Transcript The chemistry of the transition metals

The chemistry of the transition
metals
Chapter 22
1
Electron configurations
• Let’s write the 3d metals’ and their ions’ econfigurations
• Sc, V, Cr, Fe, Ni, Cu, Zn
2
Atomic size
• Atomic size decreases across
•
•
period & increases down a column
Not a big change to the radii
across the row
– Due to e-’s being stable in
outermost orbital (4s) while
adding to 3d
But 3rd row roughly same as 2nd,
not larger
– Due to e-’s going into 4f
– The f-subshell is ineffective at
shielding outer e-’s from
nuclear charge
• Outer e-’s held more
tightly by nucleus
– Called lanthanide
contraction
3
Ionization energy
• Increase across row
• But increase smaller than
•
for main-group elements
Also, 3rd transition row
has higher ionization E
(generally) than first 2
rows
– Runs counter to maingroup elements
• Due to outer e-’s being
held more tightly
4
Electronegativity
• Increase across row, same as main-group
• However, increase going down a group
– Conversely, main-group decrease going down a group
• 1st & 2nd rows different, but 2nd & 3rd same
– Due to small change in atomic size going down group
w/large increase in nuclear charge
• Au = 2.4 !
• Au- found to exist!
5
Oxidation states
• Variable
– Up to +8 in Os & Ru
• Re has widest range: -3  +7 !
6
Coordination compounds
• Can form complex ions
– Ion containing central metal ion bound to one
or more ligands
• Lewis base (or e- donor) that forms bond w/metal
• When complex ion combines w/counterions (non-ligands) they yield a neutral
compound
– Coordination compound
7
Its history
• Alfred Werner
• Central metal ion has 2 types of interactions:
• 1. primary valence: oxidation state on central
•
•
metal atom
2. secondary valence: # of molecules/ions
bound to metal atom, called coordination #.
Ex: CoCl3  6NH3 has primary valence of 3 and
coordination # of 6
8
More
• Can think of metalligand complex as
Lewis acid/base
adduct
– Since ligand donates epair to empty orbital
on metal
• Called coordinate
covalent bond
9
More on ligands
• Those that donate one e- pair = monodentate
•
•
(why is it CO and not OC?)
Those that donate 2 = bidentate
More than 2 = polydentate
• Latter two called chelates
• Coordinating ligand = chelating agent
10
Naming coordination compounds
1. Name ligands
– Neutral ligands named as molecules except
for H2O (aqua), NH3 (ammine), and CO
(carbonyl)
– Anionic ligands have suffix changes:
• -ide  -o
fluoride  fluoro
• -ate  -ato
sulfate  sulfato
• -ite  -ito
nitrite  nitrito
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Naming coordination compounds
2. List names of ligands in alphabetical order
before name of metal cation
–
For example: ammine comes before bromo
3. Use prefix to indicate # of ligands
–
–
–
–
–
2
3
4
5
6





