Submitted to: Dr. Ali MohsinSubmitted by: Adina Tatheer 10060607-050 BS Chemistry (V) Section ‘A’
Q no. 1: Describe the magnetic properties of complex.Magnetic properties can be determined by looking at a compounds electron configurationand the size of its atoms. Since Magnetism is created by the spin of electrons, we canlook at how many unpaired electrons are present in a specific compound and determinehow magnetic the compound is. For this purpose we will be evaluating the d-blockelements or Transition Metals* (TMs) because they tend to have a large number ofunpaired electrons.Introduction:The magnetism discussed in this article is paramagnetism. Paramagnetism occurs whenthere are one or more unpaired electrons in a compound. (The opposite, when allelectrons are paired, is called diamagnetism). Di- and para-magnetism are often affectedby the presence of coordination complexes, which the transition metals (d-block) readilyform.Singular electrons have a spin, denoted by the quantum number ms as +(1/2) or –(1/2).This spin is negated when the electron is paired with another, but creates a slightmagnetic field when the electron is unpaired. The more unpaired electrons, the morelikely paramagnetic a material is. The electron configuration of the transition metals (d-block) changes when in a compound. This is due to the repulsive forces betweenelectrons in the ligands and electrons in the compound. Depending on the strength of theligand, the compound may become paramagnetic or diamagnetic.Ferromagnetism:Some paramagnetic compounds are capable of becoming ferromagnetic. This means thatthe compound shows permanent magnetic properties rather than exhibiting them only inthe presence of a magnetic field. In a ferromagnetic element, electrons of atoms aregrouped into domains, where each domain has the same charge. In the presence of amagnetic field, these domains line up so that charges are parallel throughout the entirecompound. Whether a compound can be ferromagnetic or not depends on how manyunpaired electrons it has and on its atomic size.
Small atoms pair up too easily and their charges cancel. Large atoms are difficult to keep together, their charge interaction is too weak.Therefore, only the right sized atoms will work together to group themselves intodomains. Elements with the right size include: Fe, Co, Ni. That means that Fe, Co and Niare paramagnetic with the capability of permanent magnetism; they are alsoferromagnetic.Ligand Field Theory BackgroundAn element can have up to 10 d electrons in 5 d-orbitals, dxy, dxz, dyz, dz2, and dx2-y2. During the formation of a complex, the degeneracy (equal energy) of these orbitals isbroken and the orbitals are at different energy levels.(Assuming a 6-ligand compound)In an octahedral complex, the ligands approach along the x, y, and z axes, so therepulsion is strongest in the orbitals along these axes (dz2 and dx2-y2). As a result, thedz2 and dx2-y2 orbitals are higher in energy than the dxy, dxz, and dyz orbitals. In a tetrahedral
complex, the splitting is opposite, with the dxy, dxz, and dyz orbitals higher in energy toavoid the ligands approaching between the axes. The splitting in a square planar complexhas four levels (lowest to highest): dyz and dxz, dxy, dz2, dx2-y2.Depending on the strength of the ligand, the splitting energy between the different d-orbitals may be large or small. Ligands producing a smaller splitting energy are called‘weak field’ ligands, and those with a larger splitting energy are called ‘strong field’ligands.Filling of d-orbitals in a complex:Hunds Rule states that electrons will fill all available orbitals with single electrons beforepairing up, while maintaining parallel spins (paired electrons have opposing spins). For aset of degenerated d-orbitals (not in a complex), electrons fill all orbitals before pairing toconserve the pairing energy, otherwise needed. With the addition of ligands, the situationbecomes more complicated. The splitting energy between the d-orbitals means thatadditional energy is required to place single electrons into the higher-energy orbitals.Once the lower-energy orbitals have been half-filled (one electron per orbital), an
