Download CHEM 481. Transition Metal Complexes. Assignment 7. Answers

CHEM 481. Transition Metal Complexes. Assignment 7. Answers
1.
What feature of electronic structure distinguishes transition elements, lanthanide elements, and actinide elements from the
other elements and from one another?
The presence of partly filled d or f subshells; thus we consider those ions with d n configurations from d 1 to d 9 to be transition
metal, and from f1 to f13 to be lanthanide and actinide ions.
2.
Define the following terms: (a) transition element, (b) coordination compound, (c) complex ion, (d) ligand, (e) chelate, (f)
bidentate. Give an example to illustrate each word or phrase.
A transition element is an element from the d block of the periodic table; more precisely, the characteristic behavior of such
elements belongs to those species with d n configurations from d 1 to d 9 in at least one common oxidation state. Thus Sc3+ and
Zn2+ do not behave like typical TM, but Ti3+, Ti2+ and Ti+ , as well as Cu 2+ behave are typical.
A coordination compound is a compound formed by Lewis acid/base donor acceptor bonds. It is an all encompassing term,
although because such compounds are exceedingly common for TM ions, coordination compounds most often involve TM's.
An example of a neutral coordination compound is [Cr(NH3)3Br3].
A complex ion is a coordination compound which bears an overall negative or positive charge. Examples of the two kinds are
[Cr(NH3)6]3+ and [CrBr6]3–.
A ligand is an atom, ion, molecule or group which attaches to a transition metal atom or ion by Lewis acid/base
donor/acceptor bonds. In the previous examples, both NH3 and Br– are ligands.
A chelate is a ligand with two or more donor atoms which link to the same metal ion. The term is derived from a Greek word for
claw. Ethylenediamine, H2NCH2CH2NH2, and the oxallate anion O2CCO22– are chelating ligands.
Bidentate describes chelating ligands with two donor atoms. It literally means two-toothed. Both ethylenediamine,
H2NCH2CH2NH2, and the oxallate anion O2CCO22– are bidentate ligands. Note that geometrically only two oxygen atoms of the
four in oxallate are capable of donating to a metal ion.
O
O
O
O
M
3.
Define the terms ‘‘diamagnetic’’ and ‘‘paramagnetic.’’ What feature of electronic structure distinguishes these properties?
Diamagnetic means weakly repelled from an external magnetic field. Paramagnetic means attracted to an external magnetic
field. The attraction is significantly stronger than the diamagnetic repulsion, though it is still a weak force. Atoms, ions or
molecules with all their electrons paired up experience only a diamagnetic force, whereas the presence of one or more unpaired
electrons adds a paramagnetic force which overcomes the diamagnetic force from all the paired electrons.
4.
What are three common metal coordination numbers encountered in coordination chemistry, and what structure or
structures are possible for each?
Four coordinate complexes are common, and these may be tetrahedral or square planar in shape.
Six coordinate complexes are the most common, and are almost always octahedral in shape (the other shape occasionally
observed is trigonal biprismatic).
L
L
L
LL
L
L
L
M
M
L M L
M
L
L
L
L
L
L
L
L
L
L
tetrahedral
sq planar
octahedral
trig. prismatic
5.
Give an example of (a) geometric isomerism and (b) optical isomerism.
NH3
NH3
H2N
HN
NH3
NH3
Br
Br
NH2
Br Br 2
Cr
Cr
NH2
Cr
NH3
Br
Br
H3N
Cr
Br
NH2
NH2
Br
NH3
NH3
H2N
H2N
trans D4h
C2
C2
C2v cis
optical isomers
geometrical isomers
1
6.
According to the crystal field model, what is the origin of the splitting of metal d orbitals into two sets in an octahedral
complex?
Electrostatic repulsion occurs between the electron pairs donated by the ligands and the non-bonding d-electrons. Those d
orbitals which are oriented geometrically closer to the ligands experience more repulsion than those further away.
7.
What factors determine whether a complex is high or low spin?
The most fundamental factor is the number of d electrons in the complex. Then there is the geometry, which determines the
pattern of d orbital splitting that occurs due to the ligand field (see #6). Finally, there is the field strength of the ligand, which
determines whether strong- or weak-splitting of the d orbitals will occur. If the induced splitting is sufficient to compensate
for the electron pairing energy, the electrons will pair in the lower orbital(s), and the complex will be low spin; if the splitting is
insufficient, the electrons will fill all the d orbitals as they do in the free ion, and the complex will be high spin. If the number of
electrons is such that the two descriptions lead to the exact same electron configuration, the distinction between high- and
low-spin is not made.
8.
Give the electron configuration for each ion listed here, and tell whether it is paramagnetic or diamagnetic.
