2. 23-2
The Transition Elements and Their Coordination
Compounds
23.1 Properties of the Transition Elements
23.2 The Inner Transition Elements
23.3 Coordination Compounds
23.4 Theoretical Basis for the Bonding and Properties of
Complexes
3. 23-3
Figure 23.1 The transition elements (d block) and inner
transition elements (f block) in the periodic table.
4. 23-4
Properties of the Transition Metals
All transition metals are metals, whereas main-group
elements in each period change from metal to nonmetal.
Many transition metal compounds are colored and
paramagnetic, whereas most main-group ionic compounds
are colorless and diamagnetic.
The properties of transition metal compounds are
related to the electron configuration of the metal ion.
6. 23-6
Electron Configurations of
Transition Metals and their Ions
The d-block elements have the general condensed
ground-state configuration [noble gas]ns2(n – 1)dx where
n = 4 to 7 and x = 1 to 10.
Periods 6 and 7 elements include the f sublevel:
[noble gas]ns2(n – 2)f14(n – 1)dx where n = 6 or 7.
Transition metals form ions through the loss of the ns
electrons before the (n – 1)d electrons.
7. 23-7
Table 23.1 Orbital Occupancy of the Period 4 Transition Metals
The number of unpaired electrons increases in the first half of the
series and decreases in the second half, when pairing begins.
8. 23-8
Sample Problem 23.1 Writing Electron Configurations of
Transition Metal Atoms and Ions
PROBLEM: Write condensed electron configurations for the following:
(a) Zr; (b) V3+; (c) Mo3+. (Assume that elements in higher
periods behave like those in Period 4.)
PLAN: We locate the element in the periodic table and count its
position in the respective transition series. These elements are
in Periods 4 and 5, so the general electron configuration is
[noble gas]ns2(n – 1)dx. For the ions, we call that ns electrons
are lost first.
SOLUTION:
(a) Zr is the second element in the 4d series:
[Kr]5s24d2
9. 23-9
Sample Problem 23.1
(b) V is the third element in the 3d series, so its configuration is
[Ar]4s23d3. When it forms V3+, it loses the two 4s e- first, then
one of the 3d e-:
[Ar]3d2
(c) Mo lies below Cr in group 6B(6), so we expect the same
exception as for Cr. The configuration for Mo is therefore
[Kr]5s14d5. Formation of the Mo3+ ion occurs by loss of the
single 5s electron followed by two 4d electrons:
[Kr]4d3
10. 23-10
Trends in the Properties of
Transition Metals
Across a period the following trends are observed:
Atomic size decreases at first, then remains relatively
constant.
- The d electrons fill inner orbitals, so they shield outer electrons very
efficiently and the 4s electrons are not pulled closer by the increasing
nuclear charge.
Electronegativity and ionization energies also increase
relatively little across the transition metals of a particular
period.
12. 23-12
Trends in the Properties of
Transition Metals
Within a group the trends also differ from those observed
for main group elements.
Atomic size increases from Period 4 to 5, but not from
Period 5 to 6.
- The extra shrinkage from the increase in nuclear charge (called
the lanthanide contraction) is roughly equal to the normal size
increase due to adding an extra energy level.
- A Period 6 element has 32 more protons than its preceding Period
5 group member instead of only 18.
13. 23-13
Trends in the Properties of
Transition Metals
Electronegativity increases within a group from Period 4
to 5, then generally remains unchanged from Period 5 to
6. The heavier elements often have high EN values.
Although atomic size increases slightly down the group, nuclear
charge increases much more, leading to higher EN values.
Ionization energy values generally increase down a
transition group, also running counter to the main group
trend.
Density increases dramatically down a group since atomic
volumes change little while atomic masses increase
significiantly.
15. 23-15
Oxidation States of Transition Metals
Most transition metals have multiple oxidation states.
Elements in Groups 8B(8), 8B(9) and 8B(10) exhibit
fewer oxidation states. The higher oxidation state is less
common and never equal to the group number.
- The +2 oxidation state is common because the ns2 electrons
are readily lost.
The highest oxidation state for elements in Groups 3B(3)
through 7B(7) equals the group number.
- These states are seen when the elements combine with the highly
electronegative oxygen or fluorine.
16. 23-16
Figure 23.5 Aqueous oxoanions of transition elements.
