CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. What is spectrochemical series?
Answer:
The series in which ligands are arranged in the order of increasing field strength is called spectrochemical series. The order is
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO

Teacher's Note:
1. Remember the landmarks: halides are weak, H₂O is in the middle, NH₃/en are stronger and CN⁻, CO are the strongest.

Question 2. Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field ligand, and (ii) weak field ligand. (Atomic no. of Fe = 26) [CBSE 2019 (56/3/2)]
Answer:
(i) t_2g⁵e_g⁰
(ii) t_2g³e_g²

Teacher's Note:
1. Fe³⁺ is d⁵: strong field gives 1 unpaired electron, weak field gives 5.

Question 3. Why are low spin tetrahedral complexes not formed? [NCERT Exemplar]
Answer:
Because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.

Teacher's Note:
1. Δ_t is only about 4/9 of Δ₀, which is almost always smaller than P, so electrons stay unpaired.

Question 4. Do we call metal carbonyls as organometallics? Why?
Answer:
Yes, metal carbonyls are called organometallics because C atom of CO is linked to the metal atom. In fact, the metal–carbon bonds have both σ and π character.

Teacher's Note:
1. The test for an organometallic is a direct metal–carbon bond, which carbonyls have.

Question 5. Write IUPAC name of the complex: [CoCl₂(en)₂]⁺. [CBSE Comptt. OD 2017]
Answer:
Dichloridobis(ethane-1,2-diamine)cobalt(III) ion.

Teacher's Note:
1. Find the oxidation state first: en is neutral and each Cl is −1, so Co is +3; "bis" is used because the ligand name already contains "di".

Question 6. What is the IUPAC name of the complex [Ni(NH₃)₆]Cl₂? [CBSE Comptt. Delhi 2015]
Answer:
Hexaamminenickel (II) chloride.

Teacher's Note:
1. In a cationic complex the metal keeps its ordinary name; write "ammine" with double m for NH₃ (amine with one m is for organic amines).

Question 7. Give two examples of ligands which form coordination compounds useful in analytical chemistry. [CBSE Comptt. OD 2013]
Answer:
EDTA (Ethylene-diamine-tetraacetate) and DMG (Dimethylglyoxime).

Teacher's Note:
1. DMG gives a red precipitate with Ni²⁺, and EDTA is used to estimate Ca²⁺ and Mg²⁺ in hard water – quote these uses if asked for applications.

Question 8. Give IUPAC name of ionization isomer of [Ni(NH₃)₃NO₃] Cl. [CBSE Comptt. OD 2012]
Answer:
Ionization isomer is [Ni(NH₃)₃Cl] NO₃. IUPAC name is Triamminechloridonickel (II) nitrate.

Teacher's Note:
1. To write an ionisation isomer, just swap the ligand inside the bracket with the counter ion outside it.

Question 9. What is meant by chelate effect? [CBSE Comptt. OD 2015]
Answer:
Formation of stable complex with a polydentate ligand due to stronger bonding than the non chelate complexes is known as chelate effect.

Teacher's Note:
1. The extra stability comes mainly from the gain in entropy: one chelating ligand releases several monodentate ligands or water molecules.

Question 10. What is the denticity of coordination compound used for the treatment of lead poisoning? [CBSE SQP 2012]
Answer:
In Lead poisoning, EDTA (Ethylene diamine tetraacetate) is used as ligand which is hexadentate i.e., the denticity of EDTA is 6. EDTA binds with metal in octahedral manner by two N-atoms and four acetate oxygen atoms.

Teacher's Note:
1. EDTA⁴⁻ has six donor atoms (2 N + 4 O); its Pb complex is soluble and is excreted in urine.

Short Answer Questions (2 Marks)

Question 11. Write the IUPAC name and hybridisation of the following complexes:
(i) [Ni(CN)₄]²⁻
(ii) [Fe(H₂O)₆]²⁺
(Given: Atomic number of Ni = 28, Fe = 26) [CBSE OD Set-1 2020]
Answer:
(i) [Ni(CN)₄]²⁻ – Tetracyanonickelate (II) ion; hybridisation – dsp²
(ii) [Fe(H₂O)₆]²⁺ – Hexaaquairon (II) ion; hybridisation – sp³d²

Teacher's Note:
1. Strong field CN⁻ forces pairing in Ni²⁺ (d⁸), giving inner orbital dsp²; weak field H₂O cannot pair Fe²⁺ (d⁶), so outer 4d orbitals are used.