electron can either be placed in a higher-energy orbital (preserving Hund’s rule) or pairup with an electron in a lower-energy orbital (when the splitting energy is greater than thepairing energy). The strength of the ligands determine which option is chosen.With a strong-field ligand, the splitting energy is very large and low-spin complexes areusually formed. With a weak-field ligand, the electrons can easily enter the higher-energyorbitals before pairing (high-spin).How does this relate to magnetism?Low-spin complexes contain more paired electrons since the splitting energy is largerthan the pairing energy. These complexes, such as [Fe(CN)6]3-, are more oftendiamagnetic or weakly paramagnetic. High-spin complexes usually contain moreunpaired electrons since the pairing energy is larger than the splitting energy. With moreunpaired electrons, high-spin complexes are often paramagnetic.The unpaired electrons in paramagnetic compounds create tiny magnetic fields, similar tothe domains in ferromagnetic materials. The higher the number of unpaired electrons(often the higher-spin the complex), the stronger the paramagnetism of a coordinationcomplex. We can predict paramagnetiism and its relative strength by determiningwhether a compound is a weak field ligand or a strong field ligand. Once we havedetermined whether a compound has a weak or a strong ligand, we can predict itsmagnetic properties:
Q no. 2: Explain the CF in the octahedral and tetrahedral symmetry alsoexplain the factors affecting the magnitude of ∆.CF theory: CF theory tried to describe the effect of the electrical fieldof neighboring ions on the energies of the valence orbitals of an ion in a crystal. Crystalfield theory was developed by considering two compounds: manganese(II) oxide, MnO,and copper(I) chloride, CuCl.Octahedral Crystal Fields:Each Mn2+ ion in manganese(II) oxide is surrounded by six O2- ions arranged toward thecorners of an octahedron, as shown in the figure below. MnO is therefore a model foran octahedral complex in which a transition-metal ion is coordinated to six ligands.What happens to the energies of the 4s and 4p orbitals on an Mn2+ ion when this ion isburied in an MnO crystal? Repulsion between electrons that might be added to theseorbitals and the electrons on the six O2- ions that surround the metal ion in MnO increasethe energies of these orbitals. The three 4p orbitals are still degenerate, however. Theseorbitals still have the same energy because each 4p orbital points toward two O2- ions atthe corners of the octahedron.Repulsion between electrons on the O2- ions and electrons in the 3d orbitals on the metalion in MnO also increases the energy of these orbitals. But the five 3d orbitals on theMn2+ ion are no longer degenerate. Lets assume that the six O2-ions that surround eachMn2+ ion define an XYZ coordinate system. Two of the 3d orbitals (3dx2-y2 and 3dz2) onthe Mn2+ion point directly toward the six O2- ions, as shown in the figure below. Theother three orbitals (3dxy, 3dxz, and 3dyz) lie between the O2- ions.
The energy of the five 3d orbitals increases when the six O2- ions are brought close to theMn2+ ion. However, the energy of two of these orbitals (3dx2-y2 and 3dz2) increases muchmore than the energy of the other three (3dxy, 3dxz, and 3dyz), as shown in the figurebelow. The crystal field of the six O2- ions in MnO therefore splits the degeneracy of thefive 3d orbitals. Three of these orbitals are now lower in energy than the other two.By convention, the dxy, dxz, and dyz orbitals in an octahedral complex are calledthe t2g orbitals. The dx2-y2 and dz2 orbitals, on the other hand, are called the eg orbitals.
The easiest way to remember this convention is to note that there are three orbitals inthe t2g set. t2g: dxy, dxz, and dyz eg: dx2-y2 and dz2The difference between the energies of the t2g and eg orbitals in an octahedral complex isrepresented by the symbol o. This splitting of the energy of the d orbitals is nottrivial; o for the Ti(H2O)63+ ion, for example, is 242 kJ/mol.The magnitude of the splitting of the t2g and eg orbitals changes from one octahedralcomplex to another. It depends on the identity of the metal ion, the charge on this ion, andthe nature of the ligands coordinated to the metal ion.Tetrahedral Crystal Fields:Each Cu+ ion in copper(I) chloride is surrounded by four Cl- ions arranged toward thecorners of a tetrahedron, as shown in the figure below. CuCl is therefore a model fora tetrahedral complex in which a transition-metal ion is coordinated to four ligands.Once again, the negative ions in the crystal split the energy of the d atomic orbitals on thetransition-metal ion. The tetrahedral crystal field splits these orbitals into thesame t2g and eg sets of orbitals as does the octahedral crystal field. t2g: dxy, dxz, and dyz eg: dx2-y2 and dz2But the two orbitals in the eg set are now lower in energy than the three orbitals inthe t2g set, as shown in the figure below.
