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Unit 9- Coordination Compounds
Some important terms related to coordination compounds
(i) Coordination entity: A complex compound that constitutes a central metal (atom or ion) linked with a
fixed number of ions or molecules. For example, [Ni(CO)4], [PtCl2(NH3)2], [Fe(CN)6]4–,
[Co(NH3)6]3+., etc.
(ii) Central atom/ion: The atom or ion to which a fixed number of ions/groups are bound in a certain
geometrical arrangement around it. Since it accepts a lone pair of electrons for the formation of
coordinate bond, it is also referred to as Lewis acids. For example, Fe3+ and Ni2+ are the central ions in
the coordination compounds [Fe(CN)6]3– and NiCl2.6H2O respectively.
(iii) Ligands: The atoms, ions or molecules which donate a pair of electrons to the metal atom to form a
coordinate
bond, are called ligands. For example, NH3, H2O, Cl─, CN─, CO etc.
Depending on the number of donor atoms, a ligand can be of following types:
(a) Unidentate or Monodentale ligand: It contains only one donor atom. For example, , NH3, H2O, Cl─,
CN─, CO
in which N,O, Cl, C are the donor atoms which bind with metal atom or ion.
(b) Didentate or Bidentate ligand: When a ligand has two donor atoms, for example, ethane-1,2-diamine
(H2NCH2CH2NH2 ), in which the two nitrogen atoms of the amino group act as donor atoms.
(c) Polydentate or Multidentate ligand: When several donor atoms are present in a single ligand, for
example (EDTA4–) (Ethylenediaminetetraacetate), is an important hexadentate ligand which can bind
through two nitrogen and four oxygen donor atoms to a central metal ion.
(d) Chelate ligand: A di- or polydentate ligand is said to be a chelate ligand when it uses its two or more
donor atoms to bind a single metal ion. The number of such ligating groups is called the denticity of the
ligand.
A complex compound in which the donor atoms are attached to the metal so that the metal becomes a part
of the heterocyclic ring, is called chelate complex.
(v) Coordination number (CN): The number of unidentate ligands directly bonded to the central metal
atom/ion is known as the coordination number of that metal ion/atom.
For example, in the complex ions, [ Ag(NH3)2]2+ , [Zn(CN)4]2─ , & [Ni(NH3)6]2+ the coordination
number of Ag, Zn and Ni are 2, 4 and 6 respectively.
When the bonded ligands are didentate the coordination number is double the number of ligands because
the number of bonds linked to the central metal becomes double. For example, the coordination number
of Fe in [Fe(C2O4)3] 3─is 6, because C2O42─is a didentate ligand.
(vi) Coordination polyhedron: It describes the spatial arrangement of the ligand atoms which are directly
attached to the central atom/ion. For example, the coordination polyhedra of following complexes
are tetrahedral, octahedral and square planar respectively.
(vii) Coordination sphere: The coordination complex which constitutes the central atom/ion and the ligands,
are represented in a square bracket, collectively termed as coordination sphere. The ionisable groups are
written outside the bracket, called counter ions.
For example, in the complex K4[Fe(CN)6] the coordination sphere is[Fe(CN)6]4–, and the counter ion is
K +.
(viii) Homoleptic and Heteroleptic complexes: Complexes in which a metal is bound to only one kind of
donor groups, e.g., [Co(NH3)6]3+, are known as homoleptic. Complexes in which a metal is bound to
more than one kind of donor groups,e.g., [Co(NH3)4Cl2]+, are known as heteroleptic.
(ix) Charge on a complex ion: The charge carried by a complex ion is the algebraic sum of charges carried
by the central metal ion and the coordinated groups or ions. For example,
[Fe(CN)6]3–, Fe3+ = +3 charge, 6 CN─ = 6 × (─1) = ─ 6 Charge
Thus charge o [Fe(CN)6] = +3 – 6 = − 3
(x) Types of complex ion:
(i) Cationic complex ; [ Ag(NH3)2]2+ , [Ni(NH3)6]2+
(ii) Anionic complex ; [Fe(CN)6]3–, [Zn(CN)4]2─
(iii) Neutral complex ; [Co(NH3)3Cl3],[ Ni(CO)4 ] etc
Werner’s theory of coordination compounds
The main postulates of Werner’s theory (proposed by Werner in 1898), are as follows:
a. In coordination compounds metals show two types of linkages (valancies), primary and secondary.
b. The primary valancies are normally ionisable, non-directional and are satisfied by negative ions.
c. The secondary valancies are non-ionisable, directional which are satisfied by negative ions or neutral
molecules. The secondary
valency is equal to the coordination number and is fixed for a metal.
d. The ions/groups bound by the secondry linkages to the metal have characteristic spatial arrangements
corresponding to different coordination numbers.
