CCC 1 CO-ORDINATION COMPOUNDS C1 Double Salts and Co-ordination Compounds Double Salts : are those which loose their identity in solution as alum. [K 2 S.Al 2 (S.24H 2 O Co-ordination Compounds (Complexes) : Those which retain their identity in solution (complexes) as potassium ferrocyanide, K 4 [Fe(CN. A complex (or co-ordination compound) is a compound consisting either complex ions with other ions of opposite charge. K 4 [Fe(CN complex anion [Pt(en Cl 4 complex cation [Pt(en neutral complex K + is other ion Cl is other ion Ligands : A Ligand is a species that is capable of donating an electron pair(s) to a central ion. It is a Lewis base. In accepting electron pairs, the central ion acts as a Lewis acid. Ligand can be unidentate bidentate, tridentate, tetradentate etc. Ambidient Unidentate Ligand : Ligand is said to be unidentate if it has only one pair of electrons that it can donate. Polydentate Ligand : Ligand is said to be bidentate, tridentate, tetradentate etc. depending on the number of electron pairs that it can donate. :NH 3 ; one electron pair available for donation oxidentate. two electron pairs available for donation bidendate Some common Multidentate Ligands (Chelating Agents) Abbrevation Multi Name Formulae Charge en bidentate ethylenediame 0 ox bidentate oxalate 2 gly bidentate glycinate 1
DMG bidentate dimethyl glyoximate 1 CCC 2 EDTA hexadentate ethylenediamine tetraacetate 4 Chelates : Some ligands are capable of donating more than a single electron pair, from different atoms in the ligand and to different sites in the geometric structure of a complex. These are called as multidentate ligands. When the bonding of a multidentate ligand to metal ion produces a ring (usually five or six membered) we refer to the complex as a chelate. The multidentate ligand is called a chelating agent and the process of chelate formation is called chelation Nickel (II) dimethylglyoximate (chelate) [bis-(dimethylglyoximato) nickel (II) Co-ordination Number : The co-ordination number of a metal atom in a complex is the total number of bonds the metal atom forms with Ligands. Metal ion Co-ordination Number (C.N.) of Metal Ions Ag + 2 C.N. Cu + 2, 4 Cu 2+ 4, 6 Au + 2, 6 Ca 2+ 6 Fe 2+, Fe 3+ 6 Co 2+ 4, 6 Co 3+ 6 Ni 2+ 4, 6 Zn 2+ 4
Al 3+ 4, 6 CCC 3 Sc 3+ 6 Cr 3+ 6 Pt 2+ 4 Pt 4+ 6 Complex C.N. [Ag(NH 3 + 2 [HgI 3 3 [PtCl 4, [Ni(CO 4 [CO(NH 3 3+ 6 Practice Problems : 1. FeS solution mixed with (NH 4 S solution in 1 : 1 molar ratio gives the test of Fe 2+ ion but CuS solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu 2+ ion. Explain why? 2. What is meany by unidentate, didentate and ambidentate ligands? Give two examples for each. 3. Specify the oxidation numbers of the metals in the following coordination entities : (i) [Co(H 2 O)(CN)(en 2+ (ii) [CoBr 2 (en + (iii) [PtCl 4 (iv) K 3 [Fe(CN (v) [Cr(NH 3 Cl 3 4. What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes. 5. What is meany by the chelate effect? Given an example. 6. How many ions are produced from the complex Co(NH 3 in solution? (i (ii (iii (iv [Answers : (1) FeS solution mixed with (NH 4 S solution forms a double salt, which ionises in the solution to give Fe 2+ ions. CuS solution mixed with aqueous ammonia forms a complex salt, does not ionise to give Cu 2+ ions (3) (i) +3 (ii) +2 (iii) +3 (iv) +3 (v) +3 (5) a didentate or a polydentate ligand coordinates with the central metal ion forming a five or a six membered ring, the effect is called chelate effect (6) (iii) is correct C2A Nomenclature : Following rules are adopted for naming a complex ion. cations are named before anions. oxidation state (O.S.) of the central metal ion is denoted by Roman numeral Cation O.S. Anion CuC 2 FeCl 3 Copper (II) chloride Iron (III) chloride The names of ligands are given first followed by the name of central metal ion. The names of ligands that are anions and ending with ide changed to o
