Coordination Chemistry

Question 1

How is orbital energy related to ‘effective’ nuclear charge? Give equation

Answer 1

E(orbital) is directly proportional to (Zeff/n)^2
where Zeff = ‘effective’ nuclear charge (captures electron-electron interactions within atom)
n = quantum number (“shell number”)

Question 2

What are radial distribution functions and how are they expressed graphically? 1s vs 2s

Answer 2

• Quantum mechanical description of electron location
• Pretend orbitals don’t have shape (instead look at average electron density as you go out from nucleus in spherical shell)
• Expressed as number of nodes/”bumps”
• plot R^2dr (shell of thickness dr) against r (distance from nucleus), get one bump (1s distribution) or 2 bumps (2s distribution)

Question 3

What is the general rule for radial distribution functions? (Example for s-orbitals)

Answer 3

n - l (lowercase L not capital i)
where n = 1
l = 0 for s-orbital

Question 4

2s vs 2p radial distribution functions

Answer 4

• 2s has 2 bumps, 2p has 1 bump
• Similar max. radius probability (likely to find them at same radial extent as they’re both in 2nd shell)
• 2s tunnels inside/penetrates 2p electrons (so 2s energy lower than 2p)
• Expect 2s orbital to experience higher nuclear charge than 2p orbitals

Question 5

3d vs 5d radial distribution functions

Answer 5

• 1 bump for 3d
• 3 bumps for 5d

Question 6

what are the n and l quantum numbers for s, p and d orbitals?

Answer 6

n represents electron shell/energy level (i.e. 1 for 1s, 2 for 2p, etc)
l = 0 for s orbitals
l = 1 for p orbitals
l = 2 for d orbitals etc.

Question 7

What are the shapes of the d-orbitals?

Answer 7

• Imagine z is vertical axis, y is horizontal axis, x is axis coming out of plane (towards you)
• dyz, dxy and dxz orbitals point between the axes (orbitals point between planes given, not directly on top of axis lines)
• dz^2 and d(x^2-y^2) point directly at axes (z^2 points directly at z axis + doughnut shaped at y axis, x^2-y^2 is directly along x and y axes)

Question 8

What is exchange energy?

Answer 8

Stabilisation effect of 2 electrons in separate orbitals pointing the same way (same/parallel spin)

Question 9

What is pairing energy?

Answer 9

Destabilisation of 2 electrons in the same orbital with opposite spins

Question 10

What is Aufbau filling? Give an example

Answer 10

Filling from bottom energy orbital to top energy orbital
E.g. Vanadium ([Ar] 4s2 3d3)

Question 11

Give an example of non-Aufbau filling. What are the disadvantages/advantages?

Answer 11

• E.g. Chromium ([Ar] 4s1 3d5)
• Cost: putting electrons in higher energy orbitals
• Gains: Avoided pairing energy + gained exchange energy

Question 12

What is the general trend in ionic radius of M3+ ions across d-block period with respect to Zeff?

Answer 12

• Radius shrinks as Zeff increases
• This is because outermost electron experiences bit more positive charge each time
• Due to inner electrons shielding getting poorer as nuclear charge increases

Question 13

What is the trend in radius down the groups (3d,4d,5d)?

Answer 13

• As n increases, r increases
• 3d–>4d a lot bigger
• 4d–>5d not a massive increase
• Explained by lanthanide contraction:
• Compare radial distribution of 4f and 5d:
• 5d penetrates inside 4f orbitals as 4f are poorly shielding (only 1 bump)
• So 5d experiences higher nuclear charge than expected so smaller radius than expected

Question 14

What is the ionisation trend across 3d period? (Sc –> Zn)(3 factors)

Answer 14

1. General increase explained by Zeff (easier to ionise–> harder to ionise)
2. Exchange energy effect for adding e- based on e- configurations (up to Mn, adding e- increases exchange energy)
3. Pairing energy starts from d6 (iron) onwards (bonus pairing destabilisation)

Question 15

What is the ionisation energy trend down group 6 (Cr, Mo, W)?

Answer 15

• IE starts off the same for 3d and 4d/5d (as ratio of Zeff:n ~1)
• Start to diverge after 3rd IE (Cr begins to experience higher nuclear charges at higher ox. numbers)

Question 16

What are the 2 types of geometrical isomers for MX3Y3 complexes?

Answer 16

Facial and Meridional
Facial = all X groups adjacent (90 degrees from each other)
Meridional = 2X groups 180 degrees apart, other X group in the middle (90 degrees from each)

Question 17

What is exchange isomerism?