Teacher's Note:
1. In Ni(CO)₄, Ni is in the 0 state; strong field CO pushes the two 4s electrons into 3d, leaving 4s and 4p free for sp³.

Question 13. Out of [CoF₆]³⁻ and [Co(en)₃]³⁺, which one complex is,
(i) Paramagnetic,
(ii) More stable,
(iii) Inner orbital complex and
(iv) High spin complex
(Atomic number of Co = 27) [CBSE Delhi Set 1 2019]
Answer:
(i) [CoF₆]³⁻
(ii) [Co(en)₃]³⁺
(iii) [Co(en)₃]³⁺
(iv) [CoF₆]³⁻

Teacher's Note:
1. Both have Co³⁺ (d⁶): weak field F⁻ gives high spin, outer orbital, paramagnetic (4 unpaired); en gives low spin, inner orbital, diamagnetic and extra stability from chelation.

Teacher's Note:
1. Five ligands means dsp³ (trigonal bipyramidal); six ligands means d²sp³ or sp³d² – count the ligands before choosing the hybridisation.

Question 15. Define the following terms with a suitable example of each:
(a) Chelate complex
(b) Ambidentate ligand
Answer:
(a) A complex formed by bi or polydentate ligands with metal. Example- [Co(en)₃]³⁺.
(b) A ligand which can ligate through two different donor atoms. Example-SCN⁻.
Detailed Answer:
(a) Chelate complex is the complex formed when a di– or polydentate ligand uses its two or more donor atoms to bind a single metal atom.
(b) Ligand which can ligate through two different atoms is called ambidentate ligand.

Teacher's Note:
1. For SCN⁻, mention both modes: thiocyanato-S (through S) and thiocyanato-N or isothiocyanato (through N).

Question 16. Using IUPAC norms, write the formula for the following complexes:
(a) Tetraamminediaquacobalt(III) chloride
(b) Dibromidobis(ethane-1,2-diamine)platinum(IV) nitrate [CBSE OD Set-2 2019]
Answer:
(a) [Co(NH₃)₄(H₂O)₂]Cl₃
(b) [PtBr₂(en)₂](NO₃)₂

Teacher's Note:
1. Pt(IV) with two Br⁻ and two neutral en gives a +2 complex ion, so two NO₃⁻ are needed.

Teacher's Note:
1. Δ_o > P means pairing is cheaper than promotion, so all six electrons pair up in t_2g before any enter e_g.

Question 18. (a) Although both [NiCl₄]²⁻ and [Ni(CO)₄] have sp³ hybridisation yet [NiCl₄]²⁻ is paramagnetic and [Ni(CO)₄] is diamagnetic. Give reason. (Atomic no. of Ni = 28)
(b) Write the electronic configuration of d⁵ on the basis of crystal field theory when:
(i) Δ_o < P and (ii) Δ_o > P [CBSE OD Set-3 2019]
Answer:
(a) In [NiCl₄]²⁻, Cl⁻ is a weak field ligand due to which there are two unpaired electrons in 3d orbital whereas in [Ni(CO)₄], CO is a strong field ligand due to which pairing leads to no unpaired electron in 3d-orbital/ or structural representation.
(b) (i) t_2g³e_g²   (ii) t_2g⁵e_g⁰
Detailed Answer:
(a) In [NiCl₄]²⁻ Ni is in +2 oxidation state and each Cl⁻ donates a pair of electron. So, Cl⁻ acts as a weak ligand and does not cause any forced pairing. Thus, electrons remains unpaired making it paramagnetic. In [Ni(CO)₄], Ni is in zero oxidation state and CO acts as a strong ligand causing forced pairing. Thus, no electron remains unpaired making it diamagnetic.

Teacher's Note:
1. Note the different oxidation states: Ni²⁺ is 3d⁸ (two unpaired in sp³), while Ni(0) is 3d⁸4s², which becomes 3d¹⁰ with CO.