(e) (e) (f) ite changed to ito ate changed to ato Neutral Some Common Unidentate Ligands Anions (replacing ide by o ) CCC 4 Formula Names as Ligand Formula Name as Ligand H 2 O aqua F (Fluoride) Fluorido NH 3 ammine Cl Chlorido CO Carbonyl S 2 sulphido NO nitrosyl H hydrido CH 3 NH 2 methyl amine O 2 (oxide) oxo C 5 H 5 N or Py Pyridine OH (hydroxide) hydroxo (C 2 H 5 N Triethylamine CN (cyanide) cyano* (C 6 H 5 N Tripphenyl amine NC isocyano* [* marked are ambident) Positive groups end in-ium NH 2 NH 3 + hydrazinium S S 2 O 3 CO 3 NO 2 ONO SCN NCS Replacing e by o sulphato thiosulphato carbonato nitro* nitrito* thiocyanato* isothiocyanato* When there are several ligands of the same kind we normally use the prefix di, tri, tetra soon. If the name of ligand includes a number e.g. ethylenediamine (en). To avoid confusion in such cases, bis, trix and tetrakis are used. e.g. bis (ethylenediammine) (g) If anion is a complex then metal ends with ate [Ni(CN tetracyanonickelate (II) ion lead plumbate gold Aurate tin stannate silver argentate iron ferrate copper cuprate
(h) (i) CCC 5 If the complex contain two or more metal atoms, it is termed polynuclear. The bridging ligands which link the two metal atoms together are indicated by the prefix Ambidient Ligands may be attached through different atoms M NO 2 M ONO (NO 2 ) joined to metal M through N; it is nitro (NO 2 joined to metal M through O it is nitro) Similarly the SCN group may bond M-SCN (thiocyanato) or (M NCS) (isothiocyanato). They may be named as thiocyanato S, Practice Problems : thiocyanato N 1. Write the IUPAC names of the following coordination compounds : (i) [Co(NH 3 Cl 3 (ii) [Co(NH 3 Cl (iii) K 3 [Fe(CN (iv) K 3 [Fe(C 2 (v) K 2 [PdCl 4 (vi) [Pt(NH 3 Cl(NH 2 CH 3 )Cl 2. Using IUPAC norms, write the systematic names of the following : (i) [Co(NH 3 Cl 3 (ii) [Pt(NH 3 Cl(NH 2 CH 3 )Cl (iii) [Ti(H 2 O 3+ (iv) [Co(NH 3 Cl(NO 2 )Cl (v) [Mn(H 2 O 2+ (vi) [NiCl 4 (vii) [Ni(NH 3 (viii) [Co(en 3+ (ix) [Ni(CO [Answers : (1) (i) hexaamminecobalt(iii) chloride (ii) pentaamminechloridocobalt(iii) chloride (iii) potassium hexacyanoferrate(iii) (iv) potassium trioxalatoferrate(iii) (v) potassium tetrachloridopalladate(ii) (vi) diamminechlorido (methylamine) platinum(ii) chloride (2) (i) hexaamminecobalt(iii) chloride (ii) diamminechlorido(methyl amine)platinum(ii) chloride (iii) hexaaquatitanium(iii) ion (iv) tetraamminechloridonitrito-n-cobalt(iii) chloride (v) hexaaquamanganese(ii) ion (vi) tatrachloridonickelate(ii) ion (vii) hexaamminenickel(ii) chloride (viii) tris(ethane-1, 2-diamine)cobalt(III) ion (ix) tetracarbonylnickel(0) C2B When writing (not naming) the formulae of the complex complex ion should be enclosed by square brackets. Ligands are place after metal in the alphabetical order but first negative ligands, then neutral then positive. [Pt (H 2 O (NH 3 wrong aqua ammine [Pt(NH 3 (H 2 O correct [Cr (CN) 4 (NH ) correct negative 3 2 neutral
Example :- CCC 6 1. K [Fe(CN) 4 6 cation Anion Potassium hexacyanoferrate(ii) cation anion anion is complex hence metal ends with ate. Ligands are named before metal ion of complex. 