Download Coordination Compounds

CCC – 1
CO-ORDINATION COMPOUNDS
C1
Double Salts and Co-ordination Compounds
Double Salts : are those which loose their identity in solution as alum.
[K2SO4.Al2(SO4)3.24H2O]
Co-ordination Compounds (Complexes) : Those which retain their identity in solution
(complexes) as potassium ferrocyanide, K4[Fe(CN)6].
A complex (or co-ordination compound) is a compound consisting either complex ions
with other ions of opposite charge.
K4[Fe(CN)6]
complex anion
K+ is other ion
[Pt(en)3]Cl4
complex cation
Cl– is other ion
[Pt(en)2Cl2]
neutral complex
Ligands : A Ligand is a species that is capable of donating an electron pair(s) to a central
ion. It is a Lewis base. In accepting electron pairs, the central ion acts as a Lewis acid.
Ligand can be (a)
unidentate
(b)
bidentate, tridentate, tetradentate etc.
(c)
Ambidient
Unidentate Ligand : Ligand is said to be unidentate if it has only one pair of electrons
that it can donate.
Polydentate Ligand : Ligand is said to be bidentate, tridentate, tetradentate etc.
depending on the number of electron pairs that it can donate.
(a)
:NH3 ; one electron pair available for donation oxidentate.
(b)
two electron pairs available for donation bidendate
Some common Multidentate Ligands (Chelating Agents)
Abbrevation
Multi
Name
en
bidentate
ethylenediame
0
ox
bidentate
oxalate
–2
gly
bidentate
glycinate
–1
Einstein Classes,
Formulae
Charge
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CCC – 2
DMG
bidentate
dimethyl glyoximate
EDTA
hexadentate
ethylenediamine
–1
tetraacetate
–4
Chelates : Some ligands are capable of donating more than a single electron pair, from
different atoms in the ligand and to different sites in the geometric structure of a complex.
These are called as multidentate ligands. When the bonding of a multidentate ligand to
metal ion produces a ring (usually five or six membered) we refer to the complex as a
chelate. The multidentate ligand is called a chelating agent and the process of chelate
formation is called chelation
Nickel (II) dimethylglyoximate (chelate)
[bis-(dimethylglyoximato) nickel (II)]
Co-ordination Number : The co-ordination number of a metal atom in a complex is the
total number of bonds the metal atom forms with Ligands.
Co-ordination Number (C.N.) of Metal Ions
Metal ion
C.N.
Ag+
2
Cu+
2, 4
Cu2+
4, 6
Au+
2, 6
Ca2+
6
Fe2+, Fe3+
6
Co2+
4, 6
Co3+
6
Ni2+
4, 6
Zn2+
4
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CCC – 3
Al3+
4, 6
Sc3+
6
Cr3+
6
Pt2+
4
Pt4+
6
Complex
C.N.
[Ag(NH3)2]+
2
[HgI3]–
3
[PtCl4]2–, [Ni(CO)4]
4
[CO(NH3)6]3+
6
Practice Problems :
1.
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 ?
2.
What is meany by unidentate, didentate and ambidentate ligands ? Give two examples for each.
3.
Specify the oxidation numbers of the metals in the following coordination entities :
(i)
[Co(H2O)(CN)(en)2]2+
2–
(iii)
[PtCl4]
(v)
[Cr(NH3)3Cl3]
(ii)
[CoBr2(en)2]+
(iv)
K3[Fe(CN)6]
4.
What is meant by stability of a coordination compound in solution ? State the factors which govern
stability of complexes.
5.
What is meany by the chelate effect ? Given an example.
6.
How many ions are produced from the complex Co(NH3)6Cl2 in solution ?
(i)
6
(ii)
4
(iii)
3
(iv)
2
[Answers : (1) FeSO4 solution mixed with (NH4)2SO4 solution forms a double salt, which ionises in
the solution to give Fe2+ ions. CuSO4 solution mixed with aqueous ammonia forms a complex salt,
does not ionise to give Cu2+ ions (3) (i) +3 (ii) +2 (iii) +3 (iv) +3 (v) +3 (5) a didentate or a polydentate
