Unit 9- Coordination Compounds

Transcription

1 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, Fe 3+ and Ni 2+ are the central ions in the coordination compounds [Fe(CN)6] 3 and NiCl 2.6H 2 O 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, NH 3, H 2 O, 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,, NH 3, H 2 O, 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 (EDTA 4 ) (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(NH 3 ) 2 ] 2+, [Zn(CN) 4 ] 2, & [Ni(NH 3 ) 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 C 2 O 4 2 is a didentate ligand.

2 (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, Fe 3+ = +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(NH 3 ) 2 ] 2+, [Ni(NH 3 ) 6 ] 2+ (ii) Anionic complex ; [Fe(CN)6] 3, [Zn(CN) 4 ] 2 (iii) Neutral complex ; [Co(NH3) 3 Cl3],[ 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(NH 3 ) 2 ] 2+, [Ni(NH 3 ) 6 ] 2+ [Fe(CN)6] 3, etc. (g) The charge of cation(s) is balanced by the charge of anion(s).

3 (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 H 2 O, ammine for NH 3, 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, [NiCl 2 (PPh 3 ) 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)Br 5 ] 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(NH 3) 2Cl(NO 2)] (iii) Tetraammineaquachloridocobalt(III) chloride : [Co(NH 3) 4(H 2O)Cl] (iv) Potassium trioxalatoaluminate(iii) : K 3[Al(C 2O 4) 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.

4 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.

5 (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 [CoCl 2 (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(NH 3 )(NO 2 )]Cl 2, 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(NH 3 ) 5 SO 4 ]Br (red complex) & [Co(NH 3 ) 5 Br] SO 4 (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(NH 3 ) 6 ] [Cr(CN) 6 ] and [Cr(NH 3 ) 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(H 2 O) 6 ]Cl 3 (violet) and, [Cr(H 2 O) 5 Cl]Cl 2.H 2 O (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.

6 Number of orbitals and types of Hybridisations Coordination Type of hybridisation Distribution of hybrid orbitals in Type of Complex number space 4 sp 3 Tetrahedral Outer orbital/ high spin 4 dsp 2 Square planar Inner orbital/ low spin 5 sp 3 d Trigonal bipyramidal Outer orbital/ high spin 6 sp 3 d 2 Octahedral Outer orbital/ high spin 6 d 2 sp 3 Octahedral Inner orbital/ low spin Application of Valence Bond Treatment to Some Complexes Ion/ Complex Central metal ion Confi-guration of metal ion Hybridi-zation of metal ion involved Geometry of the complex Number of unpaired electrons Magnetic behaviour d 2 sp 3 Octa-hedral 1 Para-magnetic d 2 sp 3 Octa-hedral 2 Para-magnetic d 2 sp 3 Octa-hedral 3 Para-magnetic d 2 sp 3 Octa-hedral 3 Para-magnetic sp 3 d 2 Octa-hedral 4 Para-magnetic d 2 sp 3 Octa-hedral 2 Para-magnetic sp 3 Tetra-hedral 5 Para-magnetic sp 3 d 2 Octa-hedral 5 Para-magnetic sp 3 d 2 Octa-hedral 5 Para-magnetic d 2 sp 3 Octa-hedral 1 Para-magnetic d 2 sp 3 Octa-hedral 0 Dia-magnetic sp 3 Tetra-hedral 4 Para-magnetic d 2 sp 3 Octa-hedral 0 Dia-magnetic sp 3 d 2 Octa-hedral 4 Para-magnetic sp 3 Tetra-hedral 0 Dia-magnetic dsp 2 Square planar 0 Dia-magnetic sp 3 Tetra-hedral 2 Para-magnetic sp 3 d 2 Octa-hedral 2 Para-magnetic sp 3 Tetra-hedral 1 Para-magnetic sp 3 Tetra-hedral 0 Dia-magnetic dsp 2 Square planar 0 Dia-magnetic