2. [CO(NH3) Cl cation 6 3 anion Hexaammine cobalt (III) chloride 3. [CO(en Cl 3 Tris (ethylenediammine) cobalt (III) chloride 4. [Pt(Cl (NH 3 Tetraamminedichloroplatinum (IV) chloride [NH 3 (ammine) and Cl(chloro) are written in alphabetical order 5. In the following examples complex ion exist as cation :- [Fe(NH 3 Cl 3 Hexaammine iron (III) chloride [CoCl(NH 3 2+ Pentaamminechlorocobalt (III) ion [CoS (NH 3 NO 3 Tetraamminesulphatocobalt (III) nitrate 6. Complex ion exist as anions :- [ZnCl 4 Tetrachlorozincate (II) ion [AlH 4 Tetrahydridoaluminate (III) ion Na 3 [Ag(S 2 O 3 Sodium bis (thiosulphato) argentate (I) Na 2 [OSCl 5 N sodium pentachloronitrido osmate (VI) (e) K 2 [Cr(CN O 2 (O 2 )NH 3 Potassium amminedicyanodioxo peroxochromate (VI) 7. Organic Ligands have been used :- [Pt(Py [Pt Cl 4 Tetrapyridineplatinum (II) tetrachloroplatinate (II) [Cr(C 6 H 6 Bis (benzene) chromium (O) [Fe(C 5 H 5 Bis (cyclopentadienyl) iron (II) 8. In the following examples, bridging groups are used :-
Tetrakin (ethylenediammine)-µ-imido-µ-hydroxo dicobalt (III) ion. CCC 7 Trans-bis (chlortiphenylphosphine-µ-chloropalladium (II) or Chlorotriphenylphosphinepallodium (II)-µ-dichlorochlorotriphenylphosphinepalladium (II) Practice Problems : 1. Write the formulas for the following coordination compounds : (i) (ii) (iii) (iv) (v) (vi) Tetraamminediaquacobalt(III) chloride Potassium tetracyanonickelate(ii) Tris(ethane-1, 2-diamine) chromium(iii) chloride Amminebromidochloridonitrito-N-platinate(II) Dichloridobis(ethane-1, 2-diamine)platinum (IV) nitrate Iron (III) hexacyanoferrate (II) 2. Using IUPAC norms write the formulas for the following : (i) Tetrahydroxozincate(II) (ii) Potassium tetrachloridopalladate(ii) (iii) Diamminedichloridoplatinum(II) (iv) Potassium tetracyanonickelate(ii) (v) Pentaaminenitrito-O-cobalt(III) (vi) Hexaamminecobalt(III) sulphate (vii) Potassium trioxalatochromate(iii) (viii) Hexaammineplatinum(IV) (ix) Tetrabromidocuprate(II) (x) Pentaamminenitrito-N-cobalt(III) [Answers : (1) (i) [Co(NH 3 (H 2 O Cl 3 (ii) K 2 [Ni(CN (iii) [Cr(en Cl 3 (iv) [PtBrClNO 2 (NH 3 ) (v) [Pt (en (NO 3 (vi) Fe 4 [Fe(CN 3 (2) (i) [Zn(OH (ii) K 2 [PdCl 4 (iii) [Pt(NH 3 (iv) K 2 [Ni(CN (v) [Co(NH 3 (ONO) 2+ (vi) [Co(NH 3 2 (S (vii) K 3 [Cr(C 2 (viii) [Pt(NH 3 4+ (ix) [CuBr 4 (x) [Co(NH 3 (NO 2 ) 2+ C2C (EAN) Effective Atomic Number : Each ligand donates an electron pair to the metal ion, thus forming a co-ordinate bond EAN = Z O.N. + 2 (C.N.) ON oxidation number or C.N. co-ordination number EAN = Z O.N. + 2 (Ligands) Example :- (assuming that each ligand is unidentate) [Cr(CO EAN = 24 O + 2 6 = 36 [Fe(CN 2 EAN = 26 2 + 6 2 = 36 [CO(en 3+ EAN = 27 3 + 6 2 = 36 (en) bidentate
Practice Problems : CCC 8 1. Explain with two examples each of the following : coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic. C3 Isomerism in Complexes :- Compounds having same chemical formulae but differ structural arrangements are called isomers. Structural isomerism :- Differ from each other. How the atoms are joined together that is in the order in which the atoms are bonded to each other. H N = C = O and N C O H are structural isomer. Ionisation Isomerism :- This type of isomerism occurs when there is an interchange of groups between the co-ordination sphere of the metal ion and the ions outside this sphere. Examples are : [CO(NH 3 (S Br [CO(NH 3 Br S [give yellow ppt. with Ag + [give white ppt. with Ba 2+ 2 as S being ion as Br outside the bracket ionisable is ionisable [Pt(NH 3 Co-ordination isomerism :- Br [Pt(NH 2 3 Br 2 ionisable ionisable When both positive and negative ions are complex ions. Isomerism may be caused due to interchange of ligands between anion and cation. This arises only when cation and anion have same C.N. and charge [CO(NH 3 3+ [Cr(CN 3 same charge & [Cr(NH 3 3+ [CO(CN 3 Linkage Isomerism :- same O.N. (i) [CO(NH 3 ONO and [CO(NH 3 NO 2 (ii) [Mn(CO (SCN) + [Mn(CO (NCS) + Co-ordination Position Isomerism : In polynuclear complexes an interchange of ligands between the different metal nuclei give rise to positional isomerism e.g.
