Download 9. Coordination Compounds.

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1. Write the formulae for the following coordination compounds:
(i) Tetraamminediaquacobalt(III) chloride
(ii) Potassium tetracyanonickelate(II)
(iii) Tris(ethane–1,2–diamine) chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane–1,2–diamine)platinum(IV) nitrate
(vi) Iron(III) hexacyanoferrate(II).
•
Solution:
(i) [Co(NH3)4(H2O)2]Cl3
(ii) K2[Ni(CN)4]
(iii) [Cr(en)3]Cl3
(iv) [Pt(NH3)BrCl(NO2)]
(v) [PtCl2(en)2](NO3)2
(vi) Fe4[Fe(CN)6]3-
2. Write the IUPAC names of the following coordination compounds:
(i) [Co(NH3)6]Cl3
(ii) [Co(NH3)5Cl]Cl2
(iii) K3[Fe(CN)6]
(v) K2[PdCl4]
in
(iv) K3[Fe(C2O4)3]
(vi) [Pt(NH3)2Cl(NH2CH3)]Cl.
•
Solution:
(i) Hexaammine cobalt(II) chloride
(ii) Pentaammine chlorido cobalt(III) chloride
(iii) Potassium hexacyano ferrate(III)
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(iv) Potassium tris(oxalato) ferrate(III)
(v) Potassium tetrachlorido palladate(II)
(vi) Diammine chlorido(methylamine) platinum(II)chloride.
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3. Indicate the types of isomerism exhibited by the following complexes
and draw the structures for these isomers:
(i) K[Cr(H2O)2(C2O4)2]
(ii) [Co(en)3]Cl3
(iii) [Co(NH3)5(NO2)](NO3)2
(iv) [Pt(NH3)(H2O)Cl2].
•
Solution:
(i) Both geometrical (Cis, Trans) and optical isomers for Cis can exist.
(ii) Two optical isomers can exist
(iii) There are 10 possible isomers. There are geometrical, ionisation and linkage
isomers possible.
(iv) Geometrical (Cis, Trans) isomers can exist.
4. Give evidence that [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br are
ionisation isomers.
•
Solution:
The ionisation isomers dissolve in water to yield different ions and thus react
differently to various reagents:
in
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5. Explain on the basis of valence bond theory that [Ni(CN)4]2– ion with
square planar structure is diamagnetic and the [NiCl4]2– ion with tetrahedral
geometry is paramagnetic.
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Solution:
a. [Ni(CN)4]2-
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-
dsp2 hybridisation
CN will cause pairing of electrons. It is diamagnetic in nature due to the unpaired
electron.
b. [Ni(Cl4)]2-
[NiCl4]2 ion
Four pairs of electrons from
Cl ions (ligands)
It is paramagnetic and has two unpaired electrons.
6. [NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both
are tetrahedral. Why?
in
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Solution:
In Ni(CO)4, Ni is the zero oxidation state whereas in [NiCl4]2-, is in +2 oxidation state.
In the presence of CO ligand, the unpaired d electrons of Ni pair up bu Cl- being a
weak ligand is unable to pair up the unpaired electrons.
7. [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3– is weakly
paramagnetic. Explain.
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•
Solution:
In presence of CN-, (a strong ligand) the 3d electrons pair up leaving only one
unpaired electron. The hybridization of d2sp3 forming inner orbital complex. In the
presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridization is
sp3d2 forming an outer orbital complex containing five unpaired electrons, it is strongly
paramagnetic.
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8. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+
is an outer orbital complex.
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Solution:
In the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be
involved in d2sp3 hybridisation forming inner orbial complex in case of [Co (NH3)6]3+.
In [Ni(NH3)6]3+, Ni is in +2 oxidation state and has d8 configuration, the hybridization
involved in sp3d2 forming outer orbital complex.
9. Predict the number of unpaired electrons in the square planar
[Pt(CN)4]2– ion.
•
Solution:
For square planar shape, the hybridization is dsp2. Hence the unpaired electrons in 5d
orbital pair up to make one d orbital empty for dsp3hybridisation. Thus there is no
unpaired electron.
10. The hexaquo manganese(II) ion contains five unpaired electrons,
while the hexacyanoion contains only one unpaired electron. Explain using
Crystal Field Theory.
•
in
Solution:
Hexaaqua manganese II ion
Mn2+
No. of unpaired electrons = 5
Hexacyano-ion
[Mn(CN)6]4-
No. of unpaired electrons = 1.
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Aqua ligand is weaker when compared to CN- which is stronger. Hence according to
CFT greater splitting occurs in the former than the latter.