To understand the splitting of d orbitals in a tetrahedral crystal field, imagine four ligandslying at alternating corners of a cube to form a tetrahedral geometry, as shown in thefigure below. The dx2-y2 and dz2 orbitals on the metal ion at the center of the cube liebetween the ligands, and the dxy, dxz, and dyz orbitals point toward the ligands. As aresult, the splitting observed in a tetrahedral crystal field is the opposite of the splitting inan octahedral complex.Because a tetrahedral complex has fewer ligands, the magnitude of the splitting issmaller. The difference between the energies of the t2g and eg orbitals in a tetrahedralcomplex ( t) is slightly less than half as large as the splitting in analogous octahedralcomplexes ( o). t = 4/9 oFactors effecting the magnitude of Δ:
The size of the gap Δ between the two or more sets of orbitals depends on several factors,including the ligands and geometry of the complex. Some ligands always produce a smallvalue of Δ, while others always give a large splitting. The reasons behind this can beexplained by ligand field theory. The spectrochemical series is an empirically-derived listof ligands ordered by the size of the splitting Δ that they produce (small Δ to large Δ; seealso this table): - I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH < C2O42− < H2O < NCS− < CH3CN < py < NH3 < en < 2,2- bipyridine < phen <NO2− < PPh3 < CN− < COIt is useful to note that the ligands producing the most splitting are those that can engagein metal to ligand back-bonding.The oxidation state of the metal also contributes to the size of Δ between the high andlow energy levels. As the oxidation state increases for a given metal, the magnitude of Δincreases. A V3+ complex will have a larger Δ than a V2+ complex for a given set ofligands, as the difference in charge density allows the ligands to be closer to a V3+ ionthan to a V2+ ion. The smaller distance between the ligand and the metal ion results in alarger Δ, because the ligand and metal electrons are closer together and therefore repelmore.Crystal field stabilization energy:The crystal field stabilization energy (CFSE) is the stability that results from placing atransition metal ion in the crystal field generated by a set of ligands. It arises due to thefact that when the d-orbitals are split in a ligand field (as described above), some of thembecome lower in energy than before with respect to a spherical field known as thebarycenter in which all five d-orbitals are degenerate. For example, in an octahedral case,the t2g set becomes lower in energy than the orbitals in the barycenter. As a result of this,if there are any electrons occupying these orbitals, the metal ion is more stable in theligand field relative to the barycenter by an amount known as the CFSE. Conversely,the eg orbitals (in the octahedral case) are higher in energy than in the barycenter, soputting electrons in these reduces the amount of CFSE.