Mn2+ MnO4
2−
MnO4
−
+2 +6 +7
The highest oxidation state for
Mn equals its group number.
VO4
3− Cr2O7
2−
MnO4
−
+5 +6 +7
Transition metal ions are
often highly colored.
18. 23-18
Metallic Behavior of Transition Metals
The lower the oxidation state of the transition metal, the
more metallic its behavior.
Ionic bonding is more prevalent for the lower oxidation
states, whereas covalent bonding occurs more
frequently for higher oxidation states.
Metal oxides become less basic (more acidic) as the
oxidation state increases.
A metal atom in a positive oxidation state has a greater attraction
for bonded electrons, and therefore a greater effective
electronegativity, or valence-state electronegativity, than in the
zero oxidation state. This effect increases as its oxidation state
increases.
19. 23-19
Table 23.3 Standard Electrode Potentials of Period 4 M2+ Ions
Half-Reaction E°(
V)
Ti2+(aq) + 2e− Ti(s)
V2+(aq) + 2e− V(s)
Cr2+(aq) + 2e− Cr(s)
Co2+(aq) + 2e− Co(s)
Fe2+(aq) + 2e− Fe(s)
Mn2+(aq) + 2e− Mn(s)
Ni2+(aq) + 2e− Ni(s)
Cu2+(aq) + 2e− Cu(s)
Zn2+(aq) + 2e− Zn(s)
-1.63
-1.19
-0.91
-0.76
0.34
-0.28
-0.25
-0.44
-1.18 In general, reducing strength
decreases across the series.
20. 23-20
Color and Magnetic Behavior
Most main-group ionic compounds are colorless and
diamagnetic because the metal ion has no unpaired
electrons.
Many transition metal ionic compounds are highly
colored and paramagnetic because the metal ion has
one or more unpaired electrons.
Transition metal ions with a d0 or d10 configuration are
also colorless and diamagnetic.
21. 23-21
Figure 23.6 Colors of representative compounds of the Period 4
transition metals.
titanium(IV) oxide sodium chromate
potassium
ferricyanide
nickel(II) nitrate
hexahydrate
zinc sulfate
heptahydrate
scandium oxide vanadyl sulfate
dihydrate
manganese(II)
chloride
tetrahydrate
cobalt(II)
chloride
hexahydrate
copper(II) sulfate
pentahydrate
22. 23-22
Table 23.4 Some Properties of Group 6B(6) Elements
Element
Atomic Radius
(pm) IE1(kJ/mol)
E° (V) for
M3+(aq)/M(s)
Cr 128 653 -0.74
Mo 139 685 -0.20
W 139 770 -0.11
IE1 increases down the group, so reactivity decreases. This
trend is opposite to that seen in main-group elements.
23. 23-23
Lanthanides and Actinides
The lanthanides are also called the rare earth elements.
The atomic properties of the lanthanides vary little
across the period, and their chemical properties are
also very similar.
Most lanthanides have the ground-state electron
configuation [Xe]6s24fx5d0.
All actinides are radioactive, and have very similar physical
and chemical properties.
The +3 oxidation state is common for both lanthanides
and actinides.
24. 23-24
Sm is the eighth element after Xe. Two electrons go into the 6s
sublevel. In general, the 4f sublevel fills before the 5d, so the
remaining six electrons go into the 4f sublevel.
Sample Problem 23.2
SOLUTION:
Finding the Number of Unpaired Electrons
PROBLEM: The alloy SmCo5 forms a permanent magnet because
both samarium and cobalt have unpaired electrons. How
many unpaired electrons are in Sm (Z = 62)?
PLAN: We write the condensed electron configuration of Sm and
then, using Hund’s rule and the aufbau principle, place
electrons into a partial orbital diagram and count the
unpaired electrons.
The condensed configuration of Sm is [Xe]6s24f6.
25. 23-25
Sample Problem 23.2
5d 6p
↑
4f
↑ ↑ ↑ ↑ ↑
6s
↑↓
Sm has six unpaired electrons.
The partial orbital diagram is:
26. 23-26
Coordination Compounds
A coordination compound contains at least one complex
ion, which consists of a central metal cation bonded to
molecules and/or anions called ligands.
The complex ion is associated with counter ions of
opposite charge.
