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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , PREVIOUS HSE QUESTIONS AND ANSWERS FOR THE CHAPTER “CO-ORDINATION COMPOUNDS”, 1) A list of co-ordination compounds are given below: [PtCl2(NH3)2], [PtCl2(NH3)4]Br2, [Cr(H2O)6]Cl3., Which type of isomerism do these compounds exhibit? (3), [March 2008], Ans: [PtCl2(NH3)2] – Geometrical isomerism, [PtCl2(NH3)4]Br2 – Ionisation isomerism and Geometrical isomerism, [Cr(H2O)6]Cl3 – Hydrate (Solvate) isomerism, 2) Teacher asked two students to write the electronic configuration of d 4 system using CFT., Student I: t2g3 eg1, Student II: t2g4 eg0, a) Suggest which student gives correct configuration. Justify your answer. (2), b) Draw figure to show splitting of degenerate‘d’ orbitals in an octahedral crystal field. (1) *SAY 2008+, Ans: a) Both of the configurations are correct depending on the strength of the ligand. If the ligand is weak, field, student I is correct. If the ligand is strong field, student II is correct., b), , 3) The central metal ion Co3+ with co-ordination number 6 can form a series of complexes in which both Cland NH3 are acting as ligands., a) Give the formulae of each complex molecule (three molecules). (1½), b) Give the IUPAC names of the above complexes., (1½) [March 2009], Ans: a) [Co(NH3)5Cl]Cl2, [Co(NH3)4Cl2]Cl and [Co(NH3)3Cl3], b) [Co(NH3)5Cl]Cl2 – Pentaamminechloridocobalt(III)chloride, [Co(NH3)4Cl2]Cl – Tetraamminedichloridocobalt(III)chloride, [Co(NH3)3Cl3] – Triamminetrichloridocobalt(III), 4) [Cr(NH3)4Cl2]Br is a co-ordination compound., a) Identify the central metal ion of the above compound. (½ ), b) Name the ligands present in it. (1), c) What is its co-ordination number?, (½), d) Write the IUPAC name. (1), e) Write the ionisation isomer of the above compound. (1), [March 2010], 3+, Ans: a) Cr, b) NH3 and Cl–, c) 6, d) Tetraamminedichloridochromium(III)bromide, e) [Cr(NH3)4ClBr]Cl, Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 1

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Join Telegram Channel: https://t.me/hsslive, , 5) Na2EDTA is used in the estimation of hardness of water., a) Draw the structure of EDTA4-., (1), b) What is its denticity?, (½), c) What are the donor atoms in it? (½), d) Why is it called a chelating ligand?, (1), [March 2010], Ans: a), , 6), , 7), , 8), , c), , 9), , Downloaded from www.Hsslive.in ®, , b) Denticity = 6, c) Here the donor atoms are Oxygen and Nitrogen., d) It can form ring complexes with the central atom. So it is a chelating ligand., a) NO2- and ONO- constitute ambidentate ligands. Give another set of ambidentate ligands., (1), 4b) EDTA is a chelating ligand. Give two other examples. (1), c) Give the denticity of NO2- and NH3., (1), [March 2010], –, –, Ans: a) SCN and NCS, b) Ethane-1,2-diamine (en) and oxalate ion (C2O42–), c) 1, The central ion Ag+ with co-ordination number 2 forms a positive complex ion with NH3 ligand. Also Ag+, forms a negative complex with CN- ligand., a) Write the formulae of the above positive and negative complex ions. Give the IUPAC name of each., (2), b) Give the denticity of NH3 and CN- ligands., (1), c) Write the formula and name of a hexadentate ligand. (1), [SAY 2011], +, –, Ans: a) [Ag(NH3)2] - Diamminesilver(I) and [Ag(CN)2] - Dicyanidoargentate(I), b) 1, c) EDTA4– (Ethylenediaminetetraacetate ion), Consider the co-ordination compound [Co(NH3)5SO4]Br., a) Write the IUPAC name of the above co-ordination compound. (1), b) What are the primary and secondary valencies of the central metal cobalt in the above co-ordination, compound? (1), Which type of structural isomerism is exhibited by the above co-ordination compound? (1) *March ‘12+, Ans: a), Pentaamminesulphatocobalt(III)bromide, b), Primary valency = 3 and secondary valency = 6, c), Ionisation isomerism, [Cr(NH3)5CO3]Cl is a co-ordination compound., a) Name the central metal ion of the above compound. (1), b) What is the IUPAC name? (1), c) Name the ligands present in the above compound., (1), d) Whether the ligands present in the above compound are ambidentate ligands? Why?, (1), , Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 2