Werner further postulated that the most common geometrical shapes of coordination compounds are
octahedral, tetrahedral and square planar.
IUPAC Nomenclature of coordination compounds
(i) Writing the formulas of mononuclear (containing single central metal atom) coordination
compounds:
(a) The central atom is listed first.
(b) The ligands are then listed in alphabetical order without considering their charge.
(c) Polydentate ligands are also written alphabetically. In case of abbreviated ligand, the first letter of
abbreviation is used to determine the position of ligand in alphabetical order.
(d) The formula of the entire coordination entity, whether charged or uncharged, is enclosed in square
brackets. When ligands are polyatomic, their formulas are enclosed in parantheses. Ligand abbreviations
are also enclosed in parantheses.
(e) There should be no space between the ligands and the metal within a coordination sphere.
(f) When the formula of a charged coordination entity is to be written without that of the counter ion, the
charge is indicated outside the square brackets as a right superscript with the number before the sign.
For example, [Ag(NH3)2]2+ , [Ni(NH3)6]2+ [Fe(CN)6]3–, etc.
(g) The charge of cation(s) is balanced by the charge of anion(s).
(ii) Writing the name of coordination compounds:
(a) The name of cation is written first in both positively and negatively charged coordination entities
followed by the naming of anion.
(b) The legands are named in an alphabetical order before the name of central atom/ion. (This procedure
is opposite to that in writing formula).
(c) Names of anionic legands and in – O, those of cationic and neutral ligands are the same except aqua
for
H2O, ammine for NH3, carbonyl for CO and nitrosyl for NO. These are placed within closing
marks [ ].
Note: IUPAC recommendations (2004) The anion endings 'ide', 'ate' and 'ite' (cf.
Section IR-5.3.3) are changed to 'ido', 'ato' and 'ito', respectively, when generating the
prefix for the central atom
(d) Prefixes mono, di, tri, etc., are used to indicate the number of the individual ligands in the
coordination entity.
When the names of the ligands include a numerical prefix, then the terms, bis, tris, tetrakis are used,
the ligand to which they refer being placed in parenthesis. For example,
[NiCl2(PPh3)2] is named as dichlorobis(triphenylphosphine) nickel (II).
(e) Oxidation state of the metal in cation, anion or neutral coordination entity is indicated by roman
numerical in parenthesis.
(f) If the complex ion is a cation, the metal is named same as the element. For example, Co in a complex
cation is called cobalt and Pt is called platinum.
(g) If the complex ion is an anion, the name of the metal ends with the suffix- ate. For example, Co in a
complex
[Co(NH3)Br5]2− anion, is called cobaltate. For some metals, the latin names are used in the
complex anions, e.g., ferrate for Fe.
(g) The neutral complex molecule is named similar to that of the complex cation.
Example 1. Write the IUPAC name of the following coordination compounds:
1. [Cr(NH3)3(H2O)3]Cl3
: triamminetriaquachromium(III) chloride
2. [Co(H2NCH2CH2NH2)3]2(SO4)3
: tris(ethane-1,2–diammine)cobalt(III)
sulphate.
