Reactivity in Chemistry

Ligand Binding

 in Coordination Complexes and Organometallic Compounds

CC13. More Solutions to Selected Problems

Problem CC5.2.

a)  The resulting anion has aromatic stability.  It is cyclic, fully conjugated, flat and has an odd number of electron pairs.

b)  Cp anion could bind to a metal through just one pair or through two pairs, but in most cases it will bind via three pairs of electrons.

d)  Valence count on metal:  8

     Count on metal, correcting for +2 charge:  6

      Donated from ligands:  2 x 6 = 12

      Total:  18 

Problem CC5.3.

This problem deals with the "bite angle" of the ligand.  Remember, a chain of atoms becomes more flexible the longer it gets, because of the possibility for rotation around each bond along the chain.   As the two double bonds move further apart from each other (one bond apart in (a), two bonds apart in (b) and three bonds apart in (c)), the chain can "open up" and bind with a more optimal overlap with the metal.

 

 

Problem CC7.1.

a)  Zn(II), because it is smaller and less polarizable.

b)  K+, because it is less electronegative.

c)  Fe(III), because of the higher charge.

Problem CC7.2.

a)  Me3P, because phosphorus is larger and more polarizable than nitrogen.

b)  Iodide, which is larger and more polarizable than chloride.

c)  Azide, which has a more polarizable, delocalized pi bonding system.

Problem CC7.3.

a)  ZrO2

b)  CdS

c)  WO3

d)  ZnS

e)  Cu2S

Problem CC7.4.

a)  Hg(I) and Hg(II) are both large, polarizable ions.  They are soft cations and should bind well to soft donors. 

b)  The most common soft donor is a sulfur atom or sulfide ion; in amino acids, that suggests cysteine or methionine.

Problem CC7.5.

Fe(III) is a hard cation and should bind well to oxygen donors.  Enterobactin has several oxygen donors it could provide to the iron.  In fact, there is a pair of OH groups on each of the benzene rings in enterobactin.  These benzene rings with two OH groups next to each other are called "catechols".  Because there are three of these groups in enterobactin, and there is enough space in between for the groups to fold around a central atom, enterobactin is a chelating (hexadentate) donor with a high binding constant.

Problem CC8.1.

Problem CC8.2.

The three orbitals shown above interact a little more strongly with the ligands.  The three orbitals shown below interact a little more weakly.

The reason for the difference in the interaction has to do with how close the nearest lobe of a d orbital comes to a ligand.  There are really two possible positions:  the face of a cube or the edge of a cube.  If the ligands are at alternating corners of the cube, then the orbitals pointing at the edges are a little closer than those pointing at the faces of the cube.

Problem CC8.3.

The ligands do not overlap with the d orbitals as well in tetrahedral complexes as they do in octahedral complexes.  Thus, there is a weaker bonding interaction in the tetrahedral case.  That means the antibonding orbital involving the d electrons is not raised as high in energy, so the splitting between the two d levels is smaller.

Problem CC8.4.

 

Problem CC8.5.

Because the d orbital splitting is much smaller in the tetrahedral case, it is likely that the energy required to pair two electrons in the same orbital will be grester than the energy requied to promote an electron to the next energy level.  In most cases, the complex will be high spin.

Problem CC8.6.

The orbitals are shown in order of energy.

Problem CC8.7.

a. Fe+2

d6 low spin
SE = [-0.4(6)+ 0.6(0)] Δo + 3PE
= [-2.4] Δo + 3PE

d6 high spin
SE = [-0.4(4)+ 0.6(2)] Δo + 1PE
= [-.4] Δo + 1PE


b. Co+2

d7 low spin
SE = [-0.4(6)+ 0.6(1)] Δo + 3PE
= [-1.8] Δo + 3PE

d7 high spin
SE = [-0.4(5)+ 0.6(2)] Δo + 2PE
= [-.8] Δo + 1PE

c. Co+3
d6 so this looks the same as Fe+2

d. Mn+2

d5 low spin
SE = [-0.4(5)+ 0.6(0)] Δo + 2PE
= [-2.0] Δo + 2PE

d5 high spin
SE = [-0.4(3)+ 0.6(2)] Δo + 0PE
= [0] Δo

e. Ti+3
d1 so there is no possibility of low spin or high spin

Problem CC8.8.

