Coordination Compounds

9.Coordination Compound Coordination Compounds are the backbone of modern inorganic and bio–inorganic chemistry and chemical industry.In the previous Unit we learnt that the transition metals form a large number of complex compounds in which the metal atoms are bound to a number of anions or neutral molecules. In modern terminology such compounds are called coordination compounds. The chemistry of coordination compounds is an important and challenging area of modern inorganic chemistry. New concepts of chemical bonding and molecular structure have provided insights into the functioning of vital components of biological systems. Chlorophyll, haemoglobin and vitamin B12 are coordination compounds of magnesium, iron and cobalt respectively. Variety of metallurgical processes, industrial catalysts and analytical reagents involve the use of coordination compounds. Coordination compounds also find many applications in electroplating, textile dyeing and medicinal chemistry. 9.1 Werner’s Theory of Coordination Compounds Alfred Werner (1866-1919), a Swiss chemist was the first to formulate his ideas about the structures of coordination compounds. He prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour by simple experimental techniques. Werner proposed the concept of a primary valence and a secondary valence for a metal ion. Binary compounds such as CrCl3, CoCl2 or PdCl2have primary valence of 3, 2 and 2 respectively. In a series of compounds of cobalt(III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess silver nitrate solution in cold but some remained in solution. 1 mol CoCl3.6NH3 (Yellow) gave 3 mol AgCl 1 mol CoCl3.5NH3 (Purple) gave 2 mol AgCl 1 mol CoCl3.4NH3 (Green) gave 1 mol AgCl 1 mol CoCl3.4NH3 (Violet) gave 1 mol AgCl These observations, together with the results of conductivity measurements in solution can be explained if (i) six groups in all, either chloride ions or ammonia molecules or both, remain bonded to the cobalt ion during the reaction and (ii) the compounds are formulated as shown in Table 9.1, where the atoms within the square brackets form a single entity which does not dissociate under the reaction conditions. Werner proposed the term secondary valence for the number of groups bound directly to the metal ion; in each of these examples the secondary valences are six. SOLUTION CONDUCTIVIT Y COLOUR FORMULA CORRESPONDS TO Yellow [Co(NH3)6]3+3ClPurple [CoCl(NH3)6]2+2ClGreen [CoCl2(NH3)6]+ClViolet [CoCl2(NH3)4]+2Cl- 1:3 electrolyte 1:2 electrolyte 1:1 electrolyte 1:1 electrolyte Table 9.1: Formulation of Cobalt(III) Chloride-Ammonia Complexes Note that the last two compounds in Table 9.1 have identical empirical formula, CoCl3.4NH3, but distinct properties. Such compounds are termed as isomers. Werner in 1898, propounded his theory of coordination compounds. The main postulates are: 1. In coordination compounds metals show two types of linkages (valences)-primary and secondary. 2. The primary valences are normally ionisable and are satisfied by negative ions. 3. The secondary valences are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal. 4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers. In modern formulations, such spatial arrangements are called coordination polyhedra. The species within the square bracket are coordination entities or complexes and the ions outside the square bracket are called counter ions. He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. Thus, [Co(NH3)6]3+, [CoCl(NH3)5]2+ and [CoCl2(NH3)4]+ are octahedral entities, while [Ni(CO)4] and [PtCl4]2−are tetrahedral and square planar, respectively. Example 9.1 On the basis of the following observations made with aqueous solutions, assign secondary valences to metals in the following compounds: dissociate into simple ions completely when dissolved in water.MgCl2. etc. However.6H2O.6H2O 2 (iii)PtCl4.6H2O. Mohr’s salt.4NH3 1 (v)PtCl2.Formula Moles of AgCl precipitated per mole of the compounds with excess AgNO3 MOLES OF AGCL PRECIP ITATED PER MOLE OF THE FORMULA COMPOUNDS WITH EXCESS AGNO3 (i) PdCl2. do not dissociate into Fe2+ and CN− ions. potash alum. complex ions such as [Fe(CN)6]4− of K4Fe(CN)6.2HCl 0 (iv)CoCl3.4NH3 2 (ii) NiCl2.12H2O. However. 1866.2NH3 0 Solution (i) Secondary 4 (ii) Secondary 6 (iii) Secondary 6 (iv) Secondary 6 (v) Secondary 4 Difference between a double salt and a complex Both double salts as well as complexes are formed by the combination of two or more stable compounds in stoichiometric ratio.(NH4)2SO4. KAl(SO4)2. they differ in the fact that double salts such as carnallite. FeSO4. a small . in Mülhouse. KCl. Werner was born on December 12. His accomplishments included the development of the theory of coordination compounds. Other examples are [Ni(CO)4]. This theory. [Fe(CN)6]4− . at the age of 29 years became a full professor at Technische Hochschule in Zurich in 1895. from 1890 to 1893. in which Werner proposed evolutionary ideas about how atoms and molecules are linked together. In fact. was formulated in a span of only three years. For example. 9. [CoCl3(NH3)3] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions. Werner was the first to discover optical activity in certain coordination compounds. He. His study of chemistry began in Karlsruhe (Germany) and continued in Zurich (Switzerland). [PtCl2(NH3)2]. Alfred Werner was a chemist and educationist.2 Definitions of Some Important Terms Pertaining to Coordination Compounds (a) Coordination entity A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. he explained the difference in properties of certain nitrogen containing organic substances on the basis of isomerism. where in his doctoral thesis in 1890.community in the French province of Alsace. Werner became the first Swiss chemist to win the Nobel Prize in 1913 for his work on the linkage of atoms and the coordination theory. He extended vant Hoff’s theory of tetrahedral carbon atom and modified it for nitrogen. Wer ner showed optical and electrical differences between complex compounds based on physical measurements. [Co(NH3)6]3+ . . The remainder of his career was spent gathering the experimental support required to validate his new ideas. the atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement around it. [CoCl(NH3)5]2+ and [Fe(CN)6]3– are Ni2+.2-diamine) or C2O42− (oxalate). When a di. respectively. the ligand is said to be unidentate. When a ligand is bound to a metal ion through a single donor atom. the ligand is said to be didentate and when several donor atoms are present in a single ligand as in N(CH2CH2NH2)3. called chelate complexes tend to be . Such complexes. The number of such ligating groups is called the denticity of the ligand. is called the central atom or ion. Co3+ and Fe3+. larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules. These may be simple ions such as Cl− . as with Cl− . it is said to be a chelate ligand. such as proteins. When a ligand can bind through two donor atoms as in H2NCH2CH2NH2 (ethane-1. It can bind through two nitrogen and four oxygen atoms to a central metal ion. the central atom/ion in the coordination entities: [NiCl2(H2O)4]. Ethylenediaminetetraacetate ion (EDTA ) is an important hexadentate ligand. H2O or NH3. For example.