Question 19. (i) Using crystal field theory, write the electronic configuration of iron ion in the following complex ion. Also predict its magnetic behaviour: [Fe(H₂O)₆]²⁺
(ii) Write the IUPAC name of the coordination complex: [CoCl₂(en)₂]NO₃
Answer:
(i) t_2g⁴e_g², Paramagnetic
(ii) Dichloridobis(ethane-1,2-diamine)cobalt(III) nitrate

Teacher's Note:
1. Fe²⁺ is d⁶; H₂O is a weak field ligand (Δ_o < P), so the configuration is high spin with 4 unpaired electrons.

Question 20. (i) Predict the geometry of [Ni(CN)₄]²⁻.
(ii) Calculate the spin only magnetic moment of [Cu(NH₃)₄]²⁺ ion. [CBSE SQP 2021]
Answer:
(i) Square planar.
(ii) In Cu²⁺, electronic configuration (3d⁹) has one unpaired electron so, magnetic moment
= \( \sqrt{n(n+2)} \) = \( \sqrt{1(1+2)} \) = \( \sqrt{3} \) = 1.73 BM

Teacher's Note:
1. Use μ = √[n(n + 2)] BM; memorise the values for n = 1 to 5: 1.73, 2.83, 3.87, 4.90, 5.92 BM.

Question 21. (i) Write the IUPAC name of the isomer of the following complex: [Pt(NH₃)₂Cl₂]
(ii) Write the formula for the following: Tetraammineaquachloridocobalt(III) nitrate [CBSE (F) 2017]
Answer:
(i) cis-diamminedichloridoplatinum(II) and trans-diamminedichloridoplatinum(II)
(ii) [Co(NH₃)₄(H₂O)Cl](NO₃)₂

Teacher's Note:
1. Square planar MA₂B₂ complexes show cis-trans isomerism; prefix the name with cis- or trans-.
2. Charge check for (ii): +3 + 0 + 0 − 1 = +2, so two NO₃⁻ are needed.

Question 22. A chloride of fourth group cation in qualitative analysis gives a green coloured complex [A] in aqueous solution which when treated with ethane-1,2-diamine (en) gives pale-yellow solution [B] which on subsequent addition of ethane-1,2-diamine turns to blue/purple [C] and finally to violet [D]. Write the structures of complexes [A], [B], [C] and [D]. [CBSE Sample Paper 2017]
Answer:
A = [Ni(H₂O)₆]²⁺    B = [Ni(H₂O)₄(en)]²⁺
C = [Ni(H₂O)₂(en)₂]²⁺    D = [Ni(en)₃]²⁺

Teacher's Note:
1. Each en replaces two H₂O molecules; as stronger en ligands replace water, Δ₀ increases and the colour shifts.

Question 23. Write IUPAC name of the complex [Co(en)₂(NO₂)Cl]⁺. What type of structural isomerism is shown by this complex? [CBSE 2019 (56/2/1)]
Answer:
Chloridobis(ethane-1,2-diamine)nitrito-N-cobalt(III) ion.
This complex shows linkage isomerism.

Teacher's Note:
1. For ligands with numbers in their names, use bis-, tris- with brackets; alphabetical order goes by the ligand name: chlorido, ethane-1,2-diamine, nitrito.

Question 24. Why do compounds having similar geometry have different magnetic moment? [NCERT Exemplar]
Answer:
It is due to the presence of weak and strong ligands in complexes, if CFSE is high, the complex will show low value of magnetic moment and vice versa, e.g., [CoF₆]³⁻ and [Co(NH₃)₆]³⁺, the former is paramagnetic and the latter is diamagnetic.

Teacher's Note:
1. Both complexes are octahedral, yet F⁻ (weak) leaves 4 unpaired electrons while NH₃ (strong) pairs all six – quote such a pair as your example.