ligand coordinates with the central metal ion forming a five or a six membered ring, the effect is
called chelate effect (6) (iii) is correct]
C2A Nomenclature : Following rules are adopted for naming a complex ion.
(a) cations are named before anions.
(b)
oxidation state (O.S.) of the central metal ion is denoted by Roman numeral
Cation O.S. Anion
CuC2
Copper (II) chloride
FeCl3
Iron (III) chloride
(c)
The names of ligands are given first followed by the name of central metal ion.
(d)
The names of ligands that are anions and ending with
‘ide’ changed to ‘o’
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CCC – 4
‘ite’ changed to ‘ito’
‘ate’ changed to ‘ato’
(e)
Some Common Unidentate Ligands
Neutral
Anions (replacing ‘ide’ by ‘o’)
Formula
Names as Ligand
Formula
Name as Ligand
H2O
aqua
F–(Fluoride)
Fluorido
NH3
ammine
Cl–
Chlorido
CO
Carbonyl
S–2
sulphido
NO
nitrosyl
H–
hydrido
CH3NH2 methyl amine
O–2 (oxide)
oxo
C5H5N or Py
Pyridine
OH– (hydroxide)
hydroxo
(C2H5)3N
Triethylamine
CN–(cyanide)
cyano*
(C6H5)3N
Tripphenyl amine
NC
isocyano*
[* marked are ambident)
Replacing ‘e’ by ‘o’
(e)
SO42–
sulphato
S2O32–
thiosulphato
CO32–
carbonato
NO2–
nitro*
ONO–
nitrito*
SCN–
thiocyanato*
NCS–
isothiocyanato*
Positive groups end in-ium
NH2 – NH3+ hydrazinium
(f)
When there are several ligands of the same kind we normally use the prefix di, tri, tetra
soon. If the name of ligand includes a number e.g. ethylenediamine (en). To avoid
confusion in such cases, bis, trix and tetrakis are used.
e.g.
(g)
bis (ethylenediammine)
If anion is a complex then metal ends with ‘ate’
[Ni(CN)4]2–
tetracyanonickelate (II) ion
Einstein Classes,
lead – plumbate
gold – Aurate
tin – stannate
silver – argentate
iron – ferrate
copper – cuprate
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CCC – 5
(h)
If the complex contain two or more metal atoms, it is termed polynuclear. The bridging
ligands which link the two metal atoms together are indicated by the prefix
(i)
Ambidient Ligands may be attached through different atoms
M – NO2
(NO2) joined to metal M through N; it is nitro
M – ONO
(NO2 joined to metal M through O it is nitro)
Similarly the SCN group may bond M-SCN (thiocyanato) or (M – NCS) (isothiocyanato).
They may be named as
thiocyanato – S,
thiocyanato – N
Practice Problems :
1.
2.
Write the IUPAC names of the following coordination compounds :
(i)
[Co(NH3)6]Cl3
(ii)
[Co(NH3)5Cl]Cl2
(iv)
K3[Fe(C2O4)3]
(v)
K2[PdCl4]
(vi)
[Pt(NH3)2Cl(NH2CH3)]Cl
(iii)
K3[Fe(CN)6]
Using IUPAC norms, write the systematic names of the following :
(i)
[Co(NH3)6]Cl3
(iii)
[Ti(H2O)6]
3+
2+
(v)
[Mn(H2O)6]
(vii)
[Ni(NH3)6]Cl2
(ix)
[Ni(CO)4]
(ii)
[Pt(NH3)2Cl(NH2CH3)]Cl
(iv)
[Co(NH3)4Cl(NO2)]Cl
(vi)
[NiCl4]2–
(viii)
[Co(en)3]3+
[Answers : (1) (i) hexaamminecobalt(III) chloride (ii) pentaamminechloridocobalt(III) chloride (iii)
potassium hexacyanoferrate(III) (iv) potassium trioxalatoferrate(III) (v) potassium
tetrachloridopalladate(II) (vi) diamminechlorido (methylamine) platinum(II) chloride
(2) (i) hexaamminecobalt(III) chloride (ii) diamminechlorido(methyl amine)platinum(II) chloride
(iii) hexaaquatitanium(III) ion (iv) tetraamminechloridonitrito-N-cobalt(III) chloride (v)
hexaaquamanganese(II) ion (vi) tatrachloridonickelate(II) ion (vii) hexaamminenickel(II)
chloride (viii) tris(ethane-1, 2-diamine)cobalt(III) ion (ix) tetracarbonylnickel(0)]
C2B When writing (not naming) the formulae of the complex

