Chemistry of Transition Metals Part 2. Theories/ Concepts Bonding - PowerPoint PPT Presentation

Chemistry of Transition Metals Part 2. Theories/ Concepts

Bonding in transition metal compounds Theories : (i) Werner Coordination Theory (ii) 18 electron rule/ EAN (iii) Valence Bond Theory (iv) Crystal field theory (v) Molecular orbital approach Consequences : (i) High spin - low spin complexes (ii) Spectrochemical series (iii) Crystal Field Stabilization Energy (CFSE) (iv) Jahn-Teller distortions (v) Spinels

CFT- Octahedral Field

d xy d xz d yz d z2 d x2-y2

d xy d xz d yz d z2 d x2-y2

d xy d xz d yz d z2 d x2-y2

d xy d xz d yz d z2 d x2-y2

d xy d xz d yz d z 2 d x 2 -y 2

Octahedral Field • Interaction of d-orbitals with six point charges at +x, -x, +y, -y, +z and -z axes are not same . • The orbitals lying along the axes (i.e. x 2 -y 2 , z 2 ) will be destabilized more than the orbitals lying in- between the axes (i.e. xy, xz, yz).

CFT-Octahedral Complexes For O h point group x 2 -y 2 , z 2 orbitals: e g xy, xz, yz orbitals: t 2g  Difference between t 2g and e g = Δ 0 or 10 Dq.  Conservation of barycenter from a spherical field to octahedral field indicates t 2g set must be stabilized as much as the e g set is destabilized .

CFSE: Octahedral complex

Dependence: Δ o • Nature of the ligands • The charge on the metal ion • Whether the metal is a 3d , 4d , or 5d element Ligands : Weak field ligands ; small splitting ( Δ ο ~7000 – 30000 cm -1 ) Strong field ligands ; large splitting ( Δ ο > 30000 cm -1 ) M 2+ < M 3+ < M 4+ 3d < 4d < 5d

Tetrahedral field

Δ t = 4/9 Δ o

Octahedral vs Tetrahedral Field

Spinels- Use of CFSE

Spinels- Use of CFSE

Spinels- Use of CFSE

Spinels- Use of CFSE

Special case of d 8 Octahedral

Jahn-Teller Distortion

Jahn-Teller Distortion Non-linear unsymmetrical molecule: Higher energy J. T. distortion Lower degeneracy/ lower energy.

Tetrahedral, Octahedral and Square Planer

Magnetism CH CH-105

Magnetism Why do we need it? Familiar world How does it work???

Magnetism Why do we need it? Outer world Life Weather on and Earth? Sky?

Magnetism Magnetism is everywhere!!!! Origin: Lets move inside……

Chemistry Magnetism Transition & Lanthanide ions and their complexes

Magnetism Origin: Paired & unpaired electron spins How ? Spinning of electron Paired electron – mutual neutralization Elements with unpaired electron – no cancellation They are magnets (Fe, Co, Ni)

N 2 diamagnetic Dia.M. ex. H 2 O, KCl organic ligands, etc. Bismuth metal (most diamagnetic of all metals) 36

O 2 paramagnetic when dioxygen is in its ground state it is a triplet (spin S=1) and its reactivity is weak. Liquid O2 37

Magnetic Levitation (Suspension) : Property of diamagnetic molecules

Magnetochemistry Electron spin: An electron has two intrinsic spin states, which are referred as up and down or alpha and beta. Electron orbital motion: A magnetic field is generated due to the electron moving around the nucleus. Nuclear spin: Some nuclei, such as hydrogen, have a net spin, which generates a magnetic field. Mutual interaction also and with external magnetic field Shows effect strong/weak and negligible.

Magnetism

Magnetism Molar Susceptibility Type: Mass (gram) susceptibility, χ g Volume susceptibility, κ Molar susceptibility, X m Interrelation: Χ g = κ/ρ where ρ is density X m = X g x M.Wt. Where, M. Wt. is molecular weight of the sample Measurable quatity (X m ) - related to atomic properties Summary: Volume  mass  molar SUSCEPTIBILITY

Magnetism Magnetic moment (µ) from susceptibility (  )

Magnetism

µ total µ orbital µ spin e -

Magnetism

Magnetism Conditions of orbital angular momentum (µ L ) The orbitals should be degenerate (t 2g or e g ) Similar in shape and size Interconvertible by rotation eg: t 2g orbitals into each other by 90 o rotation. Such transformation is not possible with the orbitals of e g . The orbitals must not contain electrons of identical spin during this transformation and the movement of electron These conditions are fulfilled only when one or two orbitals contain partially filled electrons in t 2g and NOT in e g

