CHAPTER 15 THE CHEMISTRY OF TRANSITION METALS

Transcription

1 HAPTER 15 THE HEMISTRY F TRANSITIN METAS 15.1 The efinition of transition metal aopte by the International Union of Pure an Applie hemists (IUPA) is that Transition elements (transition metals) either have incompletely fille subshells or reaily give rise to cations that have incompletely fille subshells. Zn oes not have incompletely fille subshells an oes not reaily give rise to cations that have incompletely fille subshells. Zn forms a cation by losing its outer s electrons. 15. Going from scanium to copper the number of protons in the nucleus increase, increasing the strength of the electrostatic interaction between the nucleus an the electrons an reucing the sie of the atoms Sc: [Ar] 4s 3 1 Ti: [Ar] 4s 3 V: [Ar] 4s 3 3 r: [Ar] 4s Mn: [Ar] 4s 3 5 Fe: [Ar] 4s 3 6 : [Ar] 4s 3 7 Ni [Ar] 4s 3 8 u: [Ar] 4s a half-fille subshell is a little extra stable a full subshell is a little extra stable 15.4 V 5+ : [Ar] r 3+ : [Ar] 3 3 Mn + : [Ar] 3 5 Fe + : [Ar] 3 6 u + : [Ar] 3 9 Sc 3+ : [Ar] Ti 4+ : [Ar] 15.5 The transition metals ten to have many outer electrons (s an ) that are in states with relatively similar energies. Hence, a transition metal may lose one, two, three, or many electrons, leaing to multiple possible oxiation states While chromium metal has a negative stanar reuction potential, inicating that it shoul be oxiie easily, chromium metal oes not appear very reactive because chromium oxie forms on its surface. This oxie layer is not very reactive an protects the chromium metal (a) En is the abbreviation for ethyleneiamine (HNHHNH). () The oxiation number of is 3. (Why?) The coorination number of is six. (Why isn't this the same as the number of ligans?) Ethyleneiamine (en) is a bientate ligan. ul cyanie ion be a bientate ligan? Ask your instructor (a) The oxiation number of r is 3. The coorination number of r is 6. xalate ion (4 ) is a bientate ligan.

2 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS (a) The net charge of the complex ion is the sum of the charges of the ligans an the central metal ion. In this case the complex ion has a 3 charge. (Potassium is always 1. Why?) Since the six cyanies are 1 each, the Fe must be 3. The complex ion has a 3 charge. Each oxalate ion has a charge (Table 15.3 of the text). Therefore, the r must be 3. Since cyanie ion has a 1 charge, Ni must have a charge to make the complex ion carry a net charge Strategy: The oxiation number of the metal atom is equal to its charge. First we look for known charges in the species. Recall that alkali metals are 1 an alkaline earth metals are. Also etermine if the ligan is a charge or neutral species. From the known charges, we can euce the net charge of the metal an hence its oxiation number. Solution: (a) Since soium is always 1 an the oxygens are, Mo must have an oxiation number of 6. Magnesium is an oxygen ; therefore W is 6. ligans are neutral species, so the iron atom bears no net charge. The oxiation number of Fe is (a) tetraammineichlorocobalt(iii) ibromobis(ethyleneiamine)cobalt(iii) triamminetrichlorochromium(iii) () hexaamminecobalt(iii) chlorie 15.1 Strategy: We follow the proceure for naming coorination compouns outline in Section 15. of the text an refer to Tables 15.4 an 15.5 of the text for names of ligans an anions containing metal atoms. Solution: (a) Ethyleneiamine is a neutral ligan, an each chlorie has a 1 charge. Therefore, cobalt has a oxiation number of 3. The correct name for the ion is cisichlorobis(ethyleneiammine)cobalt(iii). The prefix bis means two; we use this instea of i because i alreay appears in the name ethyleneiamine. There are four chlories each with a 1 charge; therefore, has a 4 charge. The correct name for the compoun is pentaamminechloroplatinum(iv) chlorie. There are three chlories each with a 1 charge; therefore, has a 3 charge. The correct name for the compoun is pentaamminechlorocobalt(iii) chlorie The formulas are: (a) [Zn(H)4] [r(h)5] [ubr4] () [Fe(EDTA)] In, why two chlorie ions at the en of the formula? In (), oes the "(II)" following ferrate refer to the charge of the complex ion or the charge of the iron atom? Strategy: We follow the proceure in Section 15. of the text an refer to Tables 15.4 an 15.5 of the text for names of ligans an anions containing metal atoms. Solution: (a) There are two ethyleneiamine ligans an two chlorie ligans. The correct formula is [r(en)]. There are five carbonyl () ligans. The correct formula is Fe()5.