complex ion should be enclosed by square brackets.

Ligands are place after metal in the alphabetical order but first negative ligands,
then neutral then positive.
[Pt (H 2 O) 4 (NH 3 ) 2 ]
–
wrong
[Pt(NH3)2 (H2O)4]
–
correct
[Cr (CN ) 4 ( NH 3 ) 2 ]
–
correct
|
aqua

negative
Einstein Classes,
|
ammine

neutral
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CCC – 6
Example :1.
K 4 [Fe(CN) 6 ]
cation
Anion
Potassium hexacyanof errate(II)
cation
2.
anion

anion is complex hence metal ends with ate.

Ligands are named before metal ion of complex.
[CO( NH 3 ) 6 ] Cl 3
cation
anion
Hexaammine cobalt (III) chloride
3.
[CO(en)3]Cl3
Tris (ethylenediammine) cobalt (III) chloride
4.
[Pt(Cl)2(NH3)4]Cl2
Tetraamminedichloroplatinum (IV) chloride
[NH3(ammine) and Cl(chloro) are written in alphabetical order]
5.
6.
7.
8.
In the following examples complex ion exist as cation :(a)
[Fe(NH3)6]Cl3 Hexaammine iron (III) chloride
(b)
[CoCl(NH3)5]2+ Pentaamminechlorocobalt (III) ion
(c)
[CoSO4(NH3)4]NO3 Tetraamminesulphatocobalt (III) nitrate
Complex ion exist as anions :(a)
[ZnCl4]2– Tetrachlorozincate (II) ion
(b)
[AlH4]— Tetrahydridoaluminate (III) ion
(c)
Na3[Ag(S2O3)2] Sodium bis (thiosulphato) argentate (I)
(d)
Na2 [OSCl5N] sodium pentachloronitrido osmate (VI)
(e)
K2[Cr(CN)2O2(O2)NH3] Potassium amminedicyanodioxo peroxochromate (VI)
Organic Ligands have been used :(a)
[Pt(Py)4] [Pt Cl4] Tetrapyridineplatinum (II) tetrachloroplatinate (II)
(b)
[Cr(C6H6)2] Bis (benzene) chromium (O)
(c)
[Fe(C5H5)2] Bis (cyclopentadienyl) iron (II)
In the following examples, bridging groups are used :-
(a)
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CCC – 7
Tetrakin (ethylenediammine)-µ-imido-µ-hydroxo dicobalt (III) ion.
Trans-bis (chlortiphenylphosphine-µ-chloropalladium (II)
or
Chlorotriphenylphosphinepallodium (II)-µ-dichlorochlorotriphenylphosphinepalladium (II)
Practice Problems :
1.
2.
Write the formulas 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)
Using IUPAC norms write the formulas for the following :
(i)
Tetrahydroxozincate(II)
(ii)
Potassium tetrachloridopalladate(II)
(iii)
Diamminedichloridoplatinum(II)
(iv)
Potassium tetracyanonickelate(Ii)
(v)
Pentaaminenitrito-O-cobalt(III)
(vi)
Hexaamminecobalt(III) sulphate
(vii)
Potassium trioxalatochromate(III)
(viii)
Hexaammineplatinum(IV)
(ix)
Tetrabromidocuprate(II)
(x)
Pentaamminenitrito-N-cobalt(III)
[Answers : (1) (i) [Co(NH3)4(H2O)2]Cl3 (ii) K2[Ni(CN)4] (iii) [Cr(en)3]Cl3 (iv) [PtBrClNO2(NH3)]–
(v) [PtCl2(en)2](NO3)2 (vi) Fe 4[Fe(CN) 6] 3 (2) (i) [Zn(OH) 4] 2– (ii) K 2[PdCl 4] (iii) [Pt(NH 3 ) 2 Cl 2 ]
(iv) K2[Ni(CN)4] (v) [Co(NH3)5(ONO)]2+ (vi) [Co(NH3)6]2(SO4)3 (vii) K3[Cr(C2O4)3] (viii) [Pt(NH3)6]4+
(ix) [CuBr4]2– (x) [Co(NH3)5(NO2)]2+]
C2C (EAN) Effective Atomic Number :
Each ligand donates an electron pair to the metal ion, thus forming a co-ordinate bond
EAN
=
Z – O.N. + 2 (C.N.)
ON  oxidation number
EAN
or
C.N.  co-ordination number
=
Z – O.N. + 2 (Ligands)
(assuming that each ligand is unidentate)
Example :(a)
[Cr(CO)6]  EAN = 24 – O + 2 ×6 = 36
(b)
[Fe(CN)6]–2  EAN = 26 – 2 + 6 × 2 = 36
(c)
[CO(en)3]3+  EAN = 27 – 3 + 6 × 2 = 36
(en)  bidentate
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CCC – 8
Practice Problems :
1.
Explain with two examples each of the following : coordination entity, ligand, coordination number,
coordination polyhedron, homoleptic and heteroleptic.
C3
Isomerism in Complexes :Compounds having same chemical formulae but differ structural arrangements are called
isomers.