11. Calculate the overall complex dissociation equilibrium constant for the
Cu(NH3)42+ ion, given that B4 for this complex is 2.1 x 1013.
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Solution:
The overall dissociation constant is the reciprocal of overall stability constant i.e
1/B4 = 4.7 x 10-14.
12. Explain the bonding in coordination compounds in terms of Werner’s
postulates.
•
Solution:
(1) Metals exert two types of linkages (i) the primary or ionisable bonds which are
satisfied by negative ions and equal to the oxidation state of the metal.
(2) The secondary or non ionisable bonds which can be satisfied by neutral or negative
ions / groups. The secondary linkages equal the coordination number of central metal
atom / ion. This number is fixed for a metal.
(3) The ions / groups bound by the secondary linkages have characteristic spatial
arrangements corresponding to different coordination numbers. In the modern
terminology such spatial arrangements are called Coordination Polyhedra.
In the modern formulations, such spatial arrangements are called coordination
polyhedra. The species with the square brackets are coordination entities or complexes
and the ions outside the square bracket are called counter ions.
•
in
13. FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives
the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4
molar ratio does not give the test of Cu2+ ion. Explain why?
Solution:
FeSO4 does not form any complex with (NH4)2SO4. Instead, if forms a double salt
FeSO4.(NH4)2SO4.6H2O which dissociates completely into ions. CuSO4 when mixed with
NH3 forms a complex [Cu (NH3)4]SO4 in which the complex ion Cu(NH3)42+ does not
dissociate to give Cu2+ ion.
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14. Explain with two examples each of the following: coordination entity,
ligand, coordination number, coordination polyhedron, homoleptic and
heteroleptic.
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Solution:
Coordination Entity
A coordination entity constitutes a central atom/ions, usually of a metal, to which are
attached a fixed number of other atoms of groups each of which is called a ligand. It
may be neutral or charged.
Examples: [Co(NH3)6]3+, [PtCl4]2-
Ligand
A ligand is an ion or a small molecule having at least one lone pair of electrons and
capable of forming a coordinate bond with central atom/ion in the coordination entity.
Examples are Cl-, OH-, and NH3 etc., A ligand may be neutral or charged species. It
always acts as a Lewis base.
Coordination Number
Coordination number of the central atom / ion is determined by the number of sigma
bonds between the ligands and the central atom / ion. Eg: [Co (NH3)6]3+. The
coordination number is six.
Coordination Polyhedron
The spatial arrangement of the ligands which are direcly attached to the central
atom/ion defines a coordination polyhedron about the central atom. For example,
[Co(NH3)6]3+ is octahedral.
Homoleptic and Heteroleptic
Complexes in which a metal is bound to only one kind of donor groups are known as
homoleptic . For example, [Co(NH3)6]3+.
Complexes in which a metal is bound to more than one kind of donor groups are
known as heteroleptic. For example [Co(NH3)4Cl2]+
15. What is meant by unidentate, didentate and ambidentate ligands? Give
two examples for each.
Solution:
Ligand on the basis of their ligating ability can be classified as:
in
•
Unidentate Ligand
When the ligand can donate the pair of electrons from one atom, it is called
monodentate ligand, e.g.,NH3 , CN- etc.
Bidentate Ligand
When the ligand can donate the pair of electrons through two atoms of the ligands, it
is called didentate ligand, e.g., ethylene diamine
6
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Ambidentate Ligand
Ligands which can attach themselves (called ligating) through two different atoms of
the same molecule are called ambident ligands or group. E.g., SCN-, No2-… They
introduce linkage isomerism in the complexes.
16. Specify the oxidation numbers of the metals in the following
coordination entities:
(i) [Co(H2O)(CN)(en)2]2+.
(ii) Ca2 [Fe(CN)6]
•
Solution:
x-6= -4
x=2
(i) x + 0 + (-1) + 0 = +2
x=3
Oxidation no. of Co = 3.
17. Using IUPAC norms write the formulas for the following:
(a) Tetra hydroxozincate (II)
(b) Hexammine Cobalt (III) Sulphate
(c) Potassium tetrachlorido palladate (II)
(d) Potassium tris (oxalato) chromate (III)
•
Solution:
(a) [Zn (OH)4]2 –
in
(e) Diammine di chloro platinum (II).
(b) [Co (NH3)6]+ SO4
(c) K2 [Pd (Cl)4]
(d) K3 [Cr (C2O4)3]
(e) [Pt (NH3)2 (Cl2)].
7
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18. Using IUPAC norms write the formulas for the following:
(a) Hexaamine platinum (IV)
(b) Potassium tetra cyano nickelate (II)
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(c) Tetra bromido Cuprate (II)
(d) Penta ammine nitrito – O – Cobalt (III)
(e) Pentaammine nitro – N – Cobalt (III).