CCC 9 Hydrate Isomerism :- Hydrate isomers of a complex that differ in the placement of water molecules in the complex CrCl 3.6H 2 O can be written as [Cr(H 2 O Cl 3 violet (anhydrous) [Cr(H 2 O Cl.H 2 O light green (monohydrate) [Cr(H 2 O Cl.2H 2 O Dart green (dihydrate) Stereo Isomers :- 1. Geometrical 2. Optical Geometrical :- are isomers in which the atoms are joined to one another in the same way but differ because some atom occupy different relative position in space. [Pt(NH 3 Geometrical Isomers of C.N. 4 [CO(NH 3 + Geometrical isomers of C.N. 6
CCC 10 [Pt(NH 3 ) Br 2 [Pt(NH 3 Cl 3 Optical Isomerism :- If a molecule is asymetric that it cannot superimposed on its mirror image. The two forms have the same type of symmetry shown by the left and right hands and are called enantiomeric pair. The two forms are called optical isomers. They are called either dextro or leavo (dor l). Optical isomerism is common in octahedral complexes involving bidentate ligands. [CO(en exists as cis and trans but cis-form can have optical isomerism.
CCC 11 Practice Problems : 1. Give evidence that [Co(NH 3 ClS and [Co(NH 3 S Cl are ionisation isomers. 2. List various types of isomerism possible for coordination compounds, giving an example of each. 3. How many geometrical isomers are possible in the following coordination entities? [Cr(C 2 3 [Co(NH 3 Cl 3 4. Draw the structures of optical isomers of : [Cr(C 2 3 [Pt (en 2+ [Cr(NH 3 (en) + 5. Draw all the isomers (geometrical and optical) of : (i) [Co (en + (ii) [Co(NH 3 )Cl(en 2+ (iii) [Co(NH 3 (en) + 6. Write all the geometrical isomers of [Pt(NH 3 )(Br)(Cl)(py) and how many of these will exhibit optical isomers?
CCC 12 [Answers : (1) On dissolving in water, the two compounds will give different ions in the solution which can be tested by adding AgNO 3 solution and Ba solution (3) no possibility two geometrical isomers (6) Three isomers are possible C4A Werner s theory of co-ordination compounds : 1. There are two types of valency shown by central metal atom/metal ion in a compound Primary Secondary Primary Valency corresponds to oxidation number and secondary valency corresponds to co-ordination number. In the complex [CO(NH 3 Cl 3, primary valency is satisfied by three Cl, secondary valency is six. In [CO(NH 3 (H 2 O S, four NH 3 and two H 2 O (ligands) satisfy secondary valency. One S 2 satisfy primary valency. Primary valency is ionisable. The secondary valencies are directional and are responsible for isomerism in complexes. The primary valencies are non-directional and is represented by... and secondary valency by In all cases, metal or metal ions should satisfy primary and secondary valency both. Some negative ions can satisfy primary as well as secondary valency both. e.g. [Cr(NH 3 Cl satisfy both primary and secondary valency. Practice Problems : 1. Explain the bonding in coordination compounds in terms of Werner s postulates. C4B Bonding in complexes The metal m loses required no. of electrons to form cation. No. of electrons lost correspond to O.N. The metal ion makes available a no. of empty orbitals equal ot its co-ordinates number for the formation of co-ordinates bonds with the Ligand Orbitals. A weak ligand like (H 2 O, halide) will not affect the electronic configuration of the metal/ metal ion. A strong ligand like (NH 3, CN, CO) affect the electronic configuration of the metal/ metalion. They make unpaired electrons paired. The metal ion orbitals hybridise to form a new set of equivalent hybridised orbitals. If there are unpaired electrons in the complex it is called paramagnetic. Magnetic moment = N(N 2) B.M. N no. of unpaired electrons.
