Previous year questions: Coordination Compounds
Coordination Compounds30 min readLesson 6 of 6
Q1 (CBSE Board 2023, 3 marks)
Answer any three of the following. (a) Explain the type of hybridisation in [Fe(CN)6]3- on the basis of valence bond theory. (b) Sketch the geometrical forms of [PtCl2(en)2]2+. (c) Both [NiCl4]2- and [Ni(CO)4] have tetrahedral shapes, yet one is paramagnetic and the other diamagnetic. Account for this difference. (d) Name the type of isomerism shown by an ambidentate ligand and give one example of such a ligand.
Answer: (a) Fe3+ is d5. Cyanide is a strong field ligand, so the electrons pair in the t2g set, leaving one inner d orbital free. The hybridisation is d2sp3, the geometry is octahedral, and the complex has one unpaired electron. (b) The two geometrical isomers are cis and trans: in the cis form the two chloride ligands are adjacent and in the trans form they are opposite, with the two bidentate en ligands completing the octahedron. (c) In [NiCl4]2- nickel is +2 with a d8 configuration; chloride is a weak field ligand, so the complex is tetrahedral with two unpaired electrons and is paramagnetic. In [Ni(CO)4] nickel is in the zero oxidation state with a d10 configuration, so all electrons are paired and the complex is diamagnetic. (d) The isomerism is linkage isomerism, and the nitrite ion is an ambidentate ligand that can bind through nitrogen or through oxygen.
Q2 (CBSE Board 2023, 4 marks)
A case study explains Werner's primary and secondary valences, valence bond theory and crystal field theory. (i) NiCl2·6H2O gives 2 moles of AgCl per mole of the compound. Write the structural formula and the secondary valency of nickel. (ii) Write the IUPAC name of the ionisation isomer of [Co(NH3)5(SO4)]Cl. (iii) Using valence bond theory, predict the geometry and magnetic nature of [Ni(CO)4] and [Fe(CN)6]3-.
Answer: (i) Two chloride ions are ionisable and four remain coordinated along with six water molecules, but since only two chlorides precipitate, the structure is [Ni(H2O)6]Cl2 and the secondary valency of nickel is six. (ii) The ionisation isomer has the chloride inside the coordination sphere and the sulphate outside: [Co(NH3)5Cl]SO4, which is pentaamminechloridocobalt(III) sulphate. (iii) [Ni(CO)4] has nickel in the zero oxidation state with d10, so it is sp3 hybridised, tetrahedral and diamagnetic. [Fe(CN)6]3- has Fe3+ with d5 and cyanide as a strong field ligand, so it is d2sp3 hybridised, octahedral and has one unpaired electron, which makes it paramagnetic with a moment near 1.73 BM.
Q3 (CBSE Board 2024, 3 marks)
(a) Write the formula of potassium tetrahydroxidozincate(II). (b) Arrange [Cr(NH3)5Cl]Cl2, [Cr(NH3)3Cl3] and [Cr(NH3)6]Cl3 in increasing order of the conductivity of their solutions. (c) Identify the type of isomerism shown by [Co(NH3)5NO2]2+ and by [Co(en)3]Cl3.
Answer: (a) The complex ion is [Zn(OH)4]2- with potassium as counter ion, so the formula is K2[Zn(OH)4]. (b) Conductivity depends on the number of ions produced in solution. [Cr(NH3)3Cl3] is a neutral complex and gives no ions, [Cr(NH3)5Cl]Cl2 gives three ions and [Cr(NH3)6]Cl3 gives four ions, so the increasing order is [Cr(NH3)3Cl3], then [Cr(NH3)5Cl]Cl2, then [Cr(NH3)6]Cl3. (c) [Co(NH3)5NO2]2+ shows linkage isomerism because the nitrite ligand can bind through nitrogen or oxygen, while [Co(en)3]Cl3 shows optical isomerism because the three bidentate en ligands give two non-superimposable mirror image forms.
Q4 (CBSE Board 2024, 5 marks)
Answer any five of the following. (a) How is the crystal field splitting energy of an octahedral complex related to that of a tetrahedral complex? (b) Write the IUPAC name of PtCl2(en)22. (c) Write the geometry and magnetic behaviour of [Ni(CO)4] on the basis of valence bond theory. (d) What type of isomerism is shown by [Co(NH3)6][Cr(CN)6]? (e) For a d4 ion in a crystal field with Δo less than P, write the electronic configuration and state whether the complex is high spin or low spin. (f) Out of [Co(NH3)6]3+ and [Co(NH3)4Cl2]+, which is heteroleptic and why? (g) Draw the structures of the optical isomers of [PtCl2(en)2]2+.
Answer: (a) The tetrahedral splitting is four ninths of the octahedral splitting for the same ligand set, so Δt = (4/9)Δo. (b) The complex ion is dichloridobis(ethane-1,2-diamine)platinum(IV), so the name is dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate. (c) Nickel is in the zero oxidation state with d10, so the hybridisation is sp3, the geometry is tetrahedral and the complex is diamagnetic. (d) The compound shows coordination isomerism because cobalt and chromium are both in complex ions and the ligands can be distributed differently between them. (e) With Δo less than P the configuration is t2g3 eg1, so the complex is high spin with four unpaired electrons. (f) [Co(NH3)4Cl2]+ is heteroleptic because it contains two different kinds of ligand, ammonia and chloride, while [Co(NH3)6]3+ has only ammonia and is homoleptic. (g) Only the cis isomer is optically active: it exists as two non-superimposable mirror images, while the trans isomer has a plane of symmetry and is optically inactive.