a. [Fe(py)6]2+ 3d metal, M+2, pi acceptor ligand → low spin

b. [Fe(H2O)6]2+ 3d metal, M+2, pi donor ligand → high spin

c. [FeBr6]3- 3d metal, M+3, pi donor ligand → high spin

d. [Co(NH3)6]3+ 3d metal, M+3, sigma donor ligand → low spin

e. [Cu(NH3)6]2+ 3d metal, M+2, sigma donor ligand → low spin

f. [Rh(CO)6]3+ 4d metal, M+3 → low spin

g. [Cr(CO)6]3+ 3d metal, M+3, pi acceptor ligand → low spin

f. [PtCl6]2- 5d metal, M+4 → low spin

Problem CC8.9.

a. [Zn(NH3)4] 2+ 3d metal, d10, sigma donor ligand → tetrahedral

b. [NiCl4] 2+ 3d metal, d8, pi donor ligand → tetrahedral

c. [Ni(CN)4] 2- 3d metal, d8, pi acceptor ligand → square planar

d. [Ir(CO)(OH)(PCy3)2] 2+ 5d metal, d8 → square planar

e. [Ag(dppb)2]1+ 4d metal, d10, sigma donor ligand → tetrahedral

f. [PtCl2(NH3)2] 5d metal, d8 → square planar

g. [PdCl2(NH3)2] 4d metal, d8, M+2, sigma donor ligand → square planar

i. [CoCl4] 2 3d metal, d7, sigma donor ligand → tetrahedral

f. [Rh(PPh3)3Cl] 5d metal,  d8 → square planar

Problem CC9.1.

The weaker donors include halides and oxygen donors.  Nitrogen atom donors are mostly a little stronger than that.  The strongest donors include carbon donors, especially with pi bonds.

Problem CC9.2.

 

Problem CC9.3.

Pi donation raises the t2g electrons (the d electrons of proper symmetry for pi overlap with the ligands in an octahedral geometry).

Problem CC9.6.

Because chloride is a weaker ligand than hydroxide, the d orbital splitting gets smaller.  A d-d transition would involve less energy, so it would move to longer wavelength (red shift).  

 

Problem CC9.7.

a) The [Cu(OH2)6]2+ ion appears blue-green, so it absorbs a reddish orange, about 650 nm; whereas the [Cu(NH3)2(OH2)4]2+ ion appears indigo-violet, so it absorbs yellow, about 600 nm.

b) Shorter wavelength is higher energy, according to the Planck-Einstein relation: E = hc/λ (in which h = Planck's constant, c = speed of light, λ = wavelength of photon).  The [Cu(NH3)2(OH2)4]2+ ion absorbs the shorter wavelength, higher energy photon.

c)  The [Cu(NH3)2(OH2)4]2+ ion has the greater d-d gap.

d)  The ammonia must be a stronger field ligand than water.  In general terms we might think of that as a result of ammonia being only a sigma-donor, whereas water is also a pi-donor.

 

Problem CC9.8.

a) The [Cu(NH3)6]3+ ion appears yellow, so it absorbs violet, about 400 nm; whereas the [Cu(NH3)5Cl]2+ ion appears pink, so it absorbs green, about 550 nm.

b) Shorter wavelength is higher energy, according to the Planck-Einstein relation: E = hc/λ (in which h = Planck's constant, c = speed of light, λ = wavelength of photon).  The  [Cu(NH3)6]3+ ion absorbs the shorter wavelength, higher energy photon.

c)  The  [Cu(NH3)6]3+  ion has the greater d-d gap.

d)  The ammonia must be a stronger field ligand than chloride.  In general terms we might think of that as a result of ammonia being only a sigma-donor, whereas chloride is also a pi-donor.

 

Problem CC9.9.

The green compound absorbs red; the violet compound absorbs yellow; the yellow compound absorbs violet.

 In terms of energy these absorbances can be ranked:

 (high energy) violet photon > yellow photon > red photon (low energy)

or in terms of d-d gap:

 large gap > middle gap > small gap

In the spectrochemical series, we would expect the order of corresponding ligands to be:

-CN > H2O > F-

The cyano compound absorbs violet and appears yellow; the aquo compound absorbs yellow and appears violet; the fluoro compound absords red and appears green.

