GENERAL CHEMISTRY I
GENERAL CHEMISTRY I CHEM 121
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This 21 page Class Notes was uploaded by Jay O'Keefe on Monday October 19, 2015. The Class Notes belongs to CHEM 121 at Rice University taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/225048/chem-121-rice-university in Chemistry at Rice University.
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Date Created: 10/19/15
Exceptions to the octet rule 0 too few electrons B Be sometimes Try BeHg 0 too many ligands hexa uoride pentachloride Try XeF4 0 odd number of electrons free radicals N has 5 valence electrons Try NO N02 Bond Properties 0 Bond Order the number of bonidng electron pairs shared by two atoms in a molecule lf resonance structures exist the bond order may be fractional Consider carbonate anoin C032 0 Bond Length the distance between the nuclei of two bonded atoms Because atom size varies smoothly with position in the periodic table predictionsof trends in bond length can be quickly made Let s try to rank the following 1 H71 H7C1 HiBr H71 2 oio oio CiF CiN 3 ozo oio oo Example Draw resonance structures NO What is the NO bond order in this ion The N70 bond length is 136 pm the NO bond length is 115 pm Compare these values with the NO bond length is N021 124 pm Account for any differences Charge Distribution in Covalent Compounds We can ne tune the Lewis structures to get a more precise description of the distribution of the electrons in a molecule The valence electrons are usually not shared evenly between bonded species some atoms may have a slight negative change and others a slight postive charge The result is that some atoms have partial charges and the way they are distributed in the molecule is called a charge distribution The charge distribution can have a large effect on the way molecule line up in solids and solution The location of the charge can also in uence the site at which a reaction occurs Will H4r attach to the oxygen or chlorine of OCl 2 Formal Charge 0 The formal charge of a molecule or ion is equal to the group number of the atom minus the number of lone pair electrons minus one half the number of bonding electrons o This represents a difference between the number of valence electrons of the lone atom and the number of electrons assigned to the atom in the Lewis structure 0 The sum of the formal charges on the atoms must equal the net charge Examples 1 CN 2 SCO 3 803 Bond Polarity and Electronegativity Covalent bonds indicate the atoms share the electron pair equally and only occur between identical atoms while ionic bonds indicate that one atom donates its electron entirely to the other atom These are the two extreme cases In the middle are polar covalent bonds polar covalent bond a bond in which the two atoms have residual or partial charges Not all atoms hang on to their valence electrons with the same force nor do they take on additional electrons with the same ease Recall that atoms have different effective charges and electron af nities Consider NaCl HF and F2 Linus Pailing proposed a parameter called the electronegativity x to allow us to decide if a bond is polar which atom on the bond is negative and if one bond is more polar than another It is de ned as the measure of the ability of an atom in a molecule to attract electrons to itself Example For each of the following bond pairs decide which is the more polar and indicate the postive and negative poles 1 LiiF and LiiCl 2 SiiO and FiF 3 CO and CS Combining Formal Charge and Bond Polarity Consider BF Does it make sense that boron has a formal charge of 1 while the more electronegative uorine has a formal charge of 0 The bonds in this molecules are very polar AX 20 indicating that boron has a partially positive charge and uorine has a partially negative charge These two considerations give opposing answers the book rationalizes this by saying the charge is spread out 7 over the entire molecule The electroneutrality principle Pauling states that electrons in a molecule are distributed in such a way that the charges on the atoms are as close to zero as possible Consider C02 Where would an electron attack OCN Valence Shell ElectroniPair Repulsion VSEPR Model I Z llzrf mm mm 6 WW sz 1mm 180 Ber 002 Ax3 tzigonal 120 BFg co2 AX ce ah ml 109 5 0mm3 Ax6 txigonalbipyzamidal 90 120 PF6 AXE octahedzal 