•
Solution:
(a) [Pt (NH3)6]4+
(b) k2[Ni (CN)4]
(c) [Cu (Br)4]2-
(d) [Co (NH3)5 ONO]2+
(e) [Co (NH3)5 NO2]2+.
19. Using IUPAC norms write the systematic names of the following:
(a) [Co(NH3)6]Cl3
(b) [CoCl(NO2)(NH3)4]Cl
(c) [Ni(NH3)6]Cl2
(d) [Pt Cl(NH2CH3)(NH3)2]Cl
(e) [Mn(H2O)6]2+.
•
Solution:
(a) Hexaammine cobalt (III) chloride
(c) Hexaammine nickel (II) chloride.
in
(b) Tetraammine chloro nitro cobalt (III) chloride
(d) Diammine chloro methylamine platinum (II) chloride.
(e) Hexa aqua manganese (II) ion.
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20. Using IUPAC norms write the systematic names of the following:
(a) [Co(en)3]3+
(b) [Ti(H2O)6]3+
(c) [NiCl4]2(d) [Ni(CO)4].
Solution:
(a) Tris (ethane 1,2-diamine) cobalt (III) ion.
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(b) Hexa aqua titanium (III) ion.
(c) Tetra chlorido nickelate (II) ion.
(d) Tetra carbonyl nickel (O).
21. List various types of isomerism possible for coordination compounds.
•
Solution:
(i) Ionisation isomerism
(ii) Coordination isomerism
(iii) Linkage isomerism
(iv) Geometrical isomerism
(v) Optical isomerism.
22. How many geometrical isomers are possible in the following
coordination entities?
[Co(NH3)3Cl3].
Solution:
The two entities are represented as:
in
•
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23. Draw the structures of optical isomers of:
[Cr(C2O4)3]3–.
•
Solution:
in
24. Draw all the isomers (geometrical and optical) of:
[CoCl2(en)2]+.
•
Solution:
(i) Two geometrical isomers cis- and trans- forms are possible.
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Cis form
d- form
Trans form
l-form.
25. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how
many of these will exhibit optical isomers?
•
Solution:
Geometrical Isomerism
in
Optical Isomerism
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26. Aqueous copper sulphate solution (blue in colour) gives a green
precipitate with aqueous potassium fluoride.
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Solution:
Aqueous copper sulphate contains coordination entities [Cu(H2O)4]2+ which are blue in
colour. Water molecule is a weaker ligand than Cl- and F-.
On addition of aqueous KF solution, a new complex entity is formed which is of green
colour.
Green ppt.
27. What is the coordination entity formed when excess of aqueous KCN is
added to an aqueous solution of copper sulphate? Why is it that no
precipitate of copper sulphide is obtained when H2S(g) is passed through this
solution?
•
Solution:
Aqueous solution of copper sulphate contains Cu2+ ions in form of complex entity,
[Cu(H2O)4]2+ and H2O ligand is a weak ligand. When excess of KCN is added, a new
coordination entity, [Cu (CN)4]2- is formed due to following reaction:
[Cu(H2O)4]2+ + 4CN-
[Cu (CN)4]2- + 4H2O
in
Cyanide ligand CN- is a strong field ligand and stability constant of [Cu (CN)4]2- is quite
large and thus practically no Cu2+ ions are left in solution.
28. Draw a figure to show the splitting of d orbitals in an octahedral
crystal field.
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Solution:
To explain the splitting of ‘d’ orbitals in the presence of ligands, let us assume that the
six ligands are positioned symmetrically along the cartesian axes, with the metal atom
at the origin. As the ligands approach there is an increase in the energy of ‘d’ orbitals
relative to that of the free ion just as would be the case in a spherical field. The
orbitals lying along the axes (dz2 and dx2-y2) get repelled more strongly than dxy,
dyz and dxz orbitals, which have lobes directly between the axes. The dz2 and dx2y2 orbitals get raised in energy and dxy, dyz and dxz orbitals are lowered in energy
relative to the average energy in the spherical crystal field. Thus the degenerate set of
d orbitals get split into two sets. The lower energy orbital set. T2g and the higher
energy eg set. The energy separation is denoted by ∆0 (the subscript `O’ is for
octahedral)
Consider the complex [Ti (H2O)6]3+ formed in aqueous solutions of Ti3+ (d1) ion the
single d electron occupies one of the lower energy t2g orbitals. In d2 and d3 coordination entities the d electrons occupy the t2g orbitals singly in accordance with the
Hund’s rule. For d4 ions, two possible patterns of electron distribution arise (i) the
fourth electron may enter an eg orbital or it may pair an electron in the t2g level. The
actual configuration adopted is decided by the relative values of ∆0 and P. P represents
the energy required for electron pairing in a single orbital.
