Homework #5 Chapter 19 Transition Metals and Coordination Chemistry 6. a) b) b) Ti = [Ar]4s23d2 Re = [Xe]6s24f145d5 Ir = [Xe]6s24f145d7 Ti2+ = [Ar]3d2 Re2+ = [Xe]4f145d5 Ir2+ = [Xe]4f145d7 Ti4+ = [Ar] Re3+ = [Xe]4f145d4 Ir3+ = [Xe]4f145d6 a) b) c) Cr = [Ar]4s13d5 Cu = [Ar]4s13d10 V = [Ar]4s23d3 Cr2+ = [Ar]3d4 Cu+ = [Ar]3d10 V2+ = [Ar]3d3 Cr3+ = [Ar]3d3 Cu2+ = [Ar]3d9 V3+ = [Ar]3d2 7. 16. 22. 23. 24. Coordination Compound: A compound composed of a complex ion and counter ion sufficient to give no net charge. Complex Ion: A charged species consisting of a metal ion surrounded by ligands. Counter Ion: Anions or cations that balance the charge on the complex ion in a coordination complex. Coordination Number: The number of bonds formed between the metal ion and the ligands in a complex ion. Ligand: A group attached to the central metal ion in a complex. Chelate: A complex containing at least one polydentate ligand that forms a ring of atoms including the central metal atom. Bidentate: A ligand that can form two bonds to a metal ion. a) hexaamminecobalt(II) chloride c) potassium tetrachloroplatinate(II) e) pentaamminechlorocobalt(III) chloride b) hexaquacobalt(III) iodide d) potassium hexachloroplatinate(II) f) triamminetrinitrocobalt(III) a) pentaamminechlororuthenium(III) ion b) hexacyanoferrate(II) ion c) tris(ethylenediamine)manganese(II) ion d) pentaamminenitrocobalt(III) ion a) pentaaquabromochromium(III) bromide b) sodium hexacyanocobaltate(III) b) bis(ethylenediamine)dinitroiron(III) chloride d) tetraamminediiodoplatinum(IV) tetraiodoplatinate(II) Note: The oxidation numbers of platinum where determined by looking at the possible oxidation numbers for platinum, these can be found in slide 7 of your class notes. Platinum is known to have oxidation numbers of +2,+4, +5, and +6 with +2 and +4 being the most common. If the platinum in the cation ([PtI2(NH3)]+x) had a oxidation state of +2 then the overall charge of the ion would be 0. All of these compounds are ionically bound together. In order for this to happen you need to have positive and negative species. Ionic bonds cannot form if there is not a transfer of electrons therefore, the oxidation state cannot be +2 on platinum in the cation. If the platinum in the cation ([PtI2(NH3)]+x) had a oxidation state of +4 then the overall charge of the ion would be +2. This would make the oxidation state of the platinum in the anion ([PtI4]‐y) be +2 giving 1 the tottal anion a ch
harge of ‐2. Since +2 is alsoo a commonly observed o
oxidation num
mber for plaatinum, this gives us the oxxidation num bers of both the platinum atoms. The oxidation number o
of the platinum in the catioon is +4 and iin the anion is +2. If you d
do the math ffor when the oxidation number of the ccation +5 or ++6 (other observed oxidatiion numbe
ers of platinum) you will se
ee that you c annot get an oxidation staate for the platinu
um in the anio
on that is obsserved. 25. 26. 27. 31. a)) K2[CoCl4] c)) Na3[Fe(CN)2(C2O4)2] b) [PPt(H2O)(CO)3]Br2 d) [CCrCl(NH3)3(NH
H2CH2CH2NH2))]I2 a)) [FeCl4]‐ c)) [Cr(OH)2(CO
O)4]+ b) [R
Ru(NH3)5(H2O))]3+ d) [PPtCl3(NH3)]‐ l2 is added to
When the BaC
W
o the solution,, no precipitaate forms. Th
herefore, the sulfate ion m
must be bound to th
he cobalt because BaSO4 iss not soluble and would haave generated a precipitatte if not bound to tthe metal. When AgNO
W
he solution a p
precipitate dooes form. Therefore, Cl‐ m
must be a counter 3 is added to th
io
on allowing AggCl2 to form. SO
O42‐ must be aa bidentate ligand in order for the met al to have a ccoordination number of 6. Th
