Chem 2210 Worksheet on Transition Metals Chemistry 2210 - Dr. Clase - Winter 1999

Solutions to Worksheet on Transition metal Complexes

| Solutions Index | 2210 Index |

(Oh = octahedral Td = tetrahedral D4h = square planar)

1. How many unpaired electrons are there in: 

         a. High spin Oh d5            b. Low spin Oh d5
         c. High spin Td d6            d. Low spin Td d6
         e. Large delta Oh Cr3+        f. Small delta Oh V2+
         g. Fe(CN)64-                  h. CoCl42-
         i. Mn(OH2)62+                 j. CuCl2-
         k. D4h Ni(CN)42-              l. Cu(NH3)42+
         m. Ni(CN)44-                  n. FeF63-
      
      
         a. 5                          b. 1      
         c. 4                          d. 2      
         e. 3                          f. 3      
         g. 0 (low spin)               h. 3 (Td) 
         i. 5 (high spin)              j. 0 (d10)          
         k. 0 (d8 - eg split)          l. 1 (d9) 
         m. 0 (d10)                    n. 5 (high spin)

2. How many isomers are possible for each of the following? (en = the bidentate ligand NH2CH2CH2NH2) 

      
         a. Co(NH3)3Cl3                b. [Fe(OH2)5SCN]2+
         c. [CrCl2en2]+Cl-             d. [Co(NH3)5Cl](NO2)2
         e. Co(NH3)63+.Cr((C2O4)3)3-   f. Td Ni(PR3)Br2
         g. Planar PdCl2en

a. 2 ("Fac" - all cis and "mer" - two trans with 1 cis to both) 

      
                        Cl                       Cl
                    Cl  |  Cl                Cl  |  NH3
                      \ | /                    \ | /
                        M                        M
                      / | \                    / | \
                   H3N  |  NH3               H3N |  NH3
                        NH3                      Cl
      
                     facial                  meridional
  • b. No other isomer (all positions equivalent in Oh)
  • c. 3 (cis with 2 optical isomers + 1 trans isomer)
  • d. 2 (The other is [Co(NH3)5NO2]ClNO2.)
  • e. 8 Theoretically at any rate there are also the following:
  • Co(NH3)4C2O4+.Cr(NH3)2(C2O4)2-
  • Cr(NH3)4C2O4+.Co(NH3)2(C2O4)2-
  • Cr(NH3)63+.Co(C2O4)33-

    • All the anions are chiral and have optical isomers.

  • f. No other form possible for tetrahedral.
  • g. 2 (The other is Pden22+.PdCl42- - en must be cis, so there is no trans isomer.)
    •

    3. Roadmaps - identify the lettered species and write balanced equations for the reactions.

    a. A is a colourless liquid that fumes in moist air. When A is treated with water a white precipitate B and a water soluble acidic gas C are produced. A dissolves in a concentrated aqueous solution of C and when zinc metal is added to the solution a purple coloured ion, D is formed. 

          
         TiCl4(l) + 2H2O(l)  ---> TiO2(s) + HCl(g)
           A                        B        C 
      
         TiCl4(l) + HCl(aq) --->  TiCl6-2(aq)
      
         2TiCl6-2(aq) + Zn(s) ---> 2Ti+3(aq) + Zn+2(aq) + 12Cl-(aq)
                                    D 
      
      OR 2TiCl4(l) + Zn(s) ---> 2Ti+3(aq) + Zn+2(aq) + 4Cl-(aq)
                                 D

    b. E is a blue crystalline solid readily soluble in water. When barium nitrate solution is added to a solution of E a white precipitate, F, is produced. F is not soluble in 6 mol.L-1 nitric acid.

    When a limited amount of potassium iodide solution is added to a solution of E two precipitates are formed, one, G, is white and the other, H, is blackish and soluble in carbon tetrachloride giving a purple solution. Both G & H dissolve in excess potassium iodide to give complex anions I & J respectively. I is colourless and J is brown.

    When aqueous ammonia is added carefully to a solution of E a pale blue precipitate, K, forms which dissolves in excess ammonia to give a solution containing the deep blue ion L.

    (Sorry! There was an error in the original printing:- E appeared as F in the second and third paragraphs.) 

          
         CuSO4(aq) + Ba(NO3)2(aq) ---> Cu+2(aq) + 2NO3-(aq) + BaSO4(s)
           E                                                    F 
      
      Net ionic:-
      
         2Cu+2(aq) + 4I-(aq) ---> 2CuI(s) + I2(s)
           E                        G       H 
      
         CuI(s) + I-(aq) ---> CuI2-(aq)
                               I
      
         I2(s) + I-(aq) ---> I3-(aq)
                             J
      
      Aqueous ammonia acts both as a base and a source of ammonia ligands:
      
         NH3(aq) + H2O(aq) <===> NH4+(aq) + OH-(aq)
      
      
         Cu+2(aq) + 2OH- ---> Cu(OH)2(s)
                                K  
      
         Cu(OH)2(s) + 4NH3(aq) ---> Cu(NH3)4+2(aq) + 2OH-(aq)     
                                       L

    (c) 1998, 1999 H.J. Clase - Updated 1999.04.12

    | Top | Solutions Index | 2210 Index |