Download A2 CHEMISTRY TRANSITION METALS FAHAD

A2 CHEMISTRY
TRANSITION METALS
TRANSITION METALS
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Atoms that can form stable ions which have
incomplete 3d orbitals
o Zn does not qualify as transition
elements as Zn+2 = 1s2, 2s2, 2p6, 3s2, 3p6,
3d10, 4s0
o Sc does not qualify as transition
elements as Sc+3 = 1s2, 2s2, 2p6, 3s2, 3p6,
3d0, 4s0
Chemical Properties
o Variable Oxidation States
o Form colored compounds
o Form Metal Complexes
o Catalysts
Electronic Configuration
Atomic
Electronic
Number
Configuration
Sc
21
[Ar], 4s2, 3d1
Ti
22
[Ar], 4s2, 3d2
V
23
[Ar], 4s2, 3d3
Cr
24
[Ar], 4s1, 3d5
Mn
25
[Ar], 4s2, 3d5
Fe
26
[Ar], 4s2, 3d6
Co
27
[Ar], 4s2, 3d7
Ni
28
[Ar], 4s2, 3d8
Cu
29
[Ar], 4s1, 3d10
Zn
30
[Ar], 4s2, 3d10
 Cr and Cu promote electrons from lower
energy orbital (4s) to higher energy orbital
3d to form more stable half-filled and
completely filled orbitals.

Sc
+3

Element

Electrons are lost from the outer most 4s
electrons first when ions are formed.

Fe+3 = [Ar], 3d5, 4s0
Transition elements can have variable oxidation
states. 4s and 3d orbitals exist at approximately
the same energy levels, hence there is very little
difference in successive ionization energies.
Ti
+1
+2
+3
+4
V
+1
+2
+3
+4
+5
Cr
+1
+2
+3
+4
+5
+6
Mn Fe Co Ni
+1 +1 +1 +1
+2 +2 +2 +2
+3 +3 +3 +3
+4 +4 +4 +4
+5 +5 +5
+6 +6
+7
Underlined Oxidation States are more
stable than the rest
Cu
+1
+2
+3
From left-to-right period 2 (transition
metals), proton number increases which
would suggest IE should increase and
Atomic Radii should decrease. But, each
new electron gets added in the inner 3d
orbital which increases shielding and
decreases IE and increases Atomic Radii as
well as the outer 4s electrons are now held
less tightly
Transition Metals vs Calcium

Oxidation States
Why can’t Na have variable oxidation
states.
o Na = 1s2, 2s2, 2p6, 3s1 Na+1 can be
formed by removing 1 electron from Na,
but removing the second electron
becomes extremely difficult from the
lower energy and less shielded 2p orbital.
Invariant Atomic Radii and 1st Ionization
Energies
Transition Metal Ions

FAHAD H.AHMAD
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Melting Point: Higher MP for Transition
Metals due to Stronger Metallic Bonding as
more Electrons get delocalized.
Density: Atomic Radii of transition elements
are smaller than Ca and masses of elements
increase, hence transition elements have a
much higher density.
Atomic Radius: Atomic and Ionic Radius of
Transition elements are smaller due to
higher proton number, plus the increasing
Zn
+2
A2 CHEMISTRY

TRANSITION METALS
nuclear charge is poorly shielded by the
inner incomplete 3d orbitals
Conductivity: Conductivity increases for
transition metals as more electrons get
delocalized.
LIGAND: Ion or Molecule that has a lone pair of
electrons which can form a coordinate bond
with a metal.
List of Ligands
Degenerate-Orbitals (D-Orbitals)