3. [Ag(NH3)2][Ag(CN)2]
: diamminesilver(I) dicyanoargentate(I)
4. Hg[Co(SCN)4]
: Mercury tetrathiocyanatocobaltate(III)
5. [CoCl2(en)2]Cl
: Dichloridobis(ethane-1,2-diamine)cobalt(III)
chloride
Example 2. Write the molecular formulas of the following coordination compounds:
(i) Potassium tetrahydroxidoozincate(II)
: K 2[Zn (OH) 4 ]
(ii) Diamminechloridonitrito-N-platinum(II)
: [Pt(NH3)2Cl(NO2)]
(iii) Tetraammineaquachloridocobalt(III) chloride
: [Co(NH3)4(H2O)Cl]
(iv) Potassium trioxalatoaluminate(III)
: K3[Al(C2O4)3]
Isomerism In Coordination Compounds
Isomers are those compounds which have the same chemical formula but different structural
arrangements of their atoms. Different arrangement of atoms due to their different structures are
responsible for their different physical or chemical properties.
1. Stereoisomerism: Stereoisomers have the same chemical formula and chemical bonds but they have
different special arrangements.. They are further classified as follows:
(i) Geometrical isomerism: It arises in heteroleptic coordination complexes due to different possible
geometric arrangements of the ligands. When similar ligands are adjacent to each other, they form cis
isomer and when they are opposite to each other a trans isomer is formed. Geometrical isomerism is very
common in complexes with coordination number 4 and 6.For example, platinum ammine complexes are
geometrical isomers, as described below:
(A) Square Planar Complexes :-
(B) Octrahedral Complexes :(i) Cis – Trans :
(ii) facial (fac) o and meridional (mer) isomer
Note : Tetrahedral complexes do not show geometrical isomerism because the
relative positions of the unidentate ligands attached to the central metal atom
are the same with respect to each other.
(ii) Optical isomerism: It arises due to absence of elements of symmetry (plane of symmetry or axis
of symmetry) in the complex.
Optical isomers or enantiomers are the mirror images that cannot be superimposed on one another. The
molecules or ions that cannot be superimposed are called chiral.
A chiral molecule is an optically active and has the property of rotating the plane of polarized light either
to its left (called laevo) or to its right (called dextro). If polarized light remains undeflected, the
compound is inactive or racemic (i.e., mixture of 50% laevo and 50% dextro).
Optical isomerism is common in octahedral complexes involving didentate ligands.
oo
In a coordination compound of the type [CoCl2(en)2]2+ only the cis-isomer shows optical activity (see
the figure above).
2. Structural isomerism: same chemical formula but possess different types of bonds , differ in the
extent of ionization, position of ligands, etc. These are further classified as follows:
(i) Linkage isomerism: It arises in the coordination compounds containing ambidentate ligands. An
ambidentate ligand can link with the metal atom/ion in two different ways. So two types of structures are
formed, called linkage isomers. For example, in the complex [Co(NH3)(NO2)]Cl2, nitrite ligand is bound
to the metal in two different ways a red form, in which the nitrite ligand is bound through oxygen (–ONO),
and as the yellow form, in which the nitrite ligand is bound through nitrogen (–NO2).
(ii) Ionisation isomerism: When the counter ion in a complex salt acts as a ligand and the ligand of the
complex becomes counter ion (i.e., a mutual exchange between counter ion and ligand), the two forms of
the complex are called ionisation isomers and the process is called ionisation isomerism. For example,
[Co(NH3)5SO4]Br (red complex) & [Co(NH3)5 Br] SO4 (violet complex)
(iii) Coordination isomerism: When there is an interchange between cationic and anionic species of
different metal ions and the ligands present in a complex, this type of isomerism arises. For example,
[Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]
(iv) Solvate isomerism: It is known as hydrate isomerism when water is the solvent. In solvate isomers
solvent molecules are either directly bound to the metal ion or may be present as free solvent molecules in
the crystal lattice. For example,
[Cr(H2O)6]Cl3 (violet) and , [Cr(H2O)5Cl]Cl2.H2O (grey
green)
Bonding in coordination compounds
(i) Valency Bond Theory:
· Empty Metal orbitals hybridise to form equal number of hybrid orbitals.
· The hybrid metal orbitals then overlap with those ligand orbitals that can donate an electron pair for
bonding. In this way a bond is formed between metal ion and the ligand’s donor atom.
· The resulting complex will be diamagnetic if all the electrons are paired. If unpaired electrons are
present then the complex will be paramagnetic.
