Lecture 23

Lecture 23 - %34#%56789#+4#(:;835#

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Unformatted text preview: %34#%56789#+4#(:;835# !"#$%&'$"%()*&+$,("**,$&,(&-#"$.%& '/$"01,$&,(&12"&3,45$&6"7,4)1/,#*&+$,8"01& '/$"01,$&,(&12"&-#/&3,45$&9$,#1/"$*&:"#1"$& ;5**502)*"11*&<#*1/1)1"&,(&="02#,4,.%& >/44&*?"5@&,# 3/10/10 <283=:6597>8?#$683@A#B:3#C#DE#F#G7997:6HI# 1#":9JK7:6#K:#KL8#+9:M59#$683@A#!L59986@8<# # In both [Fe(H2O)6]2+ and [Fe(CN)6]4– ions, the iron is present as Fe(II). However one is paramagnetic and the other is diamagnetic. Which is which? Fe(II): d6 A)  [Fe(H2O)6]2+ = paramagnetic; [Fe(CN)6]4– = diamagnetic B) [Fe(H2O)6]2+ = diamagnetic; [Fe(CN)6]4– = paramagnetic C) Insufficient information CN- > NO2- > en > NH3 > SCN- > H2O > ONO- > ox2- > OH- > F- > SCN- > Cl- > Br- >I- [Fe(H2O)6] # # !"!##$%&'()'(##%*")*(+,*"-$%##.$!),/$##"$/*$"## & *(#!-$&*")/0##1(%##"$$##2/$"$()1)*'(# 2+ weaker-field ligands (! will be smaller) —— paramagnetic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e(CN)6]4- %34#%56789#+4#(:;835# & & & ——— stronger-field ligands — —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diamagnetic (! will be larger) '/$"01,$&,(&12"&3,45$&6"7,4)1/,#*&+$,8"01& 54 '/$"01,$&,(&12"&-#/&3,45$&9$,#1/"$*&:"#1"$& ———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ii) Colour !"!##$%&'()'(##%*")*(+,*"-$%##.$!),/$##"$/*$"## *(#!-$&*")/0##1(%##"$$##2/$"$()1)*'(# 1G?@35;@H&=2"&?,**/A/4/1%&,(&."#"$51/#.& 1"$5>511*&,(&05$A,#B($""&"#"$.%C&5#D&12)*&?$,7/D/#.&*,0/"1%&>/12&The colour of & /1*& E,*1& D/$"01& ?512& 1,& $"54/F/#.& 5& 4,>& .$""#2,)*"& .5*& ()1)$"C& E5%& A"& $"54/F"D& A%& E5@/#.& *,45$& +-&visible light causing electrons to be promoted 575/45A4"&1,&12"&G&A/44/,#&#">&"#"$.%&)*"$*&A%&2/.2&12$,).2?)1&E5#)(501)$/#.H&I,1>/12*15#D/#.C&0)$$"#1& ,?1/,#*& 1,& 25$#"**& 5#D& *1,$"& *,45$& "#"$.%& 51& 12"& /#D/7/D)54& 4"7"4& 5$"& 1,,& "J?"#*/7"& 1,& A"& /E?4"E"#1"DC&lower level to the higher level "*?"0/544%& /#& 5& #,#B4".50%& >,$4DH& =2"& /E?"$51/7"& 1,& *0/"#0"& /*& 1,& D"7"4,?& #">& E51"$/54*C& $"501/,#*& 5#D& !"#$%&'$"%()*&+$,("**,$&,(&-#"$.%& A solution of [Ti(H2O)6]3+ is violet. Why? ?$,0"**"*&1251&"#5A4"&?"$*,#54/F"D&*,45$&"#"$.%&1,&A"& *)((/0/"#14%&/#"J?"#*/7"&1,&?"#"1$51"&.4,A54&"#"$.%& '/$"01,$&,(&12"&3,45$&6"7,4)1/,#*&+$,8"01& E5$@"1*&5#D&"*?"0/544%&12"&#,#B4".50%&>,$4DH&&+"$*,#54/F"D&"#"$.%&51&4,>&0,*1&?$"*"#1*&#">&A5*/0&$"*"5$02& '/$"01,$&,(&12"&-#/&3,45$&9$,#1/"$*&:"#1"$& —— —— ;5**502)*"11*&<#*1/1)1"&,(&="02#,4,.