CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. What is the name of the bidentate ligand 'dmg' ? (HP Board 2011)
Answer:
dimethylglyoxime.

Teacher's Note:
1. dmg is used to test Ni²⁺ ions, which give a red precipitate of [Ni(dmg)₂].

Question 2. Give an example of a hexadentate ligand. (Raj. Board 2011)
Answer:
EDTA (ethylenediaminetetraacetate ion).

Teacher's Note:
1. EDTA⁴⁻ binds through 2 nitrogen and 4 oxygen atoms.

Question 3. Give one example of chelate complex.

Answer:
[Cu(en)₂]SO₄ :

Teacher's Note:
1. A chelate needs a polydentate ligand that forms a ring with the metal; chelate complexes are more stable than similar complexes with monodentate ligands (chelate effect).

Question 4. Give one example of each :
(a) Neutral ligand of carbon
(b) Positive ligand of nitrogen. (Raj. Board 2011)
Answer:
(a) CO (carbonyl)
(b) NO⁺ (nitrosonium) or NO₂⁺ (nitronium).

Teacher's Note:
1. Positive ligands are rare; NO⁺ (in nitroprusside) is the standard example to remember.

Question 5. Write the formula for dichlorotetraammine platinum (IV) ion.
Answer:
[Pt(NH₃)₄Cl₂]²⁺

Teacher's Note:
1. Charge = +4 (Pt) + 0 (4 NH₃) + 2(−1) = +2.

Question 6. Write the chemical formula of nitropentaamminecobalt (III) chloride.
Answer:
[Co(NH₃)₅(NO₂)]Cl₂

Teacher's Note:
1. Complex ion charge = +3 − 1 = +2, so two Cl⁻ counter ions are needed.

Question 7. Write the formula of
(i) hexaaquairon (II) sulphate
(ii) potassium hexacyanoferrate (III)
(iii) hexaammine platinum (IV) chloride
(iv) potassium trioxalatoaluminate (III).
Answer:
(i) [Fe(H₂O)₆]SO₄
(ii) K₃[Fe(CN)₆]
(iii) [Pt(NH₃)₆]Cl₄
(iv) K₃[Al(C₂O₄)₃]

Teacher's Note:
1. Work out the charge on the complex ion first; the number of counter ions follows from it (e.g. Al³⁺ + 3 C₂O₄²⁻ = −3, so 3 K⁺).

Question 8. Write IUPAC name for K₃[Co(NO₃)₆].
Answer:
Potassium hexanitratocobaltate (III)

Teacher's Note:
1. The suffix '-ate' is added to the metal name because the complex ion is an anion.

Question 9. Write the IUPAC name of
(i) Zn₂[Fe(CN)₆] (Karnataka Board 2012)
(ii) Pt[Cl₂(NH₃)₂] (Assam Board 2012)
Answer:
(i) zinc hexacyanoferrate (II)
(ii) diamminedichloridoplatinum (II).

Teacher's Note:
1. Ligands are named in alphabetical order ignoring prefixes: ammine (a) before chlorido (c).

Question 10. Write IUPAC name of (i) Na₃[Co(NO₂)₆] (ii) K₃[Fe(CN)₅NO]
Answer:
(i) sodium hexanitritocobaltate (III)
(ii) potassium pentacyanonitrosylferrate (II)

Teacher's Note:
1. The NO ligand is named 'nitrosyl'; NO₂⁻ bonded through N is 'nitrito-N' (older name 'nitro').

Question 11. Write IUPAC name of (i) Na₃[Cr(OH)₂F₄] (ii) [Cr(H₂O)₅Cl]Cl₂
Answer:
(i) sodium tetrafluoridodihydroxidochromate (III)
(ii) pentaaquachloridochromium (III) chloride.

Teacher's Note:
1. Check the oxidation state: in (i) x − 2 − 4 = −3 gives x = +3.

Short Answer Questions (2 Marks)

Question 12. How many ions are produced from the complex Co(NH₃)₅ Cl₃ in the solution ? (CBSE 2020)
Answer:
Coordination number of cobalt is 6. Hence, formula of the complex will be [Co(NH₃)₅Cl] Cl₂.
It ionizes to produce one [Co(NH₃)₅Cl]²⁺ ion and 2 Cl⁻ ions. Hence, total ions produced = 3.

Teacher's Note:
1. Write the complex formula first (fill C.N. 6 with NH₃, then Cl); only then count the ions.