Question 25. Write the state of hybridisation, shape and IUPAC name of the complex [CoF₆]³⁻. (Atomic no. of Co = 27) [CBSE (F) 2014]
OR
Write IUPAC name and hybridisation of the complex: [CoF₆]³⁻. (Atomic number of Co = 27) [CBSE 2020 (56/4/2)]
Answer:
[CoF₆]³⁻: hexafluoridocobaltate(III); Co³⁺ = [Ar]3d⁶

Hybridisation = sp³d², Shape = Octahedral

Teacher's Note:
1. F⁻ is weak field, so the 3d electrons are not paired and the outer 4d orbitals are used – an outer orbital, high spin complex with 4 unpaired electrons.

Short Answer Questions (3 Marks)

Question 26. A metal complex having composition Cr(NH₃)₄Cl₂Br has been isolated in two forms A and B. The form A reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
(i) Write the formulae of isomers A and B.
(ii) State the hybridisation of chromium in each of them.
(iii) Calculate the magnetic moment (spin only value) of the isomer A.
Answer:
(i) Isomer A: [Cr(NH₃)₄BrCl]Cl
Isomer B: [Cr(NH₃)₄Cl₂]Br
(ii) Hybridisation of Cr in isomer A and B is d²sp³.
(iii) Number of unpaired electrons in Cr³⁺(3d³) is 3
Magnetic moment = \( \sqrt{n(n+2)} \)
= \( \sqrt{3(3+2)} \)
= 3.87 BM
(deduct half mark for wrong unit/unit not written)

Teacher's Note:
1. White, dilute-NH₃-soluble precipitate = AgCl (Cl⁻ outside); pale yellow, conc.-NH₃-soluble = AgBr (Br⁻ outside). These two forms are ionisation isomers.

Question 27. Write IUPAC name for each of the following complexes:
(i) [Ni(NH₃)₆]Cl₂
(ii) K₃[Fe(CN)₆]
(iii) [Co(en)₃]³⁺ [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) Hexaamminenickel (II) chloride
(ii) Potassium hexacyanidoferrate (III)
(iii) Tris(ethane-1,2-diamine)cobalt (III) ion

Teacher's Note:
1. Name the cation first and the anion second, even when the complex part is the anion, as in potassium hexacyanidoferrate(III).

Question 28. (a) Write the formula of the following coordination compound:
Iron(III) hexacyanoferrate(II)
(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅Cl]SO₄?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF₆]³⁻. (Atomic No. of Co = 27)
Answer:
(a) Fe₄[Fe(CN)₆]₃
(b) Ionisation isomerism
(c) sp³d², octahedral complex with 4 unpaired electrons

Teacher's Note:
1. For Fe₄[Fe(CN)₆]₃ (Prussian blue), balance charges: four Fe³⁺ (+12) against three [Fe(CN)₆]⁴⁻ (−12).

Teacher's Note:
1. [Cr(NH₃)₃Cl₃] is neutral, so its name has no counter ion and no "ion" at the end.

Question 30. (i) What type of isomerism is shown by the complex [Cr(H₂O)₆]Cl₃?
(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ_o > P.
(iii) Write the hybridisation and shape of [CoF₆]³⁻. (Atomic number of Co = 27) [CBSE OD 2015]
Answer:
(i) Hydrate isomerism
(ii) Electronic configuration is t_2g⁴e_g⁰ or by diagram.
(iii) Hybridisation is sp³d² and shape is octahedral.

Teacher's Note:
1. Hydrate (solvate) 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).

Teacher's Note:
1. Count ions per formula unit: 0, 2, 3 and 4 ions respectively; the neutral complex [Co(NH₃)₃Cl₃] is practically a non-electrolyte.

Teacher's Note:
1. A ligand must have at least one lone pair to donate; H₂O, NH₃ and CO all have lone pairs, but the C in CH₄ uses all its electrons in bonds.