Answer 17

E.g. hydration isomers of CrCl3
[CrCl2(H2O)4]+ (+ H2O)–> [CrCl(H2O)5]2+

Question 18

What is linkage isomerism?

Answer 18

E.g. isomers of the M(NCS) and M(NO2) groups
M<–:N(triple bond)C-S^- and M<– -S-C(triple bond)N
M-N(=O)-O and M-O-N=O
atom linked to metal is different

Question 19

Describe the 2 metal bonding models

Answer 19

• Crystal Field model: start by modelling bonding as ionic, then add some covalency (good fit for 3d metals in mid-high ox. states)
• Ligand Field model: start by modelling as covalent then add ionicity
• Both models useful as they emphasise different truths about metal-ligand bonding
• Neither is right nor wrong

Question 20

Why are complexes of transition metals normally very colourful?

Answer 20

Promotion of electron from lower energy orbital to higher energy orbital through energy gap delta requires light, “left behind” frequencies are visible
E.g. Co2+ appears pale red as blue light absorbed

Question 21

how do delta (tetrahedral) and delta (octahedral) relate to each other via an equation?

Answer 21

delta(tet) ~ 4/9 delta(oct)

Question 22

how does square planar splitting derive from octahedral?

Answer 22

Ligands on z axis pulled away
Orbitals with z component fall in energy as they aren’t clashing with ligands any more

Question 23

What affects size of splitting (changes delta)?

Answer 23

1. Geometry (esp. oct/tet)
2. ox. state (larger ox. state means bigger delta as shorter M-L distance causes L to clash with d-orbitals more)
3. Metal identity (esp. 3d/4d/5d)(1 group down means bigger delta as 4d orbital reaches over to the ligand)
4. Ligand identity (can’t be explained by cft)(Stronger field ligand means bigger delta)

Question 24

What is the structure of a spinel?

Answer 24

Oxide (sometimes sulphide) anions form a face-centered cubic array
Metal ions find ways to fit in (octahedral or tetrahedral)
Octahedral holes in anion lattice, 4 oct holes per unit cell (6 on face, 1 in centre)
Also tetrahedral holes in anion lattice (O2-)

Question 25

What mnemonic can be used to describe normal spinels?

Answer 25

BOAT: B in Octahedral sites A in Tetrahedral sites But B fills up 1/2 octahedral sites and 1/2 tetrahedral sites AB2O4 where A2+ and B3+

Question 26

What 2 factors cause strong ionic bonding?

Answer 26

- Big charge - Small size/radius

Question 27

Why is Mg in the tetrahedral site in the structure MgAl2O4? (no d electrons)

Answer 27

- Mg2+ has weaker ionic bonds (smaller charge, bigger radius) - Al3+ has stronger ionic bonds (big charge, small radius) - Energy of structure minimised and bond strength maximised with Al in octahedral site (site with 6 bonds) (Al = 6 coordinated compared to Mg's 4)

Question 28

Why does Fe3O4 (Fe(II)Fe(III)2O4) adopt inverse spinel structure with Fe(III) in tetrahedral sites? (d-electrons)

Answer 28

Fe(II) (octahedral) has d6 electrons, 4 in t2g, 2 in eg so 2/5 delta (oct) overall Fe(II) (tetrahedral) has d6 electrons, 3 in t2g, 3 in eg so 3/5 delta (tet) overall Different splitting so use delta(tet) = 4/9 delta (oct) equation: 3/5 delta tet = 3/5 x 4/9 x delta oct = 12/45 delta oct 2/5 delta oct = 18/45 delta oct Therefore, octahedral preference of Fe(II) Fe(III) has no preference as delta tet and delta oct both = 0 (d5) no net stabilisation

Question 29

What in paramagnetism in analysis? What can be assumed for this model?

Answer 29

- Weigh sample then weigh sample in magnetic field - If mass changes, sample is magnetically susceptible (influenced by magnets) - In simple model of paramagnetism in transition metal complexes, assume only electronic spin contributes to magnetic behaviour

Question 30

What is the spin only formula?

Answer 30

mu (spin only) = sqrt(n(n+2)) where n = no. unpaired electrons Do not use quadratic formula to solve, use trial and error to find closest value

Question 31

Why do octahedral Fe(II) complexes have magnetic moments of 1.7 BM and 0BM?