3 64 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS () There are four cyanie ligans each with a 1 charge. Therefore, the complex ion has a charge, an two K ions are neee to balance the charge of the anion. The correct formula is K[u(N)4]. There are four NH3 ligans an two H ligans. Two chlorie ions are neee to balance the charge of the complex ion. The correct formula is [(NH3)4(H)] The isomers are: Br Br Ni N N N Br Ni Br N cis trans (a) In general for any MAB4 octaheral molecule, only two geometric isomers are possible. The only real istinction is whether the two Aligans are cis or trans. In Figure of the text, (a) an are the same compoun ( atoms cis in both), an an () are ientical ( atoms trans in both). A moel or a careful rawing is very helpful to unerstan the MA3B3 octaheral structure. There are only two possible geometric isomers. The first has all A's (an all B's) cis; this is calle the facial isomer. The secon has two A's (an two B's) at opposite ens of the molecule (trans). Try to make or raw other possibilities. What happens? (a) All six ligans are ientical in this octaheral complex. There are no geometric or optical isomers. NH 3 H 3 N NH 3 H 3 N NH 3 NH 3 Again there are no geometric or optical isomers. To have cis an trans isomers there woul have to be two chlorine ligans. H 3 N NH 3 H 3 N NH 3 NH 3 There are two optical isomers. They are like Figure 15.1 of the text with the two chlorine atoms replace by one more bientate ligan. The three bientate oxalate ligans are represente by the curve lines.

4 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS 65 mirror (a) There are cis an trans geometric isomers (See Problem 15.16). No optical isomers. H 3 N NH 3 H 3 N H 3 N NH 3 H 3 N NH 3 NH 3 trans cis There are two optical isomers. See Figure 15.1 of the text. The three bientate en ligans are represente by the curve lines. mirror x y xy xy x y x y x y [Ni(N)4] [Ni4]

5 66 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS 15.0 When a substance appears to be yellow, it is absorbing light from the blue-violet, high energy en of the visible spectrum. ften this absorption is just the tail of a strong absorption in the ultraviolet. Substances that appear green or blue to the eye are absorbing light from the lower energy re or orange part of the spectrum. yanie ion is a very strong fiel ligan. It causes a larger crystal fiel splitting than water, resulting in the absorption of higher energy (shorter wavelength) raiation when a electron is excite to a higher energy orbital (a) Each cyanie ligan has a 1 charge, so the oxiation state of r is. The electron configuration of r is [Ar]3 4. yanie is a strong fiel ligan (see Problem 15.0). The four 3 electrons shoul occupy the three lower orbitals as shown; there shoul be two unpaire electrons. x y xy x y [r(n)6] 4 Water is a weak fiel ligan. The four 3 electrons shoul occupy the five orbitals as shown. There shoul be four unpaire electrons. x y xy x y [r(h)6] 15. (a) Wavelengths of 470 nm fall between blue an blue-green, corresponing to an observe color in the orange part of the spectrum. We convert wavelength to photon energy using the Planck relationship. E hc ( J s)( m s 1 ) m J J 1 photon photons 1 mol 1 kj 1000 J 55 kj mol Recall the wavelength an energy are inversely proportional. Thus, absorption of longer wavelength raiation correspons to a lower energy transition. ower energy correspons to a smaller crystal fiel splitting. (a) H is a weaker fiel ligan than NH3. Therefore, the crystal fiel splitting will be smaller for the aquo complex, an it will absorb at longer wavelengths. Fluorie is a weaker fiel ligan than cyanie. The fluoro complex will absorb at longer wavelengths. hlorie is a weaker fiel ligan than NH3. The chloro complex will absorb at longer wavelengths.

