Practice Problems 1

 

Look at all the examples in the appendix located in the resources section.  Use those examples for nomenclature practice.  Here are some more, some are a bit tricky and we will cover isomers mer and fac next lecture, but give them a try.  The answers are in green (except for the structures, which are in black). The alternate approach to the 18 electron rule is in red.

 

1)      Draw the complex:

 

tri-m-carbonyl-bis{tricarbonyliron(0)}

 

 

cis-(dithiocyanato)bis(triphenylphosphine)platinum(II)

 

I didn’t specify which linkage isomer I wanted, so there are two answers:

 

 

trans-bis(dimethylsulfoxide-k-S)tetrachlororuthenium(III)

 

 

2)      Give the name and formula

 

 

mer-triamminetrichlorocobalt (III)  Remember, for the name of the complex all the ligands are written in alphabetical order (not anions, then neutrals).

mer-[CoCl₃(NH₃)_­3]                       In the formula, write the anion ligands first, then the neutral ligands.

 

 

 

bis(ethylenediamine)platinum (II)

[Pt(en)₂]²⁺

 

 

bis(acetylacetonate)diaquacopper (II)    Since acetylacetonate is a complex ligand, use bis, tris etc.

[Cu(acac)₂(OH₂)₂]

 

 

silver tetrachloroplatinate(II) or silver tetrachloroplatinum(II)

Ag₂[PtCl₄]

 

3)      Do these complexes obey the 18 electron rule?

 

[Cr(NH₃)₅Cl]²⁺

 

d³ (Cr³⁺) + (5´2) + 2 = 15; no

d⁶ (neutral Cr) + (5´2) +1-2 (because of overall +2 charge)=15; no

 

[Ni(CO)₄]

 

d¹⁰ + (2´4)=18; yes

d¹⁰ + (2´4) =18;yes (notice that both methods are the same in this case)

 

[Ru(bipy)₂(SCN)₂]  (remember that bipy is a bidentate ligand)

 

d⁶ + (4´2) + (2´2) = 18; yes

d⁸ (neutral Ru) + (4´2) + (2´1) =18; yes

 

[Cu(en)₃]²⁺ (en is also a bidentate ligand)

 

d⁹ + (3´4) =21; no

d¹¹ (neutral Cu) + (3´4) -2 (because of overall +2 charge) = 21; no

 

     [Pd(NH₃)₄Cl₂]²⁺

 

     d⁶ + (4´2) + (2´2)= 18; yes

     d¹⁰ (neutral Pd) + (4´2) + (2´1) -2 (because of overall +2 charge) = 18; yes.