Number of orbitals and types of Hybridisations
Coordination Type of hybridisation
number
Distribution of hybrid orbitals in
space
Type of Complex
4
4
sp3
dsp2
Tetrahedral
Square planar
Outer orbital/ high spin
Inner orbital/ low spin
5
sp3d
Trigonal bipyramidal
Outer orbital/ high spin
6
3 2
Octahedral
Outer orbital/ high spin
3
Octahedral
Inner orbital/ low spin
6
sp d
2
d sp
Application of Valence Bond Treatment to Some Complexes
Ion/
Complex
Central
metal
ion
Confi-guration
of metal ion
Hybridi-zation of Geometry of the
metal ion involved
complex
d2sp3
Octa-hedral
Number of
unpaired
electrons
1
Magnetic
behaviour
d2sp3
Octa-hedral
2
Para-magnetic
d2sp3
Octa-hedral
3
Para-magnetic
d2sp3
Octa-hedral
3
Para-magnetic
sp3d2
Octa-hedral
4
Para-magnetic
d2sp3
Octa-hedral
2
Para-magnetic
sp3
Tetra-hedral
5
Para-magnetic
sp3d2
Octa-hedral
5
Para-magnetic
sp3d2
Octa-hedral
5
Para-magnetic
d2sp3
Octa-hedral
1
Para-magnetic
d2sp3
Octa-hedral
0
Dia-magnetic
sp3
Tetra-hedral
4
Para-magnetic
d2sp3
Octa-hedral
0
Dia-magnetic
sp3d2
Octa-hedral
4
Para-magnetic
sp3
Tetra-hedral
0
Dia-magnetic
dsp2
Square planar
0
Dia-magnetic
sp3
Tetra-hedral
2
Para-magnetic
sp3d2
Octa-hedral
2
Para-magnetic
sp3
Tetra-hedral
1
Para-magnetic
sp3
Tetra-hedral
0
Dia-magnetic
dsp2
Square planar
0
Dia-magnetic
Para-magnetic
(ii) Crystal Field Theory:
 According to crystal field theory, the bonding between a central metal ion and a ligand is purely
electrostatic.
 In an octahedral field s-orbital (because of no degeneracy) and p-orbitals (because of their shape)
are not affected, but the degeneracy of d-orbitals is lifted because all d-orbitals are not spatially
equivalent.
 The valence electrons of metal are repelled by the negatively charge ligands, so that they occupy
those d-orbitals which have their lobes away from the direction of ligands.
 The effect of ligands is particularly marked on d-electrons and it depends on the number of
electrons.
Crystal Field Splitting of d-orbitals.
The five d-orbitals can be classified into two sets as follows:
Three of d-orbitals i.e., dxy, dyz and dzx which are oriented in between the co-ordinate axes are called t2g
-orbitals.
The other two d-orbitals i.e., dx2-y2 and dz2 oriented along the axes are called eg orbitals.
· In the case of free metal ions, all the five d-orbitals degenerate, i.e., they have equal energy. But their
interactions from the one pair of ligands and their energies also become deficit. This splitting of five dorbitals of metal ions under the influence of approaching ligands is called crystal field splitting. It is
designated by and is called crystal field splitting energy.
· The ligands which cause greater crystal field splitting are termed as strong ligands while those causing
lesser crystal field splitting are weak ligands. The decreasing order of field strength among some of the
ligands are:
-
-
4-
-
-
2-
-
-
-
-
-
CO>CN > NO > en > py > NH > EDTA > SCN > H O >ONO > ox > OH > F > SCN > Cl > Br >
-
2
3
2
Crystal Field Splitting in Octahedral Complexes
3
5
2
∆o
5
∆o
In octahedral complexes the six ligands approach the central metal ion along the co-ordinate axe dx2-y2
and dz2orbitals. Consequently, the eg set of orbitals has higher energy than t2g of orbitals.
Electron Configuration in d-Orbitals
4
Δ > P , low spin d
4
Δ < P , high spin d
(i) If ∆o < P, the fourth electron enters one of the eg orbitals giving the configuration 1 . Ligands for
which ∆o < P
are known as weak field ligands and form high spin complexes.