Question 13. Write the correct formulae for the following coordination compounds ;
(a) CrCl₃ · 6 H₂O (violet, with 3 chloride ions/unit formula)
(b) CrCl₃ · 6 H₂O (light green colour with 2 chloride ions/unit formula)
(c) CrCl₃ · 6 H₂O (dark green colour with 1 chloride ion/unit formula)
Answer:
(a) [Cr (H₂O)₆] Cl₃
(b) [Cr (H₂O)₅ Cl] Cl₂ · H₂O
(c) [Cr (H₂O)₄Cl₂] Cl·2 H₂O
Note that coordination number of Cr is 6 in each case.

Teacher's Note:
1. These three are hydrate isomers; the water not needed to complete C.N. 6 is written as water of crystallisation.

Question 14. Write the formulae of the following :
(i) bis (acetylacetonato) oxovanadium (IV)
(ii) dichloridoplatinum (IV) tetrachloridoplatinate (II)
Answer:
(i) [V (acac)₂ O]
(ii) [PtCl₂] [PtCl₄]

Teacher's Note:
1. Check charges: Pt(IV) with two Cl⁻ gives [PtCl₂]²⁺, and Pt(II) with four Cl⁻ gives [PtCl₄]²⁻, so they combine 1 : 1.

Question 15. Give the IUPAC names of the complexes
(i) K [BF₄] (ii) [Fe (C₅ H₅)₂] (iii) [Mn₃ (CO)₁₂]
Answer:
(i) potassium tetrafluoridoborate (III)
(ii) bis (cyclopentadienyl) iron
(iii) dodecacarbonyl trimanganese (0).

Teacher's Note:
1. For an anionic complex the element name ends in –ate (borate); neutral complexes keep the plain element name.

Question 16. Write the IUPAC name of the following complex:

Answer:
bis(ethane-1, 2-diamine) cobalt(II)-μ-imido-μ-carbonylbis(ethane-1, 2-diamine)cobalt (II) nitrate.

Teacher's Note:
1. A bridging ligand is shown by the prefix μ-; list the bridges after one metal unit, and give the oxidation state of each metal.

Question 17. How many isomers are there for the complex [Co(NH₃)₄Cl₂] Cl ?

Answer:
No structural isomers because coordination number of cobalt is 6 but two geometrical isomers, viz. cis and trans (the two Cl at adjacent or at opposite positions).
No optical isomerism because no didentate ligand is present.

Teacher's Note:
1. [MA₄B₂] always gives just two isomers, cis and trans, and both have a plane of symmetry.

Question 18. Draw cis and trans isomers of [CoCl₂(NH₃)₄]⁺ ion.

Answer:
In the cis isomer the two Cl are at adjacent positions (90° apart); in the trans isomer they are opposite to each other (180° apart).

Teacher's Note:
1. The cis isomer is violet and the trans isomer is green, which is why they can be told apart.

Question 19. Write all isomers of [Co(NH₃)₅(SCN)] Cl.
Answer:
[Co(NH₃)₅(SCN)] Cl, [Co(NH₃)₅(NCS)] Cl, [Co(NH₃)₅Cl] SCN, [Co(NH₃)₅Cl] NCS
MA₅B type complexes do not show geometrical or optical isomerism.

Teacher's Note:
1. SCN⁻ is ambidentate (gives linkage isomers) and can swap places with Cl⁻ (gives ionisation isomers).

Question 20. Write the name (IUPAC norms) and draw the possible optical isomers of [CrCl₂ (en) (NH₃)₂]⁺.

Answer:
diamminedichlorido(ethane-1, 2-diamine)chromium (III) ion.
Optical isomers: only the isomer in which the two Cl are cis to each other and the two NH₃ are cis to each other is optically active. It exists as a pair of non-superimposable mirror images (d and l forms), shown in the figure.
The isomers with the two Cl trans or the two NH₃ trans have a plane of symmetry and are optically inactive.

Teacher's Note:
1. To check chirality, look for a plane of symmetry; if none exists, draw the mirror image and confirm it cannot be superimposed.

Question 21. Write IUPAC name of the complex [Pt(en)₂Cl₂]. Draw geometrical isomers for this complex. (CBSE 2019)

Answer:
dichlorobis(ethane-1, 2-diamine) platinum (IV)
Geometrical isomers: cis (two Cl at adjacent positions) and trans (two Cl opposite to each other), as shown in the figure.

