Coordination compounds (12th Maharashtra state board)

Coordination compunds
Presented by: Freya Cardozo
1
Presented by
Freya Cardozo

Central
metal atom
or ion
Coordination
compound
Ligand
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Coordination compunds
• Donates electrons = Ligand and
• Accepts the donated electrons= Central metal atom or ion
• Eg. Cis platin (cancer drug) here,
• Pt is Central metal atom
• Cl- and N are donors of lone pair
• A formation of coordinate bond occurs when the shared
electron pair is contributed by ligands.
• A coordinate bond is conveniently represented by an arrow →,
where the arrow head points to electron acceptor.Presented by: Freya Cardozo
3

Ligands
• The ligands are linked directly to central metal ion through coordinate
bonds.
• A formation of coordinate bond occurs when the shared electron pair is
contributed by ligands.
• Classification of ligands based on the number of donor atoms
1. Monodentate ligands
2. Polydentate ligands – Bidentate ligand and hexadentate ligands
3. Ambidentate ligands
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• A monodentate ligand is the one where a single donor atom shares an
electron pair to form a coordinate bond with the central metal ion.
Eg. Cl-, H2O, NH3
MONODENTATE LIGANDS
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Bidentate ligands
• Bidentate ligands are those that have 2 donor atoms in a single
molecule.
• Eg. Oxalate and ethylene diamine.
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Hexadentate Ligands
• Hexadentate L- six. Eg. EDTA (ethylene diamine tertraacetate)
• Four oxygen and 2 nitrogen donor atoms
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Ambidentate ligands
• Ambidentate ligands : The ligands which have two donor atoms
and use the electron pair of either donor atoms to form a
coordinate bond with the metal ion, are ambidentate ligands.
• For example, the ligand NO2 links to metal ion through nitrogen
or oxygen.
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Difference between
Double salts
• Dissociates completely into ions
when dissolved in water
• Eg. Mohr’s salt ans Carnalite
Coordination compounds
• Dissociates with atleast one complex
ion and counter ions
• Eg. K4[Fe(CN)6]
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Werners postulates
1. Metal has two types of valancies
• Primary valency -ionizable valency
• Secondary valency – non ionizable valency
2. The ionizable sphere consists of entities which satisfy the primary
valency of the metal. Primary valencies are generally satisfied by anions.
3. The secondary coordination sphere consists of entities which satisfy the
secondary valencies and are non ionizable. The secondary valencies for a
metal ion are fixed and satisfied by either anions or neutral ligands.
Number of secondary valencies is equal to the coordination number.
4. The secondary valencies have a fixed spatial arrangement around the
metal ion
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Example
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Coordination sphere
• The central metal ion and ligands linked to it are enclosed in a square
bracket. This is called a coordination sphere, which is a discrete
structural unit. When the coordination sphere comprising central
metal ion and the surrounding ligands together carry a net charge, it
is called the complex ion.
• The ionisable groups shown outside the bracket are the counter ions.
• For example, the compound K4[Fe(CN)6] has [Fe(CN)6]4-coordination
sphere with the ionisable K⊕ ions representing counter ions
CS=
Ligand+Central
metal atom/ionPresented by: Freya Cardozo
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O.S and C.N
• PO – primary valency oxidation state
• [Fe(CN)6]4 has charge number of -4. It can be utilised to calculate O.S. of Fe
• C.N. of metal ion in a complex is the number of ligand donor atoms directly
attached to it or the number of electron pairs involved in the coordinate
bond.
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Solve
One mole of a purple coloured complex CoCl3 and NH3 on treatment
with excess AgNO3 produces two moles AgCl. Write the formula of
the complex if the coordination number of Co is 6.
• If Cl ions are free only
then they will reacts with
AgNO3 and get
precipitated
• If NH3 is not evolved
means it is tightly bound
• Coordination
Number=No.of ligands
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Classifications
Classification on the basis of types of ligands
1. Homoleptic Complexes
2. Heteroleptic complexes
• Classification on the basis of charge on the complex
1. Anionic complexes
2. Cationic complexes
3. Neutral Complexes
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O.S you must know
• F, Cl, Br, I = -1
• NO3, NO2, ONO =-1
• CN, SCN,NSC = -1
• SO4, CO3,C2O4 =-2
• Na, K = 1
• NH3, H2O, en, CO= 0
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IUPAC nomenclature
• For cationic and neutral Complexes always name the ligand first followed by the Central metal
atom or ion
• For anionic complexes name the counterion first then give a space n name ligands
• For anionic complexes the Metal atom name should end with suffix –ate
• The names of anionic ligands are obtained by changing the ending -ide to -o and -ate to -ato.
• If there are more than one ligands write acc to alphabetical order
• if One ligand is Present more than two give prefix di,tri,tetra
• The oxidation of metal atom should be written at the end in () using roman numerals
• If the name of ligand itself contains numerical prefix then display number by prefixes with bis for
2, tris for 3, tetrakis for 4 and so forth. Put the ligand name in parentheses. Forexample,
(ethylenediamine)3 or (en3) would appear as tris(ethylenediamine) or tris(ethane-1, 2-diamine).
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18