The complex ion [Cr(NH3)6]3+ has a central Cr3+ ion bonded to six
NH3 ligands. The complex ion behaves like a polyatomic ion in
solution.
27. 23-27
Coordination Number
The coordination number is the number of ligand atoms
bonded directly to the central metal ion.
Coordination number is specific for a given metal ion in a
particular oxidation state and compound.
The most common coordination number in complex ions is
6, but 2 and 4 are often seen.
- [Cr(NH3)6]3+ has a coordination number of 6.
28. 23-28
Figure 23.7 Components of a coordination compound.
[Co(NH3)6]Cl3 dissolves in water. The six
ligands remain bound to the complex ion.
[Pt(NH3)4]Br2 has four
NH3 ligands and two Br-
counter ions.
30. 23-30
Ligands
The ligands of a complex ion are molecules or anions
with one or more donor atoms.
Each donor atom donates a lone pair of electrons to
the metal ion to form a covalent bond.
Ligands are classified in terms of their number of donor
atoms, or “teeth”:
- Monodentate ligands bond through a single donor atom.
- Bidentate ligands have two donor atoms, each of which bonds to
the metal ion.
- Polydentate ligands have more than two donor atoms.
32. 23-32
Chelates
Bidentate and polydentate ligands give rise to rings in
the complex ion.
A complex ion containing this type of structure is called a
chelate because the ligand seems to grab the metal ion
like claws.
EDTA has six donor atoms and forms very stable complexes
with metal ions.
33. 23-33
Formulas of Coordination Compounds
• A coordination compound may consist of
– a complex cation with simple anionic counterions,
– a complex anion with simple cationic counterions, or
– a complex cation with complex anion as counterion.
• When writing the formula for a coordination compound
– the cation is written before the anion,
– the charge of the cation(s) is/are balanced by the charge of the
anion(s), and
– neutral ligands are written before anionic ligands, and the
formula of the whole complex ion is placed in square brackets.
34. 23-34
Determining the Charge of the Metal Ion
The charge of the cation(s) is/are balanced by the charge
of the anion(s).
K2[Co(NH3)2Cl4] contains a complex anion.
The charge of the anion is balanced by the two K+ counter ions, so
the anion must be [Co(NH3)2Cl4]2-.
There are two neutral NH3 ligands and four Cl- ligands. To have an
overall charge of 2-, the metal ion must have a charge of 2+.
Charge of complex ion = charge of metal ion + total charge of ligands
2- = charge of metal ion + [(2 x 0) + (4 x -1)]
Charge of metal ion = (-2) – (-4) = +2 or 2+
The metal ion in this complex anion is Co2+.
35. 23-35
[Co(NH3)4Cl2]Cl contains a complex cation.
The charge of the cation is balanced by the Cl- counter ion, so the
cation must be [Co(NH3)4Cl2]+.
There are four neutral NH3 ligands and two Cl- ligands. To have an
overall charge of 1+, the metal ion must have a charge of 3+.
Charge of complex ion = charge of metal ion + total charge of ligands
1+ = charge of metal ion + [(4 x 0) + (2 x 1-)]
Charge of metal ion = (+1) – (2-) = +3 or 3+
The metal ion in this complex cation is Co3+.
36. 23-36
Naming Coordination Compounds
• The cation is named before the anion.
• Within the complex ion, the ligands are named in alphabetical order
before the metal ion.
– Anionic ligands drop the –ide and add –o after the root name.
• A numerical prefix is used to indicate the number of ligands of a
particular type.
– Prefixes do not affect the alphabetical order of ligand names.
– Ligands that include a numerical prefix in the name use the
prefixes bis (2), tris (3), or tetrakis (4) to indicate their number.
• A Roman numeral is used to indicate the oxidation state for a metal
that can have more than one state.
• If the complex ion is an anion, we drop the ending of the metal name
and add –ate.
37. 23-37
Table 23.7 Names of Some Neutral and Anionic Ligands
Neutral Anionic
Name Formula Name Formula
Aqua H2O Fluoro F-
Ammine NH3 Chloro Cl-
Carbonyl CO Bromo Br-
Nitrosyl NO Iodo I-
Hydroxo OH-
Cyano CN-
38. 23-38
Table 23.8 Names of Some Metal Ions in Complex Anions
Metal Name in Anion
Iron Ferrate
Copper Cuprate
Lead Plumbate
Silver Argentate
Gold Aurate
Tin Stannate
39. 23-39
Sample Problem 23.3
PLAN: We use the rules for writing formulas and names of
coordination compounds.