Teacher's Note:
1. Find the oxidation state before naming: en is neutral and two Cl⁻ give −2, so Pt in [Pt(en)₂Cl₂]²⁺ is +4.

Question 22. Draw one of the geometrical isomers of [Pt(en)₂Cl₂]²⁺ which is optically inactive. Also write the name of this entity according to the IUPAC nomenclature. (CBSE Sample Paper 2019)

Answer:
trans isomer is optically inactive (it has a plane of symmetry). Its structure is shown in the figure (similar to trans-[Co(en)₂Cl₂]⁺).
The name is : dichlorobis(ethane-1, 2-diamine) platinum (IV).

Teacher's Note:
1. The cis isomer of [M(AA)₂B₂] is optically active, so only the trans form can be drawn here.

Question 23. [Co(CN)₆]³⁻ and [CoF₆]³⁻ both are octahedral complexes. Then what is the difference between the two ?
Answer:
[Co(CN)₆]³⁻ is an inner orbital or low spin complex involving the hybridisation d²sp³ while [CoF₆]³⁻ is an outer orbital or high spin complex involving sp³d² hybridisation.

Teacher's Note:
1. Add the magnetic consequence in exams: [Co(CN)₆]³⁻ is diamagnetic, [CoF₆]³⁻ has four unpaired electrons.

Question 24. Account for the fact that [Ni(CO)₄] has tetrahedral geometry whereas [Ni(CN)₄]²⁻ has square planar geometry.
Answer:
In [Ni(CO)₄], Ni is in zero oxidation state (3d⁸4s²). CO is a strong ligand, so the 4s electrons are shifted to 3d (3d¹⁰) and the empty 4s and three 4p orbitals give sp³ hybridisation: tetrahedral.
In [Ni(CN)₄]²⁻, Ni²⁺ is 3d⁸. The strong ligand CN⁻ pairs the two unpaired electrons, leaving one 3d orbital empty, so it is dsp² hybridised: square planar.

Teacher's Note:
1. The key difference is the oxidation state: Ni(0) has no free 3d orbital after pairing, Ni(II) has one.

Teacher's Note:
1. Both cases use one empty 3d, the 4s and two 4p orbitals; what differs is where the two electrons end up.

Question 26. Out of tetracarbonyl nickel (0) and tetraamminecopper (II) sulphate, which is paramagnetic and why ?
Answer:
Tetraamminecopper (II) sulphate is paramagnetic whereas tetracarbonyl nickel (0) is diamagnetic.
In [Cu(NH₃)₄]²⁺, Cu²⁺ is 3d⁹ and always has one unpaired electron. In [Ni(CO)₄], the strong ligand CO pairs all the electrons of Ni (3d¹⁰), so there is no unpaired electron.

Teacher's Note:
1. Any ion with an odd number of electrons (like Cu²⁺, d⁹) is paramagnetic.

Short Answer Questions (3 Marks)

Question 27. The molar conductivity of the complex CoCl₃·4NH₃·2H₂O is found to be same as that of 3 : 1 electrolyte. What is the structural formula, name and number of geometrical isomers of the complex?
Answer:
As coordination number of Co is 6 and complex should be A₃B or AB₃ type, formula is [Co(NH₃)₄(H₂O)₂]Cl₃.
Name: Tetraamminediaquacobalt(III) chloride
Geometrical isomers: 2 (cis and trans)

Teacher's Note:
1. A 3 : 1 electrolyte gives 4 ions, so all three Cl⁻ must be outside the coordination sphere.

Question 28. [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless. Why?
Answer:
In [Ti(H₂O)₆]³⁺, Ti³⁺ (3d¹) has one unpaired electron which can undergo d-d transition, while Sc³⁺ (3d⁰) does not have any d electron, so no d-d transition is possible.

Teacher's Note:
1. d⁰ and d¹⁰ ions (Sc³⁺, Zn²⁺) are colourless because no d-d transition is possible.

Question 29. Describe with an example of each, the role of coordination compounds in:
(i) Biological system   (ii) Analytical chemistry   (iii) Medicinal chemistry
Answer:
(i) Vitamin B₁₂ (a cobalt complex) is the antipernicious anaemia factor; haemoglobin (Fe) carries oxygen.
(ii) Detection and estimation of metal ions, e.g., hardness of water is estimated by titration with Na₂EDTA.
(iii) EDTA is used in lead poisoning; cisplatin is used in cancer treatment.