7 (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., dx 2 -y 2 and dz 2 oriented along the axes are called e g 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 d- orbitals 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: - CO>CN > NO > en > py > NH 3 > EDTA > SCN > H O >ONO > ox 2 > OH > F > SCN > Cl > Br > Crystal Field Splitting in Octahedral Complexes 3 5 o 2 5 o In octahedral complexes the six ligands approach the central metal ion along the co-ordinate axe d x 2 -y 2 and d z 2 orbitals. Consequently, the e g set of orbitals has higher energy than t 2g of orbitals. Electron Configuration in d-orbitals 4 Δ > P, low spin d 4 Δ < P, high spin d

8 (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 t 2g set of orbital lie relatively nearer to the approaching ligands and therefore t 2g set of d-orbitals have higher energy than, e g set of orbitals. Relationship between t and o is t = 4 9 o Colour of transition metal complexes: Colour associated with the transition metal complexes is due to the transition of electrons between d-orbitals (from t 2g to e g in octahedral complexes and from e g to t 2g 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 of complex compounds: Complexes with unpaired e are paramagnetic. The no. of unpaired e depends upon electronic structure of d n+ 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 Fe 3+ in [Fe(CN)6] 3 : d 5 Fe 3+ in [FeF6] 3 : d 5 t 2g e g

9 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 Cu 2+ is based on the formation of a blue complex with NH 3 :- CuCl 2 + 4NH [Cu(NH 3 ) 4 ]Cl 2 and that of Fe 3+ on the formation of a red complex with KSCN :- FeCl 3 + KSCN K 3 [Fe(SCN) 6 ]+ 3KCl Similarly, Ni 2+ 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, vitaminb 12,cisplatinare complexes. (v) Hardness of water is estimated by complexometric titration of Ca 2+ and Mg 2+ with ethylene diaminetetraacetic acid (EDTA). VERY SHORT ANSWER TYPE QUESTION (1 marks) Q.1- Write the IUPAC name of ionization isomer of [ Co (NH 3 ) 5 Br] SO 4. A.1- Ionization isomer is [Co (NH 3 ) 5 SO 4 ] 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 (NO 2 ) 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(NH 3 ) 4 (H 2 O) 2 ]Cl 3 (ii) Fe 4 [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.

10 Q.5-What is the coordination number of central atom in followings:- (i) [Cr (en) 2 (NO 2 ) 2 ] Br. (ii) [Co (NH 3 ) 3 SO 4 ] Br. A.5-(I) 6, because en is bidentate ligand. (ii) 4. Q.6- Write the IUPAC name of Na 3 [ Cr (OH) 2 F 4 ]. 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. NO 2 - & ONO - Q.8- Which complex is used in the treatment of cancer? A.8- cis-platin or cis- [Pt (NH 3 ) 2 Cl 2 ] Q.9- Write the geometry of coordination compound in which metal atom is present in (i) sp 3 (ii) dsp 2 hybridization state. A.9- sp 3 tetrahedral dsp 2 square planar. Q.10- What is crystal field splitting energy? A.10- The energy difference between t 2g & e g 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 ] 3- Ans.-The oxidation state of Fe in this complex is +3.i.e. Fe + [Ar] 3d 6 4s 2 Fe 3+ [Ar] 3d 5 Fe 3+ in complex state:- CN - strong field ligand and thus filling takes place against Hund rule. Therefore hybridization state is d 2 sp 3 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.

11 (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 Mg 2+ &Fe 3+ respectively. (ii) cis-platin or cis- [Pt (NH 3 ) 2 Cl 2 ] is used in the treatment of cancer. (iii)wilkinson catalyst [(Ph 3 ) 3 RhCl] 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(NH 3 )(Br)(Cl)(py)]and how many of these will exhibit optical isomers? Ans)[Pt(NH 3 )(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) :- 3d 8,4s 2 Ni 2+ :- 3d 8 3d 4s 4p

12 Ni 2+ in complex state: - since CN - is strong field ligand, therefore filling takes against Hund s rule. 3d 4s 4p Thus, hybridization state is dsp 2, therefore geometry is square planar.since all electrons are paired, therefore complex is Diamagnetic. Q.7- Draw facial and meridional isomer of [Co(NH 3 ) 3 (NO 2 ) 3 ] A.7- Q.8-Draw optical isomers of [Co(en) 3 ] 2+ A.8. Q.9-Draw cis-trans isomer of [Pt(en) 2 Cl 2 ] 2+ A.9.-