in
If ∆0 is less than P (∆0 < P), we have the so called weak field, high spin situation, and
the fourth electron enters one of the eg. orbitals giving the configuration t2g3 eg1. If
now a fifth electron is added to a weak field co-ordination entity, the configuration
t2g3 eg2. When ∆0 > P, we have the strong field, low spin situation, the pairing will
occur in the t2g level with the eg. level remaining unoccupied entities of d1 to d6 ions.
29. What is spectrochemical series? Explain the difference between a
weak field ligand and a strong field ligand.
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•
Solution:
Spectro-chemical series is a seris in which the ligands have been arranged in order of
increasing magnitude of splitting they produce. The order is
I-< Br- <S2- < Cl- < F- OH- < C2O42- < H2O < NH3 <CN- < CO
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Weak field
coordination entity
Strong field coordination
entity
1. They are formed when
the crystal field
stabilisation energy in
octahedral complexes is
less that the energy
required for an electron
pairing in a single orbital
(p)
They are formed when the
crystal field stabilisation
energy is greater than the
p.
2. They are also called
high spin complexes.
They are called low spin
complexes.
3. They are mostly
paramagnetic in nature
complex.
They are mostly
diamagnetic or less
paramagnetic than weak
field.
4. Never formed by CHligands
Formed by CN- like ligands.
in
30. What is crystal field splitting energy? How does the magnitude
of ∆o decide the actual configuration of d orbitals in a coordination entity?
•
Solution:
∆0 is the energy difference between the t2g and eg set of d- orbitlas and is called
crystal field stabilization energy is octahedral complex.
14
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In d1 coordination entity, d1 electron occupies t2g orbitals. In d2 and d3 also, electrons
occupy t2g orbitals singly in accordance with the Hund’s rule. For d4 ins two possibilities
exist. If ∆0 is less than P energy required for electron pairing in a single orbital, we
have weak field, high spin situation and the fourth electron enters one of the
eg orbitals giving teg3 eg1. Fifth electron also enters eg orbitals, i.e., teg3 eg1
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When ∆0>P, we have strong field, low spin situation and pairing will occur in the
t2g level with eg orbitals remain unoccupied in d1 to d8 ions. Greater the ∆0 value
greater the chances of pairing.
31. Discuss the nature of bonding in metal carbonyls.
•
Solution:
Carbonyls are formed by the transitional elements because of the presence of vacant
d-orbitals. In carbonyl the oxidation state of the metal is zero.
During the formation of carbonyls the vacant d-orbitals of metal are over-lapped by
the filled carbon σ-bond between metal and C of CO due to the donatin of electron pair
of carbon in to the vacant d-orbitals of metal.
On the other hand a π-overlap takes place by a back donation of electrons from a filld
d-orbitals of metal into the vacany antibonding π*p-orbitals of CO and a π M C is
formed.
32. Give the oxidation state, d orbital occupation and coordination number
of the central metal ion in the following complexes:
K3[Co(C2O4)3].
•
Solution:
Oxidation state = +3
Coordination number = 6
d- orbital occupation is t2g6 eg0.
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33. Write down the IUPAC name for each of the following complexes and
indicate the oxidation state, electronic configuration and coordination
number. Also give stereochemistry and magnetic moment of the complex:
K[Cr(H2O)2(C2O4)2].3H2O.
•
Solution:
Systematic name:
Potassium diaqua bis(oxalato)chromate(III) trihydrate
Oxidation state = +3
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Electronic configuration = 3d5
Coordination number = 6
Magnetic moment = 4.80.
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34. What is meant by stability of a coordination compound in solution?
State the factors which govern stability of complexes.
•
Solution:
The stability of a complex in solution refers to the degree of association between the
two species involved in the state of equilibrium. The magnitude of the equilibrium
constant for the association, quantitatively express the stability.
The stability of the coordination compound depends on:
i. Nature of the ligand: Chelating ligands form strong and more stable complexes than
the monodentate ligands.
ii. Nature of the metal atom/ion: Small, highly charge metal ions form more stable
complexes than large size, lowly charged metal ions.
35. What is meant by the chelate effect? Give an example.
•
Solution:
The complexes which are formed by chelating ligands like ethylene diamine(en), EDTA
etc. are more stable than those formed by monodentate ligands such as H2O or NH3.
This enhanced stability of complexes containing chelating ligand is called chelate
effect.
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