Q5 (CBSE Board 2025, 2 marks)
Write the IUPAC name of [Pt(en)2Cl2]2+ and draw the structure of the geometrical isomer of this complex that is optically inactive.
Answer: The complex contains two bidentate ethane-1,2-diamine ligands and two chloride ligands around platinum(IV), so the name is dichloridobis(ethane-1,2-diamine)platinum(IV) ion. The optically inactive geometrical isomer is the trans form, in which the two chloride ligands are opposite each other and the two en ligands lie in perpendicular planes. The trans form has a plane of symmetry, so its mirror image is the same molecule. The cis form has the two chlorides adjacent and exists as two optically active enantiomers.
Q6 (CBSE Board 2026, 3 marks)
(a) CrCl3·6H2O is treated with excess AgNO3 solution and three moles of AgCl are precipitated per mole of the compound. Write the structural formula of the compound. (b) Write the oxidation state and hybridisation of the central metal in [Cr(NH3)6]3+. (c) Why is [Sc(H2O)6]3+ colourless?
Answer: (a) Three ionisable chloride ions mean all three chlorides are outside the coordination sphere and six water molecules occupy the coordination sites, so the formula is [Cr(H2O)6]Cl3. (b) Chromium is in the +3 oxidation state, so it is d3. Ammonia is a strong field ligand, the three d electrons occupy all three t2g orbitals singly, and one inner d orbital is free for d2sp3 hybridisation, giving an octahedral inner orbital complex. (c) Scandium in [Sc(H2O)6]3+ is in the +3 state with a d0 configuration. With no d electrons there is no d-d transition in the visible region and no charge transfer in this case, so the complex is colourless.
Q7 (CBSE SQP 2022-23, 1 mark)
The CFSE of [CoCl6]3- is 18000 cm^-1. What is the CFSE of [CoCl4]-? Options: 18000 cm^-1, 8000 cm^-1, 2000 cm^-1, 16000 cm^-1.
Answer: The tetrahedral splitting is four ninths of the octahedral splitting for the same ligand, so the CFSE of the tetrahedral complex is four ninths of 18000, which is 8000 cm^-1. The correct option is 8000 cm^-1. The small tetrahedral splitting is also why low spin tetrahedral complexes are rarely formed: the gap is too small to force pairing, so the electrons stay unpaired.
Q8 (CBSE SQP 2022-23, 3 marks)
Using valence bond theory, explain the following for the paramagnetic complex [Mn(CN)6]3-: (a) the type of hybridisation, (b) the magnetic moment value, (c) whether the complex is inner or outer orbital.
Answer: Manganese in [Mn(CN)6]3- is in the +3 oxidation state, so the ion is d4. Cyanide is a strong field ligand that pairs the fourth electron, giving a t2g4 configuration with two unpaired electrons. One inner d orbital is freed, so the hybridisation is d2sp3 and the geometry is octahedral. The magnetic moment is the square root of 2 times 4, which is the square root of 8, or 2.83 BM. Because the inner d orbital is used, the complex is an inner orbital complex and also a low spin complex.
Q9 (CBSE SQP 2023-24, 3 marks)
(a) Write the formula of bis(ethane-1,2-diamine)dihydroxidochromium(III) chloride. (b) Does an ionisation isomer exist for Hg[Co(SCN)4]? Justify your answer. (c) Is the central metal atom in a coordination complex a Lewis acid or a Lewis base? Explain.
Answer: (a) Two en ligands are neutral, two hydroxide ligands give minus two, and chromium is +3, so the complex ion has a charge of +1 and the formula is [Cr(en)2(OH)2]Cl. (b) No. Ionisation isomerism needs an exchangeable ligand and counter ion; here the cation Hg2+ carries no ligands and the complex anion [Co(SCN)4]2- has its ligands tightly bound, so there is no pair of ions that can be interchanged. (c) The central metal ion acts as a Lewis acid because it accepts electron pairs from the ligands to form coordinate bonds, and the ligands act as Lewis bases because they donate those pairs.
Q10 (CBSE SQP 2024-25, 2 marks)
(a) Write the IUPAC name of K[Cr(H2O)2(C2O4)2]·H2O. (b) Name the metal present in the coordination compound of haemoglobin and of vitamin B12.
Answer: (a) The complex ion is an anion, so chromium becomes chromate. The ligands are two water molecules and two bidentate oxalate ions, giving diaqua and dioxalato, and the potassium is the counter cation, with one water of crystallisation. The name is potassium diaquadioxalatochromate(III) monohydrate. (b) Haemoglobin contains iron in a porphyrin ring and carries oxygen in the blood. Vitamin B12, also called cobalamin, contains cobalt at the centre of a corrin ring.