(c) Rh 3+ [Kr]4d 6 (d) V2+ [Ar]3d 3 (e) Ce 4+ [Xe]
(f) U 4+ [Rn]5f2
(a) Y3+ [Kr] (b) Pt 2+ [Xe]4f145d 8
9.
Identify two transition metal ions with the following electron configurations:
(c) [Ar] 3d 5 Fe3+, Mn2+
(d) [Ar] 3d 8 Co + , Ni2+
(a) [Ar] 3d 6 Co 3+, Fe2+ (b) [Ar] 3d 10 Cu + , Zn2+
10. Which of the following ligands is expected to be monodentate, which polydentate?.
(a) CH3NH2 mono; (b) C2O42– poly; (c) Br– mono; d) CH3CN mono; (e) en poly; (f) phen poly; (g) N3– mono
Note that (b), (e) and (f) are all examples of bidentate ligands
11. Give the oxidation number of the metal ion in each of the following compounds:
(a) [Mn(NH3) 6]SO4 +2 (since ammonia is neutral, sulfate is -2)
(b) K3[Co(CN) 6] +3 (since cyanide are -1 each, and potassium 1+ each)
(c) [Co(NH3) 4Cl2]Cl +3 (since ammonia is neutral, chloride -1 each).
(d) Mn(en) 2Cl2 +2 (since ethylenediamine are neutral, cloride -1 each)
12. Write formulas for the following ions or compounds:
(a) dichlorobis(ethylenediamine)nickel(II)
[NiCl2(en)]
(b) potassium tetrachloroplatinate(II)
K2[PtCl4]
(c) potassium dicyanocuprate(I)
K[Cu(CN)2]
(d) diaquatetraammineiron(II)
[Fe(NH3)4(OH2)2]2+
13. Draw all possible geometric isomers of:
NH3
NH3
NH3
NH3
Cl
Cl
Fe
Fe
NH3
Cl
Cl
H3N
NH3
NH3 D
C
NH3
NH3
Pt NH3
Br
Br
C s NCS
cis
Pt NCS
NH3
trans
C2v
cis
trans
(a) [Fe(NH3)4Cl2]
(b) [Pt(NH3)2(NCS)(Br)] Hint: NCS is bonded to Pt 2+ through N
NH3
NH3
Cl
O2N
O2N
NH3
Cl
NH2
NH3
Co
Co
Co
NH
NO
NH
O2N
O2N
Cl
3
2
2
NO2
NH3
Cl
C3v
C2v
C2v
fac
mer
no other isomers!
(c) [Co(NH3)3(NO2)3] Hint: NO2 is bonded to Co 3+ through N
(d) [Co(en)Cl4]–
2v
4h
14. Four isomers are possible for [Co(en)(NH3) 2(H2O)Cl] 2+ . (Two of the four are optical isomers and are nonsuperimposable
mirror images.) Draw the structures of the four isomers.
NH3
NH2
Co
NH2
H2 O
NH3
Cs
Cl
OH2
NH2
Co
NH2
H3N
Cl
Cs
H3N
Cl
H3N
C1
NH3
OH2
NH2
HN
NH2
Co 3
Co
NH2
NH2
Cl
OH2
NH3
C1
optical isomers
2
15. Are the following complexes chiral? Draw their structures:
(b) fac-[Co(en)(H2O)Cl3]+
(a) [Fe(en)3]2+
+
(c) cis-[Co(en)2Br2]
(d) [Pt(NH3)(H2O)(Cl)(NO2)] Hint: square-planar Pt
H2N
NH 2
NH 2
Fe
NH 2
H2N
Cl
Cl
NH 2
D3, ∴ chiral
OH2
NH 2
Co
NH 2
Cl
Cs , ∴ not chiral
NH 3
H 2N
Br
NH 2
Co
NH 2
Br
H 2N
Cl
C2, ∴ chiral
C s, ∴ not chiral
Pt NO 2
OH 2
16. What d electron configurations exhibit both high and low spin in octahedral complexes?
↑
↑ ↑ ↑
d4
↑ ↑
↑ ↑
↑ ↑
↑ ↑ ↑
↑↓ ↑ ↑
↑↓ ↑↓ ↑
d5
d6
"high spin" forms
↑
↑↓ ↑ ↑
d7
↑↓ ↑↓ ↑
d4
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
d5
d6
"low spin" forms
d7
17. Can [Ni(NH3) 6] 2+ form both low-spin and high-spin complexes? Explain your answer.
This complex contains Ni2+, which has a d 8 configuration. This configuration for an octahedral (CN=6) complex does not have
the option of being high- or low-spin.