90 SFE What is the stzuctuxe of CH20127 How do lone pairs affect the geometry Lone pairs of electrons on the ventral atom occupy sparial positions even though their location is not included in the verbal description of the shape of the molecule or ion Consider ammonia Consider water lone pairi lone pair gt lone pairibond pair gt bond pairibond pair AXg linear 1800 Bng AXg trigonal 120O BFg AXQE bent N02 AX4 tetrahedral 1095O CH4 AXgE trigonal pyramidal NHg AXQEQ bent H20 AX5 trigonal bipyramidal 90quot 1200 PF5 AX4E seeisavv SF4 AXgEQ Tishaped Cng AXQEg linear 1 AX6 octahedral 90O SF6 AX5E square pyramidal BrF5 AX4E2 square planar XeF4 lulml xu 5 6 Tvuvy xnulckul ndvnen 7 modelu VSEPR Ndzev warn Zn zarn nf varu mulekuly OA3 3 3 b Z When there are several different ligands7 it is possible to have isomers of the complex which may behave quite different Chemically Draw the structures of all possible isomers for the following complepes Indicate which iso mers are mirror image pairs enantiomers Z diamminebromochloroplatinumU square planar 2 diaquachlorotricyanocolbaltateU octahedral 3 trioccalatovanadateIU octahedral The octahedral structure is not the only possible sipicoordinate structure Other possibilities include a planar heccagonal structure and a triangular prism structure In the latter the ligands are arranged in two parallel triangles one lying above the metal atom and the other below the metal atom with its corners directly in line with the corners of the rst triangle Show that the eccistence of two and only two isomers of C39oll134Clngr provides evidence against both of these possible structures What is the nature of bonding in coordination complexes Why does PtlV form only octahedral complexes7 While Ptll forms square planar ones A simple but quite useful model for bonding in coordination compounds is crystal eld theory Which starts from an ionic description ofthe metaliligand bonds Crystal eld theory considers the response ofthe metal7s diorbitals coming into contact With lone pairs of ligands EDD D D D D D In In dv DEEDS dorbilal energies l Average energy a dorbilals b I orbilals c d orbilals In quotfreequot metal of 1 orbitals in in tetrahedral in octahedral in square am or ion the presence of complex complex planar 39 complex Ezperr rnertts cart measure rtot ortlg whether a compourtd ts paramagnetr c but also the humber of urtpar red electrons It ts fourtd that the octahedral complex tort FeCN63 has few WW patred electrorts thart the octahedral cornpr tort FeH2 O63 How Marty urtpar red electrorts are presertt m each spectes G39r39oe the dielectrort cort guratr ort of each spectes Luw eld splitLing 00000 Higm new 5p lining xv dn Fxgure 5 The two cases ofcrystal eld sphmng forthe octahedral geometry thl the coorch39rtatr ort compourtd CoNHg6Cb be dramagrtetr c or paramagnettc Many of the colors of octahedral transition metal compounds arise from excitation of an electron from an occupied bgg level to an empty e9 level The frequency of light that is capable of inducing such a transition is related to the crystali led splitting energy The larger AD the higher the frequency of light absorbed most strongly7 and the lower bluer its wavelength The ligands tune the crystal eld splitting according to the spectrochernical series Field strength Strong Weak CN gt N02 gt an gt NH3 gt H20 gt OH gt F gt 11 gt Br gt T dLevel 7 39 splitting A Small The chromiumUH ion in aqueous solution has a blueiuiolet color 1 What is the color complementary to bluewiolet 2 Estimate the wavelength of maaimum absorption for CrNOg3 solution 5 Will the wavelength of maximum absorption increase or decrease if cyano ligands are substituted for the coordinated water Explain ULI RAVIDLET P A Violet I w E u m 5 450 g I Blue 4 m 5 500 m g 5 Green 4 a u 570 m lt g 590 g lt em w u g z gt Each color that makes up visible lrght corresponds to a specmc wavelength and energy range Ruby and Emerald Rqu ALEXANDRHk MM Vlbgycr Vibgycr absorhu an absorbu on EMERALD Ihe Crw H ruby alexandrne and emerald is H octahedral coonllnatmn absorbu on vibgycr The deep red of ruby results from the substitution of a small number of chromium atom impurities into an aluminum oxide corundum crystal Corundum A1203 is colorless it 7s an insulatorl but when it is combined with about 1 of the green solid 01 203 the deepired hue of ruby results