Transition Metals Flashcards

Question

identifying the type of ligand

Answer 1

π - donor ligands ~ more than one donor pair on the same atom ~ have 2 active MOs, both occupied ~ VSEPR shows two double dots (lone pairs) ~ late in periodic table (group 6,7) π - acceptor ligands ~ have an empty π* MO on the donor atom ~ look at MO to identify examples: CO (cabonyl complexes) PR3 (phosphine) C=C organic ligands σ-only ligands ~ only have one active MO, the donor pair ~ no available π/π* MO examples: (group 5) NH3, NR3, PH3, PR3, AsH3

Answer 2

1. O more electronegative than C - lower in energy 2. symmetry of AOs (relative to bond direction, z) rotate around z axis: - no change in sign = σ - change in sign = π 2s - σ symmetry 2px - π symmetry 2py - π symmetry 2pz - σ symmetry 3. sp hybrids (spz) (s + pz) (s - pz) 4. frontier MOs HOMO = σ3 (lone pair on C) - acts as a Lewis σ base (an electron-pair donor) LUMO = 2π (π*) - acts as a Lewis π acid (an electron pair acceptor) ~ accepts electron density from filled metal d orbitals both frontier MOs are on C - CO always binds from C side (not electronegative O)

Answer 3

π backbonding σ bond from ligand to metal (1) π bond from metal to ligand (2) CO is not nucleophilic - ∴ σ bonding is weak but d-metal carbonyl compounds are very stable ∴ π backbonding is very strong bonding is synergic (mutually enhancing): π backbonding from metal to CO increases electron density on CO, this increases ability of CO to form σ bond to metal

Answer 4

in isolated CO molecule, ν = 2143 cm-1 (stretching frequency) CO stretching frequency is very intense in IR (large change in dipole) - easy to monitor σ donation has little effect on stretching frequency π backdonation - electron density goes into π* of CO - weakens C-O bond (greater C-O bond length) - stretching frequency decreases

Answer 5

depends on: - orbital overlap - energy match between M d and CO π* orbitals (ΔE) - electron density on M influenced by: - type and charge of M - other ligands (competition for the same M-d electrons) M-C≡O with no backbonding M=C=O with complete backbonding

Answer 6

decrease in oxidation state: more electron density available for π backbonding - greater weakening of C≡O bond d10 metal has more electrons for π backbonding than d6 - greater weakening of C≡O bond Td: only 3 other CO ligands comepeting for π backbonding electrons - greater weakening of C≡O bond

Answer 7

CO can bridge 2/3 metal atoms CO stretching frequencies generally follow the order: MCO > M2CO > M3CO - increasing occupation of π* orbital as the CO molecule bonds to more metal atoms - more electron density from metals enters the CO π* orbitals M3(μ3-CO): the symbol μ3 indicates that CO bridges 3 metals

Answer 8

C-based π acceptors stable as 18 electron complexes Cr(0): d6 (each CO brings 1 electron pair (2 electrons) 6 CO needed (6 x 2 = 12) for there to be 18 electrons Cr(CO)6 Fe(0): d8 - 5 CO needed (5 x 2 = 10) for there to be 10 electrons Fe(CO)5 Ni(0): d10 - 4 CO needed (4 x 2 = 8) for there to be 18 electrons Ni(CO)4 Fe(2-) d10 - 4 CO needed (4 x 2 = 8) for there to be 18 electrons [Fe(CO)4]2-

Answer 9

Mn(CO)5 Mn(0): d7 + (2 x 5) = 17 electrons ~ does not meet 18 electron rule Mn(CO)5 ~ gain electron → [Mn(CO)5]- (18e- stable) Mn(CO)5 ~ lose electron → [Mn(CO)6]+ (18e- stable) Mn(CO)5 ~ react with methyl radical → Mn(CO)5(CH3) (18e- stable) Mn(CO)5 ~ share with another complex → Mn2(CO)10 (18e- with Mn-Mn single bond. dimer of Mn(CO)5 ~ reactive)