(b) Central atom/ion In a coordination entity.or polydentate ligand uses its two or more donor atoms to bind a single metal ion. small molecules such as H2O or NH3. the ligand is said to be polydentate. (c) Ligands The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These central atoms/ions are also referred to as Lewis acids. NO2−ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. (d) Coordination number The coordination number (CN) of a metal ion in a complex can be defined as the number of ligand donor atoms to which the metal is directly bonded. in the complex ions. The ionisable groups are written outside the bracket and . are not counted for this purpose. if formed between the ligand and the central atom/ion. It is important to note here that coordination number of the central atom/ion is determined only by the number of sigma bonds formed by the ligand with the central atom/ion. Pi bonds.2-diamine) are didentate ligands. the coordination number of Pt and Ni are 6 and 4 respectively. is 6 because C2O42– and en (ethane-1. SCN− ion can coordinate through the sulphur or nitrogen atom. Similarly. [PtCl6]2– and [Ni(NH3)2+] . Fe and Co. in the complex ions. Examples of such ligands are the NO2− and SCN− ions.8). Ligand which can ligate through two different atoms is called ambidentate ligand. Similarly.more stable than similar complexes containing unidentate ligands (for reasons see Section 9. [Fe(C2O4)3]3– and [Co(en)3]3+ . (e) Coordination sphere The central atom/ion and the ligands attached to it are enclosed in square bracket and is collectively termed as the coordination sphere. the coordination number of both. For example. Complexes in which a metal is bound to more than one kind of donor groups. For example. 9. (g) Oxidation number of central atom The oxidation number of the central atom in a complex is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. the coordination sphere is [Fe(CN)6]4– and the counter ion is K+ ..1 shows the shapes of different coordination polyhedra. e. For example. For example.g. [Co(NH3)6]3+ is octahedral. e. are known as heteroleptic. are known as homoleptic. (f)Coordination polyhedron The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom.g.are called counter ions. The most common coordination polyhedra are octahedral..3 Nomenclature of Coodination Compounds . [Co(NH3)4Cl2]+ . Fig. oxidation number of copper in [Cu(CN)4]3– is +1 and it is written as Cu(I). 9. [Ni(CO)4] is tetrahedral and [PtCl4]2− is square planar. [Co(NH3)6]3+ . (h) Homoleptic and heteroleptic complexes Complexes in which a metal is bound to only one kind of donor groups. in the complex K4[Fe(CN)6]. square planar and tetrahedral. The oxidation number is represented by a Roman numeral in parenthesis following the name of the coordination entity. (v) There should be no space between the ligan ds and the metal within a coordination sphere. Ligand abbreviations are also enclosed in parentheses. 9. their formulas are enclosed in parentheses.3. For example. The placement of a ligand in the list does not depend on its charge. (vii) The charge of the cation(s) is balanced by the charge of the anion(s). [Co(CN)63−] . the charge is indicated outside the square brackets as a right superscript with the number before the sign. When ligands are polyatomic. The formulas and names adopted for coordination entities are based on the recommendations of the International Union of Pure and Applied Chemistry (IUPAC). (vi) When the formula of a charged coordination entity is to be written without that of the counter ion. (iii) Polydentate ligands are also listed alphabetically. . (ii) The ligands are then listed in alphabetical order. the first letter of the abbreviation is used to determine the position of the ligand in the alphabetical order. etc.1 Formulas of Mononuclear Coordination Entities Information about the constitution of the compound in a concise and convenient manner. [Cr(H2O)6]3+ . (iv) The formula for the entire coordination entity. particularly when dealing with isomers. In case of abbreviated ligand. is enclosed in square brackets. Mononuclear coordination entities contain a single central metal atom. whether charged or not. The following rules are applied while writing the formulas: (i) The central atom is listed first.Nomenclature is important in Coordination Chemistry because of the need to have an unambiguous method of describing formulas and writing systematic names. When the names of the ligands include a numerical prefix. (v) Oxidation state of the metal in cation. If the complex ion is an anion. Co in a complex anion. (ii) The ligands are named in an alphabetical order before the name of the central atom/ion. those of neutral and cationic ligands are the same except aqua for H2O. For example. the ligand to which they refer being placed in parentheses. then the terms. The following rules are used when naming coordination compounds: (i) The cation is named first in both positively and negatively charged coordination entities. They are listed as prefixes to the name of the central atom along with any appropriate multipliers. (iv) Prefixes mono. These are placed within enclosing marks ( ).. anion or neutral coordination entity is indicated by Roman numeral in parenthesis. For some metals. are used to indicate the number of the individual ligands in the coordination entity. the metal is named same as the element. tris.3. For example. the Latin names are used in the complex anions. tri. Co in a complex cation is called cobalt and Pt is called platinum. (This procedure is reversed from writing formula). . Thus. (iii) Names of the anionic ligands end in –o. the groups that surround the central atom must be identified in the name. [NiCl2(PPh3)2] is named as dichlorobis(triphenylphosphine)nickel(II).2 Naming of Mononuclear Coordination Compounds The names of coordination compounds are derived by following the principles of additive nomenclature. [Co (SCN)4]2− is called cobaltate. For example. carbonyl for CO and nitrosyl for NO. bis. etc.9. the name of the metal ends with the suffix – ate. tetrakis are used. di. (vi) If the complex ion is a cation. ammine for NH3. 2–diammine)cobalt(III) sulphate Explanation: The sulphate is the counter anion in this molecule. The amine ligands are named before the aqua ligands according to alphabetical order.2 Write the formulas for the following coordination compounds: (i) Tetraammineaquachloridocobalt(III) chloride .2– diamine is a neutral molecule. Since there are three chloride ions in the compound. 2. the charge on each complex cation must be +3. all the ligands are neutral molecules. In this example.g. the oxidation number of chromium must be the same as the charge of the complex ion. 3. [Co(H2NCH2CH2NH2)3]2(SO4)3 is named as: tris(ethane-1. (vii) The neutral complex molecule is named similar to that of the complex cation.e. we can calculate the oxidation number of the metal. ethane-1. so the oxidation number of cobalt in the complex ion must be +3. Therefore. +3. Remember that you never have to indicate the number of cations and anions in the name of an ionic compound.. which is a cation. [Ag(NH3)2][Ag(CN)2] is named as: diamminesilver(I) dicyanoargentate(I) Example 9. ferrate for Fe. From the charge on the complex ion and the charge on the ligands. Since it takes 3 sulphates to bond with two complex cations. 