Teacher's Note:
1. One clear named example per part is enough to earn the mark in each.

Question 30. Write the type of isomerism exhibited by the following complexes:
(i) [Co(NH₃)₅Cl]SO₄   (ii) [Co(en)₃]³⁺   (iii) [Co(NH₃)₆][Cr(CN)₆]
Answer:
(i) Ionisation isomerism
(ii) Optical isomerism
(iii) Coordination isomerism

Teacher's Note:
1. Ionisation isomer of (i) is [Co(NH₃)₅SO₄]Cl; coordination isomer of (iii) is [Cr(NH₃)₆][Co(CN)₆].

Question 31. Explain the following:
(i) CO is stronger ligand than NH₃.
(ii) Low spin octahedral complexes of nickel are not known.
(iii) Aqueous solution of [Ti(H₂O)₆]³⁺ is coloured.
Answer:
(i) CO has higher value of crystal field splitting energy than NH₃, because besides the σ bond it forms a π bond by accepting electrons from the metal (back bonding).
(ii) Ni²⁺ has d⁸ configuration which is not affected by field strength of ligand; in an octahedral field it is always t_2g⁶e_g² with two unpaired electrons.
(iii) In this case, Ti³⁺ has t_2g¹e_g⁰ configuration. It can perform d-d transition by absorbing visible light, so the solution is coloured (violet).

Teacher's Note:
1. For d⁸ octahedral there is only one way to fill the orbitals, so high and low spin cannot be distinguished.

Question 32. Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(Py)] and how many of these will exhibit optical isomerism? Here, Py = Pyridine.

Answer:
3 isomers are possible (shown in the figure): in (I) NH₃ is trans to Br, in (II) NH₃ is trans to Cl and in (III) NH₃ is trans to Py.
None of these will exhibit optical isomerism, as monodentate ligands in a square planar complex do not show optical isomerism (the plane of the molecule is a plane of symmetry).

Teacher's Note:
1. For [MABCD] square planar, fix one ligand and change the ligand trans to it – that gives exactly three isomers.

Question 33. How will you distinguish between the following isomer pairs ?
(a) (i) [Cr(H₂O)₆] Cl₃ and (ii) [Cr (H₂O)₅ Cl] Cl₂H₂O
(b) (i) cis [Pt (NH₃)₂ Cl₂] and (ii) trans [Pt (NH₃)₂ Cl₂]
(c) The two enantiomers of [CoCl₂(en)₂]⁺
Answer:
(a) The water molecules in isomer (i) are not lost easily, being coordinated to the central atom whereas the water molecule in isomer (ii) is lost easily on heating.
(b) cis isomer (i) has dipole moment, the trans isomer (ii) does not.
(c) one isomer is dextro-rotatory whereas the other is laevo-rotatory.

Teacher's Note:
1. For (a) a second test also works: with AgNO₃, (i) gives 3 mol and (ii) gives 2 mol of AgCl per mole.

Question 34. Draw the structures of (a) cis-dichlorido-tetracyanochromate (III)
(b) Mer-triamminetrichloridocobalt (III)
(c) Fac-triaquatrinitro-N-cobalt (III)

Answer:
(a) [CrCl₂(CN)₄]³⁻: the two Cl at adjacent (cis) positions.
(b) [Co(NH₃)₃Cl₃]: the three NH₃ (and the three Cl) lie in one plane passing through the metal (meridional).
(c) [Co(NO₂)₃(H₂O)₃]: the three NO₂ (and the three H₂O) occupy one triangular face of the octahedron (facial).
The structures are shown in the figure.

Teacher's Note:
1. In a mer isomer two of the three identical ligands are trans to each other; in a fac isomer all three are cis to one another.

Question 35. On the basis of valence bond theory, explain geometry, nature of hybridisation, magnetic property and optical isomerism of [Co(ox)₃]³⁻ and [CoF₆]³⁻.
Answer:
[Co(ox)₃]³⁻ : Geometry = octahedral, Hybridisation = d²sp³ (the six 3d electrons of Co³⁺ pair up), Magnetic properties = diamagnetic. Two optical isomers (d and l), as the three chelating oxalate groups leave no plane of symmetry.
[CoF₆]³⁻ : Geometry = octahedral, Hybridisation = sp³d², Magnetic properties = paramagnetic (4 unpaired electrons, as F⁻ is a weak ligand). No optical isomerism, as the ion is symmetrical.

