EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 9781305856745
Author: DECOSTE
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 19, Problem 84AE
Interpretation Introduction

Interpretation: The structure of three salts formed needs to be determined if 0.27 g of octahedral complex of Cr reacts with strong dehydrating agent and lost mass of 0.036 g. Reaction of 270 mg of second salt with same dehydrating agent show mass loss of 18 mg. The 3rd salt did not lose any mass while treating with dehydrating agent.

  Co(en)33++  e-  Co(en)32+

Given:

  Co3++  e-  Co2+    ξ°=1.82VCo(en)33+ = Kf =  2.0×1047 Co(en)32+= Kf  = 1.5×1012

Concept Introduction:

Crystal field theory is the theory given to explain the bonding in the coordination complexes. As ligand approaches towards the metal ion, the d-orbital of metal ion divide according to the energy of metal ion. On the basis of energy and degeneracy, the d-orbital can be classified as t2g and eg orbitals. The electrons will fill according to their energy levels.

In octahedral complex, the t2g orbitals are lower energy orbitals in d-orbital splitting whereas eg orbitals are higher energy orbitals.

Expert Solution & Answer
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Answer to Problem 84AE

  EBK CHEMICAL PRINCIPLES, Chapter 19, Problem 84AE , additional homework tip  1

Explanation of Solution

Given information:

Addition of excess of aqueous silver nitrate to 100.0 mL portion of 0.100 M solution of each salt forms silver chloride,

  • 1st solution= 1430 mg AgCl
  • 2nd solution = 2870 mg AgCl
  • 3rd solution = 4300 mg AgCl

Two salts are green and one is violet here.

Compound 1:

  Moles CrCl3.6H2O  = 0.27 g × 1 mole266.5 g =1.0×10-3moles CrCl3.6H2O

Calculate moles of water of hydration:

  Moles  of H2O  = 0.036 g × 1 mole18.0 g = 0.002 moles H2O

  Moles  of H2mole compound =   0.002 moles0.001 moles =  2.0 

With 2 moles of water; the formula must be [Cr(H2O)4Cl2]Cl.2H2O .

Mass of AgCl = 1430 mg

Check the mass of Cl in [Cr(H2O)4Cl2]Cl.2H2O :

  Moles  of AgCl  =0.100 L  × 0.100 moles [Cr(H2O)4Cl2]Cl  1 L ×  1 mole  Cl 1 mole  [Cr(H2O)4Cl2]Cl× 1 mole  AgCl 1 mole  Cl-× 143.4 g 1 mole  = 1.43 g =1430 mg AgCl Compound 2:

  Moles CrCl3.6H2O  = 0.27 g × 1 mole266.5 g =1.0×10-3moles CrCl3.6H2O

Calculate moles of water of hydration:

  Moles  of H2O  = 0.018 g × 1 mole18.0 g = 0.001 moles H2O

  Moles  of H2mole compound =   0.001 moles0.001 moles =  1.0 

With 1 moles of water; the formula must be [Cr(H2O)5Cl]Cl2.H2O .

Mass of AgCl = 2870 mg

Check the mass of Cl in [Cr(H2O)5Cl]Cl2.H2O :

  Moles  of AgCl  =0.0200 moles Cl- × 1 mole AgCl   1 mole Cl- × 143.4 g AgCl 1 mole AgCl  = 2.87 g = 2870 mg AgCl

Compound 3:

  Moles CrCl3.6H2O  = 0.27 g × 1 mole266.5 g =1.0×10-3moles CrCl3.6H2O

Moles of water of hydration is 0 as no water lose is there. Thus the formula must be [Cr(H2O)6]Cl3 .

Mass of AgCl = 4300 mg

Check the mass of Cl in [Cr(H2O)6]Cl3 :

  Moles  of AgCl  =0.0300 moles Cl- × 1 mole AgCl   1 mole Cl- × 143.4 g AgCl 1 mole AgCl  = 4.30 g = 4300 mg AgCl

Hence the structure of complexes should be:

  EBK CHEMICAL PRINCIPLES, Chapter 19, Problem 84AE , additional homework tip  2

Conclusion

Thus,

  EBK CHEMICAL PRINCIPLES, Chapter 19, Problem 84AE , additional homework tip  3

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Chapter 19 Solutions

EBK CHEMICAL PRINCIPLES

Ch. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Define each of the following terms. a....Ch. 19 - Prob. 17ECh. 19 - When a metal ion has a coordination number of 2,...Ch. 19 - The compound cisplatin, Pt(NH3)2Cl2 , has been...Ch. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - For the process Co(NH3)5Cl2++Cl2Co(NH3)4Cl2++NH3...Ch. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Consider the complex ions...Ch. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - How many unpaired electrons are in the following...Ch. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - Prob. 66ECh. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69AECh. 19 - Prob. 70AECh. 19 - Prob. 71AECh. 19 - Prob. 72AECh. 19 - Prob. 73AECh. 19 - Prob. 74AECh. 19 - Prob. 75AECh. 19 - Prob. 76AECh. 19 - Prob. 77AECh. 19 - Prob. 78AECh. 19 - Prob. 79AECh. 19 - Prob. 80AECh. 19 - Prob. 81AECh. 19 - Prob. 82AECh. 19 - Prob. 83AECh. 19 - Prob. 84AECh. 19 - Prob. 85AECh. 19 - Prob. 86AECh. 19 - Prob. 87AECh. 19 - Prob. 88AECh. 19 - Prob. 89AECh. 19 - Prob. 90AECh. 19 - Prob. 91AECh. 19 - Prob. 92AECh. 19 - Prob. 93AECh. 19 - Prob. 94AECh. 19 - Prob. 95AECh. 19 - Prob. 96AECh. 19 - Prob. 97CPCh. 19 - Prob. 98CPCh. 19 - Prob. 99CPCh. 19 - Prob. 100CPCh. 19 - Prob. 101CPCh. 19 - Prob. 102CPCh. 19 - Prob. 103CPCh. 19 - Prob. 104CP
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