(ii) If ∆o > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital
with
Configuration 2. Ligands which produce this effect are known as strong field ligands
and form
low spin complexe
Crystal Field Splitting in Tetrahedral Complexes
In tetrahed ral complex, four ligands may be imagined to occupy the alternate corners of the cube and the
centre ion at the centre of the cube. In this situation, the t2g set of orbital lie relatively nearer to the
approaching ligands and therefore t2g set of d-orbitals have higher energy than , eg set of orbitals.
Relationship between ∆t and ∆o is
𝟒
∆t = ∆o
𝟗
Colour of transition metal complexes: Colour associated with the transition metal complexes is due to
the transition of electrons between d-orbitals (from t2g to eg in octahedral complexes and from eg to t2g in
tetrahedral complexes). Such transitions are called d-d transitions. Transition metal complexes absorb
some selected wavelengths of visible light and
appear coloured.
Magnetic properties
compounds:
of
complex
Complexes
with
unpaired
e
are
paramagnetic. The no. of
unpaired e
depends upon electronic structure of dn+ ion
which further depends upon extent of CF
splitting. e.g. [Fe(CN)6]3–has magnetic
moment of a single unpaired electron while
[FeF6]3– has paramagnetic moment of five .
unpaired electrons
Fe3+ in [Fe(CN)6]3– : d5
3+
Fe in [FeF6]3– : d
5
↑↓ ↑↓
t2g
↑
↑
↑
↑
eg
↑
↑
Bonding in metal carbonyls :
These are Homoleptic Complexs of transition metals with CO.
The metal-carbon bond in metal carbonyls possess both s and p
character. The M–C  bond is formed by the donation of lone
pair of electrons on the carbonyl carbon into a vacant orbital of
the metal. The M–C  bond is formed by the donation of a
pair of electrons from a filled d orbital of metal into the vacant
antibonding * orbital of carbon monoxide. The etal to ligand
bonding creates a synergic effect which strengthens the bond
between CO and the metal.
Application of coordination compounds
(i)
Complex formation is frequently encountered in analytical chemistry. For example
identification of Cu2+ is based on
the formation of a blue complex with NH3 :CuCl2 + 4NH3 ----- [Cu(NH3)4]Cl2
and that of Fe3+on the formation of a red complex with KSCN :FeCl3 + KSCN ------ K3[Fe(SCN)6]+ 3KCl
Similarly, Ni2+ is estimated as red complex with dimethyl glyoxime (DMG).
(ii) Complex formation is used in the extraction of metals from their ores. For example Ni is extracted
from its ores as volatile nickel carbonyl.
Ni + 4CO ------ Ni(CO)4 ↑
 Ni + 4CO ↑
(iii) Metal complexes of Ag, Au, Cu, etc., are used for electroplating of these metals on the desired
objects.
(iv) Many biological processes involve complex formation. For example haemoglobin, chlorophyll,
vitaminB12,cisplatinare complexes.
(v) Hardness of water is estimated by complexometric titration of Ca2+ and Mg2+ with ethylene
diaminetetraacetic acid (EDTA).
VERY SHORT ANSWER TYPE QUESTION (1 marks)
Q.1- Write the IUPAC name of ionization isomer of [ Co (NH3)5Br] SO4.
A.1- Ionization isomer is [Co (NH3)5 SO4] Br.
IUPAC name- pentaamminesulphatocobalt (iii) bromide.
Q.2- Write the IUPAC name of linkage isomer of [Cr (en)2(ONO)2]Cl.
A.2-Linkage isomer is [Cr (en)2 (NO2)2] Br.
IUPAC name- bis(ethane-1,2-diamine)dinitrito-N-chromium(iii) bromide.
Q.3- Write the formula for the following coordination compounds:(i) Tetraamminediaquacobalt(III) chloride
(ii) Iron (III) hexacyanoferrate (II).
A.3- (i) [Co(NH3)4(H2O)2]Cl3 (ii) Fe4[Fe(CN)6]3
Q.4- What is meant by the chelate effect? Give an example.
A.4- A complex in which there is a close ring of atoms caused by attachment of ligand to a metal
atom at two points is called chelate effect. e.g.
Q.5-What is the coordination number of central atom in followings:(i) [Cr (en)2 (NO2)2] Br. (ii) [Co (NH3)3 SO4] Br.
A.5-(I) 6, because en is bidentate ligand. (ii) 4.