Answer 10

- frontier MOs in CO are both polarised towards C - no polarisation of HOMO/LUMO in N2: worse overlap with M-d atomic orbitals, lower bond strength - donor pair in CO is a C-based level - N-based in N2, lower in energy ∴ higher gap from M-d, lower bond stength - N-N bond is more covalent than C-O - larger split of π/π* levels - π* at higher E in N2 than CO, N2 is a worse π acceptor

Answer 11

bridging and side-on interactions possible for N2, not for CO

Answer 12

10e- (valence e-) N2 dinitrogen CN- cyano R-NC isonitrile same number of total electrons/same electronic structure as CO 11e- (valence e-) NO nitrosyl

Answer 13

lone pair on P (HOMO: basic and nucleophilic) - σ donor empty π* orbital on P (LUMO) - large gap between σ and π* orbital (π* orbital is inactive) PH3 is σ donor only - only active orbital is 2a1 PR3 is σ donor only (high LUMO) PF3 is π acceptor (more electronegative heteroatom to P lowers LUMO)

Answer 14

electron-rich phosphines (PMe3): good σ donor but poor π acceptor - increase e- density on M - increase in backdonation to CO electron-poor phosphines (PF3): poor σ donor but good π acceptor - decrease e- density on M - decrease in backdonation to CO lewis basicity can indicate donor/acceptor ability σ only PCy3 > PEt3 > PMe3 > PPh3 > π acceptor P(OMe)3 > P(OPh)3 > PCl3 > PF3 PF3 π acceptor of similar strength to CO

Answer 15

alkenes bond side-on to a M, with both C atoms of double bond equidistant from M. groups on alkene are perpendicular to plane of M and the two C atoms alkenes have no lone pair

Answer 16

the electron density of the C=C π bond (HOMO) donated to an empty orbital on M - forms σ bond - removes e- density from π bond - weakens the C=C bond filled metal d orbital donates electron density to empty π* orbital of alkene (LUMO) - forms π bond - backdonate e- into π* - weakens C=C bond structure tends to that of a C-C singly bonded structure activation of C-C bond towards addition reaction

Answer 17

when π backbonding from metal atom increases, strength of C=C bond decreases as the electron density is located in the C=C antibonding orbital

Answer 18

number of C atoms bonded to the same metal

Answer 19

behave as polydentate ligands, in which each C=C double bond behaves like an η2 ethylene Conjugated π systems: C-based π-acceptor ligand Like CO, very large Δo

Answer 20

side-on interaction to M

Answer 21

Δo depends on the type of ligand π donors << σ only << π acceptors Δo depends on electronegativity (C > N > O > F) and period (N < P < As) of donor atom Δo depends on period of metal (3d << 4d ~ 5d) and its oxidation state (Δo increases with metal oxidation state - M will be a stronger acid)

Answer 22

high spin: electrons prefer to singly occupy each orbital before pairing (weak field) - π donors - species has a high number of parallel electron spins low spin: electrons pair in the same d orbital (strong field) π acceptors - species has a small number of parallel electron spins

Answer 23

which fraction of the splitting Δ is the electronic configuration in the complex more stable than in a spherical field - depends on high/low spin arrangement Fe3+ (d5) [Fe(H2O)6]3+ high spin configuration LFSE: (3 x 0.4)Δo - (2 x 0.6)Δo LFSE = 0 ~ no stabilisation [Fe(CN)6]3- low spin configuration LFSE: (5 x 0.4)Δo LFSE = 2Δo ~ large stabilisation

Answer 24

stable complexes with large LFSE d5 high spin (max spin multiplicity) - half filled set of MOs d6 low spin (large LFSE) - completely filled subset