18. Depict high- and low-spin configurations for each of the complexes below. Tell whether each is diamagnetic or
paramagnetic. Give the number of unpaired electrons for the paramagnetic cases. These configurations are all included
among the examples covered in #16, so I will not repeat them (but you should on an actual test.)
(a) [Fe(CN)6]4– Here the iron is +2, so it is d 6. It would have 4 u.p.e. in the high spin form, and thus paramagnetic, but 0 u.p.e.
in the low spin form. Given the position of CN– on the spectrochemical series, we would definitely expect the hexacyanoferrate
anion to be low spin.
(b) [Co(NH3)6]3+ Here the cobalt is +3, so it is d 6 as well. Again ammonia is a strong-field ligand, so we expect a diamagnetic
complex.
(c) [Fe(H2O)6]3+ Here the iron is +3, so it is d 5. If high-spin, it would have 5 u.p.e., and if low-spin one u.p.e., both
paramagnetic. With the rather weak field water ligand, we would probably expect a high-spin complex.
(d) [CrF6]4– Here the chromium is +2, so it is d 4. If high-spin, it would have 4 u.p.e., and if low-spin 2 u.p.e., both paramagnetic.
With the very weak field fluoride ligand, we would definitely expect a high-spin complex.
19. From experiment we know that [Mn(H2O) 6] 2+ has five unpaired electrons, whereas [Mn(CN)6] 4– has one unpaired electron.
Using the crystal field model, depict the electron configuration for each ion. What can you conclude about the effect of the
ligand on the magnitude of ∆o?
Both complexes contain manganese in the +2 form, so it is d 5. The diagrams are again in #16, and indeed high-spin d 5 has five
u.p.e., while the low-spin form has a single u.p.e. This implies that water is a weak-field ligand, resulting in a small ∆o less than
the electron pairing energy, while cyanide is a strong-field ligand, causing ∆o to be sufficiently large that the electrons are
forced to pair up in the lower d orbitals.
20. Comment on the fact that, although an aqueous solution of cobalt(III) sulfate is diamagnetic, the solution becomes
paramagnetic when a large excess of fluoride ion is added.
Cobalt is +3, so it is d 6, which can be diamagnetic in the low-spin form, and paramagnetic with 4 u.p.e. in the high-spin form.
The implication of this experiment is that water in the aqua ion has sufficient field strength to cause low-spin, but when the
water is displaced by the excess fluoride, the lower ligand field strength of fluoride causes ∆o to shrink, and a high-spin state is
induced. Note that water is not particularly high on the spectrochemical series, but evidently high-enough for cobalt +3 to be
low-spin.
[ Co (OH ) ]
3+
2 6
+
6F −
→
[CoF ]
3−
6
+
6 H2 O
21. The complex ion [Co(CO3) 3] 3– , an octahedral complex with bidentate carbonate ions as ligands, has one absorption in the
visible region of the spectrum at 640 nm. From this information
(a) Predict the color of this complex, and explain your reasoning. 640 nm is orange light (see spectrum in notes). Using a
colour wheel such as on p.1074 of Kotz and Treichel, 3rd ed. the complement of orange is blue, so we expect a blue complex
(b) Place the carbonate ion in the proper place in the spectrochemi cal series. Carbonate fits between halides and oxallate
on the spectrochemical series (see Table 23.4, p. 1076 of Kotz and Treichel).
3
(c) Predict whether [Co(CO3) 3] 3– will be paramagnetic or diamagnetic. This is another d 6 complex, which when high spin
has a single d-d band, and low spin has three. Thus the single band spectrum signifies high-spin, which is diamagnetic.
22. Why are the colored f-element aqua ions very pale?
The f-f bands are of low probability, and are also extremely narrow. This means that their integrated intensity is low, and hence
the colours are very pale relative to the concentration of the ions.
23. For the d-elements, it is essential to know what the coordination number of the ligands is before setting out to interpret the
electronic or magnetic properties. Why is this less important for the f-elements?
In the f element ions, the f orbitals are buried deeply, and are little affected by the ligand environment. The 7 f orbitals are close
enough to the free-ion form that the spectra resemble those of the free ions. The environment has little influence, and the
number of u.p.e. comes directly from the ion electron configuration.
24. The glycinate ion, H2NCH2CO2– (formed by deprotonation of the amino acid glycine), can function as a bidentate ligand,
binding through N and one O. Draw all of the isomers that can be formed by the complex ion [Cu(H2NCH2CO2) 2(H2O) 2].
OH 2
OH 2
O
H 2N
O
O
NH 2 H 2N
O
NH 2
O
NH 2
H 2O
H 2O
H 2O
Cu
Cu
Cu
Cu
Cu
O
O
NH 2
NH 2
H 2O
H 2O
H 2O
O
O
NH 2
H 2N
H 2N
O
OH 2
OH 2
4