CCC 13 The complex with so many unpaired electrons is high spin complex and that with paired electrons is low spin complex. If inner d-orbital is used for hybridisation it is called inner d complex. e.g. d 2 sp 3 [Fe(CN 4 If outer d-orbital is used is called outer d-complex. sp 3 d 2 Weak field ligand [Fe(H 2 O 3+ I < Br < S < Cl < NO 3 < F < oxalate < H 2 O < EDTA < NH 3 < en < NO 2 < CN < CO Examples : [Fe(CN 4 O.N. = +2 C.N. = 6 [six empty orbitals are required by six CN ligands t 2 g d xy, d yz, d zx e.g. dx 2 y 2, dz 2 Complex ion is d 2 sp 3 hybridised Inned d-complex Diamagnetic Magnetic moment = O Octahedral geometry [Fe(H 2 O 3+ (H 2 O is weak ligand) Hybridisation sp 3 d 2 Outer-d-complex Paramagnetic (High spin complex) Octahedral geometry
Commom type of hybridisation and Geometry of complexes Hybridisation Geometry Examples CCC 14 sp Linear [Ag(NH 3 + sp 2 Trigonal Planal [HgI 3, [CuP(Me + sp 3 Tetrahedral [V 3, [Mn, [NiCl 4 2, [Ni(CO, [Zn(Cl 4 ) 2 dsp 2 Square planar [Cu(NH 3 2+, [Ni(CN 2, [Pt(NH 3 2+ sp 3 d or dsp 3 Trigonal [Cu(Cl 5, [Ni(CN 3 d 2 sp 3 or Bipyramidal [Fe(CO sp 3 d 2 Octahedral [Fe(H 2 O 3+ Practice Problems : 1. Explain on the basis of valence bond theory that [Ni(CN ion with square planar structure is diamagnetic and the [NiCl 4 ion with tetrahedral geometry is paramagnetic. 2. [NiCl 4 is paramagnetic while [Ni(CO is diamagnetic though both are tetrahedral. Why? 3. Explain [Co(NH 3 3+ is an inner orbital complex whereas [Ni(NH 3 2+ is an outer orbital complex. 4. Predict the number of unpaired electrons in the square planar [Pt(CN ion. 5. The hexaaquo manganese (II) ion contains five unpaired electrons while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. 6. Aqueous copper sulphate solution (blue in colour) gives : (i) (ii) a green precipitate with aqueous potassium fluoride and a bright green solution with aqueous potassium chloride. Explain these experimental results. 7. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H 2 S(g) is passed through this solution? 8. Discuss the nature of bonding in the following coordination entities on the basis of Valence Bond Theory : (i) [Fe(CN 4 (ii) [FeF 6 3 (iii) [Co(C 2 (iv) [CoF 6 3 9. Draw a figure to show the splitting of d-orbitals in an octahedral crystal field. 10. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand. 11. What is crystal field splitting energy? How does the magnitude of 0 decide the actual configuration of d-orbitals in a coordination entity? 12. [Cr(NH 3 3+ is paramagnetic while [Ni(CN is diamagnetic. Explain, why? 13. A solution of [Ni(H 2 O 2+ is green but a solution of [Ni(CN is colourless. Explain. 14. [Fe(CN 4 and [Fe(H 2 O 2+ are of different colours in dilute solutions. Why? 15. Discuss the nature of bonding in metal carbonyls.