6 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS Step 1: The equation for freeing-point epression is Tf Kfm Solve this equation algebraically for molality (m), then substitute Tf an Kf into the equation to calculate the molality. m T f K f m m Step : Multiplying the molality by the mass of solvent (in kg) gives moles of unknown solute. Then, iviing the mass of solute (in g) by the moles of solute, gives the molar mass of the unknown solute mol solute? mol of unknown solute kg water mol solute 1 kg water molar mass of unknown g 117 g/mol mol The molar mass of (NH3)43 is 33.4 g mol 1, which is twice the compute molar mass. This implies issociation into two ions in solution; hence, there are two moles of ions prouce per one mole of (NH3)43. The formula must be: [(NH3)4] which contains the complex ion [(NH3)4] an a chlorie ion,. Refer to Problem (a) for a iagram of the structure of the complex ion Rust stain removal involves forming a water soluble oxalate ion complex of iron like [Fe(4)3] 3. The overall reaction is: Fe3(s) 6H4(aq) Fe(4)3 3 (aq) 3H(l) 6H (aq) Does this reaction epen on ph? 15.6 Use a raioactive label such as 14 N (in NaN). A NaN to a solution of K3Fe(N)6. Isolate some of the K3Fe(N)6 an check its raioactivity. If the complex shows raioactivity, then it must mean that the N ion has participate in the exchange reaction The green precipitate in uf. When K is ae, the bright green solution is ue to the formation of u4 : u (aq) 4 (aq) u4 (aq) 15.8 The white precipitate is copper(ii) cyanie. u (aq) N (aq) u(n)(s) This forms a soluble complex with excess cyanie. u(n)(s) N (aq) u(n)4 (aq)

7 68 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS pper(ii) sulfie is normally a very insoluble substance. In the presence of excess cyanie ion, the concentration of the copper(ii) ion is so low that us precipitation cannot occur. In other wors, the cyanie complex of copper has a very large formation constant The overall reaction is: u4 (aq) 6H(l) u(h)6 4 (aq) green blue Aition of excess water (ilution) shifts the equilibrium to the right (e hâtelier s principle) The formation constant expression (assuming that activities can be replace by concentrations) is: K f [Fe(H ) 5 NS ] [Fe(H ) 6 3 ][SN ] Notice that the original volumes of the Fe(III) an SN solutions were both 1.0 m an that the final volume is 10.0 m. This represents a tenfol ilution, an the concentrations of Fe(III) an SN become 0.00 M an M, respectively. We make a table. Fe(H)6 3 SN Fe(H)5NS H Initial (M): hange (M): Equilibrium (M): Kf (0.00)(.7 10 ) The thir ioniation energy increases rapily from left to right. Thus metals ten to form M ions rather than M 3 ions Mn 3 is 3 4 an r 3 is 3 5. Therefore, Mn 3 has a greater tenency to accept an electron an is a stronger oxiiing agent. The 3 5 electron configuration of r 3 is a stable configuration (a) Since carbon is less electronegative than oxygen, it is more likely that carbon will share electrons with Fe forming a metal-to-ligan sigma bon. The sp orbital on carbon containing the lone pair overlaps with the empty sp 3 orbital on Fe. sp hybri orbital sp 3 hybri orbital Fe sp 3 hybri orbital Ti is 3 an Fe is The three cobalt compouns woul issociate as follow:

8 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS 69 [(NH3)6]3(aq) [(NH3)6] 3 (aq) 3 (aq) [(NH3)5](aq) [(NH3)5] (aq) (aq) [(NH3)4](aq) [(NH3)4] (aq) (aq) The concentration of free ions in the three 1.00 M solutions woul be 4.00 M, 3.00 M, an.00 M, respectively. If you mae up 1.00 M solutions of Fe3, Mg, an Na, these woul serve as reference solutions in which the ion concentrations were 4.00 M, 3.00 M, an.00 M, respectively. A 1.00 M solution of [(NH3)5] woul have an electrolytic conuctivity close to that of the Mg solution, etc A g sample of hemoglobin contains 0.34 g of iron. In moles this is: 1 mol g Fe mol Fe g The amount of hemoglobin that contains one mole of iron must be: g hemoglobin mol Fe g hemoglobin per mol Fe We compare this to the actual molar mass of hemoglobin: g hemoglobin 1 mol hemoglobin 1 mol Fe g hemoglobin 4 mol Fe per 1 mol hemoglobin The iscrepancy between our minimum value an the actual value can be explaine by realiing that one hemoglobin molecule contains four iron atoms (a) Zinc(II) has a completely fille 3 subshell giving the ion greater stability. Normally the colors of transition metal ions result from transitions within incompletely fille subshells. The 3 subshell of inc(ii) ion is fille (a) [r(h)6]3, [r(h)5]h, [r(h)4]h The compouns can be ientifie by a conuctance experiment. mpare the conuctances of equal molar solutions of the three compouns with equal molar solutions of Na, Mg, an Fe3. The solution that has similar conuctance to the Na solution contains ; the solution with the conuctance similar to Mg contains ; an the solution with conuctance similar to Fe3 contains (a) Reversing the first equation: Ag(NH3) (aq) Ag (aq) NH3(aq) K Ag (aq) N (aq) Ag(N) (aq) K Ag(NH3) (aq) N (aq) Ag(N) (aq) NH3(aq) 8 K K1K ( )( )