Octahedral crystal field stabilization energyIf the splitting of the d-orbitals in an octahedral field is Δoct, the three t2g orbitals arestabilized relative to the barycenter by2/5 Δoct, and the eg orbitals are destabilizedby 3/5 Δoct. As examples, consider the twod5 configurations shown further up the page.The low-spin (top) example has five electrons in the t2g orbitals, so the total CFSE is 5x 2/5 Δoct = 2Δoct. In the high-spin (lower) example, the CFSE is (3 x 2/5Δoct) - (2x 3/5 Δoct) = 0 - in this case, the stabilization generated by the electrons in the lowerorbitals is canceled out by the destabilizing effect of the electrons in the upper orbitals.Crystal Field stabilization is applicable to transition-metal complexes of all geometries.Indeed, the reason that many d8 complexes are square-planar is the very large amount ofcrystal field stabilization that this geometry produces with this number of electrons.TetrahedralOctahedral
Q no. 3: How pi bonding is explained by MOT and give the experimentalevidence for pi bonding?Pi Bonding and MO Theory:Pi-bonding also exists for tetrahedral and square planar complexes. The MO treatmentfor these systems is very similar to what is observed for an octahedral system. First,there are four plausible L-M pi-interactions: pπ-dπ, dπ-dπ, π*-dπ, σ*-dπ.The pπ-dπ interaction involves ligand-to-metal pi donation while the other three aremetal-to-ligand pi donations. Pi-bonds will involve the t2g set, not the eg*. This isbecause the eg* orbitals point directly at the ligands and are set up for σ overlap. Thedirection of electron donation and the energy levels of ligand pi-bonding orbitals willhave a pronounced effect on molecules. We will consider a molecule with six pi-donorligands (e.g. halide ions) and then 6 pi-acceptor ligands (e.g. CO).MX6n- :The halide p orbitals are lower in energy than the metal d orbitals and they are filled,while metal d orbitals may or may not contain electrons. Thus:When the MOs form the ligand p electrons fill the t2g orbitals, thus metal t2gelectrons go into the t2g* MOs. The result of this type of interaction is a small ∆O.M(CO)6: The CO pi* orbitals are empty and are high in energy (remember CO bondenergy).
Since the CO pi* orbitals are empty, the t2g MO is filled with metal t2g electrons andpromotion is then a relatively high energy process. These diagrams explain the relativeplacements of the halides and CN-/CO in the spectrochemical series. In an electrostaticmodel, the reverse would be expected.Experimental Evidence For Pi Bonding: So what evidence is there for π-bonding (i.e. what do we look for)? Webegin by asking what would the interaction look like without π-bonding? Then whathappens with full π-bonding: M-L → M=LSince the bonding between metal and ligand changes between these forms, bondingwithin the ligand must change. If electron density is fed into a pi or σ orbital on themetal, a bond within the ligand will be weakened. The strongest evidence for pi- bondingcomes from metal-carbonyl complexes. Crystallography - The greater the extent of pi-back bonding, the more M=C character there will be and the more C≡O will resembleC=O. The difference in C≡O and C=O bond lengths is about 0.1 Å and should be useablefor quantification. Unfortunately, this has not been observed. In contrast, M-C bondlengths do change. Consider the complexes Cr(CO)6 and Cr(CO)5(PR3). In the absenceof π-backbonding the Cr-C bond lengths should be the same. If it does occur, then thebond lengths should be shorter in Cr(CO)5(PMe3). Why? Two reasons: PMe3 is at besta very poor pi-acceptor so only 5 COs are competing for electron density from the metal,not 6; and PR3 is a very good σ donor, CO is not. Thus, the Cr has more electron densityto share with fewer acceptors. One other trend is expected. The Cr-C(O) bond trans toPR3 should be shorter than those cis. This is because the trans CO will bind to the samed-orbital as the PR3 and the effect will be greatest there.Infrared Spectroscopy:
Evidence for C=O character is most clearly seen in IR spectroscopy. ν(CO) forC≡O is about 2150 cm-1, while in R2C=O ν(C=O) is about 1700 cm-1. Thus, the greaterthe extent of backbonding the lower the expected ν(C≡O). This is seen dramatically fortwo series of complexes M(CO)6n+/- and M(CO)4n+/- .This can also be seen when a CO issubstituted for by another ligand as seen in crystallography. The only problem with usingthis technique is that the CO stretching band is almost always split into severalcomponents making interpretation difficult.Q no. 4:Explain the Electroneutrality and BackBonding with suitableexamples?Electroneutrality:It is assumed; that is, that there is no measurable charge excess in any side of themembrane. So, although there is an electric potential across the membrane due to chargeseparation, there is no actual measurable difference in the global concentration of positiveand negative ions across the membrane, that is, there is no actual measurable chargeexcess in either side. That occurs because the effect of charge on electrochemicalpotential is hugely greater than the effect of concentration so an undetectable change inconcentration creates a great change on electric potential.Example:Nature seems to strongly discourage any process that would lead to an excess of positiveor negative charge in matter. Suppose, for example, that we immerse a piece of zincmetal in pure water. A small number of zinc atoms go into solution as Zn2+ ions, leavingtheir electrons behind in the metal: Zn(s) → Zn2+ + 2e–As this process goes on, the electrons which remain in the zinc cause a negativecharge to build up within the metal which makes it increasingly difficult for additionalpositive ions to leave the metallic phase. A similar buildup of positive charge in theliquid phase adds to this inhibition. Very soon, therefore, the process comes to a halt,resulting in a solution in which the concentration of Zn2+ is still too low (around 10–10 M)to be detected by ordinary chemical means.