CCC 15 16. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex : (i) K[Cr(H 2 O (C 2.3H 2 O (ii) CrCl 3 (py (iii) [Co(NH 3 Cl (iv) [Cs[FeCl 4 17. Discuss briefly giving an example in each case the role of coordination compounds in : (i) biological systems (ii) medicinal chemistry (iii) analytical chemistry and (iv) extraction/metallurgy of metals 18. Amongst the following ions, which one has the highest magnetic moment value? (i) [Cr(H 2 O 3+ (ii) [Fe(H 2 O 2+ (iii) [Zn(H 2 O 2+ 19. The oxidation number of cobalt in K[Co(CO is (i) +1 (ii) +3 (iii) 1 (iv) 3 20. Amongst the following, the most stable complex is : (i) [Fe(H 2 O 3+ (ii) [Fe(NH 3 3+ (iii) [Fe(C 2 3 (iv) [FeCl 6 3 21. What will be the correct order for the wavelengths of absorption in the visible region for the following : [Ni(NO 2 4, [Ni(NH 3 2+, [Ni(H 2 O 2+ [Answers : (2) Cl is weak ligand. It cannot pair up the electrons in 3d orbitals. Presence of strong CO ligand and 4s electrons shift to 3d to pair up with 3d electrons (3) Presence of NH 3 pairing of 3d electrons takes place leaving two d-orbitals empty. In [Ni(NH 3 2+, Ni is in +2 state with the configuration 3d 8. The NH 3 can not pair up 3d electrons (4) No unpaired electron (6) (i) weak H 2 O ligands are replaced by F ligands forming [CuF 4 ions which is a green precipitate (ii) weak H 2 O ligands are replaced by Cl ligands forming [CuCl 4 ion which has bright green colour (7) The complex ion is highly stable and does not dissociate to give Cu 2+ ions. Hence, no precipitate with H 2 S is obtained (8) (i) d 2 sp 3 (ii) sp 3 d 2 (iii) d 2 sp 3 (iv) sp 3 d 2 (10) ligands in order of their increasing field strengths (11) The difference of energy between the two sets of orbitals is called crystal field splitting energy. If 0 < P, then 4th electron pairsup in one of the t 2g orbitals giving the configuration 4 0 g g thereby forming low spin complexes (13) In [Ni(H 2 O 2+, it has two unpaired electrons. In [Ni(CN, no unpaired electrons (14) ligands H 2 O and CN possess different crystal field splitting energy ( 0 ) (18) (ii) having greatest value of n has highest magnetic moment (19) (iii) is the correct option (20) (iii) is correct option (21) Ni(H 2 O 2+ > [Ni(NH 3 2+ > [Ni(NO 2 4 t 2 e
Complex State of Hybridisation and Magnetic Behaviour of some co-ordination complex Ion Configuration of metal ion Hybridisation of metal ion orbitals for ligand bonds No. of unpaired electrons Magnetic Behaviour [V(H 2 O 3+ V 3+ d 2 sp 3 d 2 2 Paramagnetic [Cr(NH 3 3+ Cr 3+ d 3 d 2 sp 3 3 Paramagnetic [MnF 6 3 Mn 3+ d 4 sp 3 d 2 4 Paramagnetic [Mn(CN 3 Mn 3+ d 4 d 2 sp 3 2 Paramagnetic [Fe(CN 3 Fe 3+ d 5 d 2 sp 3 1 Paramagnetic [Fe(F 6 Fe 3+ d 5 sp 3 d 2 5 Paramagnetic [Fe(Cl 4 ) 2 Fe 2+ d 6 sp 3 4 Paramagnetic [COF 6 3 CO 3+ d 6 sp 3 d 2 4 Paramagnetic [CO(CN 3 CO 3+ d 6 d 2 sp 3 0 Diamagnetic [Ni(Cl 2 Ni 2+ d 8 sp 3 2 Paramagnetic [Ni(CN 2 Ni 2+ d 8 dsp 2 0 Diamagnetic [COCl 4 2 CO 2+ d 9 sp 3 1 Diamagnetic [Zn(NH 3 2+ Zn 2+ d 10 sp 3 0 Diamagnetic CCC 16 Metal Metal