1. the charge on the complex ion must be +3 (since the compound is electrically neutral). [Cr(NH3)3(H2O)3]Cl3 is named as: triamminetriaquachromium(III) chloride Explanation: The complex ion is inside the square bracket. Further. The following examples illustrate the nomenclature for coordination compounds. (ii) Potassium tetrahydroxozincate(II) (iii) Potassium trioxalatoaluminate(III) (iv) Dichloridobis(ethane-1.2–diamine)platinum(IV) nitrate .1 Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Potassium tetracyanonickelate(II) (iii) Tris(ethane–1.2-diamine)cobalt(III) (v) Tetracarbonylnickel(0) Solution (i)[Co(NH3)4(H2O)Cl]Cl2 (ii)K2[Zn(OH)4] (iii)K3[Al(C2O4)3] (iv)[CoCl2(en)2] (v)[Ni(CO)4] Example 9.3 Write the IUPAC names of the following coordination compounds: (i) [Pt(NH3)2Cl(NO2)] (ii) K3[Cr(C2O4)3] (iii) [CoCl2(en)2]Cl (iv) [Co(NH3)5(CO3)]Cl (v) Hg[Co(SCN)4] Solution (i) Diamminechloridonitrito-N-platinum(II) (ii) Potassium trioxalatochromate(III) (iii) Dichloridobis(ethane-1.2–diamine) chromium(III) chloride (iv) Amminebromidochloridonitrito-N-platinate(II) (v) Dichloridobis(ethane–1.2-diamine)cobalt(III) chloride (iv) Pentaamminecarbonatocobalt(III) chloride (v) Mercury tetrathiocyanatocobaltate(III) Intext Questions 9. they differ in one or more physical or chemical properties. Two principal types of isomerism are known among coordination compounds.1 Geometric Isomerism . A detailed account of these isomers are given below.4.2 Write the IUPAC names of the following coordination compounds: (i) [Co(NH3)6]Cl3 (ii) [Co(NH3)5Cl]Cl2 (iii) K3[Fe(CN)6] (iv) K3[Fe(C2O4)3] (v) K2[PdCl4] (vi) [Pt(NH3)2Cl(NH2CH3)]Cl 9. Structural isomers have different bonds. Each of which can be further subdivided. 9. (a) Stereoisomerism (i) Geometrical isomerism (ii) Optical isomerism (b) Structural isomerism (i) Linkage isomerism (ii) Coordination isomerism (iii) Ionisation isomerism (iv) Solvate isomerism Stereoisomers have the same chemical formula and chemical bonds but they have different spatial arrangement.(vi) Iron(III) hexacyanoferrate(II) 9.4 Isomerism in Coordination Compounds Isomers are two or more compounds that have the same chemical formula but a different arrangement of atoms. Because of the different arrangement of atoms. X. You may attempt to draw these structures.3). or opposite to each other in a trans isomer as depicted in Fig. the two ligands X may be arranged adjacent to each other in a cis isomer. NH2 CH2 CH2 NH2(en)] are present in complexes of . B. Other square planar complex of the type MABXL (where A.. L are unidentates) shows three isomers-two cis and one trans. Important examples of this behaviour are found with coordination numbers 4 and 6. This type of isomerism also arises when didentate ligands L – L [e. 9.This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.g. Such isomerism is not possible for a tetrahedral geometry but similar behaviour is possible in octahedral complexes of formula [MX2L4] in which the two ligands X may be oriented cis or trans to each other (Fig.2. In a square planar complex of formula [MX2L2] (X and L are unidentate). 9. 9.4).5). When the positions are around the meridian of the octahedron. The molecules or ions that cannot be superimposed are called chiral.2 Optical Isomerism Optical isomers are mirror images that cannot be superimposed on one another. These are called as enantiomers. Example 9. we get the meridional (mer) isomer (Fig. The two forms are called dextro (d) and laevo (l) depending upon the direction they rotate the plane of polarised light in a polarimeter . 9.4. Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Ma3b3] like [Co(NH3)3(NO2)3].4 Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion ? Solution 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. we have the facial (fac) isomer.formula [MX2(L – L)2] (Fig. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face. 9. In a coordination entity of the type [PtCl2(en)2]2+ . Optical isomerism is common in octahedral complexes involving didentate ligands (Fig.7). only the cisisomer shows optical activity (Fig.(d rotates to the right. 9.5 Draw structures of geometrical isomers of [Fe(NH3)2(CN)4]– Solution Example 9.6). l to the left). 9. Example 9.6 Out of the following two coordination entities which is chiral (optically active)? (a) cis-[CrCl2(ox)2]3– . (b) trans-[CrCl2(ox)2]3– Solution The two entities are represented as Out of the two, (a) cis – [CrCl2(ox)2] is chiral (optically active). 9.4.3 Linkage Isomerism Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS–, which may bind through the nitrogen to give M–NCS or through sulphur to give M–SCN. Jørgensen discovered such behaviour in the complex [Co(NH3)5(NO2)]Cl2, which is obtained as the 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). 9.4.4 Coordination Isomerism This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH3)6][Cr(CN)6], in which the NH3 ligands are bound to Co3+ and the CN– ligands to Cr3+ . In its coordination isomer [Cr(NH3)6][Co(CN)6], the NH3 ligands are bound to Cr3+and the CN– ligands to Co3+ . 9.4.5 Ionisation Isomerism This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4. 9.4.6 Solvate Isomerism This form of isomerism is known as ‘hydrate isomerism’ in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H2O)6]Cl3(violet) and its solvate isomer [Cr(H2O)5Cl]Cl2.H2O (grey-green). Intext Questions 9.3 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] 9.4 Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionisation isomers. 9.5 Bonding in coordination Compounds Werner was the first to describe the bonding features in coordination compounds. But his theory could not answer basic questions like: (i) Why only certain elements possess the remarkable property of forming coordination compounds? (ii) Why the bonds in coordination compounds have directional properties? (iii) Why coordination compounds have characteristic magnetic and optical properties? Many approaches have been put forth to explain the nature of bonding in coordination compounds viz. Valence Bond Theory (VBT),Crystal Field Theory (CFT), Ligand Field Theory (LFT) and Molecular Orbital Theory (MOT). We shall focus our attention on elementary treatment of the application of VBT and CFT to coordination compounds. 9.5.1 Valence Bond Theory According to this theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on (Table 9.2). These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. This is illustrated by the following examples. DISTRIBUTION OF HYBRID COORDINATION TYPE OF NUMBER 4 4 5 6 6 sp3+ dsp2+ sp d sp3d2 d2sp3 3 ORBITALS IN Tetrahedral Square planar Trigonal bipyramidal Octahedral Octahedral HYBRIDISATION SPACE Thus. It is thus called outer orbital or high spin or spin free complex. one from each NH3 molecule. In the formation of this complex. The paramagnetic octahedral complex. since the inner d orbital (3d) is used in hybridisation. the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex.2: Number of Orbitals and Types of Hybridisations It is usually possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of the valence bond theory. The hybridisation scheme is as shown in diagram. the complex. the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6. [CoF6]3− uses outer orbital (4d ) in hybridisation (sp3d ).[Co(NH3)6]3+ .Table 9. Thus: . In the diamagnetic octahedral complex. Six pairs of electrons. occupy the six hybrid orbitals. Here nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. Here nickel is in +2 oxidation state and has the electronic configuration 3d8. In the square planar complexes. An example is [Ni(CN)4]2–.In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. Similarly. the hybridisation involved is dsp2. [Ni(CO)4] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron. The compound is paramagnetic since it contains two unpaired electrons. The hybridisation scheme is as shown in diagram: .ion donates a pair of electrons. This is ill-ustrated below for [NiCl42−]. Each Cl. The hybridisation scheme is as shown in diagram. like Ti3+(d1). In fact. When more than three 3d electrons are present. 