CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. A coordination compound with the molecular formula CrCl₃.4H₂O precipitates AgCl with AgNO₃ solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound? [HOTS]
Answer:
[Cr(H₂O)₄Cl₂]Cl; tetraaquadichloridochromium(III) chloride

Teacher's Note:
1. Two ions means one ionisable Cl⁻; the other two chlorides and four water molecules complete coordination number 6.

Question 2. Arrange the following complexes in the order of increasing electrical conductivity:
[Co(NH₃)₃Cl₃], [Co(NH₃)₅Cl]Cl₂, [Co(NH₃)₆]Cl₃, [Co(NH₃)₄Cl₂]Cl [HOTS]
Answer:
[Co(NH₃)₃Cl₃] < [Co(NH₃)₄Cl₂]Cl < [Co(NH₃)₅Cl]Cl₂ < [Co(NH₃)₆]Cl₃

Teacher's Note:
1. Count ions: 0, 2, 3 and 4 respectively – more ions, higher conductivity.

Question 3. Which of the following is more stable complex and why? [Co(NH₃)₆]³⁺ and [Co(en)₃]³⁺ [CBSE Delhi 2014]
Answer:
[Co(en)₃]³⁺ is more stable complex because of chelation.

Teacher's Note:
1. Always name the reason – “chelate effect” – to get the full mark; en is a bidentate ligand forming five-membered rings.

Question 4. Why is CO a stronger ligand than Cl⁻? [CBSE (F) 2011]
Answer:
CO is a stronger ligand than Cl⁻ as it is a π-acceptor ligand.

Teacher's Note:
1. Back bonding from metal d-orbitals into empty π* orbitals of CO increases Δ₀; Cl⁻ instead donates π-electrons and lowers it.

Question 5. Square planar complexes with coordination number of four exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands attached to the central metal atom are same with respect to each other.

Teacher's Note:
1. In a tetrahedron every position is adjacent to every other, so “cis” and “trans” positions do not exist.

Question 6. How many isomers are there for octahedral complex [CoCl₂(en)(NH₃)₂]⁺? [NCERT Exemplar]
Answer:
There will be three isomers: cis and trans isomers. Cis will also show optical isomerism.

Teacher's Note:
1. Total = trans + the two optically active forms (d and l) of the cis isomer = 3.

Question 7. A complex of the type [M(AA)₂X₂]ⁿ⁺ is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex. [NCERT Exemplar] [HOTS]
Answer:
An optically active complex of the type [M(AA)₂X₂]ⁿ⁺ indicates cis-octahedral structure, e.g., cis-[Pt(en)₂Cl₂]²⁺ or cis-[Cr(en)₂Cl₂]⁺.

Teacher's Note:
1. The trans form has a plane of symmetry and is optically inactive; only the cis form is chiral.

Question 8. Give IUPAC name of linkage isomer of [Co(NH₃)₅(NO₂)]²⁺. [HOTS]
Answer:
The linkage isomer is [Co(NH₃)₅(ONO)]²⁺. Its IUPAC name is pentaamminenitrito-O-cobalt(III) ion.

Teacher's Note:
1. Write -N or -O after “nitrito” to show the donor atom; NO₂ bonded through N is nitrito-N.

Question 9. What type of isomerism is shown by the following complex: [Co(NH₃)₆][Cr(CN)₆] [CBSE (F) 2014]
Answer:
Coordination isomerism

Teacher's Note:
1. Coordination isomerism needs both cation and anion to be complexes, so that ligands can be exchanged between the two metals.

Question 10. Write IUPAC name of coordination isomer of [Co(NH₃)₆][Cr(CN)₆]. [HOTS]
Answer:
The coordination isomer is [Cr(NH₃)₆][Co(CN)₆], hexaamminechromium(III) hexacyanidocobaltate(III).

Teacher's Note:
1. In the anion, the metal name takes “-ate”: cobalt becomes cobaltate.