Teacher's Note:
1. [CoF₆]³⁻ is the standard example of a high spin, outer orbital Co(III) complex; most other Co(III) complexes are low spin.

Question 36. Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ_o < P. (CBSE Foreign 2017)
Answer:
In a free transition metal ion, all the five d-orbitals are degenerate but when it is involved in a complex formation, the degeneracy is split into two sets of d-orbitals. This is called crystal field splitting.
The difference of energy between the two sets of d-orbitals is called crystal field splitting energy (CFSE).
The lower energy set of three d-orbitals is called t_2g and the higher energy set of two d-orbitals is called e_g. The three electrons first enter into t_2g, one in each orbital. If Δ_o < P, the 4th electron will enter e_g. Hence, the configuration will be t_2g³ e_g¹.

Teacher's Note:
1. Δ_o < P (weak field) → electrons spread out (high spin); Δ_o > P (strong field) → electrons pair in t_2g (low spin).

Question 37. Draw a sketch to show the splitting of d-orbitals in an octahedral crystal field. State for a d⁶ configuration, how the actual configuration of the split d-orbitals in an octahedral crystal field is decided by the relative values of Δ_o and P ?

Answer:
Splitting of d-orbitals in an octahedral field: the five degenerate d-orbitals split into a lower set of three t_2g orbitals (d_xy, d_yz, d_xz), at −0.4Δ_o, and a higher set of two e_g orbitals (d_x²−y², d_z²), at +0.6Δ_o, as shown in the figure.
For d⁶ configuration, three electrons will first enter into t_2g. Now, if P < Δ_o, electrons will pair up in t_2g orbitals, giving the configuration t_2g⁶ and if Δ_o < P, two electrons will first enter into e_g orbitals and one will pair up in t_2g giving the configuration t_2g⁴ e_g².

Teacher's Note:
1. In the sketch, the ligands point at the e_g orbitals (along the axes), which is why e_g is raised in an octahedral field.

Long Answer Questions (5 Marks)

Question 38. Name various types of isomerism shown by coordination compounds. Explain each type with an example.

Answer:
Coordination compounds show two main types of isomerism : A. Structural isomerism and B. Stereoisomerism.
A. Structural isomerism : Isomers have different bonds (different ligands in the coordination sphere or different donor atoms).
(i) Ionisation isomerism : A counter ion is itself a potential ligand and exchanges place with a ligand, so the isomers give different ions in solution, e.g., [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄.
(ii) Hydrate (solvate) isomerism : Isomers differ in whether water is a ligand or water of crystallisation, e.g., [Cr(H₂O)₆]Cl₃ (violet) and [Cr(H₂O)₅Cl]Cl₂.H₂O (grey-green).
(iii) Linkage isomerism : Arises with ambidentate ligands bonded through different donor atoms, e.g., [Co(NH₃)₅(NO₂)]Cl₂ (yellow) and [Co(NH₃)₅(ONO)]Cl₂ (red).
(iv) Coordination isomerism : Arises from interchange of ligands between cationic and anionic entities of different metal ions, e.g., [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆].
B. Stereoisomerism : Isomers have the same chemical formula and chemical bonds but different spatial arrangement.
(v) Geometrical isomerism : Arises in heteroleptic complexes due to different possible arrangements of ligands. In square planar [MX₂L₂], e.g., [Pt(NH₃)₂Cl₂], the two Cl may be adjacent (cis) or opposite (trans). Octahedral [MX₂L₄], e.g., [Co(NH₃)₄Cl₂]⁺, also shows cis-trans isomerism, and [Ma₃b₃], e.g., [Co(NH₃)₃Cl₃], shows fac and mer isomerism (shown below). Tetrahedral complexes do not show geometrical isomerism.
(vi) Optical isomerism : Optical isomers are mirror images that cannot be superimposed on one another (enantiomers); they rotate the plane of polarised light equally but in opposite directions (dextro, d and laevo, l). It is common in octahedral complexes with bidentate ligands, e.g., [Co(en)₃]³⁺ and cis-[PtCl₂(en)₂]²⁺.

Teacher's Note:
1. For each type, an exam answer needs three things: the name, a one-line definition and a correct pair of formulae.

Teacher's Note:
1. Werner's theory explains the formulae and isomers, but not why only certain metals form complexes or their magnetic and colour properties - that needs VBT/CFT.