Structural isomerism :Differ from each other. How the atoms are joined together that is in the order in which the
atoms are bonded to each other.
H – N = C = O and N  C – O – H are structural isomer.
Ionisation Isomerism :This type of isomerism occurs when there is an interchange of groups between the
co-ordination sphere of the metal ion and the ions outside this sphere. Examples are :
(a)
[CO(NH3)5 (SO4]Br
[CO(NH3)5Br] SO4
[give yellow ppt. with Ag+]
[give white ppt. with Ba 2+ as SO 4 2— being
ion as Br– outside the bracket ionisable]
is ionisable]
(b)
[Pt(NH3)4Cl2] Br2
[Pt(NH3)4Br2] Cl 2
ionisable
ionisable
Co-ordination isomerism :When both positive and negative ions are complex ions. Isomerism may be caused due to
interchange of ligands between anion and cation. This arises only when cation and anion
have same C.N. and charge
(a)
[CO(NH3)6]3+ [Cr(CN)6]3—
same charge
&
[Cr(NH3)6]3+ [CO(CN)6]3—
same O.N.
Linkage Isomerism :(i)
[CO(NH3)5ONO]Cl2 and [CO(NH3)5NO2]Cl2
(ii)
[Mn(CO)5 (SCN)]+
[Mn(CO)5(NCS)]+
Co-ordination Position Isomerism :
In polynuclear complexes an interchange of ligands between the different metal nuclei
give rise to positional isomerism e.g.
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CCC – 9
Hydrate Isomerism :Hydrate isomers of a complex that differ in the placement of water molecules in the complex CrCl3.6H2O can be written as
[Cr(H2O)6]Cl3
—
violet (anhydrous)
[Cr(H2O)5Cl]Cl2.H2O
—
light green (monohydrate)
[Cr(H2O)4Cl2]Cl.2H2O
—
Dart green (dihydrate)
Stereo Isomers :1.
Geometrical
2.
Optical
Geometrical :- are isomers in which the atoms are joined to one another in the same way
but differ because some atom occupy different relative position in space.
Geometrical Isomers of C.N. – 4
(a)
[Pt(NH3)2Cl2]
(b)
Geometrical isomers of C.N. – 6
(a)
[CO(NH3)4Cl2]+
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CCC – 10
(b)
[Pt(NH3)Cl2Br2]
(c)
[Pt(NH3)3Cl3]
Optical Isomerism :- If a molecule is asymetric that it cannot superimposed on its mirror
image. The two forms have the same type of symmetry shown by the left and right hands
and are called enantiomeric pair. The two forms are called optical isomers. They are called
either dextro or leavo (dor l).
Optical isomerism is common in octahedral complexes involving bidentate ligands.
[CO(en)2Cl2] exists as cis and trans but cis-form can have optical isomerism.
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CCC – 11
Practice Problems :
1.
Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionisation isomers.
2.
List various types of isomerism possible for coordination compounds, giving an example of each.
3.
How many geometrical isomers are possible in the following coordination entities ?
(a)
4.
5.
6.
[Cr(C2O4)3]3–
(b)
[Co(NH3)3Cl3]
(b)
[PtCl2(en)2]2+
Draw the structures of optical isomers of :
(a)
[Cr(C2O4)3]3–
(c)
[Cr(NH3)2Cl2(en)]+
Draw all the isomers (geometrical and optical) of :
(i)
[CoCl2(en)2]+
(iii)
[Co(NH3)2Cl2(en)]+
(ii)
[Co(NH3)Cl(en)2]2+
Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit
optical isomers ?
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CCC – 12
[Answers : (1) On dissolving in water, the two compounds will give different ions in the solution
which can be tested by adding AgNO3 solution and BaCl2 solution (3) (a) no possibility (b) two
geometrical isomers (6) Three isomers are possible]
C4A Werner’s theory of co-ordination compounds :
1.
There are two types of valency shown by central metal atom/metal ion in a compound
(a)
Primary
(b)
Secondary