Question 11. What type of isomerism is exhibited by the following complex: [Co(NH₃)₅SO₄]Cl [CBSE (F) 2014]
Answer:
Ionisation isomerism

Teacher's Note:
1. Its ionisation isomer is [Co(NH₃)₅Cl]SO₄; the two give different ions in solution (Cl⁻ vs SO₄²⁻).

Short Answer Questions (2 Marks)

Question 12. (i) A coordination compound with molecular formula CrCl₃.4H₂O precipitates one mole of AgCl with AgNO₃ solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound?
(ii) On the basis of crystal field theory, write the electronic configuration of d⁶ in terms of t_2g and e_g in an octahedral field when Δ_o < P.
Answer:
(i) [Cr(H₂O)₄Cl₂]Cl
Tetraaquadichloridochromium (III) chloride.
(ii) t_2g⁴e_g².

Teacher's Note:
1. When Δ_o < P, fill all five orbitals singly first (t_2g³e_g²) and only then pair; the sixth electron goes into t_2g.

Question 13. Using IUPAC norms write the formulae for the following:
(i) Sodium dicyanidoaurate (I)
(ii) Tetraamminechloridonitrito-N-platinum (IV) sulphate [CBSE OD Set-1 2017]
Answer:
(i) Na[Au(CN)₂]
(ii) [Pt(NH₃)₄Cl(NO₂)]SO₄

Teacher's Note:
1. Check the charge: Pt(IV) + 4(0) + (−1) + (−1) = +2, which is balanced by one SO₄²⁻.

Question 14. Using IUPAC norms write the formulae for the following:
(i) Tris(ethane-1,2-diamine)chromium(III) chloride
(ii) Potassium tetrahydroxozincate(II) [CBSE OD Set-2 2017]
Answer:
(i) [Cr(en)₃]Cl₃
(ii) K₂[Zn(OH)₄]

Teacher's Note:
1. The prefix "tris" (not "tri") is used because the ligand name ethane-1,2-diamine already contains a numerical prefix.

Question 15. Using IUPAC norms write the formulae for the following:
(i) Potassium trioxalatoaluminate (III)
(ii) Dichloridobis(ethane-1,2-diamine)cobalt(III) [CBSE OD Set-3 2017]
Answer:
(i) K₃[Al(C₂O₄)₃]
(ii) [CoCl₂(en)₂]⁺

Teacher's Note:
1. Three oxalate ions carry −6 and Al is +3, so the complex ion is −3 and needs three K⁺.

Question 16. (i) Write the formula for the following:
Tetraammineaquachloridocobalt (III) nitrate
(ii) Write the formula for the following:
Diamminechloridonitrito-N-platinum(II)
Answer:
(i) [Co(NH₃)₄(H₂O)Cl](NO₃)₂
(ii) [Pt(NH₃)₂Cl(NO₂)]

Teacher's Note:
1. In the formula, write the metal first and then the ligands in alphabetical order of their names; keep brackets around polyatomic ligands such as (H₂O) and (NO₂).

Question 17. (i) Write the IUPAC name of the following complex: [Co(NH₃)₄Cl(NO₂)]Cl
(ii) Write the formula for the following:
Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride [CBSE Foreign Set-3 2017]
Answer:
(i) Tetraamminechloridonitrito-N-cobalt (III) chloride.
(ii) [CoCl₂(en)₂]Cl

Teacher's Note:
1. Ligands are named in alphabetical order ignoring prefixes: ammine, chlorido, nitrito-N.

Question 18. (i) Write down the IUPAC name of the following complex:
[Cr(NH₃)₂Cl₂(en)]Cl (en = ethylenediamine)
(ii) Write the formula for the following complex:
Pentaamminenitrito-O-Cobalt (III). [CBSE Delhi 2015]
Answer:
(i) Diamminedichlorido(ethane-1,2-diamine)chromium (III) chloride
(ii) [Co(NH₃)₅(ONO)]²⁺

Teacher's Note:
1. Check the coordination number: 2 NH₃ + 2 Cl + one bidentate en = 6, which suits Cr(III).