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , e) What is the ionisation isomer of the above mentioned co-ordination compound?, (1) [SAY 2012], 3+, Ans: a) Cr, b) Pentaamminecarbonatochromium(III)chloride, c) NH3 and CO32–, d) Since they do not contain more than one donor atoms, they are not ambidentate ligands., e) [Cr(NH3)5Cl]CO3, 10) The magnetic behaviour of a complex can be explained on the basis of Valence Bond (V.B) theory., a) [Co(NH3)6]3+ is a diamagnetic complex and [CoF6]3- is a paramagnetic complex. Substantiate the, above statement using V.B theory. (3), b) Classify the above mentioned complexes into inner orbital and outer orbital complexes. (1), [March 2011 & 2013], 3+, Ans: a) In [Co(NH3)6] , the electrons get paired in presence of the strong field ligand NH3. So there is no, unpaired electron and hence it is diamagnetic. But in [CoF6]3 –, F – is a weak field ligand and hence electrons, do not get paired. So there is unpaired electron and hence it is paramagnetic., b) [Co(NH3)6]3+ is an inner orbital complex, while [CoF6]3 – is an outer orbital complex., 11) Many theories have been put forth to explain the nature of bonding in co-ordination compounds., a) On the basis of valence bond theory account for the diamagnetic behaviour of [Ni(CN) 4]2-. (1½), b) What is the shape of the above complex?, (½ ), c) Arrange the following ligands in the increasing order of their field strengths (as in spectrochemical, series)., Cl-, CO, H2O, OH(1), [SAY 2013], 2–, –, Ans: a) In [Ni(CN)4] , CN is a strong field ligand and hence all the electrons get paired. So there is no, unpaired electron and hence the complex is diamagnetic., b) Square planar, c) Cl – < OH – < H2O < CO, 12) [Co(NH3)5SO4]Cl is an octahedral co-ordination compound., a) Write the IUPAC name of the above compound., (1), b) Write the formula of the ionization isomer of the above compound., (1), c) How do d - orbitals split in an octahedral crystal field?, (1), d) Draw the diagram which indicates the splitting of d-orbitals in tetrahedral field. (1), [March 2014], Ans: a) Pentaamminesulphatocobalt(III)chloride, b) [Co(NH3)5Cl] SO4, c) In octahedral field, d-orbitals split into two – the lower energy t2g levels and the higher energy eg levels., d), , Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 3

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , 13) a) Valence Bond Theory (VBT) can explain the magnetic behaviour and shape of complexes. Using VBT, explain the diamagnetism and square planar shape of [Ni(CN)4]2-., (2), b) i) Suggest the shape of the following complexes – [Ni(CO)4] and [CoF6]3- (1), ii) The central ion Co3+ with co-ordination number 6 is bonded to the ligands NH3 and Br- to form a, dipositive complex ion. Write the formula or IUPAC name of the complex ion., (1), [SAY 2014], 2–, 2, Ans: a) In [Ni(CN)4] , the central atom Ni is in dsp hybridisation. So it has a square planar geometry., Due to the absence of unpaired electrons, the complex is diamagnetic., b) i) [Ni(CO)4] is tetrahedral complex and [CoF6]3- is octahedral complex., ii) [Co(NH3)5Br]2+ [Pentaamminebromidocobalt(III)], 14) Co-ordination compounds contain central metal atom/ion and ligands., a) Primary valency of central metal atom/ion in [Co(NH3)6]Cl3 is:, i) 3 ii) 6, iii) 4 iv) 9, (1), b) i) What are the postulates of Werner’s theory?, (2), ii) Write the IUPAC names of K3[Fe(CN)6] and [Co(NH3)6]Cl3., (1), [March 2015], Ans: a) 3, b) i) The important postulates of Werner’s theory are:, 1. Every metal has two types of valencies – primary (10) valency and secondary (20) valency., Primary valency is ionisable, while secondary valency is non-ionisable., 2. Primary valency is denoted by doted lines, while secondary valency is denoted by thick lines., 3. Primary valency gives the oxidation state of the metal, while secondary valency gives the coordination number of the metal., 4. Primary valency is always satisfied by –ve ions, while secondary valency may be satisfied by –ve, ions or neutral molecules., 5. Every metal has a fixed number of secondary valencies. In order to satisfy this requirement,, some –ve ions may perform dual character – i.e., they act as primary and secondary valencies, simultaneously., 6. The primary valencies are non-directional, while the secondary valencies are directional. i.e., they are directed to some fixed positions in space., 7. Since secondary valencies are directional, co-ordination compounds have a definite geometry, and they show isomerism. [Here only 2 postulates are required], ii) K3[Fe(CN)6] – Potassiumhexacyanidoferrate (III) and [Co(NH3)6]Cl3 – Hexaamminecobalt(III)chloride., 15) a) Write the IUPAC name of the complex K3[Cr(C2O4)3]., (1), b) Draw the figure to show the splitting of ‘d’ orbitals in octahedral crystal field., (1), 3+, 3c) [Fe(H2O)6] is strongly paramagnetic, whereas [Fe(CN)6] is weakly paramagnetic. Write the reason. (2), [SAY 2015], Ans: a) Potassiumtrioxalatochromate (III), b) Refer the answer of the question number 2 (b)., c) In [Fe(H2O)6]3+, H2O is a weak field ligand and so electron pairing does not occur. Hence there are 5, unpaired electrons and it is strongly paramagnetic. But in [Fe(CN)6]3-, electron pairing occurs in, presence of the strong ligand CN –. So there is only one unpaired electron and in weakly paramagnetic., 16) a) Write down the ionization isomer of [Co(NH3)5Cl]SO4., (1), Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 4