Primary Valency corresponds to oxidation number and secondary valency corresponds
to co-ordination number.

In the complex [CO(NH3)6]Cl3, primary valency is satisfied by three Cl, secondary
valency is six.

In [CO(NH3)4(H2O)2]SO4, four NH3 and two H2O (ligands) satisfy secondary valency.

One SO42— satisfy primary valency.

Primary valency is ionisable.

The secondary valencies are directional and are responsible for isomerism in complexes.

The primary valencies are non-directional and is represented by ..... and secondary
valency by _______________

In all cases, metal or metal ions should satisfy primary and secondary valency both. Some
negative ions can satisfy primary as well as secondary valency both.
e.g. [Cr(NH3)5Cl]Cl2 satisfy both primary and secondary valency.
Practice Problems :
1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
C4B Bonding in complexes

The metal m loses required no. of electrons to form cation. No. of electrons lost correspond to O.N.

The metal ion makes available a no. of empty orbitals equal ot its co-ordinates number for
the formation of co-ordinates bonds with the Ligand Orbitals.

A weak ligand like (H2O, halide) will not affect the electronic configuration of the metal/
metal ion.

A strong ligand like (NH3, CN—, CO) affect the electronic configuration of the metal/
metalion. They make unpaired electrons paired.

The metal ion orbitals hybridise to form a new set of equivalent hybridised orbitals.

If there are unpaired electrons in the complex it is called paramagnetic.
Magnetic moment =
N( N  2) B.M.
N  no. of unpaired electrons.
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CCC – 13

The complex with so many unpaired electrons is high spin complex and that with paired
electrons is low spin complex.

If inner d-orbital is used for hybridisation it is called inner d complex. e.g. d2sp3 [Fe(CN)6]4–

If outer d-orbital is used is called outer d-complex. sp3d2
Weak field ligand [Fe(H2O)6]3+

I— < Br < S2– < Cl— < NO3— < F— < oxalate < H2O < EDTA < NH3 < en < NO2— < CN—
< CO
Examples : [Fe(CN)6]4—
O.N. = +2
C.N. = 6
[six empty orbitals are required by six CN— ligands] t2g  dxy, dyz, dzx e.g. dx2 – y2, dz2

Complex ion is d2sp3 hybridised

Inned d-complex

Diamagnetic

Magnetic moment = O

Octahedral geometry
[Fe(H2O)6]3+
(H2O is weak ligand)

Hybridisation – sp3d2

Outer-d-complex

Paramagnetic (High spin complex)