Question 19. (i) Write down the IUPAC name of the following complex: [Co(NH₃)₅Cl]²⁺
(ii) Write the formula for the following complex:
Potassium tetrachloridonickelate (II) [CBSE Delhi 2015]
Answer:
(i) Pentaamminechloridocobalt (III) ion
(ii) K₂[NiCl₄]

Teacher's Note:
1. For [NiCl₄]²⁻: Ni(+2) + 4(−1) = −2, so two K⁺ ions are needed.

Question 20. When a coordination compound CoCl₃.6NH₃ is mixed with AgNO₃, 3 moles of AgCl are precipitated per mole of the compound. Write
(i) Structural formula of the complex,
(ii) IUPAC name of the complex. [CBSE OD 2016]
Answer:
(i) [Co(NH₃)₆]Cl₃
(ii) IUPAC name: Hexaamminecobalt (III) chloride.

Teacher's Note:
1. 3 moles of AgCl means all three chlorides are outside the sphere, so all six positions are taken by NH₃.

Teacher's Note:
1. Show the pairing step explicitly: the strong field NH₃ pairs the four unpaired 3d electrons, emptying two 3d orbitals for d²sp³. A low spin (spin-paired) entity has no unpaired electrons.

Question 22. Using IUPAC norms write the formulae for the following:
(i) Hexaamminecobalt(III) sulphate
(ii) Potassium trioxalatochromate(III) [CBSE Delhi Set-1 2019]
Answer:
(i) [Co(NH₃)₆]₂(SO₄)₃
(ii) K₃[Cr(C₂O₄)₃]

Teacher's Note:
1. Balance charges by cross-multiplying: two [Co(NH₃)₆]³⁺ (+6) balance three SO₄²⁻ (−6).

Question 23. Define the following terms with a suitable example of each:
(a) Polydentate ligand
(b) Homoleptic complex
Answer:
(a) A ligand having several donor atoms. Example-EDTA.
(b) A complex in which a metal is bound to only one kind of donor groups / ligands. Example-[Co(NH₃)₆]³⁺.
(or any other correct example)
Detailed Answer:
(a) A ligand which contains several donor atoms in a single ligand is known as polydentate ligand. Example, N(CH₂CH₂NH₂)₃.
(b) Complexes in which a metal is bound to only one kind of donor groups are known as homoleptic complexes. Example, [Co(NH₃)₆]³⁺.

Teacher's Note:
1. The opposite term is heteroleptic: a complex with more than one kind of ligand, such as [Co(NH₃)₄Cl₂]⁺.

Question 24. Using IUPAC norms, write the formulae for the following complexes:
(a) Potassium tri(oxalato)chromate(III)
(b) Hexaaquamanganese(II) sulphate [CBSE Delhi Set-3 2019]
Answer:
(a) K₃[Cr(C₂O₄)₃]
(b) [Mn(H₂O)₆]SO₄

Teacher's Note:
1. A name starting with a cation such as "Potassium" tells you the complex part is an anion, so the metal name ends in "-ate".

Question 25. Using IUPAC norms, write the formulae for the following complexes:
(a) Hexaaquachromium(III) chloride
(b) Sodium trioxalatoferrate(III)
Answer:
(a) [Cr(H₂O)₆]Cl₃
(b) Na₃[Fe(ox)₃]

Teacher's Note:
1. ox stands for oxalate, C₂O₄²⁻; writing Na₃[Fe(C₂O₄)₃] is equally correct.

Teacher's Note:
1. An ambidentate ligand uses only one of its donor atoms at a time, while a chelating ligand uses two or more donor atoms at the same time to form a ring.

Short Answer Questions (3 Marks)

Teacher's Note:
1. Inner vs outer orbital depends on whether the ligand can pair the 3d electrons, not on the metal’s oxidation state, which is +3 in both.

Teacher's Note:
1. Cr³⁺ (d³) is always paramagnetic with three unpaired electrons, whatever the ligand.