9. Thus. two vacant d orbitals are available for octahedral hybridisation with 4s and 4p orbitals. However. with species containing d4 and d5ions there are complications. d5(Mn2+. The magnetic data agree with maximum spin pairing in many cases. V3+(d2). the required pair of 3d orbitals for octahedral hybridisation is not directly available (as a consequence of Hund’s rule). It is important to note that the hybrid orbitals do not actually exist. Mn3+). A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. especially with coordination compounds containing d6 ions. hybridisation is a mathematical manipulation of wave equation for the atomic orbitals involved. d6(Fe2+. one and zero unpaired electrons.5. Cr3+(d3). for d (Cr2+.Each of the hybridised orbitals receives a pair of electrons from a cyanide ion. The results can be used to obtain information about the structures adopted by metal complexes. a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two. The magnetic behaviour of these free ions and their coordination entities is similar. The compound is diamagnetic as evident from the absence of unpaired electron.Co3+) cases. Fe3+). For metal ions with upto three electrons in the d orbitals. [Mn(CN)6]3– has magnetic moment of two .2 Magnetic Properties of Coordination Compounds The magnetic moment of coordination compounds can be measured by the magnetic susceptibility experiments. respectively. [FeF6]3– and [CoF6-]3– are outer orbital complexes involving sp3d2 hybridisation and are paramagnetic corresponding to four. the former two complexes are paramagnetic and the latter diamagnetic. structures and magnetic behaviour of coordination compounds. it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d orbitals. (ii) It does not give quantitative interpretation of magnetic data.9 BM. it will be either tetrahedral (sp3 hybridisation) or square planar (dsp2 hybridisation). [MnCl6]3– .5. to a larger extent. . 9. But the fact that the magnetic moment of the complex ion is 5.7 The spin only magnetic moment of [MnBr4]2– is 5.3 Limitations of Valence Bond Theory While the VB theory. [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3hybridisation. On the other hand.unpaired electrons while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities. Example 9. five and four unpaired electrons. (iii) It does not explain the colour exhibited by coordination compounds.9 BM. Predict the geometry of the complexion ? Solution Since the coordination number of Mn2+ ion in the complex ion is 4. [Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons. explains the formation. [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3− is diamagnetic. [Mn(CN)6]3– . it suffers from the following shortcomings: (i) It involves a number of assumptions. (vi) It does not distinguish between weak and strong ligands. the dx2 − y2 and dz2 orbitals which point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy.4 Crystal Field Theory The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand.5. However.(iv) It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds. It results in splitting of the d orbitals. there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands.e. Let us explain this splitting in different crystal fields. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. and the dxy. i. The pattern of splitting depends upon the nature of the crystal field. they are degenerate. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. it becomes asymmetrical and the degeneracy of the d orbitals is lifted. 9. Such a repulsion is more when the metal d orbital is directed towards the ligand than when it is away from the ligand. when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex. (v) It does not make exact predictions regarding the tetrahedral and square planar structures of 4-coordinate complexes.. Thus. dyz and dxz orbitals . (a) Crystal field splitting in octahedral coordination entities In an octahedral coordination entity with six ligands surrounding the metal atom/ion. The five d orbitals in an isolated gaseous metal atom/ion have same energy. The crystal field splitting. t2g set and two orbitals of higher energy. the degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy. depends upon the field d orbitals produced by the ligand and charge on the metal ion. eg set. Thus.8). ligands can be arranged in a series in the order of increasing field strength as given below: . Some ligands are able to produce strong fields in which case.9. the splitting will be large whereas others produce weak fields and consequently result in small splitting of d orbitals. the energy of the two eg orbitals will increase by (3/5) Δo and that of the three t2g will decrease by (2/5)Δo.Δo. Thus. In general. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted Δo (the subscript o is for octahedral) (Fig.which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field. The two options are: (i) If Δo < P. 9. it can be shown that Δt = (4/9) Δ0. Let us assign electrons in the d orbitals of metal ion in octahedral coordination entities. Ligands for which Δo< P are known as weak field ligands and form high spin complexes. In d2 and d3 coordination entities. the single d electron occupies one of the lower energy t2g orbitals. (b) Crystal field splitting in tetrahedral coordination entities In tetrahedral coordination entity formation. It is an experimentally determined series based on the absorption of light by complexes with different ligands. . the d orbital splitting (Fig. the d electrons occupy the t2g orbitals singly in accordance with the Hund’s rule. Which of these possibilities occurs. For d4 ions. the fourth electron enters one of the eg orbitals giving the configuration t2g3e1g. two possible patterns of electron distribution arise: (i) the fourth electron could either enter the t2g level and pair with an existing electron. Δo and the pairing energy. For the same metal. depends on the relative magnitude of the crystal field splitting. Obviously. the same ligands and metal-ligand distances.I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < edta4– < NH3 < en < CN– < CO Such a series is termed as spectrochemical series. or (ii) it could avoid paying the price of the pairing energy by occupying the e g level. P (P represents the energy required for electron pairing in a single orbital). it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g4eg0 . Calculations show that d4 to d7 coordination entities are more stable for strong field as compared to weak field cases. (ii) If Δo > P. Ligands which produce this effect are known as strong field ligands and form low spin complexes.9) is inverted and is smaller as compared to the octahedral field splitting. 5 Colour in Coordination Compounds In the previous Unit. it appears red.Consequently. The complementary colour is the colour generated from the wavelength left over. if green light is absorbed by the complex.5. The colour of the complex is complementary to that which is absorbed. Table 9.3 gives the relationship of the different wavelength absorbed and the colour observed. we learnt that one of the most distinctive properties of transition metal complexes is their wide range of colours. so the light that emerges is no longer white. This means that some of the visible spectrum is being removed from white light as it passes through the sample. 9. therefore. low spin configurations are rarely observed. . the orbital splitting energies are not sufficiently large for forcing pairing and. Consequently. Similarly. It is important to note that in the absence of ligand. it would excite the electron from t2g level to the eg level (t2g1 eg0 → t2g0 eg1 ). The influence of the ligand on the colour of a complex may be illustrated by considering the [Ni(H2O)6]2+ complex. which forms when nickel(II) chloride is dissolved in water.The colour in the coordination compounds can be readily explained in terms of the crystal field theory.5H2O is 3+ blue in colour.10). the complex [Ti(H2O)6]3+. crystal field splitting does not occur and hence the substance is colourless. the complex appears violet in colour (Fig. removal of water from [Ti(HM2O)6]Cl3 on heating renders it colourless. anhydrous CuSO4 is white. ethane-1. The next higher state available for the electron is the empty eg level. The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron. Consider. 