13 Q.10- Draw optical isomers of cis-[pt(en) 2 Cl 2 ] 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 CoCl 3. 4 NH 3. 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(NH 3 ) 3 Cl 2 ]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) : K 2[Zn (OH) 4 ] (ii) Diamminechloridonitrito-N-platinum(II) : [Pt(NH 3) 2Cl(NO 2)] (iii) Tetraammineaquachloridocobalt(III) chloride : [Co(NH 3) 4(H 2O)Cl] (iv) Potassium trioxalatoaluminate(iii) : K 3[Al(C 2O 4) 3]

14 SHORT ANSWER TYPE QUESTION (3 marks) Q.1- How would account for followings:- (i) [Ti(H 2 O) 6 ] 3+ is coloured while [Sc(H 2 O) 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] 3d 1,since H2O is weak field ligand therefore filling takes place according to Hund rule. [Ti(H 2 O) 6 ] 3+ Due to presence of unpaired electron it is coloured. [Sc(H 2 O) 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- Fe in [Fe(CN) 6 ] 3- are:- 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-

15 Due to sp 3 hybridization state geometry of [Ni(CO) 4 ] is tetrahedral, Whereas due to dsp 2 hybridization state geometry is square planar. Q.2-Describe the type of hybridization, shape and magnetic property of followings:- (i)[fe(h 2 O) 6 ] 2+ (II) [Co(NH 3 )] 3+ (iii) [NiCN 4 ] 2-. A.2- (i) Fe in Fe(H 2 O) 6 ] 2+ Hybridization:- sp 3 d 2 Shape:- Octahedral Magnetic property:- Paramagnetic (ii) Co in [Co(NH 3 )] 3+ Hybridization:- d 2 sp 3 Shape:- Octahedral Magnetic property:- Diamagnetic (iii) Ni in [NiCN 4 ] 2- Hybridization:- dsp 2 Shape:- Square planar Magnetic property:- Diamagnetic. Q.3-(a) State for a d 6 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 [MA 2 B 2 ] 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 t 2g 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 t 2g and e g 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. (c) The spin only magnetic moment of [MnBr4] 2 is 5.9 BM. Predict the geometry of the complex ion

16 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(H 2 O) 2 (C 2 O 4 )] (ii) [Co(en) 3 ]Cl 3 (iii) [Co(NH 3 ) 5 (NO 2 )](NO 3 ) 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(NH 3 ) 5 (NO 2 )](NO 3 ) 2 and [Co(NH 3 ) 5 (NO 3 )](NO 3 ) (NO 2 ) Linkage isomers : [Co(NH 3 ) 5 (NO 2 )](NO 3 ) 2 and [Co(NH 3 ) 5 (ONO)](NO 3 ) 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(NH 3) 4(H 2O)Cl] : Tetraammineaquachloridocobalt(III) chloride

17 HOTS Questions 1. [NiCl 4 ] 2 is paramagnetic while [Ni(CO) 4 ] is diamagnetic though both are tetrahedral. Why? Answer Though both [NiCl 4 ] 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,[NiCl 4 ] 2 is paramagnetic. In Ni(CO) 4, Ni is in the zero oxidation state i.e., it has a configuration of 3d 8 4s 2.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) is5d 8.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(no 2 ) 6 ] 4, [Ni(NH 3 ) 6 ] 2+, [Ni(H 2 O) 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: H 2 O < NH 3 < NO Hence, the wavelengths of absorption in the visible region will be in the order: [Ni(H 2 O) 6 ] 2+ > [Ni(NH 3 ) 6 ] 2+ > [Ni(NO 2 ) 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).

18 (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 CoCl 3 and NH3. Answer i. helping in need of my friend. ii.my friend is affectionate and truthful iii.[co Cl 2 (NH 3 ) 4 ]Cl and [CoCl (NH 3 ) 5 ]Cl 2