Figure 1. Transport of zinc ions from the metal to water; the build-up of negative charge in the metal (and positive charge in the solution) soon brings the process to a halt.There would be no build-up of this opposing charge in the two phases if the excesselectrons could be removed from the metal or the positive ions consumed as theelectrode reaction proceeds. For example, we could drain off the electrons left behindin the zinc through an external circuit that forms part of a complete electrochemicalcell. Another way to remove these same electrons is to bring a good electron acceptor(that is, an oxidizing agent) into contact with the electrode. A suitable acceptor wouldbe hydrogen ions; this is why acids attack many metals. For the very active metalssuch as sodium, water itself is a sufficiently good electron acceptor.The degree of charge unbalance that is allowed produces differences in electricpotential of no more than a few volts, and corresponds to unbalances in theconcentrations of oppositely charged particles that are not chemically significant.There is nothing mysterious about this prohibition, known as the electroneutralityprinciple; it is a simple consequence of the thermodynamic work required to separateopposite charges, or to bring like charges into closer contact. The additional workraises the free energy change of the process, making it less spontaneous.The only way we can get the oxidation of the metal to continue is to couple it withsome other process that restores electroneutrality to the two phases. A simple way toaccomplish this would be to immerse the zinc in a solution of copper sulfate insteadof pure water. As you will recall if you have seen this commonly-performed
experiment carried out, the zinc metal quickly becomes covered with a black coatingof finely-divided metallic copper. The reaction is a simple oxidation-reductionprocess, a transfer of two electrons from the zinc to the copper: Zn(s) → Zn2+ + 2e– Cu2+ + 2e– → Cu(s)The dissolution of the zinc is no longer inhibited by a buildup of negative charge inthe metal, because the excess electrons are removed from the zinc by copper ions thatcome into contact with it. At the same time, the solution remains electrically neutral,since for each Zn ion introduced to the solution, one Cu ion is removed. The netreaction Zn(s) + Cu2+ → Zn2+ + Cu(s)quickly goes to completion.Back bonding:It is a type of bonding which take place between atoms in a compound ;in which oneatom has lone pair of electron and other has vacant orbital placed adjacent to eachother.A compound which posses back bonding has pi bonding character because it occursafter formation of sigma bond.Usually ; back bonding causes molecule to get stable as it completes octet.for ex. in borontriflouride ,back bonding causes decrease in bond length and increase in bond order.pi backbonding:
(Top) the HOMO and LUMO of CO. (Middle) an example of a sigma bonding orbitalin which CO donates electrons to a metals center from its HOMO. (Bottom) anexample where the metal center donates electrons through a d orbital to COs LUMO.NOTE in this depiction the y axis has no relation to energy levels.π backbonding, also called π backdonation, is a concept from chemistry in whichelectrons move from an atomic orbital on one atom to a π* anti-bonding orbital onanother atom or ligand. It is especially common in the organometallicchemistry of transition metals with multi-atomic ligands such as carbonmonoxide, ethylene or the nitrosonium cation. Electrons from the metal are used tobond to the ligand, in the process relieving the metal of excess negative charge.Compounds where π backbonding occurs include Ni(CO) and Zeisessalt. IUPAC offers the following definition for backbonding:A description of the bonding of π-conjugated ligands to a transition metal whichinvolves a