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , b) Write the IUPAC name of the above compound., (1), 2c) [Ni(CO)4] is diamagnetic while [NiCl4] is paramagnetic though both are tetrahedral. Why? (2), [March 2016], Ans: a) [Co(NH3)5SO4]Cl, b) Pentaamminechloridocobalt(III)sulphate, c) In [Ni(CO)4], there is no unpaired electron. So it is diamagnetic. But in [NiCl4]2-, there are 2 unpaired, electrons. So it is paramagnetic, 17) Consider the co-ordination compound [Co(NH3)5Cl]Cl2., a) Write the IUPAC name of the above compound., (1), b) i) What is the primary valency and secondary valency of the central metal ion in the above coordination compound?, (1), ii) Write the name of isomerism exhibited by the complex [Pt(NH3)2Cl2]. Represent the possible, isomers., (2), [SAY 2016], Ans: a) Pentaamminechloridocobalt(III)chloride, b) i) Primary valency = 3 and secondary valency = 6, ii) Geometrical isomerism., , 18) [Co(NH3)5SO4]Cl and [Co(NH3)5Cl]SO4 are co-ordination compounds., a) Identify the isomerism shown by the above compounds. (1), b) Write the IUPAC names of the above compounds. (2), c) Identify the ligands in each of the above compounds., (1), [March 2017], Ans: (a) Ionisation isomerism., (b) [Co(NH3)5SO4]Cl – Pentaamminesulphatocobalt(III)chloride, [Co(NH3)5Cl]SO4 – Pentaamminechloridocobalt(III)sulphate, (c) In the compound [Co(NH3)5SO4]Cl, the ligands are NH3 and SO42– and in [Co(NH3)5Cl]SO4 , the, ligands are NH3 and Cl –., 19) a) In which of the following, the central atom/ion is in zero oxidation state., i) [Ni(CN)4]2ii) [NiCl4]2iii) [Ni(CO)4] iv) [Ni(NH3)6]2+, (1), 2b) [Ni(CN)4] has square planar structure and it is diamagnetic., i) On the basis of valence bond theory explain why [Ni(CN)4]2- exhibit these properties. (2), ii) Identify the ligand in the above mentioned complex. (1), [SAY 2017], Ans: a) [Ni(CO)4], b) i) In [Ni(CN)4]2–, the central atom Ni is in dsp2 hybridisation. So it has a square planar geometry. Due, to the absence of unpaired electrons, the complex is diamagnetic., c) The ligand in this complex is CN –, 20) Explain how the complexes of nickel, [Ni(CN)4]2- and [Ni(CO)4] have different structures, but do not differ in, their magnetic behaviour., (Atomic no. of Ni = 28), (2), Ans: In [Ni(CN)4]2-, Ni is dsp2 hybridized and so it has square planar geometry. But in [Ni(CO)4], Ni is sp3, hybridized and so it has tetrahedral shape. Due to the absence of unpaired electrons, both are diamagnetic., Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 5

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , 21) (a) Draw the structures of geometrical isomers of [Fe(NH3)2(CN)4] – (2), (b) Write the formula of pentaamminecarbonatocobalt (lll) chloride. (1), (c) Write any two limitations of valance bond theory. (1), [March 2018], Ans: (a), , (b) [Co(NH3)5(CO3)]Cl, (c) Limitations of valence bond theory are:, (i) It involves a large number of assumptions., (ii) It does not explain the colour exhibited by co-ordination compounds., 22) Draw a diagram depicting crystal field splitting in an octahedral environment of d-orbitals. Label the, diagram properly. Calculate the crystal field stabilization energy for a d 3 configuration. (4) [SAY 2018], Ans:, , For d3 systems, the configuration is t2g3., The crystal field stabilization energy is -3 x 2∆o/5= - 6∆o/5, 23) Identify the co-ordination compound which can exhibit linkage isomerism, among the following:, (a) [Pt(NH3)2Cl2] (b) [Co(NH3)5(SO4)]Br (c) [Co(NH3)5(NO2)]Cl2 (d) [Cr(NH3)6][CoF6], (1), Ans: (c) [Co(NH3)5(NO2)]Cl2, 24) (i) With the help of a diagram give the splitting of d-orbitals of Mn2+ ion in an octahedral crystal field. (2), (ii) On the basis of crystal field theory, explain why [Mn(H2O)6]2+ contains five unpaired electrons while, [Mn(CN)6]4- contains only one unpaired electron., (2), [March 2019], Ans: (i) Refer the answer of Question no. 22, (ii) In [Mn(H2O)6]2+, H2O is a weak field ligand. Here the magnitude of ∆o is less than the energy required for, electron pairing. So the configuration is t2g3eg2. Hence it contains 5 unpaired electrons. But in [Mn(CN)6]4-,, CN- is a strong field ligand and hence ∆o > the pairing energy. So the configuration is t2g5. Hence it, contains only one unpaired electron., 25) The crystal field splitting energy for Octahedral (∆o) and Tetrahedral (∆t) complexes are related as, Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 6