Octahedral complexes The degenerate t 2g orbitals (d xy , d xz , d yz ) can be interconverted by 90 ° rotations e.g. the d xz orbital is transformed into the d yz orbital by a rotation of 90 ° about the z -axis – during this rotation the electron is orbiting the nucleus Thus, an electron in a t 2g orbital can contribute to orbital angular momentum x y 48

d yz d xz z z z x x 90 o rot. 90 o rot. x d yz y y y 49

Octahedral complexes However the e g orbitals (d z 2 and d x 2 -y 2 ) cannot be interconverted by rotation as they are different shapes Thus an electron in an e g orbital can not contribute to orbital angular momentum 50

But an e g ------> t 2g transformation is possible dxy / dx 2 -y 2 orbital motion about z axis dxz / dyz orbital motion about z axis dxz / dxy orbital motion about x axis dyz / dxy orbital motion about y axis 51

Orbital contribution to the magnetic moment high spin octahedral d n ions think of possible t 2g electron arrangements d 1 d yz d yz d yz d xz d xy d xz d xy d xz d xy Possible t 2g arrangements = 3 d 1 e.g. Ti(III) YES Orbital contribution = d 2 d yz d yz d yz d xz d xy d xz d xy d xz d xy Possible t 2g arrangements = 3 d 2 e.g. V(III) Orbital contribution = yes 52

Magnetism Orbital Contributions in Octahedral Complexes

Q01. Crystalline AgO is diamagnetic. Explain. • AgO As per formula, oxidation state is +2 Electronic configuration should be Ag: [Kr] 4d 10 5s 1 Ag(II): [Kr] 4d 9 • Mixture of Ag 2 O and Ag 2 O 3

• Has Ag(I) and Ag(III) configurations • Both are diamagnetic Linear Ag(I) (green) and square planar Ag(III) (grey)

• Q02 . Work out the hybridization and geometry for the following complexes using the valence bond approach. (a) Ni(CO) 4 ; (b) [Ni(CN) 4 ] 2- ; (c) [CoCl 4 ] 2- ; (d) OsO 4 ; (e) VOCl 3 ; (f) [Pt(NH 3 ) 4 ] 2+ ; (g) [Ag(NH 3 ) 2 ] + ; (h) [Pt(PPh 3 ) 4 ]; (i) (Cr 2 O 7 ) 2- [Ni(CN) 4 ] 2-

(c) OsO 4 Electronic configuration [Xe] 4f 14 5d 6 6s 2 Ionic approach : Oxidation state of Os(VIII) Electronic configuration: [Xe] 4f 14 5d 0 6s 0 d 3 S hybridization Tetrahedral Covalent approach : Covalent bond Electronic configuration: [Xe] 4f 14 5d 6 6s 2 d 3 S hybridization Tetrahedral The remaining four electrons form π with oxygen

Q03. While the most stable chloride of Zr is ZrCl 4 , that of Pd is PdCl 2 . Why? • The third and high I.E. of the d-block metals increases with increasing atomic number. Owing to the large Zeff making it more energetically unfavourable to attain oxidation state above +2. • Further the d-orbitals become more core-like towards the end of the series and so are less effective in stabilising higher oxidation states.

Q04. When high pressure is applied, what type of electronic configuration is favoured for a d 5 transition metal complex (Octahedral, weak field ligand)? Low spin ; because it leads to low electron density between the metal and the ligand (i.e., along the bond axis). d 5 : LS d 5 : HS

Q05. Provide reasons for the fact that a number of tetrahedral Co(II) complexes are stable, where as the corresponding Ni(II) complexes are not. The CFSE of d 7 tetrahedral complex is greater than that of d 8 tetrahedral complex. Similarly, the CFSE of d 8 octahedral complex is greater than that of the d 7 octahedral complex.

Q06. Using the crystal field stabilization energy as criterion, indicate whether you expect the following spinels to be normal or inverse: Fe 3 O 4 ; Co 3 O 4 . Spinel by definition, the 3+ ion has to go to the O h site leaving the 2+ ion in T d . Fe 3 O 4 is composed of Fe(II) Td and Fe(III) Oh ions with d6 and d5 configurations respectively. Since d 5 has no CFSE, it is more advantageous to put it in a Td environment than in Oh. In other words, by placing d 6 ions in Oh environment there is more gain in more CFSE than keeping this in Td environment. Here the Fe 3 O 4 structure is inverse spinel. Co 3 O 4 has a similar structure with d7 and d6 configurations for 2+ and 3+ ions respectively. Co(III) d6 ion is low spin because (a) high charge (even with weak ligands like oxo) and (b) maximum gain in CFSE. So the Co 3 O 4 structure is normal spinel.

Q07. By showing the details, determine the CFSE for the following complexes: (a) [FeCl 4 ] 2- ; (b) W(CO) 6 . CFSE = -0.6  t e 3 t 2 3 6 e g 0 t 2g CFSE = -2.4  o

Q08. Explain what is meant by the term “synergic bonding”? In the synergic bonding the  -donation of charge from the ligand to the metal is reinforced by π - back-donation from the metal to the ligand. In a valence bond model it may be represented by M - CO + and M=C=O. The filled CO to empty M  -donation. Filled M d- orbital to empty π* - back-donation.