Octahedral geometry
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CCC – 14
Commom type of hybridisation and Geometry of complexes
Hybridisation
Geometry
Examples
sp
Linear
[Ag(NH3)2]+
sp2
Trigonal Planal
[HgI3]—, [CuP(Me)3]+
sp3
Tetrahedral
[VO4]3—, [MnO4]—,
[NiCl4]2—, [Ni(CO)4],
[Zn(Cl4)]2—
dsp2
Square planar
[Cu(NH3)4]2+,
[Ni(CN)4]2—,
[Pt(NH3)4]2+
sp3d or dsp3
Trigonal
[Cu(Cl5)]3—, [Ni(CN)5]3—
d2sp3 or
Bipyramidal
[Fe(CO)5]
sp3d2
Octahedral
[Fe(H2O)6]3+
Practice Problems :
1.
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.
2.
[NiCl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why ?
3.
Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
4.
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2– ion.
5.
The hexaaquo manganese (II) ion contains five unpaired electrons while the hexacyanoion contains
only one unpaired electron. Explain using Crystal Field Theory.
6.
Aqueous copper sulphate solution (blue in colour) gives :
(i)
a green precipitate with aqueous potassium fluoride and
(ii)
a bright green solution with aqueous potassium chloride.
Explain these experimental results.
7.
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 ?
8.
Discuss the nature of bonding in the following coordination entities on the basis of Valence Bond
Theory :
(i)
[Fe(CN)6]4–
(ii)
[FeF6]3– (iii)
[Co(C2O4)3)]3–
(iv)
[CoF6]3–
9.
Draw a figure to show the splitting of d-orbitals in an octahedral crystal field.
10.
What is spectrochemical series ? Explain the difference between a weak field ligand and a strong
field ligand.
11.
What is crystal field splitting energy ? How does the magnitude of 0 decide the actual configuration
of d-orbitals in a coordination entity ?
12.
[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain, why ?
13.
A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.
14.
[Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why ?
15.
Discuss the nature of bonding in metal carbonyls.
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CCC – 15
16.
17.
18.
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 :
(i)
K[Cr(H2O)2(C2O4)2].3H2O
(ii)
CrCl3(py)3
(iii)
[Co(NH3)5Cl]Cl2
(iv)
[Cs[FeCl4]
Discuss briefly giving an example in each case the role of coordination compounds in :
(i)
biological systems
(ii)
medicinal chemistry
(iii)
analytical chemistry and
(iv)
extraction/metallurgy of metals
Amongst the following ions, which one has the highest magnetic moment value ?
(i)
19.
[Fe(H2O)6]2+
(iii)
[Zn(H2O)6]2+
+1
(ii)
+3
(iii)
–1
(iv)
–3
(iii)
[Fe(C2O4)3]3–
(iv)
[FeCl6]3–
Amongst the following, the most stable complex is :
(i)
21.
(ii)
The oxidation number of cobalt in K[Co(CO)4] is
(i)
20.
[Cr(H2O)6]3+
[Fe(H2O)6]3+
(ii)
[Fe(NH3)6]3+
What will be the correct order for the wavelengths of absorption in the visible region for the
following : [Ni(NO2)6]4–, [Ni(NH3)6]2+, [Ni(H2O)6]2+
[Answers : (2) Cl– is weak ligand. It cannot pair up the electrons in 3d orbitals. Presence of strong
CO ligand and 4s electrons shift to 3d to pair up with 3d electrons (3) Presence of NH3 pairing of 3d
electrons takes place leaving two d-orbitals empty. In [Ni(NH3)6]2+, Ni is in +2 state with the
configuration 3d8. The NH3 can not pair up 3d electrons (4) No unpaired electron (6) (i) weak H2O
ligands are replaced by F– ligands forming [CuF4]2– ions which is a green precipitate (ii) weak H2O
ligands are replaced by Cl– ligands forming [CuCl4]2– ion which has bright green colour (7) The
complex ion is highly stable and does not dissociate to give Cu2+ ions. Hence, no precipitate with H2S
is obtained (8) (i) d2sp3 (ii) sp3d2 (iii) d2sp3 (iv) sp3d2 (10) ligands in order of their increasing field
strengths (11) The difference of energy between the two sets of orbitals is called crystal field splitting
4 0
energy. If 0 < P, then 4th electron pairsup in one of the t2g orbitals giving the configuration t 2g e g
thereby forming low spin complexes (13) In [Ni(H 2O) 6] 2+, it has two unpaired electrons.
In [Ni(CN)4]2–, no unpaired electrons (14) ligands H2O and CN– possess different crystal field
splitting energy (0) (18) (ii) having greatest value of n has highest magnetic moment (19) (iii) is the
correct option (20) (iii) is correct option (21) Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4–]
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CCC – 16
[Fe(Cl4)]2—
[Fe(F6)]3—
[Fe(CN)6]3—
[Mn(CN)6]3—
[MnF6]3—
[Cr(NH3)6]3+