Writing Names and Formulas of
Coordination Compounds
PROBLEM:
(a) What is the systematic name of Na3[AlF6]?
(b) What is the sytematic name of [Co(en)2Cl2]NO3?
(c) What is the formula of tetraamminebromochloroplatinum(IV) chloride?
(d) What is the formula of hexaamminecobalt(III) tetrachloroferrate(III)?
SOLUTION:
(a) The complex ion is [AlF6]3-. There are six (hexa-) F- ions (fluoro)
as ligands. The complex ion is an anion, so the ending of the
metal name must be changed to –ate. Since Al has only one
oxidation state, no Roman numerals are used.
sodium hexafluoroaluminate
40. 23-40
(b) There are two ligands, Cl- (chloro) and en (ethylenediamine).
The ethylenediamine ligand already has a numerical prefix in
its name, so we indicate the two en ligands by the prefix bis
instead of di.
Sample Problem 23.3
The complex ion is a cation, so the metal name is unchanged,
but we need to specify the oxidation state of Co. The counter
ion is NO3
-, so the complex ion is [Co(en)2Cl2]+.
Charge of complex ion = charge of metal ion + total charge of ligands
1+ = charge of metal ion + [(2 x 0) + (2 x 1-)]
Charge of metal ion = (+1) – (-2) = +3 or 3+
The ligands must be named in alphabetical order:
dichlorobis(ethylenediamine)cobalt(III) nitrate
41. 23-41
Sample Problem 23.3
(c) The central metal ion is written first, followed by the neutral
ligands and then (in alphabetical order) by the negative ligands.
Charge of complex ion = charge of metal ion + total charge of ligands
= (4+) + [(4 x 0) + (1 x 1-) + (1 X 1-)]
= +4 + (-2) = +2 or 2+
We will therefore need two Cl- counter ions to balance the charge on the
complex ion.
[Pt(NH3)4BrCl]Cl2
42. 23-42
Sample Problem 23.3
(d) This compound consists of two different complex ions. In the
cation, there are six NH3 ligands and the metal ion is Co3+, so
the cation is [Co(NH3)6]3+.
The anion has four Cl- ligands and the central metal ion is Fe3+,
so the ion is [FeCl4]-.
The charge on the cation must be balanced by the charge on
the anion, so we need three anions for every one cation:
[Co(NH3)6][FeCl4]3
43. 23-43
Constitutional Isomers of Coordination
Compounds
Compounds with the same formula, but with the atoms
connected differently, are constitutional isomers.
Coordination isomers occur when the composition of
the complex ion, but not the compound, is different.
- This can occur by the exchange of a ligand and a counter ion, or be
by the exchange of ligands.
Linkage isomers occur when the composition of the
complex ion is the same but the ligand donor atom is
different.
- Some ligands can bind to the metal through either of two donor
atoms.
44. 23-44
Figure 23.8 A pair of linkage (constitutional) isomers
The nitrite ion can bind either through the N atom or either one of
the O atoms.
46. 23-46
Stereoisomers of Coordination
Compounds
Stereoisomers are compounds that have the same atomic
connections but different spatial arrangements of their
atoms.
Geometric or cis-trans isomers occur when atoms or
groups can either be arranged on the same side or on
opposite sides of the compound relative to the central
metal ion.
Optical isomers (enantiomers) are non-superimposable
mirror images of each other.
47. 23-47
Figure 23.9A Geometric (cis-trans) isomerism.
In the cis isomer, identical ligands are adjacent to each other, while in
the trans isomer they are across from each other.
The cis and trans isomers of [Pt(NH3)2Cl2].
The cis isomer (cisplatin) is an antitumor agent while the trans
isomer has no antitumor effect.
48. 23-48
Figure 23.9B Geometric (cis-trans) isomerism.
The cis and trans isomers of [Co(NH3)4Cl2]+.
Note the placement of the Cl- ligands (green spheres).
49. 23-49
Figure 23.10A Optical isomerism in an octahedral complex ion.