Q.6- Write the IUPAC name of Na3[ Cr (OH)2F4].
A.6- sodium tetrafloridodihydroxidochromate(iii).
Q.7- What is ambidentate ligand?
A.7- A monodentate ligand which has two donor atoms but attach only with one donor site is
called ambidentate ligand. e.g. NO2 - & ONOQ.8- Which complex is used in the treatment of cancer?
A.8- cis-platin or cis- [Pt (NH3)2 Cl2]
Q.9- Write the geometry of coordination compound in which metal atom is present in (i) sp3
(ii) dsp2 hybridization state.
A.9- sp3 –tetrahedral dsp2 – square planar.
Q.10- What is crystal field splitting energy?
A.10- The energy difference between t2g & eg orbitals is known as Crystal Field Splitting Energy
(CFSE) (
).
SHORT ANSWER TYPE QUESTION (2 marks)
Q.1- Deduce shape and magnetic behavior of [Fe(CN)6]3Ans.-The oxidation state of Fe in this complex is +3.i.e.
Fe+→[Ar] 3d6 4s2
Fe3+→[Ar] 3d5
Fe3+ in complex state:- CN- strong field ligand and thus filling takes place against Hund rule.
2
3
Therefore hybridization state is d sp and it is weakly paramagnetic.
Q.2- (a)What is spectrochemical series?
(b) what are inner orbital complex?
A.2- (a) It is a series in which ligands can be arranged in the order of increasing field strength
or in the order of increasing magnitude CFSE.
(b) The complex which uses its inner i.e. (n-1) d orbital for complex formation is known as
inner orbital complex. Such complex is formed by strong field ligands.
Q.3- Give an example in each case the role of coordination compound in (i) Biological systems
(ii) Medicinal chemistry (iii) Heterogeneous catalysis (iv) extraction of metals
A.3- (i) In biological system Chlorophyll & Hemoglobin is a complex of Mg2+ &Fe3+ respectively.
(ii) cis-platin or cis- [Pt (NH3)2 Cl2] is used in the treatment of cancer.
(iii)Wilkinson catalyst [(Ph3)3RhCl] is used for hydrogenation of alkenes.
(iv) In extraction of metals like gold silver complex Na [Ag(CN)2]is formed.
Q.4- Draw a sketch to show the splitting of d-orbitals in an octahedral crystal field.
A.4-
Q. 5. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)]and how many of these will
exhibit optical isomers?
Ans)[Pt(NH3)(Br)(Cl)(py)
From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely
show optical isomerization. They do so only in the presence of unsymmetrical chelating
agents.
Q.6- Using the VBT deduces the shape & magnetic character of [Ni (CN)4]2-.
A.6- Ni (28) :- 3d8 ,4s2
Ni2+ :- 3d8
↑↓ ↑↓ ↑↓ ↑
↑
3d
4s
4p
Ni2+ in complex state: - since CN- is strong field ligand, therefore filling takes against Hund’s rule.
↑↓ ↑↓ ↑↓ ↑↓
↑↓
↑↓
↑↓ ↑↓
3d
4s
2
4p
Thus, hybridization state is dsp , therefore geometry is square planar.Since all electrons are
paired, therefore complex is Diamagnetic.
Q.7- Draw facial and meridional isomer of [Co(NH3)3(NO2)3]
A.7-
Q.8-Draw optical isomers of [Co(en)3]2+
A.8.
Q.9-Draw cis-trans isomer of [Pt(en)2Cl2]2+
A.9.-
Q.10- Draw optical isomers of cis-[Pt(en)2Cl2]2+
A.10-
Q.11- (a)What is spectrochemical series?
(b) what are inner orbital complex?
A.12- (a) It is a series in which ligands can be arranged in the order of increasing field strength
or in the order of increasing magnitude CFSE.
(b) The complex which uses its inner i.e. (n-1) d orbital for complex formation is known as
inner orbital complex. Such complex is formed by strong field ligands.
Q.13-A coordination compound has the formula CoCl3.4 NH3. It does not liberate ammonia
but forms a precipitate with silver nitrate. Write the structure and IUPAC name of compound.