9 630 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS G RTln K (8.314 J K 1 mol 1)(98 K) ln ( ) J mol Zn (s) Zn (aq) e E anoe 0.76 V [u (aq) e u (aq)] E cathoe 0.15 V Zn(s) u (aq) Zn (aq) u (aq) Ecell Ecathoe E anoe 0.15 V ( 0.76 V) 0.91 V We carry aitional significant figures throughout the remainer of this calculation to minimie rouning errors. G nfe ()(96500 J V 1 mol 1 )(0.91 V) G RT ln K J mol kj mol --1 ln K G RT ( J mol 1 ) (8.314 J K 1 mol 1 )(98 K) ln K K e The half-reactions are: (aq) e (s) E cathoe 1.0 V [Ag (aq) e Ag(s)] E anoe 0.80 V Ag(s) (aq) Ag (aq) (s) E cell Ecathoe Eanoe 1.0 V 0.80 V 0.40 V Since the cell voltage is positive, proucts are favore at equilibrium. At 5, E cell V ln K n V 0.40 V ln K ln K 31.1 K e Iron is much more abunant than cobalt xyhemoglobin absorbs higher energy light than eoxyhemoglobin. xyhemoglobin is iamagnetic (low spin), while eoxyhemoglobin is paramagnetic (high spin). These ifferences occur because oxygen () is a strongfiel ligan. The crystal fiel splitting iagrams are:

10 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS 631 x y x y xy x y xy x y eoxyhemoglobin oxyhemoglobin The orbital splitting iagram below shows five unpaire electrons. The Pauli exclusion principle requires that an electron jumping to a higher energy 3 orbital woul have to change its spin. A transition that involves a change in spin state is forbien, an thus oes not occur to any appreciable extent. The colors of transition metal complexes are ue to the absorption of energy in the form of visible light when a electron is excite from the groun state into an excite state. In Mn complexes, the promotion of a electron to an excite state is spin forbien; therefore, there is little in the way of an absorption spectrum an the complexes are practically colorless. x y xy x y Mn(H) mplexes are expecte to be colore when the highest occupie orbitals have between one an nine electrons. Such complexes can therefore have transitions (that are usually in the visible part of the electromagnetic raiation spectrum). The ions V 5, a, an Sc 3 have 0 electron configurations an u, Zn, an Pb have 10 electron configurations: these complexes are colorless. The other complexes have outer electron configurations of 1 to 9 an are therefore colore exists in solution as 4 (blue) or (H)6 (pink). The equilibrium is: 4 6H (H)6 4 H < 0 blue pink ow temperature an low concentration of ions favor the formation of (H)6 ions. Aing H (more ions) favors the formation of 4. Aing Hg leas to: Hg Hg4 This reaction ecreases [ ], so the pink color is restore Dipole moment measurement. nly the cis isomer has a ipole moment Fe is 3 6 ; Fe 3 is 3 5.