Answer 25

tetrahedral arrangement is less sterically hindered square planar arrangement gives d-orbital splitting with high dx^2-y^2 - this arrangement is energetically favourable when there are 8 d-electrons and strong enough crystal field (Δ) to favour low spin - this tendency is enhanced with large 4d and 5d metal ion - larger ligand field splitting and lower pairing energy than 3d metal ions (3d metal has smaller, less diffuse orbitals ∴ poorer overlap with ligand) - forms low-spin square-planar metal complexes in this case, electronic stabilisation energy can more than compensate for unfavourable steric interactions ligand high on spectrochemical series also results in large LFSE - square-planar complex

Answer 26

when a non-linear molecule is in a degenerate electronic state, it distorts in such a way as to remove the degeneracy

Answer 27

a set of MOs at the same energy, but occupied by a different number of electrons (unevenly filled) associated with d-elements as it is common to have degenerate orbitals due to symmetry

Answer 28

if the ground electronic configuration of a nonlinear complex is orbitally degenerate, and asymmetrically filled, the complex distorts to remove the degeneracy and achieve lower energy 6-coordination d9 complex of Cu(II) departs considerably from octahedral geometry, showing pronounced tetragonal distortions - high spin d4 complex of Cr(II) or Mn(III) and low spin d7 complex of Ni(III) show similar distortion tetragonal distortion of regular octahedron: extention along z-axis and compression on the x-and y-axis - lowers energy of orbitals with z-component - raises energy of orbitals without z-component may be electronically advantageous: if 1 or 3 electrons occupy eg (such as in high spin d4, low spin d7 and d9) - degenerate level is unevenly filled - in d9 complex, 2 electrons go into dz^2 with lower energy and 1 in dx^2-y^2 with higher energy

Answer 29

large drop in log(kn) for n=5 Cu(II) d9 is Jahn-Teller distorted: has a full t2g set and degenerate eg set - complex elongates to break the degeneracy and become more stable - 4 short (stronger) and 2 long (weaker) M-L bonds amine bonds more strongly to metal centre than water (larger Δo) - 4 strong bonds form with amine - 5th amine ligand has long, weak bond replacement of the ligands with long bonds causes smaller energy change, smaller kn (drop in rate of reaction)

Answer 30

potential energy surface: plot of energy as a function of atomic coordinates - multi-dimensional surface: N-atoms = 3N-6 degrees of freedom (xi, yi, zi) choose one geometric parameter that varies along the reaction - plot energy as a function of reaction coordinate - minima: equilibrium points (reactants/products) multiple transition states and intermediates before forming product elementary reaction step: transition state to intermediate three mechanisms depending on shape of potential energy surface 1. associative (A) 2. dissociative (D) 3. interchange (I)

Answer 31

2 steps → 1 intermediate in potential energy surface two transition states and an intermediate in between - first step is slow (largest Ea): forms an intermediate with all ligands together with metal (increases coordination number of metal) - second step is fast: leaving group removed, return to original coordination number this mechanism is given: - by low coordination number transition metal complexes (e.g. square planar rather than octahedral) - low number of valence electrons (e.g. not more than d8 - to be able to accomodate the two extra electrons from Y ligand in intermediate)

Answer 32

2 steps → 1 intermediate in potential energy surface two transition states and an intermediate in between - first step: leaving group removed before incoming ligand can combine with intermediate (decreases coordination number of metal - second step: ligand combines with intermediate this mechanism is given by: - high coordination number transition metal complex - bulky ligands

Answer 33

1 step → 0 intermediate in potential energy surface one transition state this mechanism is given by: - complexes which dont fit the other two classes

Answer 34

differentiate by measuring rates at varying concentrations of reagents associative and interchange are both bimolecular - but associative has an intermediate that can be isolated - intermediate cannot be observed for interchange