A.13- The oxidation state of Co in compound is +3 and that its coordination number is +6. Since
it does not liberate ammonia and thus it is present in coordination sphere. Therefore, the
formula is [Co(NH3)3Cl2]Cl . Its IUPAC name is tetraamminedichloridocobalt(III) chloride.
Q.14- Draw a sketch to show the splitting of d-orbitals in an octahedral crystal field.
A.14-
Q15. Write the molecular formulas of the following coordination compounds:
(i) Potassium tetrahydroxidoozincate(II)
: K2[Zn (OH) 4 ]
(ii) Diamminechloridonitrito-N-platinum(II)
: [Pt(NH3)2Cl(NO2)]
(iii) Tetraammineaquachloridocobalt(III) chloride
: [Co(NH3)4(H2O)Cl]
(iv) Potassium trioxalatoaluminate(III)
: K3[Al(C2O4)3]
SHORT ANSWER TYPE QUESTION (3 marks)
Q.1- How would account for followings:(i) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.
(ii) [Fe(CN)6] 3- is weakly paramagnetic while is [Fe(CN)6] 4- diamagnetic.
(iii) [Ni(CO)4] possesses tetrahedral geometry while [Ni(CN)4]2- is square planar.
A.1-The electronic configuration of Ti 3+→[Ar] 3d1,since H2O is weak field ligand therefore filling
takes place according to Hund rule.
[Ti(H2O)6]3+
Due to presence of unpaired electron it is coloured.
[Sc(H2O)6]3+
Due to absence of unpaired electron it is Diamagnetic
(ii) The outer electronic structure of [Fe(CN)6] 3-and [Fe(CN)6] 4- are:Fe in [Fe(CN)6] 3-
Fe in [Fe(CN)6] 4-
Due to presence of one unpaired electron in 3d orbital of [Fe(CN)6] 3- it is weakly paramagnetic.
(iii) The outer electronic structure of [Ni(CO)4] and [Ni(CN)4]2- are:Ni in [Ni(CO)4]
Ni in[Ni(CN)4]2-
Due to sp3 hybridization state geometry of [Ni(CO)4] is tetrahedral, Whereas due to dsp2
hybridization state geometry is square planar.
Q.2-Describe the type of hybridization, shape and magnetic property of followings:(i)[Fe(H2O)6]2+ (II) [Co(NH3)]3+ (iii) [NiCN4]2-.
A.2- (i) Fe in Fe(H2O)6]2+
Hybridization:- sp3d2
Shape:- Octahedral
Magnetic property:- Paramagnetic
(ii) Co in [Co(NH3)]3+
Hybridization:- d2 sp3
Shape:- Octahedral
Magnetic property:- Diamagnetic
2(iii) Ni in [NiCN4]
Hybridization:- dsp2
Shape:- Square planar
Magnetic property:- Diamagnetic.
Q.3-(a) State for a d6 ion the actual configuration in split d-orbital in an octahedral crystal
field is decided by the magnitude of CFSE(∆o) and pairing energy (P).
(b) Tetrahedral complex of type [MA2B2] does not show geometrical isomerism. Why?
A.3-(a)There are two ways of placing electron, which is determined by the magnitude of ∆o and
P i.e.
(i) When ∆o > P the electrons will occupy the more stable t2g orbitals i.e. filling takes place
against Hund rule.
(ii) When ∆o < P the electrons will spread over entire set of d-orbitals. i.e t2g and eg means filling
takes place according Hund Rule.
(b) This is because the relative positions of the unidentate ligands attached to central atom are
the same with respect to each other.
Q.4-(A) Mention the factors effecting stability of complex.
(B) Draw the structure EDTA.
2–
(c) The spin only magnetic moment of [MnBr4] is 5.9 BM. Predict the geometry of the
complex ion
A.4-(A)Following factors decide stability of complex:(a)Charge on metal atom :- Higher the charge on central atom, stability of ligand will be more.
(b) Nature of ligand :- (i) Strong field ligand form stable complex than weak field ligand.
(ii) Chelating ligand form stable complex than monodentate ligand.
(B) EDTA –ethylene di amine tetra acetate.