herefore, the
e compound m
must be [Co(SSO4)(NH3)4]Cl a)) Isomer: One of two
o or more com
mpounds thatt contain the same numbeer of the samee atoms in d
different arrangements. Examp
ple: b) Structu
ural Isomer: Isomers in which the atom
ms have differrent partnerss or lie in a different orrder. (Differennt chemical fo
ormulas) Examp
ple: 2 c) d) e) f) g) 32. Stereoisomers: Isomers in which the atoms have the same partners but are in different arrangements in space. (Same chemical Formulas) Example: Coordination Isomers: Isomers that differ by the exchange of one of more ligands between cationic complex and an anionic complex. Example: [CrBr(H2O)5]Cl and [CrCl(H2O)5]Br Linkage Isomers: Isomers that differ in the identity of the atom that a ligand uses to attach to a metal ion. Example: Geometric Isomers: Isomers in which atoms or groups of atoms can assume different positions around a rigid bond. Example: Optical Isomers: Isomers in which one molecule is the mirror image of the other molecule. Example: a) 2 bonds, one to each of the lone pair of electrons on the oxygen atoms b) 3 bonds, one to each of the lone pair of electrons on the nitrogen atoms c) 4 bonds, one to each of the lone pair of electrons on the oxygen or the nitrogen atoms d) 4 bonds, one to each of the lone pair of electrons on the nitrogen atoms 3 33. a)) b) c)) d) 34. Note: a)) c)) is an abbreviation for ethyleneddiamine (NH2CH2CH2NH2) b) d) 4 e)) 37. M is a transitio
M
on metal ion 39. Liinkage Isomer: Isomers thaat differ in the identity of tthe atom thaat a ligand usees to attach to a metal ion.
In
n order to form a linkage issomer there n
needs to be aat least 2 diffeerent types of atoms in thee liggand and there needs to b
be lone pairs o
on the differeent types of aatoms. Capable of forming linkage Isomers SCN‐ (loan pair on b
both S and N) NO2‐ (loan pair on b
both O and N)) OCN‐ (loan pair on b
both O and N
N) Not Capable of forming linkkage Isomers
N3‐ (on
nly one type o
of atom) NH2CH
H2CH2NH2 (only the nitroge
en atoms havve lone pairs, therefore only 1 bonding site) I‐ (onlyy one type of atom) 40. 5 42. Th
here are 5 geometrical iso
omers (ligandss are attachedd to differentt locations (different chem
mical names)) (i‐v) Note v. (cis‐diaaquadibromo
odicyanoplatin
num(VI)) and its mirror image are both optical isomers (isomers that have the same neighboors but have d
different locattions relative to eaach other and
d are not iden
ntical to theirr mirror imagee (yet they haave the samee chemical fo
ormulas)). 45. 46. Weak‐Field Lig
W
gand: A ligand
d that produces a small ligaand field splittting (lies below NH3 in spe
ectochemicall series). Sttrong Field Liggand: A ligand that producces a large liggand field spliitting (lies abo
ove H2O in th
he spectochem
mical series). Lo
ow‐Spin Complex: A transiition metal co
omplex with tthe minimum
m number of u
unpaired electrons. High‐Spin Com
mplex: A transsition metal co
omplex with tthe maximum
m number of unpaired elecctrons. Cu = [Ar]4s13d10 (lower ene
ergy when 3d orbital is full)) 2+
9 Cu = [Ar]3d
Cu+ = [Ar]3d10 Complex appear colored be
ecause electro
ons are able tto transition b
between the split d orbitaals. +
Siince Cu has aa full d shell, n
no electrons can transitionn, therefore, Cu+ compoun
nds tend to be co
olorless. Cu2++ has space fo
or an electron
n to be promooted, therefore, Cu2+ comp
pounds are exxpected to be
e colored. Cd(NH3)4Cl2 Cd
d2+ = [Kr]4d10 Siince Cd2+ has a filled d orbital, compoun
nds formed frrom Cd2+ wou
uld not be exp
pected to havve co
olor. 