Question 29. (a) What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of d⁴ in terms of t_2g and e_g in an octahedral field when
(i) Δ_o > P   (ii) Δ_o < P
(b) Write two limitations of crystal field theory.
Answer:
(a) The difference of energy between two splitted levels of d-orbitals is called crystal field splitting energy. It is denoted by Δ or 10 Dq.
For octahedral Δ_o, for tetrahedral it is Δ_t and for square planar Δ_sp.

(i) When Δ_o > P, t_2g⁴e_g⁰   (ii) When Δ_o < P, t_2g³e_g¹
(b) (i) It assumes ligand to be point charges.
(ii) It does not take into account the covalent character of bonding between the ligand and the central atom.

Teacher's Note:
1. Another limitation often asked: CFT cannot explain why some anionic ligands (like F⁻) are weaker than neutral ones (like NH₃) in the spectrochemical series.

Question 30. [Ni(H₂O)₆]²⁺ is green and becomes violet when ethane-1, 2-diamine is added to it. Identify the observation.
Answer:
Ethane-1,2-diamine is stronger ligand than H₂O. When H₂O molecule is replaced by ethane-1,2-diamine (en) the crystal field splitting energy (Δ) increases. Complex absorbs light of higher frequency for d-d transition. This is why colour of complex changes from green to violet.
[Ni(H₂O)₆]²⁺ – Green (Δ_o small), [Ni(en)₃]²⁺ – violet (Δ_o large)

Teacher's Note:
1. Larger Δ_o shifts absorption to shorter wavelength, so the colour seen (the complement) changes.

Question 31. Write the IUPAC name of the following:
(i) [Co(NH₃)₆]Cl₃
(ii) [NiCl₄]²⁻
(iii) K₃[Fe(CN)₆] [CBSE Comptt. OD 2015]
Answer:
(i) Hexaamminecobalt (III) chloride.
(ii) Tetrachloridonickelate (II) ion.
(iii) Potassium hexacyanoferrate (III)

Teacher's Note:
1. In K₃[Fe(CN)₆], six CN⁻ (−6) and three K⁺ (+3) fix iron at +3; the anion takes the Latin-based name "ferrate".

Question 32. (i) For the complex [Fe(CN)₆]³⁻, write the hybridisation, magnetic character and spin nature of the complex. (At. number: Fe = 26).
(ii) Give the limitations of valence bond theory.
Answer:
(i) Hybridization: d²sp³
Magnetic character: Paramagnetic
Spin nature of the complex: Low spin.
(ii) (a) No explanation for magnetic properties of complex compound.
(b) No explanation for absorption spectra, difference between square planar and tetrahedral in nature.
(c) Failed to distinguish between strong and weak field ligands.

Teacher's Note:
1. [Fe(CN)₆]³⁻ is low spin yet paramagnetic: Fe³⁺ is d⁵, so one electron stays unpaired even after pairing.

Question 33. (a) (i) For the complex [Fe(CN)₆]⁴⁻, write the hybridisation, magnetic character and spin type of the complex. (At. number: Fe = 26). [CBSE OD Set-2 2016]
(b) (i) Write the hybridisation and magnetic character of [Co(C₂O₄)₃]³⁻. (At. no. of Co = 27)
Answer:
(a) (i) Hybridization: d²sp³
Magnetic character: Diamagnetic
Spin type: Low spin complex.
(b) (i) d²sp³, diamagnetic

Teacher's Note:
1. Fe²⁺ and Co³⁺ are both d⁶; with strong field ligands they become t_2g⁶, so both complexes are diamagnetic and inner orbital.

Question 34. Write the hybridisation, shape and magnetic character of [Fe(CN)₆]⁴⁻. [CBSE Comptt. Delhi 2016]
Answer:
Hybridisation: d²sp³
Shape: Octahedral
Magnetic character: Diamagnetic

Teacher's Note:
1. For each mark, give all three parts separately; d²sp³ always corresponds to an octahedral shape.

