1311fin_f05 - k = 1.3807 x 10-23 J K-1 R = 0.082057 l atm...

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M M M k = 1.3807 x 10 -23 J K -1 R = 0.082057 l atm mol -1 K -1 = 8.314 J mol -1 K -1 = 1.987 cal mol -1 K -1 = 8.314 V C mol -1 K -1 k= R / N A N A = 6.022 x 10 23 mol -1 1 kcal = 4.184 kJ c = 2.998 x 10 8 m s -1 c = λν E = -13.6(Z/n) 2 eV 1 eV = 1.6022 x 10 -19 J 1 eV = 96.485 kJ mol -1 E = h ν χ = (IE+EA)/2 Δ E = -2.18 x 10 -8 J (Z 2 ) (1/n 2 f - 1/n 2 i ) E = -me 4 Z 2 /8 ε o n 2 h 2 E= mc 2 λ = h/mv 1 nm = 10 -9 m 1 D = 10 -8 cm=100 pm ε o = 8.85 x 10 -12 J -1 C 2 m -1 4 πε o = 1.113 x 10 -10 C 2 J -1 m -1 U % N A Z + Z - e 2 M/r o Δ G = Δ H - T Δ S = -RTlnK Δ G E = -n F E E EE RT n ln [products] [reactants] =− o F
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E s s p p F F F F * B B * s s p p NaCl unit cell (n-1) d n s n p ) L orbitals F * F F F F * F * n t 2g e g * Molecular shapes linear bent trigonal planar trigonal pyramidal tetrahedral trigonal bipyramidal seesaw T-shaped square planar square pyramidal octahedral r + /r & cubic 0.732 octahedral 0.414 tetrahedral 0.225 trigonal 0.15 ccp has 2 tetrahedral holes and 1 octahedral hole per sphere coordination no. structure cation, anion M NaCl 6,6 1.748 CsCl 8,8 1.763 ZnS 4,4 1.638 CaF 2 8,4 2.519 I < Br < Cl < S CN < F , OH < NO 2 < H 2 O < SCN < NH 3 < en < N O 2 < PR 3 < P(OR) 3 , C 2 H 4 < PF 3 , CO, CN
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Name_______________________________________ Chem 1311 Final Exam 14 December 2005 In taking this examination, you are agreeing to adhere to the GT academic honor code. At a minimum this requires that you utilize only the materials supplied to you and that you do not give help to, or accept help from, others. No electronic devices, including calculators are allowed. 1. (13) Draw all diastereoisomers of the six-coordinate complex [Ir(PPh 3 ) 2 (H) 2 (CO)(Cl)] in which the two phosphorus ligands are in cis positions. Indicate which one(s) are optically active (has a non- superimposible mirror image form), if any, by enclosing the isomer in a box. Careful - points will be deducted for each duplicate structure and for each structure incorrectly indicated as being optically active. 2. (8) [Ni(PPh 3 ) 2 Br 2 ] (Ph = C 6 H 5 ) has two unpaired electrons; [Ni(PMe 3 ) 2 Br 2 ] (Me = CH 3 ) has zero unpaired electrons. Which compound is tetrahedral and which is square planar? Briefly explain the basis for your conclusion (required to get credit for the problem). Crystal field orbital splitting diagrams may be useful in illustrating your conclusion. 3. Phosphorus ligands such as P(CH 3 ) 3 and P(C 6 H 5 ) 3 are strong field ligands. a) (5) What is the definition of a strong field ligand? b) (5) Why are such phosphorus ligands strong field ligands.
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Page 2 of 7 4. (3) Give ground state electron configuration for Fe 2+ and indicate the number of unpaired electrons. Fe 2+ unpaired electrons _______ 5. (10) The salt Ca 3 P 2(s) , which is composed of Ca 2+ and P 3 & ions is formed from calcium metal and white phosphorus, P 4 . The reaction representing the standard heat of formation is 3Ca (s) + ½ P 4(s) 6 Ca 3 P 2(s) . White phosphorus has the structure shown to the right. There are five steps (reactions) leading from P 4(s) to P 3 & (g) that must be included in the thermochemical (Born-Haber) cycle needed to compute the lattice energy. Write the five steps as balanced reactions and identify the energy associated with each balanced reaction. It is not necessary to write out the entire Born-Haber cycle.
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1311fin_f05 - k = 1.3807 x 10-23 J K-1 R = 0.082057 l atm...

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