Question 33. (i) Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ_o < P.
(ii) [Ni(CN)₄]²⁻ is colourless whereas [Ni(H₂O)₆]²⁺ is green. Why? (At. no. of Ni = 28) [CBSE Foreign Set-1 2017]
Answer:
(i) It is the magnitude of difference in energy between the two sets of d orbital i.e., t_2g and e_g
t_2g³e_g¹
(ii) In [Ni(H₂O)₆]²⁺, Ni⁺²(3d⁸) has two unpaired electrons which do not pair up in the presence of weak field ligand H₂O.
Detailed Answer:
(i) The difference in energy between the two sets of d-orbital (t_2g and e_g) caused by splitting of the degenerate levels due to the presence of ligands in a definite geometry.
Electronic configuration: t_2g³e_g¹
(ii) In [Ni(H₂O)₆]²⁺, Ni is present in +2 state with the configuration 3d⁸. It has two unpaired electrons which do not pair up in the presence of the weak H₂O ligand. Therefore, it is green in colour. While undergoing d–d transition, red light is absorbed and complementary light emitted is green.
In case of [Ni(CN)₄]²⁻, Ni is in +2 state with the configuration 3d⁸ but in presence of the strong CN⁻ ligand, the two unpaired electrons in the 3d orbitals undergoes pairing. As there is no unpaired electron present, it is colourless.

Teacher's Note:
1. The colour seen is the complementary colour of the light absorbed: green [Ni(H₂O)₆]²⁺ absorbs red light.

Question 34. (i) Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ_o > P.
(ii) [Ni(CN)₄]²⁻ is diamagnetic whereas [NiCl₄]²⁻ is paramagnetic. Give reason. (At. no. of Ni = 28) [CBSE Foreign Set-2 2017]
Answer:
(i) It is the magnitude of difference in energy between the two sets of d orbital i.e. t_2g and e_g
t_2g⁴e_g⁰.
(ii) In [Ni(CN)₄]²⁻, CN⁻ is a strong field ligand and pairing takes place whereas in [NiCl₄]²⁻, due to the presence of Cl⁻, a weak field ligand no pairing occurs/diagrammatic representation.

Teacher's Note:
1. Compare with SA-II 17: the same d⁴ ion is t_2g³e_g¹ when Δ_o < P and t_2g⁴e_g⁰ when Δ_o > P.

Question 35. A metal ion Mⁿ⁺ having d⁴ valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming Δ_o > P.
(i) Write the electronic configuration of d⁴ ion.
(ii) What type of hybridisation will Mⁿ⁺ ion has?
(iii) Name the type of isomerism exhibited by this complex. [CBSE SQP 2017]
Answer:
(i) t_2g⁴e_g⁰
(ii) d²sp³
(iii) Optical isomerism

Teacher's Note:
1. With t_2g⁴e_g⁰, both e_g (3d) orbitals are empty, so inner d orbitals are used; [M(AA)₃] complexes are chiral and exist as d and l forms.

Question 36. Explain with two examples each of the following:
(i) Coordination polyhedron
(ii) Homoleptic complex
(iii) Heteroleptic complex [CBSE 2019 (56/2/3)]
Answer:
(i) Coordination Polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion is known as the coordination polyhedron around the central atom/ion. Tetrahedral, square planar, octahedral, square pyramidal and trigonal bipyramidal are common shapes of coordination polyhedra, e.g., [Co(NH₃)₆]³⁺ (octahedral) and [Ni(CO)₄] (tetrahedral).
(ii) Homoleptic Complex: The complex in which metal atom is bound to only one kind of donor groups, e.g., [Cu(CN)₄]³⁻ and [Co(NH₃)₆]³⁺.
(iii) Heteroleptic Complex: The complex in which metal atom is bound to more than one kind of donor groups, e.g., [Co(NH₃)₄Cl₂]⁺ and [Co(NH₃)₅Cl]²⁺.

Teacher's Note:
1. The question asks for two examples each – giving only one costs marks.
2. “Homo” = one kind of ligand, “hetero” = more than one kind.

Long Answer Questions (5 Marks)

Teacher's Note:
1. Parts (b) and (c) are a contrast pair: same Co³⁺, but NH₃ pairs the electrons (inner, diamagnetic) while F⁻ does not (outer, paramagnetic).
2. Experimentally, paramagnetism is shown by the substance being attracted into a magnetic field (Gouy balance); μ ≈ 4.9 B.M. confirms four unpaired electrons.

Teacher's Note:
1. Pentaammine keeps both a’s (penta + ammine); carbonato comes after ammine alphabetically.
2. The colour seen is complementary to the colour absorbed – red absorbed, green seen.