FAHAD H.AHMAD
dxy, dyz, dxz orbitals lie between axis
dx2-y2 and dz2 lie along the axis
Monodentate Ligands donate only one lone pair
to form a single coordinate bond with central
metal ion.
Metal Complexes
Fe+3 has the following electronic configuration
3d
↑
↑
↑
↑
↑
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H2O:
:NH3
:OH-1
hexa aqua metal complex
hexa ammine complex
hexa hydroxo complex
4s
4p
4d
This ion has a high Charge Density due to its
smaller size. The empty 4th shell orbitals can
therefore attract ligands and datively bond to
them. The six orbitals shaded above can be
used to dative bonding. The number of dative
bonds formed depends on the charge density
and the size of the attached ligand. These six
orbitals then hybridize, and in this case this is
known as sp3d2 hybridization.
(The above ligands are smaller in size, hence six
ligands can fit around the central metal atom
forming octahedral complexes)
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:Cl-1 tetra chloro metal complex
:Br-1 tetra bromo metal complex
:I-1 tetra iodo metal complex
A2 CHEMISTRY
TRANSITION METALS
(The above ligands are larger in size, hence only
four ligands can fit around the central metal
atom forming tetrahedral complexes)
FAHAD H.AHMAD
Following shows an example of a hexadentate
ligand forming six coordinate bonds with the
central Cu+2 metal ion
Bidentate ligands donate two lone pairs to
form two coordinate bonds with the central
metal ion
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
Ethane dioate –OOCCOO1,2-diamino ethane NH2CH2CH2NH2
Transition Metals and Colored Metal
Complexes
Following shows an example of a bidentate
ligand forming coordinate bonds with a central
metal atom
Hexadentate Ligands donates six lone pairs to
form six coordinate bonds with the central
metal ion e.g.
Increased repulsion on the inner 3d orbital due
to coordinate bond formation of the outer 4s,
4p, 4d orbitals splits the 3d orbital into two sets
of two and three orbitals.
The type of splitting depends on the structure
of the metal complex formed. For an octahedral
complex (6 ligands), dx2 and dx2-y2 split to a
A2 CHEMISTRY
TRANSITION METALS
FAHAD H.AHMAD
higher energy levels. For tetrahedral complex,
dxy, dxz, dyz split to a higher energy level.
Violet-Blue-Green-Yellow-Orange-Red
Color
Blue
Cu+2 ion has a 3d9 orbital which splits as shown
below when it forms a hexa aqua complex with
water ligands. Electrons are unevenly
distributed in the two sets of 3d orbitals.
Electrons can now be promoted from the lower
energy 3d orbital set to the higher energy 3d
orbital set. A small amount of energy is required
for this promotion which can be obtained from
the visible spectrum.
green
violet
red
orange
yellow
The above colors are arranged in a circle. If
violet, blue, green, yellow, orange get absorbed,
you will see red reflected and so on.
LIGAND COLORS
Hexa Aqua Complexes
Other metal compounds
do not have split 3d
orbitals with a small
energy gap. If a sodium ion
promotes an electron from
the outer 2p orbital to the 3s orbital then a
huge amount of energy is required and a
photon this energetic cannot be obtained from
the visible spectrum, instead it gains that
energy from a photon from a higher frequency
photon. (E
= h*frequency)
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[V (H2O)6]+2 violet
[V (H2O)6]+3 green
[Cr (H2O)6]+3 blue/violet
[Mn (H2O)6]+2 light pink
[Fe (H2O)6]+2 light green
[Fe (H2O)6]+3 red/brown
[Co (H2O)6]+2 pink
[Ni (H2O)6]+2 green
[Cu (H2O)6]+2 blue
[Zn (H2O)6]+2 colorless
The energy gap between the 3d orbital sets is
determined by the strength of the ligand field
(repulsion exerted by the ligand) and the
geometry and charge density of the complex
ion. Hence each complex ion absorbs a different
photon and form different colored compounds.
Tetra Ammine Complex
Zn+2 has complete 3d orbital, hence even orbital
splitting does not lead to colored compounds.
[Cu (H2O)6]+2 blue solution
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[Co (NH3)6]+2 brown
[Ni (NH3)6]+2 blue
[Cu (NH3)4 (H2O)2]+2 deep blue
Ammonia Ligand Exchange for Copper ions
If small amount of ammonia is added
COMPLEMENTARY COLORS
The colors that we see are the complementary
colors, A metal complex absorbs a certain
wavelegnth and we see the reflected white light
with a missing wavelength. Following is a list of
complementary colors
NH3 + H2O  NH4+1 + OH-1
OH-1 ions produced replace the H2O ligands
+2
-1
[Cu (H2O)6] + 2OH  [Cu (OH)2 (H2O)4] + 2H2O
Which gives a pale blue ppt of [Cu (OH)2 (H2O)4]
A2 CHEMISTRY
TRANSITION METALS
If ammonia is added in excess then NH3
ligands replace the OH-1 and H2O ligands
FAHAD H.AHMAD
If small amount of ammonia is added
NH3 + H2O  NH4+1 + OH-1
+2
[Cu (OH)2 (H2O)4] + NH3 ↔ [Cu (NH3)4 (H2O)2] +
2OH-1 + 2H2O
OH-1 ions produced replace the H2O ligands
+2
[Cu (NH3)4 (H2O)2]+2 gives a deep blue solution
Chloro Ligand Exchange for Cu+2 ions
When concentrated HCl is added to blue
solution of [Cu (H2O)6]+2
-1
[Ni (H2O)6] + 2OH  [Ni (OH)2 (H2O)4] + 2H2O
Which gives a grey-green ppt of [Cu (OH)2 (H2O)4]
If ammonia is added in excess then NH3
ligands replace the OH-1 and H2O ligands
+2
+2
-1
-2
[Cu (H2O)6] + Cl ↔ [CuCl4] + 6H2O
4H2O
-2
Yellow/green ppt of [CuCl4] is obtained which
changes to blue solution when excess water is
added which favours the reverse reaction
LIGAND EXCHANGE IN COBALT
[Co (NH3)6]+2 + H2O ↔ [Co (H2O)6]+2 + NH3
brown
pink
Above reversible reaction shifts to the right
or to the left depending on the
concentration of ammonia
[Co (H2O)6]+2+4Cl-1 ↔ [CoCl4]-2+6H2O
pink
blue
The above equilibrium shifts according to
the concentration of Chloride ions
Ligand Exchange for Ni+2
Green Solution of Ni+2 forms a grey-green
precipitate when a small amount of
ammonia is added, the precipitate dissolves
in excess of ammonia to give a blue violet
solution
[Ni (H2O)6]+2 green solution
-1
[Ni (OH)2 (H2O)4] + NH3 ↔ [Ni (NH3)6] + 2OH +
[Ni (NH3)6]+2 gives a blue violet solution