HWC8soln - CHM 3610 Summer 2005 HOMEWORK SOLUTIONS CHAPTER...

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CHM 3610 Summer 2005 HOMEWORK SOLUTIONS CHAPTER 8 Chapter 8. 1, 7, 10, 12(skip d), 14, 15, 16, 17, 21, 22, 29, 35, 38, 43, 44, 46, 47, 50, 55, 61, 62, 66, 70, 73 Q1 A is d(xz) t 2g B is d(z 2 ) e g C is d(x 2 y 2 ) e g D is d(xy) t 2g Q7 (a) d 4 strong field t 2g 4 e g 0 1.6 o + P 2 Mn(CN) 6 3 (b) d 7 weak field t 2g 5 e g 2 0.8 o 3C o ( H 2 O) 6 2+ (c) d 4 weak field t 2g 3 e g 1 0.6 o 4M n ( H 2 O) 6 3+ (d) d 7 strong field t 2g 6 e g 1 1.8 o + P 1 Co(CN) 6 4 Q10 (a) 2 unpaired (b) paramagnetic (c) = 2.828( T) ½ = 2.83 = 0.0034 at 298 K and 0.0100 at 100 K. Q12(skip d) (a) Ni(H 2 O) 6 2+ 2 unpaired = 2.828 = 0.00363 at 273 K (b) Co(CN) 6 3 0 unpaired = 0 = 0 (c) Co(CN) 6 4 1 unpaired = 1.73 = 0.00136 at 273 K Q14 (a) Ti(III) and Cu(II) are d 1 and d 9 ions and would have essentially 1 absorption peak. Pt(IV) is unlikely to have absorptions in the visible and will probably be colorless. Therefore, the spectra are for Ni(II). (b) t (c) A is ~8000 cm 1 , B is ~10000 cm 1 , and C is ~12000 cm 1 (d) A is green transmitted ~22000 cm 1 , B is blue to indigo transmitted is ~14000 and 24000cm 1 Q15 (a) 8130 cm 1 (b) If H 2 O is 1.0 than g = 8130 cm 1 but the value in Table 8.4 is 8600 cm 1 (c) For A f= 11500 cm 1 / 8130 cm 1 = 1.41, B = 1.56 and C = 0.79 (d) increase (e) increase (f) A = ammonia, B = CO and C = iodide. Iodide is a weak field and CO is a strong field. Q16 (a) 20000 cm 1 for MA 6 and 25000 cm 1 for NB 6 . NB 6 has only one peak so is most likely to be a d 1 (b) g for MA 6 = 20,000 cm 1 /0.80 = 25,000 cm 1 . and g for NB 6 = 20,000 cm 1 /1.80 = 13,900 cm 1 . (d) B, A, B Q17 (a) Cr 2+ “true” P = (0.78)(23,500 cm 1 ) = 18,330 cm 1 .
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HWC8soln - CHM 3610 Summer 2005 HOMEWORK SOLUTIONS CHAPTER...

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