od, vs the additional cost for pairing electrons
How do tetrahedral complexes relate to high/low spin?
Generally always high spin, as the splitting parameter is always too small to compensate for the cost of pairing
Octahedral always much larger