Structure I and its mirror image, structure II, are optical isomers
of cis-[Co(en)2Cl2]+.
50. 23-50
Figure 23.10B Optical isomerism in an octahedral complex ion.
The trans isomer of [Co(en)2Cl2]+ does not have optical isomers.
Structure I can be superimposed on its mirror image, structure II.
52. 23-52
Sample Problem 23.4 Determining the Type of Stereoisomerism
PROBLEM: Draw stereoisomers for each of the following and state the
type of isomerism:
(a) [Pt(NH3)2Br2] (square planar) (b) [Cr(en)3]3+ (en = H2NCH2CH2NH2)
PLAN: We determine the geometry around each metal ion and the
nature of the ligands. If there are different ligands that can be
placed in different positions relative to each other, geometric
(cis-trans) isomerism occurs. Then we see whether the mirror
image of an isomer is superimposable on the original. If it is
not, optical isomerism also occurs.
53. 23-53
Sample Problem 23.4
SOLUTION:
(a) The square planar Pt(II) complex has two different types of
monodentate ligands. Each pair of ligands can be next to
each other or across from each other. Thus geometric
isomerism occurs.
These are geometric isomers; they do not have optical isomers
since each compound is superimposable on its mirror image.
54. 23-54
Sample Problem 23.4
(b) Ethylenediamine (en) is a bidentate ligand. The Cr3+ ion has
a coordination number of 6 and an octahedral geometry, like
Co3+. The three bidentate ligands are identical, so there is
no geometric isomerism. However, the complex ion has a
nonsuperimposable mirror image. Thus optical isomerism
occurs.
55. 23-55
Bonding in Complex Ions
In terms of valence bond theory, the filled orbital of the
ligand overlaps with an empty orbital of the metal ion.
When a complex ion is formed, each ligand donates an
electron pair to the metal ion.
The ligand acts as a Lewis base, while the metal ion acts as a Lewis
acid.
This type of bond is called a coordinate covalent bond
since both shared e- originate from one atom in the pair.
The VB model proposes that the geometry of the complex
ion depends on the hybridization of the metal ion.
59. 23-59
Figure 23.15 An artist’s wheel.
Each color has a complementary
color; the one opposite it on the
artist’s wheel.
The color an object exhibits depends
on the wavelengths of light that it
absorbs.
An object will have a particular color because
• it reflects light of that color, or
• it absorbs light of the complementary color.
60. 23-60
Table 23.9 Relation Between Absorbed and Observed Colors
Absorbed Color λ (nm) Observed Color λ (nm)
Violet 400 Green-yellow 560
Blue 450 Yellow 600
Blue-green 490 Red 620
Yellow-green 570 Violet 410
Yellow 580 Dark blue 430
Orange 600 Blue 450
Red 650 Green 520
61. 23-61
Crystal Field Theory
Crystal field theory explains color and magnetism in
terms of the effect of the ligands on the energies of the
d-orbitals of the metal ion.
The bonding of the ligands to the metal ion cause the
energies of the metal ion d-orbitals to split.
Although the d-orbitals of the unbonded metal ion are equal in
energy, they have different shapes, and therefore different
interactions with the ligands.
The splitting of the d-orbitals depends on the relative
orientation of the ligands.
62. 23-62
Figure 23.16 The five d-orbitals in an octahedral field of ligands.
The ligands approach along the x, y and z axes. Two of the
orbitals point directly at the ligands, while the other three
point between them.
63. 23-63
Figure 23.17 Splitting of d-orbital energies in an octahedral
field of ligands.
The d orbitals split into two groups. The difference in energy
between these groups is called the crystal field splitting energy,
symbol Δ.
64. 23-64
Figure 23.18 The effect of ligands and splitting energy on
orbital occupancy.
Weak field ligands lead to
a smaller splitting energy.
Strong field ligands lead to a
larger splitting energy.
65. 23-65
Figure 23.19 The color of [Ti(H2O)6]3+.
The hydrated Ti3+ ion is purple.
Green and yellow light are absorbed
while other wavelengths are
transmitted. This gives a purple color.
66. 23-66
Figure 23.19 The color of [Ti(H2O)6]3+.