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , (a) ∆t = 3/5∆o, (b) ∆t = 5/3∆o, (c) ∆t = 4/9∆o, (d) ∆t = 9/4 ∆o, (1), Ans: (c) ∆t = 4/9∆o, 26) Write the IUPAC name of the following :, (i) [Ni(CO)4] (ii) K4[Fe(CN)6], (2), Ans: (i) Tetracarbonylnickel (0), (ii) Potassiumhexacyanidoferrate (II), 27) Draw the geometrical isomers of [PtCl2(en)2]2+. Which among the isomer is optically active? Give reason., (4), [SAY 2019], Ans:, , Pt, , Cis isomer, , Pt, , Trans isomer, , Cis isomer is optically active, since it has no plane of symmetry. [Or, trans isomer has a plane of symmetry, and hence it is optically inactive.], 28) Assign the primary valence and secondary valence of the central metal in [Ni(CO)4] (2), Ans: Primary valence = 0, Secondary valence = 4, 29) Cis isomer of [Pt(NH3)2Cl2] is used to inhibit the growth of tumors., (a) Give the IUPAC name of [pt (NH3)2CI2]., (1), (b) Give the structure of cis and trans isomers of [Pt (NH3)2Cl2]., (2), [March 2020], Ans: (a) Diamminedichloridoplatinum(II), (b), , 30) Write the IUPAC name of the following Co-ordination compounds :, (i) [Co(NH3)6]Cl3, (ii) K3[Fe(C2O4)3], (2), Ans: (i) Hexaamminecobalt(III)chloride, (ii) Potassiumtrioxalatoferrate (III), 31) The octahedral complex [Co(NH3)6]3+ is diamagnetic while [CoF6]3- is paramagnetic. Explain using V.B., Theory., (3), [SAY 2020], Ans: NH3 is a strong field ligand and hence electron pairing occurs in [Co(NH3)6]3+. So there is no unpaired, electron and hence it is diamagnetic. While in [CoF6]3-, F - is a weak field ligand and hence electron pairing, does not occur. So there is unpaired electrons and hence it is paramagnetic., 32) (i) Write the IUPAC name of K2[Zn(OH)4], (1), (ii) Metal present in chlorophyll is ………., (1), Ans: (i) Potassiumtetrahydroxidozincate(II), Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 7

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , (ii) Magnesium (Mg), 33) (i) Account for the following:, A. Zr and Hf have identical radii., (1), B. Transition metals are very good catalysts., (1), 2+, (ii) Calculate the spin only magnetic moment of M (aq) ion (Z = 27). (1), Ans: (i) A) Due to lanthanoid contraction/lanthanide contraction., B) This is due to their large surface area and the ability to show variable oxidation state., (ii) The electronic configuration of M2+ (Z = 27) is [Ar]3d7. So the no. of unpaired electrons = 3., The spin only magnetic moment, µs = √n(n+2) = √3(3+2) = √15 = 3.87 BM, 34) (i) List the various structural isomerisms possible for co-ordination compounds. (2), (ii) [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain. (2), [March 2021], Ans: (i) The different types of structural isomerism shown by co-ordination compounds are: 1. Ionisation, isomerism 2. Linkage isomerism 3. Solvate or hydrate isomerism 4. Co-ordination isomerism., (ii) H2O is a weak field ligand and hence electron pairing does not occur while CN- is a strong field ligand, and hence electron pairing occurs. So the number of unpaired electrons in [Fe(H2O)6]3+greater than that in, [Fe(CN)6]3-. Hence [Fe(H2O)6]3+ is strongly paramagnetic. [Also refer the Ans. of the Question no. 15 (c)]., ==============================================================================, , Co-ordination Compounds – Prepared by Anil Kumar K L, GHSS Ashtamudi., , Page 8