[V(H2O)6]3+
CO3+
Fe2+
Fe3+
Fe3+
Mn3+
Mn3+
Cr3+
V3+
d6
d6
d6
d5
d5
d4
d4
d3
d2
Configuration
of metal ion
sp3
d2sp3
sp3d2
sp3
sp3d2
d2sp3
d2sp3
sp3d2
d2sp3
sp3d2
Hybridisation
of metal ion
orbitals for
ligand bonds
2
0
4
4
5
1
2
4
3
2
No. of unpaired
electrons
Paramagnetic
Diamagnetic
Paramagnetic
Paramagnetic
Paramagnetic
Paramagnetic
Paramagnetic
Paramagnetic
Paramagnetic
Paramagnetic
Magnetic
Behaviour
State of Hybridisation and Magnetic Behaviour of some co-ordination complex
[COF6]3—
CO3+
d8
Metal
Ion
[CO(CN)6]3
Ni2+
Metal
Complex
[Ni(Cl)4]2—
Diamagnetic
Diamagnetic
1
Diamagnetic
0
sp3
0
dsp2
d9
sp3
d8
CO2+
d10
Ni2+
[COCl4]2—
Zn2+
[Ni(CN)4]2—
[Zn(NH3)4]2+
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E X E R C I S E (OBJECTIVE)
1.
2.
11.
The EAN of Ni in Ni(CN)42– is
(a)
34
(b)
35
(c)
36
(d)
28
Glycinato ligand is
12.
(a)
3.
4.
5.
6.
7.
8.
9.
10.
(b)
Bidentate ligand
(c)
Two donor sites N and O–
(d)
All
The compound
paramagnetism
13.
which
does
not
show
Hexafluorocobaltate (III) ion is found to be high
spin complex, the probable hybrid state of cobalt
in it is
(a)
d2sp3
(b)
sp3
(c)
sp3d
(d)
sp3d2
In sodium tetrafluorooxochromate(.....),
Na3[Cr(O)F4] the left out place should be filled with
which of the following Roman numericals
(a)
VI
(b)
III
(c)
IV
(d)
none of these
The correct name of the compound
[Cu(NH3)4](NO3)2, according to IUPAC system is
(a)
Cuprammonium nitrate
(a)
[Cu(NH3)4]Cl2
(b)
[Ag(NH3)2Cl
(b)
Tetraammine copper(II) dinitrate
(c)
NO
(d)
NO2
(c)
Tetraammine copper(II) nitrate
(d)
Tetraammine copper(I) dinitrate
Which one is bidentate ligand
(a)
C2O42–
(b)
NH2·CH2·CH2·NH2
Lithium tetrahydridoaluminate is correctly
represented as
(c)
Both
(a)
Al[LiH4]
(b)
Al2[LiH4]3
(d)
None
(c)
Li[AlH4]
(d)
Li[AlH4]2
14.
Which ion is paramagnetic
15.
The correct IUPAC name of K2[Zn(OH)4] is
(a)
[Co(NH3)4]2+
(b)
[Ni(CO)4]
(a)
Potassium tetrahydroxyzinc(II)
(c)
[Co(NH3)6]3+
(d)
[Ni(CN)4]2–
(b)
Potassium tetrahydroxozincate(II)
The shape of the complex Ag(NH3)2+ is
(c)
Potassium tetrahydroxyzincate(IV)
(a)
Octahedral
(b)
Square planar
(d)
Potassium hydroxozinc(II)
(c)
Tetrahedral
(d)
Linear
16.
K3CoF6 is high spin complex. What is the hybrid
state of Co atom in this complex
(a)
sp3d
(b)
sp3d2
(c)
d2sp3
(d)
dsp2
Which one of the following will be able to show
cis-trans isomerism
(a)
MA3B
(b)
M(A A  )2
(c)
MA2BCD
(d)
MA4
17.
The type of isomerism shown by [Co(en)2(NCS)2]Cl
and [Co(en)2(NCS)Cl]NCS is
(a)
Co-ordination
(b)
Ionisation
(c)
Linkage
(d)
All above
18.
Which of the following cations does not form an
amine complex with excess of ammonia
(a)
Ag+
(b)
Cu2+
(c)
Cd2+
(d)
Na+
Einstein Classes,
19.
The correct IUPAC name of KAl(SO4)3 · 12H2O is
(a)
Aluminium potassium sulphate-12-water
(b)
Potassium aluminium(III) sulphate-12water
(c)
Potassium aluminate(III) sulphate
hydrate
(d)
Aluminium(III) potassium sulphate
hydrate-12
Number of electrons gained by Pd in [PdCl4]–2;
(a)
4
(b)
8
(c)
10
(d)
Zero
The possible number of isomers for the complex
[MCl2Br2]SO4 is
(a)
1
(b)
2
(c)
4
(d)
5
The oxidation number of Pt in [Pt(C2H4)Cl3]– is
(a)
+1
(b)
+2
(c)
+3
(d)
+4
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CCC – 18
20.
21.
22.
Chlorophyll is a co-ordination compound having
central atom of
(b)
Presence of water molecule
(c)
Excitation of electrons from d  d
(a)
Ca
(b)
Mg
(d)
Intramolecular vibration
(c)
Na
(d)
K
2–
Among [Ni(CN)4] [NiCl4] and [Ni(CO)4]
25.
26.
27.
2
(b)
3
(c)
4
(d)
5
(b)
NiCl42– is square planar and Ni(CN)42–
and Ni(CO)4 are tetrahedral
(c)
Ni(CO)4 is square planar and Ni(CN)42–
and [NiCl4]2– are tetrahedral
(a)
Copper hydroxide is an amphoteric
substance
(d)
None
(b)
In acidic solution hydration protects
copper ions
(c)
In acidic solutions protons coordinates
with ammonia molecules forming NH4+
ions and NH3 molecules are not available
(d)
In alkaline solution insoluble Cu(OH)2 is
precipitated which is soluble in excess
of any alkali
30.
Ammonia forms the complex ion [Cu(NH3)4]2+ with
copper ions in alkaline solutions but not in acidic
solutions. What is the reason for it ?
Which one does not belong to ligand
PH3
(b)
BF3
(d)
NO+
Cl
–
Nickel metal is in highest oxidation state in
(a)
Ni(CO)4
(b)
(c)
[Ni(NH3)6](BF4)2(d)
K2NiF6
K4[Ni(CN)6]
A complex of cobalt has five ammonia molecules,
one nitro group and two chlorine atoms for each
cobalt atom. One mole of this compound produces