synergic process with donation of electrons from the filled π-orbital or loneelectron pair orbital of the ligand into an empty orbital of the metal (donor–acceptorbond), together with release (back donation) of electrons from an nd orbital of themetal (which is of π-symmetry with respect to the metal–ligand axis) into the emptyπ*-antibonding orbital of the ligand.Metal carbonyls, nitrosyls, and isocyanides:The electrons are partially transferred from a d-orbital of the metal to anti-bondingmolecular orbitals of CO (and its analogues). This electron-transfer (i) strengthens the
metal-C bond and (ii) weakens the C-O bond. The strengthening of the M-CO bond isreflected in increases of the vibrational frequencies for the M-C bond (often outside ofthe range for the usual IR spectrophotometers). Furthermore, the M-CO bond length isshortened. The weakening of the C-O bond is indicated by a decrease in the frequency ofthe νCO band(s) from that for free CO (2143 cm−1) often by more than 200 cm−1. For thisreason, IR spectroscopy is an important diagnostic technique in metal-carbonylchemistry. The article infrared spectroscopy of metal carbonyls discusses this in detail.Many ligands other than CO are strong "backbonders". Nitric oxide is an even stronger π-acceptor than is CO and νNO is a diagnostic tool in metal-nitrosyl chemistry. In the caseof isocyanide complexes, the degree of π-bonding is again indicated by shortening of theM-CNR bond and by decrease in νCN. For the isocyanides however, an additionalparameter is the MC=N-C angle, which deviates from 180° in highly electron-richsystems. Other ligands have weak π-backbonding abilities, which creates a labilizationeffect of CO, which is described by the cis effect.Metal-alkenes and alkyne complexes:As in metal-carbonyls, electrons are partially transferred from a d-orbital of the metal toanti-bonding molecular orbitals of the alkenes and alkynes. This electron-transfer (i)strengthens the metal-ligand bond and (ii) weakens the C-C bonds within the ligand. Inthe case of metal-alkenes and alkynes, the strengthening of the M-C2R4 and M-C2R2 bondis reflected in bending of the C-C-R angles which assume greater sp3 and sp2 character,respectively. Thus strong pi-backbonding causes a metal-alkene complex to assume thecharacter of a metallacyclopropane. Electronegative substituents exhibit greater pibackbonding. Thus strong pi backbonding ligandsare tetrafluoroethylene, tetracyanoethylene, and hexafluoro-2-butyne.Metal-phosphine complexes:
R3P–M σ bondingR3P–M π backbondingPhosphines accept electron density from metal p or d orbitals into combinations of P–Cσ* antibonding orbitals that have π symmetry. When phosphines bond to electron-richmetal atoms, backbonding would be expected to lengthen P–C bonds as P–C σ* orbitalsbecome populated by electrons. The expect lengthening of the P–C distance is oftenhidden by an opposing effect: as the phosphorus lone pair is donated to the metal, P(lonepair)-R(bonding pair) repulsions decrease, which acts to shorten the P–C bond. The twoeffects have been deconvoluted by comparing the structures of pairs of metal-phosphinecomplexes that differ only by one electron. Oxidation of R3P–M complexes results inlonger M–P bonds and shorter P–C bonds, consistent with π-backbonding. In early work,phosphine ligands were thought to utilize 3d orbitals to form M-P pi-bonding, but it isnow accepted that d-orbitals on phosphorus are not involved in bonding as they are toohigh in energy.
References: 1) http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Crystal_Field_Theory/Crys tal_Field_Theory/Magnetic_Properties_of_Coordination_Complexes_(CFT) 2) http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/crystal.php 3) Principles of structure and reactivity, James E Huheey, Ellen A. Keiter, Richard L Keiter, 4th edition.