 

Problem CC9.10.

a) The [V(OH2)6]3+ ion appears yellow, so it absorbs violet, about 400 nm; whereas the [V(OH2)6]2+ ion appears pink, so it absorbs green, about 550 nm.

b) Shorter wavelength is higher energy, according to the Planck-Einstein relation: E = hc/λ (in which h = Planck's constant, c = speed of light, λ = wavelength of photon).  The   [V(OH2)6]3+ ion absorbs the shorter wavelength, higher energy photon.

c)  The   [V(OH2)6]3+  ion has the greater d-d gap.

d)  One ion contains V3+, whereas the other ion contains V2+.  As charge increases on an ion, electrons contract toward the nucleus.  However, low-lying electrons, which are closer to the nucleus to begin with, are attracted even more than are high-lying electrons.  thus, as charge increases on the ion, the d-d splitting increases.

 

Problem CC9.11.

a)  [Co(NH3)6]+3     

      The metal is +3 suggesting a large Δ0

     The ligands are sigma donors, so medium field ligands

     The splitting energy is large thus low spin. 

b)  [Fe(CN)6]-4          

      The metal is first row, +2 suggesting intermediate Δ0

     The ligands are pi acceptors, so strong field ligands

     The splitting energy is large thus low spin. 

 c)  [CoF6]-4

      The metal is first row, +2 suggesting a large Δ0

     The ligands are pi donors, so weak field ligands

     The splitting energy is small thus high spin. 

d)  [Rh(CN)6]-3          

      The metal is 2nd row (4d) and +3 suggesting a large Δ0

     The ligands are pi acceptors, so strong field ligands

     The splitting energy is large thus low spin. 

e)  [V(OH2)6]+3    

      The complex is d2, so there is only one possible spin state. 

f)  [Fe(py)6]+2

       The metal is first row, +2 suggesting intermediate Δ0

     The ligands are pi acceptors, so strong field ligands

     The splitting energy is large thus low spin. 

g)  [MnCl6]-4           

     The metal is first row, +2 suggesting intermediate Δ0

     The ligands are pi donors, so weak field ligands

     The splitting energy is large thus high spin. 

h)  [Ru(NH3)6]+2

     The metal is 2nd row (4d) and +2 suggesting a large Δ0

     The ligands are sigma donors, so medium field ligands

     The splitting energy is large thus low spin. 

 

Problem CC10.1.

a)  The ions in LiF are both smaller than in KBr, so the force of attraction between the ions in LiF  is greater because of the smaller separation between the charges.

b)  Calcium has a 2+ charge in CaCl2, whereas potassium has only a + charge, so the chloride ions are more strongly attracted to the calcium than to the potassium.

 

Problem CC10.3.

a) labile (electrons in higher energy d orbital set)

b) not labile (all electrons in lower energy d orbitals)

c)  not labile (all electrons in lower energy d orbitals)

d)  labile (electrons in higher energy d orbital set)

e)  labile (electrons in higher energy d orbital set)

 

Problem CC11.1.

a) yes; ls d7 is eg1, so one level has one electron and the other has none - net decrease in energy upon distortion

b) no; d8 is eg2, so both levels have one electron -- no net change in energy upon distortion

 

Problem CC11.2.

a) Co2+ yes; d7 so eg1

b) Ag2+ yes; d9 so eg3

c)  Ni2+ no; d8 so eg2

d) Mn2+ no if hs; d5 so eg2.  However, if it were low-spin (unlikely with a first-row metal in a moderately low oxidation state and π-donors), then there would be degeneracy and distortion.

e) Ag+ no; d10 so eg4

 

Problem CC11.3.

a) [V(CN)6]3-  yes d2 t2g2        b) [Cr(CN)6]3- no d3 t2g3

c) [Mn(CN)6]3- yes d4 t2g4        d) [Fe(CN)6]3- yes d5 t2g5

e) [Cr(CN)6]4- yes d4 t2g4        f) [Mn(CN)6]4- yes d5 t2g5

g) [Fe(CN)6]4- no d6 t2g6

 

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

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This material is based upon work supported by the National Science Foundation under Grant No. 1043566.

 

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