1:1. This is an octahedral complex where the single electron (Ti3+ is a 3d1 system) in the metal d orbital is in the t2g level in the ground state of the complex.2diamine(en) is progressively added in the molar ratios en:Ni. If light corresponding to the energy of yellow-green region is absorbed by the complex. If the didentate ligand. for example. which is violet in colour. 2:1. but CuSO4. the following series of reactions and their associated colour changes occur: [Ni(H2O)6]2+ + en (aq)= [Ni(H2O)4(en)] (aq) + 2H2O . 9. 3:1. For example. 12(b)]. which are randomly distributed in positions normally occupied by Al3+ . Ruby [Fig.9.5-1% Cr3+ ions (d3). Cr3+ ions occupy octahedral sites in the mineral beryl . In emerald [Fig. 9.12(a)] is aluminium oxide (Al2O3) containing about 0. We may view these chromium(III) species as octahedral chromium(III) complexes incorporated into the alumina lattice. d–d transitions at these centres give rise to the colour.9. Colour of Some Gem Stones The colours produced by electronic transitions within the d orbitals of a transition metal ion occur frequently in everyday life.11.green pale blue [Ni(H2O)4(en)]2+(aq) + en (aq) = [Ni(H2O)2(en)2]2+(aq) + 2H2O blue/purple [Ni(H2O)2(en)2](aq) + en (aq) = [Ni(en)3]2+ (aq) violet This sequence is shown in Fig. 9.9 Predict the number of unpaired electrons in the square planar [Pt(CN)4]2−ion. it does not take into account the covalent character of bonding between the ligand and the central atom.6 [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? 9. while the hexacyanoion contains only one unpaired electron. 9. These are some of the weaknesses of CFT. Explain. Further.(Be3Al2Si6O18).10 The hexaquo manganese(II) ion contains five unpaired electrons. 9. structures. . 9.7 [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain using Crystal Field Theory. The absorption bands seen in the ruby shift to longer wavelength.6 Limitations of Crystal Field Theory The crystal field model is successful in explaining the formation. Intext Questions 9. it follows that anionic ligands should exert the greatest splitting effect. colour and magnetic properties of coordination compounds to a large extent.5. namely yellow-red and blue. However. The anionic ligands actually are found at the low end of the spectrochemical series. which are explained by ligand field theory (LFT) and molecular orbital theory which are beyond the scope of the present study. from the assumptions that the ligands are point charges.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. 9. causing emerald to transmit light in the green region.8 Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. 9. The metal-carbon bond in metal carbonyls possess both s and p character.14).13). well defined structures. Octacarbonyldicobalt(0) has a Co – Co bond bridged by two CO groups (Fig. Tetracarbonylnickel(0) is tetrahedral. The M–C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. . The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal (Fig. The M–C π bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide.9.6 Bonding in Metal Carbonyls The homoleptic carbonyls (compounds containing carbonyl ligands only) are formed by most of the transition metals. pentacarbonyliron(0) is trigonalbipyramidal while hexacarbonyl chromium(0) is octahedral. These carbonyls have simple.9. Decacarbonyldimanganese(0) is made up of two square pyramidal Mn(CO)5 units joined by a Mn – Mn bond. and be successively replaced by them.7 Stability of Coordination Compounds The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. are referred to as stepwise stability constants. we can write the overall stability constant thus: M +4L € ML4 β4 = [ML4]/[M][L]4 The stepwise and overall stability constant are therefore related as follows: β4 = K1 × K2 × K3 × K4 or more generally. if we have a reaction of the type: M + 4L € ML4 then the larger the stability constant. Free metal ions rarely exist in the solution so that M will usually be surrounded by solvent molecules which will compete with the ligand molecules. we generally ignore these solvent molecules and write four stability constants as follows: M + L€ ML K1 = [ML]/[M][L] ML + L€ ML2 K2 = [ML2]/[ML][L] ML3 + L€ ML4 K4 = [ML4]/[ML3][L] where K1. Alternatively.. For simplicity. K2. the higher the proportion of ML4 that exists in solution. L. The magnitude of the (stability or formation) equilibrium constant for the association.9. the steps involved in the formation of . Kn If we take as an example. Thus. quantitatively expresses the stability. βn = K1 × K2 × K3 × K4 ……. etc. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands). metallurgy. In this case.the cuprammonium ion. 9. as a result of formation of coordination entities. logK3 = 2. biological systems. in that the successive stability constants decrease. Also β4 = [Cu(NH3)42+]/[Cu2+][NH3)4 The addition of the four amine groups to copper shows a pattern found for most formation constants. logK2 = 3.0. industry and medicine. Intext Question 9. These are described below: • Coordination compounds find use in many qualitative and quantitative chemical analysis.0 or log β4 = 11.9 The instability constant or the dissociation constant of coordination compounds is defined as the reciprocal of the formation constant.8 Importance and Applications of Coordination Compounds The coordination compounds are of great importance. K2 are the stepwise stability constants and overall stability constant. plant and animal worlds and are known to play many important functions in the area of analytical chemistry. the four constants are: logK1 = 4. given that β4 for this complex is 2.7. These compounds are widely present in the mineral. we have the following: Cu2+ + NH3 € Cu(NH3)2+ K1 = [Cu(NH3)2+]/[Cu2+][NH3] Cu(NH3)2+ + NH3 € Cu(NH3)22+ K2 = [Cu(NH3)22+]/[Cu(NH3)][NH3] etc. form the basis for their detection and estimation by . logK4 = 2. where K1.2.11 Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ion.1 × 1013 . is a coordination compound of magnesium. Examples of such reagents include EDTA. combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2]− in aqueous solution. • Hardness of water is estimated by simple titration with Na2EDTA. make use of complex formation. cupron. the red pigment of blood which acts as oxygen carrier is a coordination compound of iron.classical and instrumental methods of analysis. like those of silver and gold. a Wilkinson catalyst. impure nickel is converted to [Ni(CO)4]. DMG (dimethylglyoxime). α– nitroso–β–naphthol. carboxypeptidase A and carbonic anhydrase (catalysts of biological systems). The pigment responsible for photosynthesis. is used for the . Gold can be separated in metallic form from this solution by the addition of zinc (Unit 6). which is decomposed to yield pure nickel. •Coordination compounds are used as catalysts for many industrial processes. chlorophyll. Haemoglobin. [(Ph3P)3RhCl]. the anti– pernicious anaemia factor. is a coordination compound of cobalt. purification of metals can be achieved through formation and subsequent decomposition of their coordination compounds. For example. Gold. • Similarly. The Ca2+ and Mg2+ions form stable complexes with EDTA. Vitamin B12. for example. Examples include rhodium complex. The selective estimation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes. etc. Among the other compounds of biological importance with coordinated metal ions are the enzymes like. •Coordination compounds are of great importance in biological systems. • Some important extraction processes of metals. cyanocobalamine. Werner. advances in this area. [Ag(S2O3)2]3− . •There is growing interest in the use of chelate therapy in medicinal chemistry. excess of copper and iron are removed by the chelating ligands D–penicillamine and desferrioxime B via the formation of coordination compounds. reactions. The first systematic attempt at explaining the formation. •In black and white photography. His theory postulated the use of two types of linkages (primary and secondary) by a metal atom/ion in a coordination compound. have provided development of new concepts and models of bonding and molecular structure. During the last fifty years. In the modern language of chemistry these linkages are recognised as the ionisable (ionic) . the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex ion. •Articles can be electroplated with silver and gold much more smoothly and evenly from solutions of the complexes. Summary The chemistry of coordination compounds is an important and challenging area of modern inorganic chemistry. Examples are: cis–platin and related compounds. Thus.hydrogenation of alkenes. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. novel breakthroughs in chemical industry and vital insights into the functioning of critical components of biological systems. structure and bonding of a coordination compound was made by A. EDTA is used in the treatment of lead poisoning. Some coordination compounds of platinum effectively inhibit the growth of tumours. [Ag(CN)2]– and [Au(CN)2]− than from a solution of simple metal ions. The treatment provides for quantitative estimations of orbital separation energies. Coordination compounds are of great importance. The β stabilisation of coordination compound due to chelation is called the chelate effect.and non-ionisable (covalent) bonds. The ligand to metal is σ bond and metal to ligand is π bond. fails to provide a quantitative interpretation of magnetic behaviour and has nothing to say about the optical properties of these compounds. The splitting of the d orbitals provides different electronic arrangements in strong and weak crystal fields. The Valence Bond Theory (VBT) explains with reasonable success. however. The Crystal Field Theory (CFT) to coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges). the formation. on the degeneracy of d orbital energies of the central metal atom/ion. It. The metal–carbon bond in metal carbonyls possesses both σ and π character. Werner predicted the geometrical shapes of a large number of coordination entities. However. magnetic behaviour and geometrical shapes of coordination compounds. The stability of coordination compounds is measured in terms of stepwise stability (or formation) constant (K) or overall stability constant (β). Using the property of isomerism. These compounds provide critical insights into the functioning and structures of vital components of biological systems. the assumption that ligands consititute point charges creates many theoretical difficulties. magnetic moments and spectral and stability parameters. This unique synergic bonding provides stability to metal carbonyls. enthalpy and entropy terms. The stability of coordination compounds is related to Gibbs energy. . respectively. 5 Specify the oxidation numbers of the metals in the following coordination entities: (i) [Co(H2O)(CN)(en)2]2+ (ii) [CoBr2(en)2]+ (iii) [PtCl4]2− (iv) K3[Fe(CN)6] (v) [Cr(NH3)3Cl3] 9. 9. 9. homoleptic and heteroleptic.3 Explain with two examples each of the following: coordination entity. coordination polyhedron. Exercises 9. didentate and ambidentate ligands? Give two examples for each.6 Using IUPAC norms write the formulas for the following: (i) Tetrahydroxozincate(II) (ii) Potassium tetrachloridopalladate(II) (iii) Diamminedichloridoplatinum(II) (iv) Potassium tetracyanonickelate(II) (v) Pentaamminenitrito-O-cobalt(III) (vi) Hexaamminecobalt(III) sulphate (vii) Potassium tri(oxalato)chromate(III) . ligand. Explain why? 9.1 Explain the bonding in coordination compounds in terms of Werner’s postulates. 9. analytical and medicinal chemistry. coordination number.4 What is meant by unidentate.2 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.Coordination compounds also find extensive applications in metallurgical processes. 9 How many geometrical isomers are possible in the following coordination entities? (i) [Cr(C2O4)3]3– (ii) [Co(NH3)3Cl3] 9.11 Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2]+ (ii) [Co(NH3)Cl(en)2]2+ (iii) [Co(NH3)2Cl2(en)]+ .8 List various types of isomerism possible for coordination compounds.10 Draw the structures of optical isomers of: (i) [Cr(C2O4)3]3– (ii) [PtCl2(en)2]2+ (iii) [Cr(NH3)2Cl2(en)]+ 9.7 Using IUPAC norms write the systematic names of the following: (i) [Co(NH3)6]Cl3 (ii) [Pt(NH3)2Cl(NH2CH3)]Cl (iii) [Ti(H2O)6]3+ (iv) [Co(NH3)4Cl(NO2)]Cl (v) [Mn(H2O)6]2+ (vi)[NiCl4]2− (vii) [Ni(NH3)6]Cl2 (viii) [Co(en)3]3+ (ix) [Ni(CO)4] 9. giving an example of each.(viii) Hexaammineplatinum(IV) (ix) Tetrabromidocuprate(II) (x) Pentaamminenitrito-N-cobalt(III) 9. 9. Explain these experimental results.9. 9.14 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? 9.17 What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.16 Draw figure to show the splitting of d orbitals in an octahedral crystal field. 9. Explain why? 9. 9. Explain.19 [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers? 9.15 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− 9.13 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.20 A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colorless.18 What is crystal field splitting energy? How does the magnitude of Δo decide the actual configuration of d orbitals in a coordination entity? 9. . 27 Discuss briefly giving an example in each case the role of coordination compounds in: (i) biological systems (ii) medicinal chemistry and (iii) analytical chemistry (iv) extraction/metallurgy of metals. d orbital occupation and coordination number of the central metal ion in the following complexes: (i) K3[Co(C2O4)3] (ii)cis-[Cr(en)2Cl2]Cl (iii)(NH4)2[CoF4] (iv) [Mn(H2O)6]SO4 9. .26 What is meant by the chelate effect? Give an example. why? 9.25 What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.21 [Fe(CN)6]4− and [Fe(H2O]2+ are of different colours in dilute solutions. 9.23 Give the oxidation state. 9.22 Discuss the nature of bonding in metal carbonyls. Also give stereochemistry and magnetic moment of the complex: (i) K[Cr(H2O)2(C2O4)2].24 Write down the IUPAC name for each of the following complexes and indicate the oxidation state.3H2O (ii) [Co(NH3)5Cl-]Cl2 (iii) CrCl3(py)3 (iv) Cs[FeCl4] (v) K4[Mn(CN)6] 9.9. 9. electronic configuration and coordination number. 29 Amongst the following ions which one has the highest magnetic moment value? (i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6]2+ (iii) [Zn(H2O)6]2+ 9. [Ni(H2O)6]2+ ? Answers to Some Intext Questions 9.32 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+ .28 How many ions are produced from the complex Co(NH3)6Cl2 in solution? (i) 6 (ii) 4 (iii) 3 (iv) 2 9.1 (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 . the most stable complex is (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+ (iii) [Fe(C2O4)3]3− (iv) [FeCl6]3− 9.9.31 Amongst the following.30 The oxidation number of cobalt in K[Co(CO)4] is (i) +1 (ii) +3 (iii) –1 (iv) –3 9. (iv) Geometrical (cis-. In the presence of CO ligand. (Hint: There are geometrical. (a strong ligand) the 3d electrons pair up leaving only one unpaired electron.8 In the presence of NH3.7 In presence of CN−. The hybridisation is sp3d2 forming an outer orbital complex containing five unpaired electrons. The hybridisation is d2sp3 forming inner orbital complex.4 The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents: [Co(NH3)5Br]SO4 + Ba2+ → BaSO4(s) [Co(NH3)5SO4]Br + Ba2+ → No reaction [Co(NH3)5Br]SO4 + Ag+ → No reaction [Co(NH3)5SO4]Br + Ag+ → AgBr (s) 9.6 In Ni(CO)4. trans-) isomers can exist. trans-) and optical isomers for cis can exist. (iii) There are 10 possible isomers.2 (i)Hexaamminecobalt(III) chloride (ii)Pentaamminechoridocobalt(III) chloride (iii)Potassium hexacyanoferrate(III) (iv)Potassium trioxalatoferrate(III) (v)Potassium tetrachloridopalladate(II) (vi)Diamminechlorido(methylamine)platinum(II) chloride 9. (a weak ligand). 