The Cr3 ions have the same oxidation number as the aluminum ions They replace Al3 ions in octahedral crystal sites7 each de ned by six oxygen ions Because CrJr has the valencewlectron con guration 3d37 its ground state in the octahedral site in the crystal has all three electrons with spins parallel in the lowerienergy tgg levels When light of the proper wavelength shines on such an ion7 either one or tow of these electrons can be excited into the higherienergy e9 level7 giving rise to two strong absorption bands in the visible region of the spectrum The lowerienergy band causes the absorption of green and yellow light and the other the absorption of violet light The two bands overlap7 so that most of the blue light is also absorbed7 but some passes through and combines with the strongly transmitted red light to give the characteristic slightly purplish red of the ruby Emeralds also have Cr3 impurities that substitute for Al3 ions in an octahedral site formed by six oxygen ions What then causes the dramatic difference between ruby red and emerald green The answer lies with a small change about 10 in the crystal eld splitting pa rameter A0 The crystal into which chromium substitutes to give emerald is not corundum but beryl7 with formula BegAlQSisolg Like corundum7 beryl is colorless in its pure form The presence of beryllium and silicon makes the bonding in this crystal weaker than in than in corundum The weaker crystal eld at the octahedral sites means that both of the Cr3 absorption bands in emerald are shifted to lower energy One absorption now overlaps the red part of the visible spectrum7 blocking transmission of red7 orange and yellow light The other absorption shifts downward from violet toward blue and also changes in shape There is now a transmission window between the two bands Green light with a trace of blue is transmitted7 giving a color that we call emerald green A subtle change in crystali eld forces has thus completely changed the appearance of the gem Structures of Coordination Compounds and Isomers Common Geometries o Complexes with the General Formula MLglin o Complexes with the General Formula ML4lin o Complexes with the General Formula MLglin Isomerisrn 1 structural isomers coordination sphere and linkage isomers butane and 2 rnethylipropane CIC13HQO6 2 stereoisomerism geometric isomerism tran572ibutene and ci5727butene Geometric isomers result when the atoms bonded directly to the metal have a different spatial arrangement Consider PtNH32Clg and Coen2Clgl mer and fac Consider CrNH33Clg optical isomerism Some molecules and their mirror images cannot be superimposed like hands and feet These molecules are called chiral and the pairs of molecules are enantiomers Consider CH3CHOHCOgH lactic acid The possibility of chirality arises for a number of coordination compounds based on octahedral geometry Consider a molecule with three bidentate ligands Square planar molecules are never chiral they are incapable of optical isomerism Bonding in Coordination Compounds The bonding in coordination complexes is usually described by either molec ular orbital theory or crystal eld theory We will approach the metaliligand bonding using metal d orbitals and ligand lone pair orbitals CFT describes coordination compound bonding electrostatically this is the theory we will be using diOrbital Energies in Coordination Compounds The diorbitals Our particular interest is in the orientation of the d orbitals with respect to the ligands in a metal complex Notice that there are two sets orbitals that point along the axes toward the ligands And orbitals that point between the axes between the ligands In an isolated atom the d orbitals are degenerate In a complex the energies shift According to CFT repusion between d electrons and the electron pairs and ligands destabilizes the d orbitals Electrons in the various orbitlas are not affected equally Splitting of the d orbitalsl Magnetic Properties What is the electron con guration of Cr What happens in an octahedral complex high spin low spin pairing energy crystal eld splitting paramagnetic diamagnetic Example Low Spin C0NH363 Example High Spin CoFglg The high low spin nature can be tuned by the ligands The spectrocheinical series Halides lt oxalate lt water lt ammonia en lt phen lt cyanide
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