hc 
 
50. Order of increa
O
asing energy absorbed  E 
770 nm
m < 440 nm <
< 290 nm Order of increa
O
asing ligand ffield strength
F‐ < NH
H3 < CN‐ Th
he stronger the field ligand
d, the greater the ligand fiield splitting.
Order of increa
O
asing ligand ffield splitting
3‐
CoF63‐ < Co(NH3)63+ < Co(CN)
<
6
CoF63‐ 770 nm 6 51. Co(NH3)63+ Co(CN)63‐ 440 nm 290 nm a) Fe2+ = [Ar]3d6 Low Spin
d z2

d xz
2+
b) Fe3+ = [Ar]3d5 High Spin

d z2
d x2  y 2

d xy

d xy

d yz
8
c) Ni = [Ar]3d 
d z2

d x2  y 2

d xz
d x2  y2

d xy

d yz
2+


d xz

d yz
10 d) Zn = [Ar]3d

d z2


d z2
d x2  y2

d xy

d xz

d x2  y2

d xy

d yz

d xz

d yz
e) Co2+ = [Ar]3d7 Low Spin

d z2
High Spin

d xy
52. 
d xz

d xy

d yz
Co(NH3)63+, therefore, Co3+ Co3+ = [Ar]3d6 (Low Spin) d z2

d z2
d x2  y 2
d x2  y 2

d x2  y 2

d xz

d yz
Fe(H2O)62+, therefore, Fe2+ Fe2+ = [Ar]3d6 (High Spin) 
d z2

d x2  y 2

d xy

d xz

d yz

d xy

d xz

d yz
No unpaired electrons (diamagnetic) Four unpaired electrons (paramagnetic) NH3 is a strong field ligand, therefore, it will form a low spin complex. Causing Co(NH3)63+ to have no unpaired electrons and be diamagnetic. H2O is a weak field ligand; therefore, it will 7 form a high spin complex. Causing Fe(H2O)62+ to have four unpaired electrons and be paramagnetic. 53. a) Ru(NH3)62+, therefore, Ru2+ Ru2+ = [Kr]4d6 (Low Spin) d z2
b) Ni(H2O)62+, therefore, Ni2+ Ni2+ = [Ar]3d8 (Low Spin) 
d z2
d x2  y 2

d x2  y 2

d xy
c) 
d xz

d yz
No unpaired electrons V(en)33+, therefore, V3+ V3+ = [Ar]3d2 (Low Spin) d z2

d xy

d xz
2 unpaired electrons 
d yz
d x2  y 2

d xy
56. 
d xz
d yz
2 unpaired electrons Fe(CN)63‐, therefore, the metal ion is Fe3+ Fe3+ = [Ar]3d5 If there is only one unpaired electron the complex must be low spin, as seen below. d z2
d x2  y 2

d xy

d xz

d yz
8 Fe(SCN)63+, therefore, Fe3+ Fe3+ = [Ar]3d5 In order to have 5 unpaired electrons the complex must be high spin, as seen below. 
d z2

d x2  y 2

d xy

d xz

d yz
Therefore, SNC‐ must be a weaker field ligand than CN‐ causing it to lie below CN‐ on the spectrochemical series. 57. Test Tube #1: Appears: Violet Absorbs: Yellow Test Tube #2: Appears: Yellow Absorbs: Violet Test Tube #3: Appears: Green Absorbs: Red Order of increasing ligand field strength Cl‐ < H2O < NH3 The stronger the field ligand, the greater the ligand field splitting. Order of increasing ligand field splitting Cr(H2O)4Cl2+ < Cr(H2O)63+ < Cr(NH3)63+ Order of increasing wavelength absorbed test tube #2 <test tube #1 < test tube #3 Order of increasing energy absorbed  E 
test tube #3 < test tube #1 < test tube #2 Cr(NH3)63+ test tube #2 Cr(H2O)63+ test tube #1 test tube #3 Cr(H2O)4Cl2+ 71. 

A will have the formula [CrIy(NH3)x]Iz What we know from the formula x  y  6 (the problem told us that A was 6 coordinated) 3   1 y     1 z  (the charge of the cation must equal opposite the charge of the anion) Calculate the moles of Cr in A (use data i) hc 
 
0.0203 g CrO 3

1mol CrO3
100.00 g

Calculate mass percent of Cr 2.03 104 mol Cr

52.00 g
1mol Cr
1mol Cr
1mol CrO3
  2.03 10
4
mol Cr   0.0106mol Cr
 m 
 0.0106 g 
mass % Cr =  Cr 100%  
100%  10.1% Cr
 0.105 g 
 mTotal 
Calculate the moles of Cr in 100 g of A 10.1g Cr

1mol Cr
52.00 g
  0.194mol Cr 9 Calculate the moles of NH3 in A (use date ii) mol
0.0329 L HCl  0.100
1 L HCl 

1mol NH 3
1mol HCl
Calculate the mass percent of NH3 0.00329mol NH 3

17.04 g
1mol NH 3
 mNH3
mass % NH3 = 
 mTotal
  0.00329mol NH
  0.0560mol NH
3
3

 0.0560 g 
100%

100%

16.4%
NH



3
 0.341g 

Calculate the moles of NH3 in 100 g of A 16.4 g NH 3

1mol NH 3
17.04 g
  0.962mol NH
3
Calculate x x
0.962 mol
 4.96  5 0.194 mol
Calculate y x y 6
y  6 x  6 5 1
Calculate z 3   1 y     1 z 
z  3  y  3 1  2
[CrI(NH3)5]I2 10