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Write IUPAC names
• [Ni(CN4)]2-
• [Fe(CN)6]4-
• Na3[Co(NO2)6]
• Na3[AlF6]
• [Cu(NH3)4]2+
• [Fe(H20)5NCS] 2+
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Write formula
• Potassium hexa cyanoferrate (II)
• Aquachlorobis(ethylenediamine)cobalt(III).
• Tetraaquadichlorochromium(III) chloride.
• Diamminedichloroplatinum(II).
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EAN(effective atomic
number)
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EAN
• Proposed by sidgwik to explain stability of complexes
• EAN equals total number of electrons around the central metal ion in the
complex
• rule states that a metal ion continues to accept electrons pairs till it attains
the electronic configuration of the next noble gas. Thus if the EAN is equal to 18
(Ar), 36 (Kr), 54 (Xe), or 86 (Rn) then the EAN rule is obeyed.
• Formula
EAN= Z-X+Y
Z= Atomic number of Central metal atom or ion
X= O.S of Central metal atom or ion
Y= number of electrons donated by ligands
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Calculate EAN
• Co[NH3]6
3⊕
• [Fe(CN)6]4-
• Mn(CO)
Co 27 Mn 25 Fe 26
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Isomerisms
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Constitutional or structural
isomers
1. linkage isomers
2. ionization isomers,
3. coordination isomers
4. solvate isomers.
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Linkage isomer
• Formed by Ambidentate ligands
• ligand with 2 do or atom.
• Eg. [Co(NH3)5(NO2)]2⊕ and [Co(NH3)5(ONO)]2⊕
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Ionisation Coordination
• Ionization isomers involve exchange
of ligands between coordination and
ionization spheres
[Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4
• Coordination isomers show interchange
of ligands between cationic and anionic
spheres of different metal ions.
[Co(NH3)6] [Cr(CN)6] and
[Cr(NH3)6] [Co(CN)6]
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Solvate isomers/ Hydrate isomers
• Both complexes differ in the number of water molecules inside
coordination sphere and hydration sphere
These are similar to
ionization isomers.
[Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2.H2O
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Stereoisomers
• Stereoisomers have the same links among constituent atoms
however the arrangements of atoms in space are different.
• These are of two types
1. Geometric isomers or distereoisomers
2. Optical isomers or enantiomers
• Enantiomers and distereoisomers are non superimposable mirror
images of each other.
• Difference is that enantiomers are chiral or optically active
whereas Di are inactive
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Geometric isomers or distereoisomers
• Seen in heteroleptic complexes
• 2 types i.e cis and trans
• same ligands on one side cis and opposite sides trans
1. Square planar complexes (CN=4)
2. Octahedral complexes (CN=6)
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Cis trans isomers in square planar complexes
• MA2B2 and MA2BC types A,B,C are Monodentate ligands
• The arrangment of cis and trans for following complex can be drawn as
follows [Pt(NH3)2Cl2]
• The cis isomer is used as an anticancer drug called cis platin. It is
physiologically active, polar with non-zero dipole Whereas trans isomer is
inactive
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Cis and trans isomers in octahedral
complexes
• octahedral complexes of the type MA4B2, MA4BC and M(AA)2B2 exist as cis and
trans isomers.
• (AA) is a bidentate ligand.
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[Fe(NH3)2(CN)4] draw geometrical isomers of
the same
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Optical isomers
• The complex molecules or ions that are nonsuperimposable mirror
images of each other are enantiomers.
• The nonsuperimposable mirror images are chiral.
• Enantiomers have identical properties however differ in their response
to the plane-polarized light. The enantiomer that rotates the plane of
plane-polarized light to right (clockwise) is called the dextro (d) isomer,
while the other that rotates the plane to left (anticlockwise) is called
laevo (l) isomer.
• Give reason : Square planar complexes do not show optical isomerism
Square planar complexes do not show enantiomers since they have mirror
plane and axis of symmetry.
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Factors affecting stability
• charge to size ratio of the metal ion
Higher the ratio greater is the stability. For the divalent metal ion
complexes their stability shows the trend : Cu2⊕ > Ni2⊕ > Co2⊕ > Fe2⊕
> Mn2⊕ > Cd2⊕. The above stability orderis called Irving-William order. In
the above list both Cu and Cd have the charge +2, however, the ionic
radius of Cu2⊕ is 69 pm and that of Cd2⊕ is 97 pm. The charge to size
ratio of Cu2⊕is greater than that of Cd2⊕ . Therefore the Cu2⊕forms
stable complexes than Cd2⊕.
• Nature of ligand
related to how easily the ligand can donate its lone pair of electrons to
the central metal ion that is, the basicity of the ligand. The ligands those
are stronger bases tend to form more stable complexes.
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38