(c) Since the coordination number of Mn2+ ion in the complex ion is 4, it will be either
tetrahedral (sp3 hybridisation) or square planar (dsp2hybridisation). However, the fact that the
magnetic moment of the complex on is 5.9 BM, it should be tetrahedral in shape rather than
square planar because of the presence of five unpaired electrons in the d orbitals.
Q.5. Indicate the types of isomerism exhibited by the following complexes and draw the structures
for these isomers :(i) K[Cr(H2O)2(C2O4)] (ii) [Co(en)3]Cl3 (iii) [Co(NH3)5(NO2)](NO3)2
Sol .(i) (a) Both geometrical (cis-trans) and optical isomers for cis can exist. (b) Optical isomers (d- and
l-) of cis
(ii) Two optical isomers can exist.
(iii) Ionisation isomers :
[Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5(NO3)](NO3) (NO2)
Linkage isomers :
[Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5(ONO)](NO3)2
Q.6. Write the IUPAC name of the following coordination compounds:
1. [Cr(NH3)3(H2O)3]Cl3
: triamminetriaquachromium(III) chloride
2. [Co(H2NCH2CH2NH2)3]2(SO4)3
: tris(ethane-1,2–diammine)cobalt(III)
sulphate.
3. [Ag(NH3)2][Ag(CN)2]
: diamminesilver(I) dicyanoargentate(I)
4. Hg[Co(SCN)4]
: Mercury tetrathiocyanatocobaltate(III)
5. [CoCl2(en)2]Cl
: Dichloridobis(ethane-1,2-diamine)cobalt(III)
Chloride
6. [Co(NH3)4(H2O)Cl]
: Tetraammineaquachloridocobalt(III) chloride
HOTS Questions
1. [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral.
Why?
Answer Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magneticcharacters are
different. This is due to a difference in the nature of ligands. Cl− is aweak field ligand and it does
not cause the pairing of unpaired 3d electrons. Hence,[NiCl4]2− is paramagnetic.
In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.But CO is a
strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons.Also, it causes the 4s
electrons to shift to the 3d orbital, thereby giving rise to sp3hybridization. Since no unpaired
electrons are present in this case, [Ni(CO)4] is diamagnetic.
2. Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion.
Answer-[Pt(CN)4]2−, In this complex, Pt is in the +2 state. It forms a square planar structure.
Thismeans that it undergoes dsp2 hybridization. Now, the electronic configuration of Pd(+2)
is5d8.CN− being a strong field ligand causes the pairing of unpaired electrons. Hence, there are
no unpaired electrons in[Pt(CN)4]2−
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+
Answer :The central metal ion in all the three complexes is the same. Therefore, absorption in
the visibleregion depends on the ligands. The order in which the CFSE values of the ligands
increases in the spectrochemical series is as follows: H2O < NH3 < NO
Hence, the wavelengths of absorption in the visible region will be in the order:
[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4−
Value Based Questions
(1) Nowadays younger generation girls are interested in wearing ornaments made of white
metal platinum
(i) Are you interested in wearing ornaments made of platinum or will you save platinum
by not wearing it? Why?
(ii) Name the life saving drug prepared from platinum
(iii) Which disease can be cured by that complex?
Answer :i.I will not wear but I will save it for medicinal use .
ii.cis platin
iii.to cure cancer
(2) Ram is a poor boy. He never brings lunch to school. Sam is his friend and Sam shares
his lunch with Ram.
(i) Which type of value Sam has?
(ii) Seeing this condition which type of chemical bond do you recall?
(iii) Can you give a complex name and formula with this type of bond?
Answer :i.co ordinate bond Sam is a kind and helping boy.
ii. coordinate bond
iii. K4[Fe(CN)6 Pottasiumhexacyanoferrate(II).
(3) When cobalt III chloride and ammonia are combined we get yellow, purple, green,
violet coloured coordination complexes. Like wise when people make friendship with
different type of people their personality changes. (i) Which type of friends are you ? (ii)
Write any one good character of your friend. (iii) Write any two complex formed by the
combination CoCl3 and NH3 .
Answer
i. helping in need of my friend.
ii.My friend is affectionate and truthful
iii.[Co Cl2 (NH3)4]Cl and [CoCl (NH3)5]Cl2