CCC 17 E X E R C I S E (OBJECTIVE) 1. The EAN of Ni in Ni(CN is 34 35 36 (d8 2. Glycinato ligand is Bidentate ligand Two donor sites N and O All 3. The compound which does not show paramagnetism [Cu(NH 3 [Ag(NH 3 Cl NO NO 2 4. Which one is bidentate ligand C 2 NH 2 CH 2 CH 2 NH 2 Both None 5. Which ion is paramagnetic [Co(NH 3 2+ [Ni(CO [Co(NH 3 3+ [Ni(CN 6. The shape of the complex Ag(NH 3 + is Octahedral Square planar Tetrahedral Linear 7. K 3 CoF 6 is high spin complex. What is the hybrid state of Co atom in this complex sp 3 d sp 3 d 2 d 2 sp 3 dsp 2 8. Which one of the following will be able to show cis-trans isomerism MA 3 B M(A A MA 2 BCD MA 4 9. The type of isomerism shown by [Co(en (NCS Cl and [Co(en (NCS)ClNCS is Co-ordination Ionisation Linkage All above 10. Which of the following cations does not form an amine complex with excess of ammonia Ag + Cu 2+ Cd 2+ Na + 11. Hexafluorocobaltate (III) ion is found to be high spin complex, the probable hybrid state of cobalt in it is d 2 sp 3 sp 3 sp 3 d sp 3 d 2 12. In sodium tetrafluorooxochromate(...), Na 3 [Cr(O)F 4 the left out place should be filled with which of the following Roman numericals VI III IV none of these 13. The correct name of the compound [Cu(NH 3 (NO 3, according to IUPAC system is Cuprammonium nitrate Tetraammine copper(ii) dinitrate Tetraammine copper(ii) nitrate Tetraammine copper(i) dinitrate 14. Lithium tetrahydridoaluminate is correctly represented as Al[LiH 4 Al 2 [LiH 4 3 Li[AlH 4 Li[AlH 4 2 15. The correct IUPAC name of K 2 [Zn(OH is Potassium tetrahydroxyzinc(ii) Potassium tetrahydroxozincate(ii) Potassium tetrahydroxyzincate(iv) Potassium hydroxozinc(ii) 16. The correct IUPAC name of KAl(S 12H 2 O is Aluminium potassium sulphate-12-water Potassium aluminium(iii) sulphate-12- water Potassium aluminate(iii) sulphate hydrate Aluminium(III) potassium sulphate hydrate-12 17. Number of electrons gained by Pd in [PdCl 4 2 ; 4 8 10 Zero 18. The possible number of isomers for the complex [M Br 2 S is 1 (b (c 5 19. The oxidation number of Pt in [Pt(C 2 H 4 )Cl 3 is +1 +2 +3 +4
20. Chlorophyll is a co-ordination compound having central atom of Ca Mg Na K 21. Among [Ni(CN [NiCl 4 and [Ni(CO Ni(CN is square planar and NiCl 4 and Ni(CO are tetrahedral NiCl 4 is square planar and Ni(CN and Ni(CO are tetrahedral Ni(CO is square planar and Ni(CN and [NiCl 4 are tetrahedral None 22. Which one does not belong to ligand PH 3 NO + BF 3 Cl 23. Nickel metal is in highest oxidation state in Ni(CO K 2 NiF 6 [Ni(NH 3 (BF 4 K 4 [Ni(CN 24. A complex of cobalt has five ammonia molecules, one nitro group and two chlorine atoms for each cobalt atom. One mole of this compound produces three moles ions in aqueous solution which on treating with excess of AgNO 3 give two moles of AgCl. The formula of the compound is [Co(NH 3 NO 2 Cl[(NH 3 )Cl [Co(NH 3 Cl[ClNO 2 [Co(NH 3 NO 2 [Co(NH 3 [(NO 2 25. K 3 [Al(C 2 is called Potassium alumino oxalate Potassium alumino (III) oxalate Potassium trioxalato aluminate Potassium trioxalato aluminate (III) 26. In hexacyanomanganate (II) ion the Mn atom assumes d 2 sp 3 hybrid state. The number of unpaired electrons in the complex is 1 (b 3 zero 27. Name the metal M which is extracted on the basis of following reactions 4M + 8CN + 2H 2 O + O 2 4[M(CN + 4OH 2[M(CN + Zn [Zn(CN + 2M Nickel Silver Copper Mercury 28. The colour of [Ti(H 2 O 3+ is due to Transfer of an electron from one Ti to another Presence of water molecule CCC 18 Excitation of electrons from d d Intramolecular vibration 29. The number of d-electrons