Question 35. (a) What type of isomerism is shown by the complex [Co(NH₃)₅(SCN)]²⁺?
(b) Why is [NiCl₄]²⁻ paramagnetic while [Ni(CN)₄]²⁻ is diamagnetic? (Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed? [CBSE OD Set-1, 2, 3 2017]
Answer:
(a) Linkage isomerism.
(b) In [NiCl₄]²⁻, due to the presence of Cl⁻, a weak field ligand no pairing occurs whereas in [Ni(CN)₄]²⁻, CN⁻ is a strong field ligand and pairing takes place.
(c) Because of very low CFSE which is not able to pair up the electrons.

Teacher's Note:
1. Linkage isomerism needs an ambidentate ligand: SCN⁻ can bind through S (thiocyanato) or through N (isothiocyanato).

Question 36. For the complex ion [CoF₆]³⁻ write the hybridisation type, magnetic character and spin nature. [Atomic number: Co = 27] [CBSE Comptt. Delhi Set-1, 2 2017]
Answer:
Hybridisation: sp³d²
Magnetic character: Paramagnetic
Spin nature: High spin

Teacher's Note:
1. F⁻ is a weak field ligand, so Co³⁺ (d⁶) keeps 4 unpaired electrons and must use the outer 4d orbitals.

Question 37. When a coordination compound CrCl₃.6H₂O is mixed with AgNO₃ solution, 3 moles of AgCl are precipitated per mole of the compound. Write:
(i) Structural formula of the complex
(ii) IUPAC name of the complex
(iii) Magnetic and spin behaviour of the complex [CBSE Comptt. OD Set-3 2017]
Answer:
(i) [Cr(H₂O)₆]Cl₃
(ii) Hexaaquachromium(III) chloride
(iii) Paramagnetic (3 unpaired electrons, t_2g³). For a d³ ion like Cr³⁺, the high-spin and low-spin arrangements are the same.

Teacher's Note:
1. Cr³⁺ is d³ (t_2g³) with 3 unpaired electrons whatever the ligand, so it is always paramagnetic (μ = 3.87 BM).

Long Answer Questions (5 Marks)

Question 38. (a) What are bidentate ligands? Explain with examples.
(b) Explain the coordination sites of polydentate ligands taking an example of EDTA.
(c) Calculate charge on the central metal in the following complexes:
(i) [Cu(NH₃)₆]²⁺ (ii) [Ag(CN)₂]⁻
Answer:
(a) Ligands which can coordinate with the central metal atom or ion through two donor atoms are known as bidentate ligands. Examples of bidentate ligands are:
Ethylene diamine (en) : H₂N̈–CH₂–CH₂–N̈H₂ (donor atoms : two N)
Oxalate anion : [O=C(–O⁻)–C(–O⁻)=O]²⁻, i.e., C₂O₄²⁻ (donor atoms : two O⁻)
Carbonate ion : O=C(O⁻)₂, i.e., CO₃²⁻ (donor atoms : two O⁻)
(b) Ligands which coordinate with the central ion through more than two donor atoms present in the molecule are called polydentate ligands. These are called tridentate (three), tetradentate (four), pentadentate (five) and hexadentate (six) ligands depending upon the number of coordinating donor atoms present in their molecules. A common example of hexadentate ligand is ethylenediamminetetraacetate ion as shown below:
(⁻OOC–CH₂)₂N–CH₂–CH₂–N(CH₂–COO⁻)₂
Structure of ethylenediamminetetraacetate ion: its six donor sites are the two N atoms (1, 2) and one O⁻ of each of the four carboxylate groups (3, 4, 5, 6).
(c) The charge carried by a complex ion is the algebraic sum of charges carried by the central ion and the ligands coordinated to it. Thus [Cu(NH₃)₆]²⁺ carries a charge of +2 and because ammonia molecule is neutral therefore, Cu²⁺ carries a charge of +2. [Ag(CN)₂]⁻, ion carries a charge of −1 and two cyanide ions coordinated to it carry a charge of −1 each. So, Ag⁺ carries a charge of +1.

Teacher's Note:
1. EDTA⁴⁻ alone fills all six positions of an octahedron, which is why it forms very stable 1 : 1 chelates with metal ions.

Teacher's Note:
1. en is bidentate, so two en + two Cl give coordination number 6 in (ii), not 4.