di
tri
tetra
penta
hexa
diammine
triiodo
tetranitrito
12
Naming coordination compounds
– If the ligand name already has a number
prefix, put () around ligand
– 2  bis
bis(ethylenediamine)
– 3  tris
tris(edta)
– 4  tetrakis
– Prefixes don’t affect order in which ligands
listed
13
Naming coordination compounds
4. Name metal
– When complex ion is cation, use name of
metal followed by oxidiation state w/Roman
numeral
• Pt2+ = platinum (II)
– If complex ion anionic, drop ending of metal
• Add –ate followed by oxid. state w/Roman
numeral
– Pt2+ = platinate(II)
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Naming coordination compounds
5. Write entire name by listing ligands first then metal
–
Chemical formula:
•
Symbol of metal first, then neutral molecules and anions last (all
one word!)
–
–
–
Hexaamminecobalt(III) ion= [Co(NH3)6]3+
Diamminetetrachloroplatinate(II) ion = [Pt(NH3)2Cl4]2-
If more than one anion or neutral molecule in ligand, list in
alphabetical order based on chemical symbol
•
•
[Mn(CO)(NH3)5]SO4 is correct
[Mn(NH3)5(CO)]SO4 is incorrect
–
SO4 is called a “counter-ion”
 There should be a space between the complex ion and the
counter-ion
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Practice naming
• [Cr(H2O)5Cl]Cl2
• K3[Fe(CN)6]
• Na2[PtCl4]
• [Mn(CO)(NH3)5]SO4
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Practice writing chemical formulae
• Tetraaquaplatinum(II)
hexachloroplatinate(IV)
• Ammonium
diaquatetrabromovanadate(III)
• Tris(ethylenediamine)cobalt(III)
trioxalatoferrate(III)
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Structure and isomerization
• 3 categories
1. Structural isomers: atoms connected in different
ways
1. Coordination isomers
2. Linkage isomers
2. Geometric isomers: ligands have different spatial
arrangement
1. Cis-trans isomers
2. Octahedral complex isomers
3. Optical isomers: nonsuperimposable mirror-images
(enantiomers)
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Structural isomerism: coordination
isomers
• Coordination isomers
– Coordination ligand exchanges places
w/uncoordinated counter-ion
• Ex: [Co(NH3)5Br]Cl vs. [Co(NH3)5Cl]Br
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Structural isomerism: linkage
isomers
• Either one of atoms in NO2- can bond to
metal
– When O, nitrito: ONO– When N, nitro: NO2-
• Different color compounds (page 1089)
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Geometric isomerism: cis-trans
isomers
• Occurs in sq-planar: MA2B2
• And octahedral complexes: MA4B2
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Geometric isomerism: octahedral
complex isomers
• MX3Y3
• Fac (facial) isomer
– Three identical ligands at corners of a
triangular face of octahedron
• Mer (meridian) isomer
– Three identical ligands at corners of a
triangular meridian (inside octahedron)
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Optical isomerism
• Nonsuperimposable mirror-images
– Enantiomers
• Have chirality/chiral center
• Rotation of plane-polarized light in
opposite directions
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Geometries
• Depend in part on coordination #
• Coordination # Shape Example
•2
linear
[Ag(NH3)2]+
• 4 d8
sq planar [PdCl4]2• 4 d10
tetrahedral [Zn(NH3)4]2+
•6
octahedral
[Fe(H2O)6]3+
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Bonding in coordination compounds
• Valence Bond Theory: hybridization of
orbitals (CHEM&141)
• Filled orbital on ligand
• Empty orbital on metal cation
– Lead to hybridized orbitals
25
Hybridizations of complex ions
Geometry
• Linear
• Tetrahedral
• Square planar
• Octahedral
Hybridization
sp
sp3
dsp2
d2sp3
26
Crystal Field Theory
• VB theory can’t explain color and magnetism of
•
coordination compounds
Basically, complex ions form due to attractions
between e- on ligands and positive charge on
metal ion
– Yet, e- on ligands repel e- in unhybridized metal dorbitals
• Crystal Field Theory (CFT) focuses on these
repulsions
27
Ligands attack on x, y, z axes
• Notice how ligands interact most strongly with top two orbitals’ lobes
– Greater interactions
• Higher repulsions
• In below three, orbitals lie between axes & have nodes directly on axes
– Less interactions
• Lower repulsions
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More…
• D-orbitals end up being split into low- and highE orbitals
– Based on spatial arrangement of ligands
• High-E orbitals are top two (see previous figure)
• Low-E orbitals are bottomw three (see previous figure)
• Diff in E between split d-orbitals
– Crystal field splitting E
– =∆
• Magnitude of ∆ depends on particular complex (mostly its
ligands)
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Octahedral geometry
30
Tetrahedral
• Based on how ligands interact with orbital
lobes
31
Square planar
• Again, based on how ligands overlap with
orbital lobes
32
Comparing all 3 ligand geometries
33
Calculate ∆ (crystal field splitting E)
• Using Ephoton = h = hc/ = ∆
• [Ti(H2O)6]3+ has max abs @ 498 nm
• What is the CFSE in kJ/mol?
• Calculate it!
34
Solution
E photon = h =
hc

=
34
8m
(6.626

10
J

s)(3.00

10
)
hc
s
=

498 109 m
-19
-19
-22
  3.99 10 J = 3.99 10 J/ion = 3.99 10 kJ/ion
3.99 10-22 kJ
 6.022 1023 ions
 240 kJ
ion
mol
mol
35
Magnitude of CFSE
• Depends largely on ligands involved
– Spectroscopic studies of various ligands
attached to same metal
• Yield Spectrochemical series
– Arranged from ligands that result in largest ∆
to smallest
36
Spectrochemical series
• CN- > NO2- > en > NH3 > (typically
strong-field ligands)
• H2O > OH- > F- > Cl-> Br-> I- (typically
weak-field ligands)
• Large ∆ = strong-field ligands
• Small ∆ = weak-field ligands
37
More
• Also, as oxidation state increases, so does
∆
– Draws ligands closer to metal, increases nodal
repulsions
38
Magnetic properties
• Magnitude of ∆
•
determines d-orbital
occupancy and # of
unpaired e-’s
Large ∆ = strong-field
ligands:
– Low-spin
complexes
• Small ∆ = weak-field
ligands:
– High-spin
complexes
39
Magnetic properties
• Low-spin complexes tend to be diamagnetic
• High-spin complexes tend to be paramagnetic
• d1, d2, d3, d8, d9, and d10 do not rely on ligand splitting
strength
• Why?
• Practice:
• How many unpaired e-’s would one expect for [FeCl6]3-?
• How many unpaired e-’s would one expect for
[Co(CN)6]4-?
40
Color of complex ions
• The color wheel: absorption of color
appears as complementary color
41
Color of complex ions
• Color in causes lower d-orbital e- to go up
to higher d-orbital state
– Specific wavelength of light kicked out
• The complement of color absorbed
• Colorless complexes are either d0 or d10
– Don’t have d-orbital e-’s to move up
42
Problem
• Two ligands form complexes with same
metal ion. The first ligand, A, complexes a
red soln, while the second ligand, B,
complexes a yellow soln.
• Which ligand produces the larger ∆?
43