in [Cr(H 2 O 3+ is (a 3 (c 5 30. Ammonia forms the complex ion [Cu(NH 3 2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it? Copper hydroxide is an amphoteric substance In acidic solution hydration protects copper ions In acidic solutions protons coordinates with ammonia molecules forming NH 4 + ions and NH 3 molecules are not available In alkaline solution insoluble Cu(OH is precipitated which is soluble in excess of any alkali 31. In the coordination compound, K 4 [Ni(CN, the oxidation state of nickel is +2 1 0 +1 32. The coordination number of a central metal atom in a complex is determined by the number of ligands around a metal ion bounded by sigma bonds the number of ligands around a metal ion bounded by pi-bonds the number of ligands around a metal ion bonded by sigma and pi-bonds both the number of only anionic ligands bonded to the metal ion. 33. Which one of the following complexes is an outer orbital complex? [Fe(CN 4 [Mn(CN 4 [Co(NH 3 3+ [Ni(NH 3 2+ [Atomic nos. : Mn = 25, Fe = 26, Co = 27, Ni = 28 34. Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statmenets about cerium is incorrect? The common oxidation states of cerium are +3 and +4 The +3 oxidation state of cerium is more stable than +4 oxidation state The +4 oxidation state of cerium is not known in solutions Cerium (IV) acts as an oxidising agent.
35. The correct order of magnetic moments (spin only values in B.M.) among is [MnCl 4 > [CoCl 4 > [Fe(CN 4 [MnCl 4 > [Fe(CN 4 > [CoCl 4 [Fe(CN 4 > [MnCl 4 > [CoCl 4 [Fe(CN 4 > [CoCl 4 > [MnCl 4 36. The IUPAC name of the coordination compound K 3 [Fe(CN is Potassium hexacyanoferrate (II) Potassium hexacyanoferrate (III) Potassium hexacyanoiron (II) Tripotassium hexacyanoiron (II) 37. Which of the following compounds shows optical isomerism? [Cu(NH 3 2+ [ZnCl 4 [Cr(C 2 3+ [Co(CN 3 38. Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour? [Cr(CN 3 [Mn(Cn 3 [Fe(CN 3 [Co(CN 3 [At. Nos : Cr = 24, Mn = 25, Fe = 26, Co = 27) 39. The value of the spin only magnetic moment for one of the following configurations is 2.84 BM. The correct one is d 4 (in strong ligand field) d 4 (in weak ligand field) d 3 (in weak as well as in strong field) d 5 (in strong ligand field) CCC 19 40. The IUPAC name for the complex [Co(NO 2 )(NH 3 is pentaamminenitrito-n-cobalt(iii) chloride nitrito-n-pentaamminecobalt(iii) chloride nitrito-n-pentaamminecobalt(ii) chloride pentaamminenitrito-n-cobalt(ii) chloride 41. Nickel (Z = 28) combines with a uninegative monodentate ligand X to form a paramagnetic complex [NiX 4. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively two, square planar one, tetrahedral two, tetrahedral one, square planar 42. In Fe(CO, the Fe-C bond possesses -character only -character only both and characters ionic character 43. The spin-only magnetic moment [in units of Bohr magneton. (µ B ) of Ni 2+ in aqueous solution would be (atomic number Ni = 28) 1.73 (b.84 (c.90 0 44. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca 2+ ion? Two Six Three One A N S W E R S 1. a 9. b 17. b 25. d 33. d 41. c 2. d 10. d 18. d 26. a 34. c 42. c 3. b 11. d 19. b 27. b 35. a 43. b 4. c 12. b 20. b 28. c 36. b 44. d 5. a 13. c 21. a 29. b 37. c 6. d 14. c 22. c 30. c 38. c 7. b 15. b 23. b 31. c 39. a 8. c 16. b 24. c 32. a 40. a