%& 15$."1*H&K"05)*"&?"$*,#54/F"D&"#"$.%&>/44&A"&?,**/A4"&,#4%&/(&*,45$&"#"$.%&/*&5&LMNO&575/45A4"&*)??4%C&12"& @"%&"#5A4"$&(,$&?"$*,#54/F"D&"#"$.%&/*&/#"J?"#*/7"&*1,$5."H&31)D/"*&/#&12"&I,0"$5&.$,)?&257"&4"D&1,&12"& light >/44&*?"5@&,# 0$"51/,#&,(&5&#">&05154%*1&1251&05?1)$"*&12"&()#01/,#54&"4"E"#1*&,(&?2,1,*%#12"*/*&5#D&/#&D,/#.&*,&?$,7/D"*& # 5& 2/.24%& E5#)(501)$5A4"& 5#D& /#"J?"#*/7"& E"12,D& 1,& "(("01& 5& 05$A,#B#")1$54& 5#D& *)*15/#5A4"& E"12,D& (,$& ——— ——— *,45$&*1,$5."&P&*,45$&()"4*&($,E&>51"$B*?4/11/#.H&K%& D"7"4,?/#.&5#&/#"J?"#*/7"& LMNO&*,45$&"#"$.%&*%*1"E& (,$&12"&/#D/7/D)54C&5&05$A,#B#")1$54&"#"$.%&*)??4%&(,$&Q&R&G&A/44/,#&A"0,E"*&575/45A4"H&&&&&&&! 3+ : d1 Ti transition metal complexes is due to from the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xcited state : t2g0eg1 Ground state : t2g1eg0 ! is inversely proportional to " (wavelength of light absorbed) transmitted light does not contain this wavelength 57 !"#$%&'()*+))(,#&-(./&0112&3)""*)&45067&0#$892&"#1&0#:(8,#;$#.&<,%+.(,#)7&=%>$8."&3##,:".$)?&& "#1&@+88"2&A8"2&45B5&!8,C$)),8&,C&0#9(#$$8(#97&D#(:$8)(.2&,C&=%>$8.") to follow March 15 event. %1)$H# )I8?D5AJ#KL#&53;=#MNKN# )*&$H# OHON#P4>4# &2.1!$H# # $).!#KENNQJ#,67R83?7@A#:B#19G83@5# & & '@=83#$R86@?H# Y283?:6597S8D#$683CA#B:3#K#TU#L#V7997:6WH#1#":9I@7:6#@:#@=8#+9:G59#$683CA#!=59986C8 Determine The & magnitude of ! for [Ti(H2O)6]3+ is 210 kJ/mol. Z&& !"#$%&'()*+))(,#&-(./&0112&3)""*)&45067&0#$892&"#1&0#:(8,#;$#.&<,%+.(,#)7&=%>$8."&3##,:".$)?&& "#1&@+88"2&A8"2&45B5&!8,C$)),8&,C&0#9(#$$8(#97&D#(:$8)(.2&,C&=%>$8.") to follow the wavelength of the light that is absorbed. March 15 event. & & ;5$02&QSC&OTUV&?EC&="4)*&QSVC&W#/7"$*/1%&,(&X4A"$15& How is the wavelength of light absorbed related to the colour we observe? Colour Wheel : 650nm 760 nm 400nm Colour observed red orange 600nm Colour absorbed " = 570 nm yellow 560nm violet blue 430nm 490nm 59 ! = 210 kJ 1mol 1000 J ! ! mol 6.022 ! 10 23 atoms kJ = 3.48 # 10-19 J != = hc hc = E ! (6.626 ! 10 "34 Js)(2.998 ! 10 8 m / s) 3.48 ! 10 "19 J green = 5.70 # 10-7 m = 570 nm 58 Know how to use the colour wheel 1 3/10/10 Demo – colors of some Ni octahedral complexes Add Ni(NO3)2 (s) to water to obtain: Ni(NO3)2 (s) + H2O ! [Ni(H2O)6]2+ (green solution) For [Ni(H2O)6]2+ : 650nm " absorbed 760 nm 400nm red 600nm yellow 560nm violet blue 430nm green Colour observed 60 Add NH3 to displace the aqua ligands to form [Ni(NH3)6]2+. What do you predict