VBT
(valence bond theory)
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Postulates of VBT
1. Metal ion provides vacant d orbitals for formation of coordinate bonds with
ligands.
2. The vacant d orbitals along with s and p orbitals of the metal ion take part in
hybridisation.
3. No.of hybrid orbitals= No of ligands= CN
4. Overlap between vacant d of metal and filled orbital of ligand forms
Coordinate bond
5. The hybrid orbitals used by the metal ion point in the direction of the ligand.
6. Complexs can be classified as inner and outer complexes depending upon
whether nd or (n-1)d orbitals are involved
7. If nd orbital used- outer spin complexes or else inner
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Geometry and magnetism
Coordination number Hybridisation Geometry of complex
2 sp Linear
4 sp3 Tetrahedral
4 dsp2 Square planar
6 Sp3d2 or spd23 Octahedral
All electrons paired= Diamagnetic
Unpaired electrons= Paramagnetic
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Strong field ligands and weak field ligands
THOSE LIGANDS THAT CAUSE PAIRING OF ELECTRONS ARE STRONG
FIELD LIGANDS
Eg. NH3, CO, CN
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Tetrahedral complexes
• Steps to follow
1. Identify central metal atom or ion and write its outer electronic
configuration
2. Calculate its O.S and then write ground state config with block diagram
3. Identify the number of ligands and electrons donated by them
4. Check if ligand is weak field or strong field(if strong-pair)
5. Calculate how many vacant orbitals you will require
6. write hybridisation and geometry
7. check for magnetism
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1. [Ni(Cl)4]2-
• Ni --- 28 [Ar] 3d8 4s2
• O.S --- +2
• [Ar] 3d8 4s0
• No of ligands=4 thus vacant
orbitals needed is 4
• Electrons donated = 8
• Hybridisation sp3
• tetrahedral
• Paramgnetic
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45

[Ni(CN)4]2-
Hint – CN is a strong filed ligand thus pairing will occur
Ni- 28 E.C
O.S
G.S now pair electrons
4 ligands thus 4 orbitals
Select vacant orbitals and hybridisation
Geometry
Magnetism
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Octahedral complexes
• Inner orbital complex ---- low spin
• Outer orbital complex ---- high spin
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48

[Co(NH3)6]3+
• Co 27
• [Ar] 3d7 4s2
• NH3 strong field ligand so pairing
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50

[CoF6]3-
• F is a weak field ligand so no pairing
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52