Answer 31

Fe(II) = d6 Can be high spin configuration [Fe(H2O)6]2+ or low spin [Fe(CN)6]4- In high spin, S = 2 (4 x 1/2 spins, 2 paired cancel out) In low spin S = 0 (all 3 paired cancel out) Use spin only formula to give values

Question 32

When do you promote or pair an electron in high/low spin configurations?

Answer 32

- In high spin, worth the cost of promoting electron - In low spin, worth the cost of pairing electron - If big delta, promoting electron is more costly than pairing so pairing preferred and vice versa

Question 33

What are the key considerations in molecular orbital theory? Give example with H2

Answer 33

- Energy match - Overlap E.g. In H2, s-orbital on each H atom taken and constructive overlap and deconstructive overlap formed (electrons stabilised as they are more stable in a molecule than in separate atoms) - H atoms identical in energy so extent to which you antibond is large

Question 34

What is the requirement for orbitals to overlap?

Answer 34

Need to have same symmetry label No overlap, no bond

Question 35

What is the trend of delta through the spectrochemical series? Give examples and reference sigma and pi donors/acceptors

Answer 35

delta increases as you go through series E.g. I- gives small delta, CO gives big delta Pi acceptors sit high in spectrochemical series - good sigma donors give big delta - pi donors give small delta - pi acceptors give big delta

Question 36

What are the atomic and molecular selection rules? Compare/contrast.

Answer 36

Atomic: delta S = 0 (cannot change spin so pairing electrons forbidden) delta L = +/- 1 (s-->p allowed but d-->d forbidden) Molecular: delta S = 0 angular momentum less important in covalent molecules (rule is weaker so can be broken)

Question 37

Give example of the consequences of the spin selection rule?

Answer 37

E.g. Mn2+ is pale pink (high spin d5) because d-->d transition breaks spin selection rule (delta S = 0) so there is a penalty to intensity

Question 38

Why is potassium permanganate such an intense colour (purple)?

Answer 38

- Electrons travelling between ligand and metal (Ligand metal charge transfer LMCT)(2 different atoms) aren't bound by selection rules - Internal redox reaction occurs that turns Mn(VII) into Mn(VI) - Also examples of metal ligand charge transfer (MLCT): - e.g. [Ru(bipy)3]2+ Ru donates e- to N atom - Useful in solar cells

Question 39

What is Irving Williams series? What is the trend? Give an example with a ligand

Answer 39

Water substituted for another ligand around M(II) Thermodynamic equilibrium shows a double hump Normally presented as variation in logK by metal identity E.g. NH3 ligand, all values of K > 1 so generally favourable Increase across period (V, Cr, Mn, Fe, Co, Ni, Cu, Zn) but peaks at Cu True for other ligands such as en and EDTA

Question 40

Explain the double hump in the Irving Williams graph (rationalise)

Answer 40

1. There is a start to finish trend reflecting metal radius (the ionic argument) 2. Double hump in middle: - First value (Sc) much lower than last value (Zn) - Sc2+ has larger radius so gives weaker ionic bonding whereas Zn2+ has smaller radius so gives stronger ionic bonding - d1-d3 increases in stability as delta increases (d3 = max) - d4-d5 decreases stability as delta decreases - then same pattern for d6-d10 (exception for Cu2+, anomaly) 3. Jahn-Teller distortion of Cu(II) (in next flashcard)

Question 41

Why is Cu2+ an anomaly in the Irving Williams graph trend?

Answer 41

All other metals assumed to be perfect octahedral Cu2+ is distorted by elongating z-axis and compressing x,y plane z orbitals become more stable energy of orbitals determined by how much ligand pushes/clashes into it so elongating ligand means it clashes with z orbitals less odd no. of electrons mean 2e- stabilised even if 1e- destabilised 4 strong short bonds even if 2 long weak bonds (substitution of ligands only favourable at strong bonds)

Question 42

What is the Nernst equation?

Answer 42

delta G = -nFE

Question 43

Do pi acceptors/sigma donors stabilise low or high ox. states?

Answer 43

- good pi acceptors (e.g. bipy) stabilise low ox. states: - When metal centre reduced, energy increases so better match with pi acceptor (t2g) orbital (gives stronger bonds) - good sigma donors (e.g. carbanion cyanide) stabilise high ox. states: - When metal oxidised, energy decreases so better match with sigma donor orbital (gives stronger bonds)

Question 44

What is the trend in standard cell potential (E standard) down the groups?

Answer 44

Gets more negative/less positive so reduction becomes less favourable Higher ox. states more stable for 4d/5d (n is big, Zeff relatively constant) Bond strength related to orbital size: 5d > 3d so stronger covalent bonding