three moles ions in aqueous solution which on
treating with excess of AgNO3 give two moles of
AgCl. The formula of the compound is
(a)
[Co(NH3)4NO2Cl][(NH3)Cl]
(b)
[Co(NH3)5Cl][ClNO2]
(c)
[Co(NH3)5NO2]Cl2
(d)
[Co(NH3)5][(NO2)2Cl2]
31.
32.
K3[Al(C2O4)3] is called
(a)
Potassium alumino oxalate
(b)
Potassium alumino (III) oxalate
(c)
Potassium trioxalato aluminate
(d)
Potassium trioxalato aluminate (III)
33.
In hexacyanomanganate (II) ion the Mn atom
assumes d2sp3 hybrid state. The number of unpaired
electrons in the complex is
(a)
1
(b)
2
(c)
3
(d)
zero
In the coordination compound, K4[Ni(CN)4], the
oxidation state of nickel is
(a)
+2
(b)
–1
(c)
0
(d)
+1
The coordination number of a central metal atom
in a complex is determined by
(a)
the number of ligands around a metal ion
bounded by sigma bonds
(b)
the number of ligands around a metal ion
bounded by pi-bonds
(c)
the number of ligands around a metal ion
bonded by sigma and pi-bonds both
(d)
the number of only anionic ligands
bonded to the metal ion.
Which one of the following complexes is an outer
orbital complex ?
(a)
[Fe(CN)6]4–
(b)
[Mn(CN)6]4–
(c)
[Co(NH3)6]3+
(d)
[Ni(NH3)6]2+
[Atomic nos. : Mn = 25, Fe = 26, Co = 27, Ni = 28]
Name the metal M which is extracted on the basis
of following reactions
Cerium (Z = 58) is an important member of the
lanthanoids. Which of the following statmenets
about cerium is incorrect ?
4M + 8CN– + 2H2O + O2  4[M(CN)2]– + 4OH–
(a)
The common oxidation states of cerium
are +3 and +4
(b)
The +3 oxidation state of cerium is more
stable than +4 oxidation state
(c)
The +4 oxidation state of cerium is not
known in solutions
(d)
Cerium (IV) acts as an oxidising agent.
2[M(CN)2]– + Zn  [Zn(CN)4]2– + 2M
28.
(a)
Ni(CN)22– is square planar and NiCl42–
and Ni(CO)4 are tetrahedral
(c)
24.
The number of d-electrons in [Cr(H2O)6]3+ is
(a)
(a)
23.
29.
2–
(a)
Nickel
(c)
Copper
(b)
Silver
(d)
Mercury
34.
3+
The colour of [Ti(H2O)6] is due to
(a)
Transfer of an electron from one Ti to
another
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35.
The correct order of magnetic moments (spin only
values in B.M.) among is
[MnCl4]2– > [CoCl4]2– > [Fe(CN)6]4–
(a)
36.
37.
38.
2–
4–
[MnCl4] > [Fe(CN)6] > [CoCl4]
(c)
[Fe(CN)6]4– > [MnCl4]2– > [CoCl4]2–
(d)
[Fe(CN)6]4– > [CoCl4]2– > [MnCl4]2–
The IUPAC name of the coordination compound
K3[Fe(CN)6] is
(a)
Potassium hexacyanoferrate (II)
(b)
Potassium hexacyanoferrate (III)
(c)
Potassium hexacyanoiron (II)
(d)
Tripotassium hexacyanoiron (II)
41.
Which of the following compounds shows optical
isomerism ?
(a)
[Cu(NH3)4]2+
(b)
[ZnCl4]2–
(c)
[Cr(C2O4)3]3+
(d)
[Co(CN)6]3–
Which one of the following cyano complexes would
exhibit the lowest value of paramagnetic
behaviour ?
42.
for
(b)
nitrito-N-pentaamminecobalt(III)
chloride
(c)
nitrito-N-pentaamminecobalt(II)
chloride
(d)
pentaamminenitrito-N-cobalt(II)
chloride
Nickel (Z = 28) combines with a uninegative
monodentate ligand X– to form a paramagnetic
complex [NiX 4 ] 2– . The number of unpaired
electron(s) in the nickel and geometry of this
complex ion are, respectively
(a)
two, square planar
(b)
one, tetrahedral
(c)
two, tetrahedral
(d)
one, square planar
In Fe(CO)5, the Fe-C bond possesses
(a)
-character only
(b)
-character only
[At. Nos : Cr = 24, Mn = 25, Fe = 26, Co = 27)
(c)
both  and  characters
The value of the ‘spin only’ magnetic moment for
one of the following configurations is 2.84 BM. The
correct one is
(d)
ionic character
(c)
[Fe(CN)6]
[Mn(Cn)6]
(d)
[Co(CN)6]
43.
complex
pentaamminenitrito-N-cobalt(III)
chloride
3–
(b)
3–
the
(a)
3–
[Cr(CN)6]
3–
The IUPAC name
[Co(NO2)(NH3)5]Cl2 is
2–
(b)
(a)
39.
40.
(a)
d4 (in strong ligand field)
The “spin-only” magnetic moment [in units of Bohr
magneton. (µB)] of Ni2+ in aqueous solution would
be (atomic number Ni = 28)
(b)
d4 (in weak ligand field)
(a)
1.73
(b)
2.84
(c)
d3 (in weak as well as in strong field)
(c)
4.90
(d)
0
(d)
d5 (in strong ligand field)
44.
How many EDTA (ethylenediaminetetraacetic acid)
molecules are required to make an octahedral
complex with a Ca2+ ion ?
(a)
Two
(b)
Six
(c)
Three
(d)
One
ANSWERS
1.
a
9.
b
17.
b
25.
d
33.
d
41.
c
2.
d
10.
d
18.
d
26.
a
34.
c
42.
c
3.
b
11.
d
19.
b
27.
b
35.
a
43.
b
4.
c
12.
b
20.
b
28.
c
36.
b
44.
d
5.
a
13.
c
21.
a
29.
b
37.
c
6.
d
14.
c
22.
c
30.
c
38.
c
7.
b
15.
b
23.
b
31.
c
39.
a
8.
c
16.
b
24.
c
32.
a
40.
a
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