11 63 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS Therefore, Fe 3 (like Mn, see Problem 15.44) is nearly colorless, so it must be light yellow in color EDTA sequesters metal ions (like a an Mg ) which are essential for growth an function, thereby epriving the bacteria of the ability to grow an multiply The Be complex exhibits optical isomerism. The u complex exhibits geometric isomerism. H 3 H F 3 Be F 3 H H 3 F 3 H H 3 Be H 3 H F 3 mirror F 3 H H 3 u F 3 H H 3 F 3 H H 3 u H 3 H F 3 cis trans The square planar complex shown in the problem has 3 geometric isomers. They are: a b c b a c Note that in the first structure a is trans to c, in the secon a is trans to, an in the thir a is trans to b. Make sure you realie that if we switch the positions of b an in structure 1, we o not obtain another geometric isomer. A 180 rotation about the ac axis gives structure 1. a b c 180 o rotation a b c a c b 15.5 Isomer I must be the cis isomer. H 3 N H 3 N The chlorines must be cis to each other for one oxalate ion to complex with.

12 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS 633 H 3 N H 3 N Isomer II must be the trans isomer. H 3 N NH 3 With the chlorines on opposite sies of the molecule, each is replace with a hyrogen oxalate ion. H NH 3 H 3 N H The reaction is: Ag (aq) N (aq) Ag(N) (aq) K f [Ag(N) ] [Ag ][N ] First, we calculate the initial concentrations of Ag an N. Then, because Kf is so large, we assume that the reaction goes to completion. This assumption will allow us to solve for the concentration of Ag at equilibrium. The initial concentrations of Ag an N are: 5.0 mol [N ] M mol [Ag ] 0.18 M 99.0 We set up a table to etermine the concentrations after complete reaction. Ag (aq) N (aq) Ag(N) (aq) Initial (M): hange (M): 0.18 ()(0.18) 0.18 Final (M): K f [Ag(N) ] [Ag ][N ]

13 634 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS M [Ag ]( M ) [Ag ] M (a) If the bons are along the -axis, the orbitals will have the lowest energy. orbital will have the highest energy. The an x y xy M x, y x y, xy If the trigonal plane is in the xy-plane, then the an orbitals will have the highest energy, an will have the lowest energy. x y xy M y x, x y x, y xy If the axial positions are along the -axis, the M orbital will have the highest energy. The trigonal plane will be in the xy-plane, an thus the an orbitals will be next highest in energy. x y xy, x y x, y xy (a) The equilibrium constant can be calculate from G. We can calculate G from the cell potential. From Table 13.1 of the text, u e u u e u E 0.34 V an G ()(96500 J V 1 mol 1 )(0.34 V) J mol 1 E 0.15 V an G (1)(96500 J V 1 mol 1 )(0.15 V) J mol 1 These two equations nee to be arrange to give the isproportionation reaction in the problem. We keep the first equation as written an reverse the secon equation an multiply by two. u e u G J mol 1 u u e G ()( J mol 1 ) u u u G J mol J mol J mol 1

14 HAPTER : TRANSITIN META HEMISTRY AND RDINATIN MPUNDS 635 We use Equation (10.1) of the text to calculate the equilibrium constant. G RT ln K K e G RT ( J mol 1 ) (8.314 J mol K e 1 K 1 )(98 K) K Free u ions are unstable in solution [as shown in part (a)]. Therefore, the only stable compouns containing u ions are insoluble (a) The secon equation shoul have a larger S because more molecules appear on the proucts sie compare to the reactants sie. nsier the equation, G H TS. The H term for both reactions shoul be approximately the same because the N bon strength is approximately the same in both complexes. Therefore, G will be preominantly epenent on the TS term. Next, consier Equation (10.1) of the text. G RT ln K A more negative G will result in a larger value of K. Therefore, the secon equation will have a larger K because it has a larger S, an hence a more negative G. This exercise shows the effect of bi- an polyentate ligans on the position of equilibrium (a) u 3 woul not be stable in solution because it can be easily reuce to the more stable u. From Figure 15.3 of the text, we see that the 3 r ioniation energy is quite high, about 3500 kj mol --1. Therefore, u 3 has a great tenency to accept an electron. Potassium hexafluorocuprate(iii). u 3 is 3 8. uf6 3 has an octaheral geometry. Accoring to Figure of the text, uf6 3 shoul be paramagnetic, containing two unpaire electrons. (Because it is 3 8, it oes not matter whether the ligan is a strong or weak-fiel ligan. See Figure of the text.) We refer to Figure 15.1 of the text. The splitting pattern is such that all the square-planar complexes of u 3 shoul be iamagnetic.