Answer 35

the extent to which ligand X weakens the bond trans to itself in the ground state of the complex - correlates with the σ donor/π acceptor ability of X - ligands trans to each other use the same orbitals on metal for bonding (compete for the same orbitals) if one ligand is a stronger σ donor, the ligand trans to it cannot donate e- to M as well, thus has a weaker interaction with M - bond in the trans position is weakened and is easier to replace ~ this is a kinetic effect ligands compete for the same orbitals of the metal ligand X trans to the leaving group in square planar complexes influence rate of substitution

Answer 36

when a ligand binds very strongly with d-AO, it polarises the d-AO and weakens the bond in the trans position ~ making it easier to replace

Answer 37

Ørsted's Law 1. an electric current (I) generates a magnetic field (B) 2. an electric current of intensity, I in a loop with area, A generates a magnetic field: μ = IA electrons in AOs/MOs are a microscopic version of this current: I ∝ e/m_e

Answer 38

total angular momentum, J orbital angular momentum, L spin angular momentum, S J = L + S in most molecules, electrons are localised in bonds ∴ L is negligible - in this case, we can use a spin-only formula where the magnetic moment only depends on S

Answer 39

S can be predicted from LFT S = 1/2(number of unpaired electrons in complex)

Answer 40

magnetic moment per unit of mass or volume in most materials, M = 0 unless in an applied magnetic field, H in an applied field: M = χH χ - magnetic susceptibility (proportionality constant) there are two different responses to the applied field, H χ < 0 - M points in opposite direction to H - diamagnetic material χ > 0 - M points in same direction as H - paramagnetic material

Answer 41

χ < 0 when placed in a magnetic field, atoms acquire an induced magnetic dipole moment that is in a direction opposite to the applied field - atom has no permanent magnetic moment ~ the resultant orbital and spin angular momentum = 0 - all paired electrons Lenz law: the applied field induces an electric current, whose associated field is opposite to the applied one

Answer 42

χ > 0 atom has a permanent magnetic dipole moment due to both electronic orbital and spin angular momentum (has unpaired electrons) in applied magnetic field, spins align parallel to the field

Answer 43

when particles are 'independent' (non-interacting) e.g. in dilute solution, magnetic susceptibility described by Curie's Law χ = C/T C - Curie constant linearised plot of Curie's Law: - plot 1/χ as a function of T - the slope of the curve is 1/C - curve passes through origin magnetisation increased with applied field M = χ H = (C/T)H - the greater the field, the greater the tendency for moments to align with the field - the susceptibility, χ = M/H is the same, irrespective of the field - magnetisation disappears when the field is removed

Answer 44

in zero applied field, moments are in random direction and fluctuating - cancel each other out to give average zero magnetisation when a magnetic field is applied - moments align with the field - net magnetic moment decreases with T

Answer 45

if magnetic moments on neighbouring sites interact with one another, the temperature dependence of χ requires an extention to Curie's Law: χ = C/(T-θ) θ - Weiss constant 1/χ = (T-θ)/C = T/C - θ/C plot 1/χ as a function of T - slope of the curve is still 1/C - curve does not pass through orgin

Answer 46

θ > 0 indicates solids where spins align in parallel fashion below a critical T (Curie temperature, Tc), all the spins align spin alignment is in the same direction as applied field and reinforces the field magnetisation increases beyond the value expected from Curie's Law

Answer 47

θ < 0 indicates solids where spins align in an antiparallel fashion below a critical T (Néel temerature, Tn), neighbour spins align in opposite directions spin alignment in opposite direction reduces the applied field magnetisation decreases below the value expected from Curie's Law

Answer 48

θ > 0 can also be obtained when there are different spins coupled in AFM way spins align in opposite directions but do not cancel out, hence behave as ferromagnetic solids - have two different ions with different magnetic moments - spins do not cancel out entirely, we have a net magnetic moment

Answer 49

the interaction between magnetic moments on neighbouring sites is called magnetic superexchange SUPEREXCHANGE: if magnetic ions have common ligands DIRECT EXCHANGE: otherwise e.g. in magnetic perovskite

Transition Metals Flashcards

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