The Transition Elements and Coordination Compounds
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1 Chapter 22 The Transition Elements and Coordination Compounds Concept Check 22.1 Another complex studied by Werner had a composition corresponding to the formula PtCl 4 2KCl. From electrical-conductance measurements, he determined that each formula unit contained three ions. He also found that silver nitrate did not give a precipitate of AgCl with this complex. Write a formula for this complex that agrees with this information. Because silver nitrate does not precipitate AgCl from the complex, you conclude that chlorine is not present as free chloride ions; the Cl is presumably present as a complex. Potassium ion is likely present as K +, which would account for two of the ions present in each formula unit. The other would be a complex ion of platinum with the six chlorine atoms (a complex of Pt with six Cl - ions). The charge on this complex ion must be -2 to counter the charges from the K + ions, so its formula is PtCl The formula of the complex then is K 2 [PtCl 6 ]. Concept Check 22.2 A complex has the composition Co(NH 3 ) 4 (H 2 O)BrCl 2. Conductance measurements show that there are three ions per formula unit, and precipitation of AgCl with silver nitrate shows that there are two Cl - ions not coordinated to cobalt. What is the structural formula of the compound? The addition of silver nitrate to the complex precipitates AgCl equivalent to two Cl - ions per formula unit. Since each formula unit consists of three ions, the complex appears to consist of two Cl - ions plus a complex ion with a charge of +2. The formula of the complex then would 177
2 178 Chapter 22 be [Co(NH 3 ) 4 (H 2 O)Br]Cl 2. The structural formula of a possible constitutional isomer is [Co(NH 3 ) 4 (H 2 O)Cl]BrCl. Concept Check 22.3 a. Which of the following molecular models of octahedral complexes are mirror images of the complex X? Keep in mind that you can rotate the models when performing comparisons. b. Which complexes are optical isomers of molecule X? c. Identify the distinct geometric isomers of the complex X. (Note that some of the models may represent the same molecule.) a. If you were to place a mirror to the right of the complex X, complex A would directly represent what you would see in the mirror, so it is a mirror image. If you were to place a mirror behind complex X, complex D would also be a mirror image. Note that complexes B and C are exactly the same, and A and E are the same, in both cases, they only differ by rotation (spin one of the complexes in each pair 180 to see this). Therefore, A, D, and E are mirror images.
3 The Transition Elements and Coordination Compounds 179 b. Optical isomers are nonsuperimposable mirror images of one another. The mirror images of molecule X, which are molecules A and D, are both superimposable mirror images, so no optical isomers of X are present. c. To answer this part, you need to rotate each of the complexes to see if they have the same bonding arrangement in space with each of the ligands. Models A & E are the same complex, neither of which has the same bonding arrangement as complex X, so the complex that they represent is a geometric isomer. Models B & C represent the same complex that also has a different bonding arrangement than complex X, so the complex that they represent is also a geometric isomer. Complex D is the same complex as complex X, so it is not a geometric isomer. Conceptual Problem A cobalt complex whose composition corresponded to the formula Co(NO 2 ) 2 Cl 4NH 3 gave an electrical conductance equivalent to two ions per formula unit. Excess silver nitrate solution immediately precipitates 1 mol AgCl per formula unit. Write a structural formula consistent with these results. Because one mole of chloride ion is precipitated per formula unit of the complex, the chlorine atoms must be present as chloride ion. All the other ligands are coordinated to the cobalt. An appropriate formula is [Co(NH 3 ) 4 (NO 2 ) 2 ]Cl. Conceptual Problem For the following coordination compounds, identify the geometric isomer(s) of compound X.
4 180 Chapter 22 In order to solve this problem, you need to check each of the possible coordination compounds (A-D) to see if they are different than compound X. Compounds B, C, and D, through rotation and flipping over, all turn out to be the same as compound X. The only compound that is different is A, so it is the only geometric isomer of compound X. Conceptual Problem Describe step by step how the name potassium hexacyanoferrate(ii) leads one to the structural formula K 4 [Fe(CN) 6 ]. Hexacyano means that there are six CN - ligands bonded to the iron cation. The Roman numeral II means that the oxidation state of the iron cation is +2, so that the overall charge of the complex ion is -4. This requires four potassium ions to counterbalance the -4 charge. Conceptual Problem Compounds A and B are known to be stereoisomers of one another. Compound A has a violet color; compound B has a green color. Are they geometric or optical isomers?
5 The Transition Elements and Coordination Compounds 181 Compounds A and B are geometric isomers because these isomers have different physical properties whereas optical isomers do not. Conceptual Problem A complex has a composition corresponding to the formula CoBr 2 Cl 4NH 3. What is the structural formula if conductance measurements show two ions per formula unit? Silver nitrate solution gives an immediate precipitate of AgCl but no AgBr. Write the structural formula of an isomer. The given compound must consist of a chloride ion (which can be precipitated with AgNO 3 solution) and a [Co(NH 3 ) 4 Br 2 ] + ion, giving [Co(NH 3 ) 4 Br 2 ]Cl. The structural formula of a possible constitutional isomer is [Co(NH 3 ) 4 BrCl]Br. Conceptual Problem For the complexes shown here, which would have the d electron distribution shown in the diagram below: [MF 6 ] 3-, [M(CN) 6 ] 3-, [MF 6 ] 4-, [M(CN) 6 ] 4-? Note that the neutral metal atom, M, in each complex is the same and has the ground state electron configuration [Ar]4s 2 3d 6. The d orbital energy level diagram contains six electrons. This indicates that the metal M must have the electron configuration [Ar]3d 6 and therefore a charge of 2+ (M 2+ ). The two complexes that have M in the 2+ oxidation state are MF 6 4- and M(CN) The d orbital energy level diagram is high spin indicating that there is a relatively small amount of crystal field splitting. According to the spectrochemical series, F - is a less strongly bonding ligand with a relatively smaller crystal field splitting, which makes the MF 6 4- the most likely answer.
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