9. In the presence of H2O. ionisation and linkage isomers possible). 9. the unpaired d electrons of Ni pair up but Cl− being a weak ligand is unable to pair up the unpaired electrons. . it is in +2 oxidation state. (ii) Two optical isomers can exist.9. 3d electrons do not pair up.3 (i) Both geometrical (cis-. it is strongly paramagnetic. Ni is in zero oxidation state whereas in NiCl42− . the 3d electrons pair up leaving two d orbitals empty to be involved in d2sp3 hybridisation forming inner orbital complex in case of [Co(NH3)6]3 . 9. 1/ β4= 4. Ni is in +2 oxidation state and has d8 configuration.2 mol of AgCl are obtained. [Co(NH3)6]3+ . What will be the correct order of absorption of wavelength of light in the visible region. Hence the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp2 hybridisation. for the complexes. The colour of the coordination compounds depends on the crystal field splitting. [Co(CN)6]3– . 9.e.7 × 10−14 I. the hybridisation involved is sp3d2 forming outer orbital complex.11 The overall dissociation constant is the reciprocal of overall stability constant i. 0.In Ni(NH3)62+ .1 mol CoCl3(NH3)5 is treated with excess of AgNO3. Multiple Choice Questions (Type-I) 1.9 For square planar shape. When 0. the hybridisation is dsp2 . 9. Which of the following complexes formed by Cu2+ ions is most stable? 2. The conductivity of solution will correspond to (i) 1:3 electrolyte (ii) 1:2 electrolyte . [Co(H2O)6]3+ (i) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+ (ii) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3– (iii) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3– (iv) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+ 3. Thus there is no unpaired electron. The stabilisation of coordination compounds due to chelation is called the chelate effect. The formula of the complex is : (i) [CrCl3(H2O)3]⋅3H2O (ii) [CrCl2(H2O)4]Cl⋅2H2O (iii) [CrCl(H2O)5]Cl2⋅H2O (iv) [Cr(H2O)6]Cl3 5.(iii) 1:1 electrolyte (iv) 3:1 electrolyte 4. Which of the following is the most stable complex species? (i) [Fe(CO)5] (ii) [Fe(CN)6]3– (iii) [Fe(C2O4)3]3– (iv) [Fe(H2O)6]3+ 7. The correct IUPAC name of [Pt(NH3)2Cl2] is (i) Diamminedichloridoplatinum (II) (ii) Diamminedichloridoplatinum (IV) (iii) Diamminedichloridoplatinum (0) (iv) Dichloridodiammineplatinum (IV) 6. 3 mol of AgCl are obtained. (i) [Cr(H2O)4Cl2]+ (ii) [Pt(NH3)3Cl] (iii) [Co(NH3)6]3+ (iv) [Co(CN)5(NC)]3– . When 1 mol CrCl3⋅6H2O is treated with excess of AgNO3. Indicate the complex ion which shows geometrical isomerism. 000 cm–1 (iv) 20.000 cm–1 9.000 cm–1. Palladium complexes of the type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are (i) linkage isomers (ii) coordination isomers (iii) ionisation isomers (iv) geometrical isomers 10. Which of the following is not a chelating agent? (i) thiosulphato (ii) oxalato (iii) glycinato (iv) ethane-1.000 cm–1 (ii) 16. The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl represent (i) linkage isomerism (ii) ionisation isomerism (iii) coordination isomerism (iv) no isomerism 11. The CFSE for tetrahedral [CoCl4]2–will be (i) 18.2-diamine .8.000 cm–1 (iii) 8. Due to the presence of ambidentate ligands coordination compounds show isomerism. The CFSE for octahedral [CoCl6]4– is 18. A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following inner orbital octahedral complex ions are diamagnetic? . Atomic number of Mn. What kind of isomerism exists between [Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2⋅H2O (greyishgreen)? (i) linkage isomerism (ii) solvate isomerism (iii) ionisation isomerism (iv) coordination isomerism 14. IUPAC name of [Pt(NH3)2Cl(NO2)] is : (i) Platinum diaminechloronitrite (ii) Chloronitrito-N-ammineplatinum (II) (iii) Diamminechloridonitrito-N-platinum (II) (iv) Diamminechloronitrito-N-platinate (II) II. Fe and Co are 25. 26 and 27 respectively. 15. Which of the following species is not expected to be a ligand? (i) NO (ii) NH4+ (iii) NH2CH2CH2NH2 (iv) CO 13. Multiple Choice Questions (Type-II) Note : In the following questions two or more options may be correct.12. (iv) tetrahedral complexes have larger crystal field splitting than octahedral complex. Atomic number of Mn.(i) [Co(NH3)6]3+ (ii) [Mn(CN)6]3– (iii) [Fe(CN)6]4– (iv) [Fe(CN)6]3– 16. (i) [Co(H2O)6]2+ is transformed into [CoCl6]4– (ii) [Co(H2O)6]2+ is transformed into [CoCl4]2– (iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes. Fe. Which of the following complexes are homoleptic? . Co and Ni are 25. Which of the following options are correct for [Fe(CN)6]3– complex? (i) d2sp3 hybridisation (ii) sp3d2 hybridisation (iii) paramagnetic (iv) diamagnetic 18. Which of the following outer orbital octahedral complexes have same number of unpaired electrons? (i) [MnCl6]3– (ii) [FeF6]3– (iii) [CoF6]3– (iv) [Ni(NH3)6]2+ 17. This is because____________. 19. 26 27 and 28 respectively. An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. (iii) It is a chelating ligand. 23. (ii) It is a didentate ligand. Identify the optically active compounds from the following : (i) [Co(en)3]3+ (ii) trans– [Co(en)2 Cl2]+ (iii) cis– [Co(en)2 Cl2]+ (iv) [Cr (NH3)5Cl] 22. (iv) It is a unidentate ligand. 2-diamine as a ligand. (i) It is a neutral ligand. Identify the correct statements for the behaviour of ethane-1. Which of the following complexes show linkage isomerism? (i) [Co(NH3)5 (NO2)]2+ (ii) [Co(H2O)5CO]3+ (iii) [Cr(NH3)5SCN]2+ (iv) [Fe(en)2Cl2]+ .(i) [Co(NH3)6]3+ (ii) [Co(NH3)4 Cl2]+ (iii) [Ni(CN)4]2– (iv) [Ni(NH3)4Cl2] 20. Which of the following complexes are heteroleptic? (i) [Cr(NH3)6]3+ (ii) [Fe(NH3)4 Cl2]+ (iii) [Mn(CN)6]4– (iv) [Co(NH3)4Cl2] 21. Short Answer Type 24.92 BM whereas [Fe(CN)6]3– has a value of only 1. 27. A coordination compound CrCl3⋅4H2O precipitates silver chloride when treated with silver nitrate. [Fe(CN)6]4– and [Cu(NH3)6]2+. 26.III. [Co(NH3)4Cl2] Cl. [Co(NH3)6]Cl3 . Explain why [Fe(H2O)6]3+ has magnetic moment value of 5. . On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands. What does this indicate about the structure of the complex? Give one example of such complex. Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory. [Cr(CN)6]3–. 29. [Cr(NH3)5Cl]Cl2 25. Write structural formula of the compound and name it. Why are low spin tetrahedral complexes not formed? 30. 28. 31. 32. The molar conductance of its solution corresponds to a total of two ions.74 BM. [Cr(NH3)6]3+. Arrange the following complexes in the increasing order of conductivity of their solution: [Co(NH3)3Cl3]. Explain giving reason. Magnetic moment of [MnCl4]2– is 5. Arrange following complex ions in increasing order of crystal field splitting energy (ΔO) : [Cr(Cl)6]3–. A complex of the type [M(AA)2X2]n+ is known to be optically active. [CoF6]3–.92 BM. Give two examples of ambidentate ligands. Match the complex ions given in Column I with the colours given in Column II and assign the correct code : Column I (Complex ion) A. 36. Pale blue 4. Why do compounds having similar geometry have different magnetic moment? 