When the ion absorbs light, electrons can move from the lower t2g
energy level to the higher eg level. The difference in energy
between the levels (Δ) determines the wavelengths of light
absorbed. The visible color is given by the combination of the
wavelengths transmitted.
67. 23-67
The Colors of Transition Metal Complexes
The color of a coordination compound is determined
by the Δ of its complex ion.
For a given ligand, the color depends on the oxidation
state of the metal ion.
For a given metal ion, the color depends on the ligand.
68. 23-68
Figure 23.20 Effects of oxidation state and ligand on color.
[V(H2O)6]2+ [V(H2O)6]3+
[Cr(NH3)6]3+ [Cr(NH3)5Cl ]2+
A change in oxidation state
causes a change in color.
Substitution of an NH3 ligand
with a Cl- ligand affects the
color of the complex ion.
69. 23-69
Figure 23.21 The spectrochemical series.
I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2
- < CN- < CO
WEAKER FIELD STRONGER FIELD
LARGER D
SMALLER D
LONGER SHORTER
As Δ increases, shorter wavelengths (higher energies) of light
must be absorbed to excite electrons. For reference H2O is
considered a weak-field ligand.
70. 23-70
Sample Problem 23.5
SOLUTION:
Ranking Crystal Field Splitting Energies (Δ)
for Complex Ions of a Metal
PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3- in
terms of Δ and of the energy of visible light absorbed.
PLAN: The formulas show that Ti has an oxidation state of +3 in all
three ions. From Figure 23.21, we tank the ligands by crystal
field strength: the stronger the ligand, the greater the
splitting, and the higher the energy of light absorbed.
The ligand field strength is CN- > NH3 > H2O, so the
relative size of Δ and energy of light absorbed will be
[Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+
71. 23-71
The Magnetic Properties of Transition
Metal Complexes
Magnetic properties are determined by the number of
unpaired electrons in the d orbitals of the metal ion.
Hund’s rule states that e- occupy orbitals of equal energy
one at a time. When all lower energy orbitals are half-
filled:
The number of unpaired e- will depend on the relative
sizes of Epairing and Δ.
- The next e- can enter a half-filled orbital and pair up by
overcoming a repulsive pairing energy, (Epairing).
- The next e- can enter an empty, higher, energy orbital by
overcoming Δ.
73. 23-73
Figure 23.23 Orbital occupancy for high-spin and low-spin
octahedral complexes of d4 through d7 metal ions.
high spin:
weak-field
ligand
low spin:
strong-field
ligand
high spin:
weak-field
ligand
low spin:
strong-field
ligand
74. 23-74
Sample Problem 23.6 Identifying High-Spin or Low-Spin Complex
Ions
PROBLEM: Iron (II) forms a complex in hemoglobin. For each of the
two octahedral complex ions [Fe(H2O)6]2+ and [Fe(CN)6]4,
draw an energy diagram showing orbital splitting, predict
the number of unpaired electrons, and identify the ion as
low spin or high spin.
PLAN: Fe2+ electron configuration shows the number of d
electrons, and the spectrochemical series shows the
relative ligand strengths. We draw energy diagrams and
separate the t2g and eg orbital sets more for the strong-field
ligand. Then we add electrons, noting that a weak-field
ligand gives the maximum number of unpaired electrons
and a high-spin complex, whereas the strong-field ligand
will give the minimum number of unpaired electrons and a
low-spin complex.
76. 23-76
Figure 23.24A Splitting of d-orbital energies by a tetrahedral
field of ligands.
The splitting of d-orbital energies is less in a tetrahedral than an
octahedral complex, and the relative d-orbital energies are
reversed. Only high-spin tetrahedral complexes are known
because Δ is small.
77. 23-77
Figure 23.24B Splitting of d-orbital energies by a square planar
field of ligands.
Square planar complexes are low-spin and usually diamagnetic
because the four pairs of d electrons fill the four lowest-energy orbitals.
78. 23-78
Figure B23.1 Hemoglobin and the octahedral complex in heme.
Chemical Connections
Hemoglobin consists of four protein chains, each with a bound
heme. In oxyhemoglobin (B), the octahedral complex in heme has
an O2 molecule as the sixth ligand for iron(II).
(Illustration by Irving Geis. Rights owned by Howard Hughes Medical Institute.
Not to be used without permission.)