the colour to be? A)  Green B) Blue C) Violet D) Yellow E) Orange CN- > NO2- > en > NH3 > SCN- > H2O > ONO- > ox2- > OH- > F- > SCN- > Cl- > Br- >I- orange ! increases " decreases " Stronger field 650nm " absorbed 760 nm 400nm red orange 600nm yellow 560nm green Colour observed 490nm 61 violet blue 430nm 490nm If less concentrated ammonia is used, only some of the aqua ligands will be replaced by ammine ligands. [Ni(NH3)4(H2O)2]2+ Colour? " absorbed 650nm [Ni(H2O)6]2+ 760 nm 400nm red orange [Ni(NH3)6]2+ 600nm yellow 560nm violet green Of the complexes [Cr(H2O)6]2+ and [CrCl6]3- , one is violet and the other is green. Which complex has which colour ? A)  [Cr(H2O)6]2+ = violet, [CrCl6]3- = green B)  [Cr(H2O)6]2+ = green, [CrCl6]3- = violet CN- > NO2- > en > NH3 > SCN- > H2O > ONO- > ox2- > OH- > F- > SCN- > Cl- > Br- >I- blue 430nm 490nm Observed Colour = light blue/dark green 62 " Stronger field Complex 1 650nm 600nm orange Shorter " Larger ! 2+ Violet = [Cr(H2O)6] red yellow 760 nm 560nm 400nm Complex 2 violet green Longer " Smaller ! blue Green = [CrCl6]3430nm 490nm 63 Explain why [Cr(H2O)6]Cl3 is violet and [Cr(H2O)4Cl2]Cl is green. See online notes Explain why [Cr(H2O)6]Cl3 is violet and [Cr(H2O)4Cl2]Cl is green. [Cr(H2O)6]Cl3 : -  only has H2O bound to metal -  counter-ion does not contribute to colour as it does not form metal-ligand bonds - this complex absorbs yellow light [Cr(H2O)4Cl2]Cl : -  now has H2O and Cl ligands -  increase number of weak-field ligands -  ! decreases compared to complex containing 6 H2O -  complex will now absorb light of longer wavelengths - red is absorbed (and is of longer wavelength than yellow) 64 65 2 3/10/10 B. Splitting of d-orbitals for tetrahedral complexes Bond angles are 109.5o. C. Splitting of d-orbitals for square planar complexes dz 2 dx 2 ! y 2 t2g eg dxy dxz dyz dz 2 dx 2 ! y 2 dxy dxz dyz Orbital energy goes down Orbital energy goes up less repulsion leads to a smaller splitting (!tet < !oct) ALL tetrahedral complexes are HIGH-SPIN 66 Orbital energy goes up Orbital energy goes down Different splitting pattern Larger splitting (!sp > !oct) ALL square planar complexes are LOW-SPIN Recall: (most) nd8 = square planar 67 Try textbook problems: 23.92, 23.96, 23.100, 23.102, 23.107 [Solutions available] Also, try: 23.97, 23.103 [Answers posted with Assignment #3] Don’t forget PS#3, Part C #4 and #5 Read textbook Sections 6.1, 6.2 68 3 ...
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