34. CuSO4. Violet 2. (Ni (H2O)4 (en)]2+ (aq) Code : (i) A (1) B (2) C (4) D (5) (ii) A (4) B (3) C (2) D (1) (iii) A (3) B (2) C (4) D (1) (iv) A (4) B (1) C (2) D (3) Column II (Colour) 1. Name the type of isomerism when ambidentate ligands are attached to central metal ion. [Ni(H2O)6]2+ D. IV. Why? 35. Yellowish orange 5. [Co(NH3)6]3+ B. Blue .5H2O is blue in colour while CuSO4 is colourless. Green 3. [Ti(H2O)6]3+ C. Matching Type Note : In the following questions match the items given in Columns I and II.33. Vitamin B12 Code : (i) A (5) B (4) C (1) D (2) (ii) A (3) B (4) C (5) D (1) (iii) A (4) B (3) C (2) D (1) (iv) A (3) B (4) C (1) D (2) 38. 5 Column II (Central metal atom) 1. Wilkinson catalyst D. dsp2. cobalt 3. magnesium . Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code : Column I (Coordination Compound) A. Chlorophyll B. sp3d2. Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code : Column I (Complex ion) A. Blood pigment C. 1 2. iron 5. number of unpaired electrons) 1. rhodium 2. [Co(CN)4]2– Column II (Hybridisation. [Cr(H2O)6]3+ B.37. calcium 4. ionisation 3. sp3. Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code : Column I (Complex species) A.C. geometrical 5. sp3d2. cis-[Co(en)2Cl2]+ C. optical 2. [Co(NH3)6][Cr(CN)6] Code : (i) A (1) B (2) C (4) D (5) (ii) A (4) B (3) C (2) D (1) (iii) A (4) B (1) C (5) D (3) (iv) A (4) B (1) C (2) D (3) Column II (Isomerism) 1. 2 Code : (i) A (3) B (1) C (5) D (2) (ii) A (4) B (3) C (2) D (1) (iii) A (3) B (2) C (4) D (1) (iv) A (4) B (1) C (2) D (3) 39. [Ni(NH3)6]2+ D. [MnF6]4– 3. 3 4. [Co(NH3)5(NO2)]Cl2 D. d2sp3. coordination 4. 4 5. linkage . [Co(NH3)4Cl2]+ B. + 1 4. + 4 B. Assertion and Reason Type Note : In the following questions a statement of assertion followed by a statement of reason is given. reason is correct explanation of assertion. [Co2(CO)8] Code : (i) A (1) B (2) C (4) D (5) (ii) A (4) B (3) C (2) D (1) (iii) A (5) B (1) C (4) D (2) (iv) A (4) B (1) C (2) D (3) V. (ii) Assertion and reason both are true but reason is not the correct explanation of assertion. (i) Assertion and reason both are true. 2. reason is false. (iii) Assertion is true. Column II (Oxidation state of Co) Column I (Compound) A. Na4[Co(S2O3)3] D. [Co(NCS)(NH3)5](SO3) 1. 0 3. + 2 5. Choose the correct answer out of the following choices. (iv) Assertion is false. reason is true. + 3 . Match the compounds given in Column I with the oxidation state of cobalt present in it (given in Column II) and assign the correct code.40. [Co(NH3)4Cl2]SO4 C. 42. Reason : Unpaired electrons are present in their d-orbitals. Reason : Because it has d2sp3 type hybridisation. Reason : Geometrical isomerism is not shown by complexes of coordination number 6. Reason : Chelate complexes tend to be more stable. [Fe(H2O)6]2+. draw energy level diagram. VI. [Co(CN)6]3– (ii) [FeF6]3–. [FeCl6]4– (i) Type of hybridisation. Using crystal field theory. Assertion : Complexes of MX6 and MX5L type (X and L are unidentate) do not show geometrical isomerism. . 43. write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following : (i) [CoF6]3–.41. [Co(H2O)6]2+ . Assertion : [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature. explain the following in relation to the complexes given below: [Mn(CN)6]3– . Assertion : Toxic metal ions are removed by the chelating ligands. [Cr(H2O)6]3+ . 44. 45. [Co(NH3)6]3+. Assertion : ([Fe(CN)6]3– ion shows magnetic moment corresponding to two unpaired electrons. [Fe(CN)6]4– 47. Long Answer Type 46. Assertion : Linkage isomerism arises in coordination compounds containing ambidentate ligand. Using valence bond theory. (ii) Inner or outer orbital complex. Reason : Ambidentate ligand has two different donor atoms. CoSO4Cl. What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex? 50. Isomer ‘A’ reacts with AgNO3to give white precipitate. (iii) 6. (iii) 22. (i) 12.5NH3 exists in two isomeric forms ‘A’ and ‘B’. (ii). but does not react with BaCl2. (i) 10. (ii) Name the type of isomerism involved. (iv) 5. (i). (ii) 4. Isomer ‘B’ gives white precipitate with BaCl2 but does not react with AgNO3. (iii) 19. (i). (iii) 20. (ii). (iii) 7. (ii) 14. (iii) 17. Short Answer Type . (iii) 18. (iii) 16. (iii) 23.(iii) Magnetic behaviour. (iii) Give the IUPAC name of ‘A’ and ‘B’. (i). (iv) 21. Multiple Choice Questions (Type-I) 1. (i). (iv) 11. Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands? ANSWERS I. 49. (i) (iii) 9. Answer the following questions. (i). 48. (i). (ii) 2. (iv) Spin only magnetic moment value. (ii). (ii) 13. (i) Identify ‘A’ and ‘B’ and write their structural formulas. (iii) III. (i). (i) 8. (iii) 3. Multiple Choice Questions (Type-II) 15. II. 32. With weak field ligands. An optically active complex of the type [M(AA)2X2]n+ indicates cisoctahedral structure. Because for tetrahedral complexes. Crystal field splitting energy increases in the order [Cr(Cl)6]3– < [Cr(NH3)6]3+ < [Cr(CN)6]3– 33. [Co(H2O)4Cl2]Cl (tetraaquadichloridocobalt(III) chloride) 26. Cu2+ (d9) t2g6 eg3. the complex will show low value of . [Fe(CN)6]3– involves d2sp3 hybridisation with one unpaired electron and [Fe(H2O)6]3+involves sp3d2 hybridisation with five unpaired electrons. Co3+(d6) t2g4 eg2. This difference is due to the presence of strong ligand CN– and weak ligand H2O in these complexes.92 BM magnetic moment value. cis[Pt(en)2Cl2]2+ or cis-[Cr(en)2Cl2]+ 27. [Cu(NH3)6]2+ . [Co(NH3)3Cl3] < [Cr(NH3)5Cl]Cl < [Co(NH3)5Cl]Cl2 < [Co(NH3)6]Cl3 25. Fe2+(d6) t2g6 eg0. the crystal field stabilisation energy is lower than pairing energy.g.92 BM corresponds to the presence of five unpaired electrons in the d-orbitals of Mn2+ ion. e. It is due to the presence of weak and strong ligands in complexes. 30. the electronic configuration of Co (III) will be t2g4 eg2and it has 4 unpaired electrons and is paramagnetic. Thus tetrahedral structure of [MnCl4]2– complex will show 5. The magnetic moment of 5. ΔO < p. Δ0 > p. [Fe(CN)6]4– .24. the electronic configuration will be t2g6g eg0. 28. if CFSE is high. With strong field ligands. [CoF6]3–. 29. It has no unpaired electrons and is diamagnetic. As a result the hybridisation involved is sp3 rather than dsp2. 31. (ii) 43. (iv) 40. (ii) 39. 35. water acts as ligand as a result it causes crystal field splitting. In CuSO4. the former is paramagnetic and the latter is diamagnetic. (i) 42.5H2O. In the anhydrous CuSO4due to the absence of water (ligand). (iv) VI. (ii) 37. e. Assertion and Reason Type 41. (i) .g. 34. Matching Type 36. [CoF6]3– and [Co(NH3)6]3+ . Long Answer Type 46.5H2O and shows colour. crystal field splitting is not possible and hence no colour. (ii) 45. (i) 38. (i) V. (i) 44. Hence d—d transition is possible in CuSO4.magnetic moment and vice versa. Linkage isomerism Examples : IV. Number of unpaired electrons = 4 (ii) Number of unpaired electron = 5 Number of unpaired electron = 4 . [Mn(CN)6]3– Mn3+ = 3d4 (i) d2sp3 (ii) Inner orbital complex (iii) Paramagnetic [Co(NH3)6]3+ Co3+ = 3d6 (i) d2sp3 (ii) Inner orbital complex (iii) Diamagnetic (iv) Zero [Cr(H2O)6]3+ Cr3+ = 3d3 .[Fe(CN)6]4– Fe2+ = 3d6 Since CN– is strong field ligand all the electrons get paired. No unpaired electrons so diamegnetic 47. Pentaamminechlorocobalt (III) sulphate.9 BM 48.87 BM [Fe(Cl)6]4– Fe2+ = 3d6 (i) sp3d2 (ii) Outer orbital complex (iii) Paramagnetic (iv) 4. Δt = (4/9) Δ0. (i) A – [Co(NH3)5SO4]Cl B – [Co(NH3)5Cl]SO4 (ii) Ionisation isomerism (iii) (A). The observed colour of complex is the colour generated from the wavelength left over. Pentaamminesulphatocobalt (III) chloride (B). Higher the crystal field splitting.(i) d2sp3 (ii) Inner orbital complex (iii) Paramagnetic (iv) 3. lower will be the wavelength absorbed by the complex. So higher wavelength is absorbed in octahedral complex than tetrahedral complex for same metal and ligands . 49. 50. When white light falls on the complex. some part of it is absorbed.