Limitations of VBT
1. It does not explain the high spin or low spin nature of the
complexes. In other words, strong and weak field nature of
ligands can not be distinguished.
2. It does not provide any explanation for the colour of coordination
compounds.
3. The structure of the complexes predicted from the VBT would
not always match necessarily with those determined from the
experiments.
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Application of C.C
• In biology
Chlorophyll in plants – complex of Mg
Hemoglobin in blood – complex of irom
• Medicine
Cis platin – cancer
EDTA – lead posining
• Determine hardness of water – using EDTA(hardness due to Ca2+ and Mg2+)
• Electroplating
When the coordination complexes are used the ligands in the complex keep the metal
atoms well separated from each other. These metal atoms tend to form a protective
layer on the surface. Certain cyanide complexes K[Ag(CN)2] and K[Au(CN)2] find
applications in the electroplating of these noble metals.
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Deleted topics
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Crystal field Theory
(CFT)
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Questions
• Give valence bond description for the bonding in the complex
[VCl4]- . Draw box diagrams for free metal ion. Which hybrid
orbitals are used by the metal ? State the number of unpaired
electrons.
57

Postulates
• The ligands are treated as point charges. The interaction between metal
ion and ligand is purely electrostatic or there are no orbital interactions
between metal and ligand
• When ligands approach the metal ion they create crystal-field area the
metal ion. If field created is symmetric then degenracy is maintained.
Mostly it is not symmetric and thus degenracy is destroyed.
• The d orbitals thus split into two sets namely, (dxy, dyz, dxz) usually
refered by t2g and ( d x2-y2 ,dz2) called as eg. These two sets of
orbitals now have different energies. A separation of energies of these
two sets of d orbitals is the crystal field splitting parameter. This is
denoted by Δo or Dq(O for octahedral).
• The Δo depends on strength of the ligands. The ligands are then
classified as (a) strong field and (b) weak field ligands
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• CN , NC , CO, NH3, EDTA, en (ethylenediammine) are considered to be strong
ligands. They cause larger splitting of d orbitals and pairing of electrons is
favoured. These ligands tend to form low spin complexes.
• Weak field ligands are those in which donor atoms are halogens, oxygen or
sulphur. For example, F , Cl , Br , I , SCN , C2O4
• In case of these ligands the Δo parameter is smaller compared to the energy
required for the pairing of electrons, which is called as electron pairing energy.
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60

Factors affecting Crystal Field Splitting
parameter (∆0)
• The magnitude of crystal field splitting depends on strength of the
ligands. The strong ligands those appear in spectrochemical series
approach closer to the central metal which results in a large crystal field
splitting.
• Oxidation state of the metal : A metal with the higher positive charge is
able to draw ligands closer to it than that with the lower one. Thus the
metal in higher oxidation state results in larger separation of t2g and eg
set of orbitals. The trivalent metal ions cause larger crystal field splitting
than corresponidng divalent ones.
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Color of octahedral complexes
• Splitting of d orbitals into t2g and eg gives this Δo
• It corresponds to a certain frequency Of electromagnetic radiation usually
in the visible region. A colour complementary to the absorbed frequency is
thus observed.
• Eg. [Ti(H2O)6]3⊕
• The absorption of the wavelength of light corresponding to Δo parameter
promotes an electron from the t2g level. Such energy gap in case of the
[Ti(H2O)6]3⊕ complex is 20,300 cm-1 (520 nm, 243 kJ/mol) and a
complimentary colour to this is imparted to the complex. A violet color of
the [Ti(H2O)6]3⊕ complex arises from such d-d transition.
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Tetrahedral complexes
64

• The dxy, dyz, dzx orbitals with their lobes lying in between the axes
point toward the ligands. On the other hand, dx2-y2 and dz2 orbitals
lie in between metal-ligand bond axes.
• The dxy, dyz and dzx orbitals experience more repulsion from the
ligands compared to that by dx2-y2 and dz2 orbitals.
• Due to larger such repulsions the dxy, dyz and dzx orbitals are of
higher energy while the dx2-y2 and dz2 orbitals are of relatively lower
energy
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Questions
• Draw qualitatively energy-level diagram showing d-orbital splitting
in the octahedral environment. Predict the number of unpaired
electrons in the complex [Fe(CN)6]
4- . Is the complex diamagnetic
or paramagnetic? Is it coloured? Explain.
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Coordination compounds (12th Maharashtra state board)

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