CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Write the chemical formula of Wilkinson’s catalyst.
Answer:
[(Ph₃P)₃RhCl]

Teacher's Note:
1. It is used for the hydrogenation of alkenes; Rh is in the +1 state.

Question 2. Which of the two is more stable – K₄[Fe(CN)₆] or K₃[Fe(CN)₆]?
Answer:
K₄[Fe(CN)₆] because Fe has d⁶ configuration in this case.

Teacher's Note:
1. With strong field CN⁻, Fe²⁺ (d⁶) gets a fully filled t_2g⁶ set, which gives maximum crystal field stabilisation.

Question 3. Which type of ligands form chelates ?
Answer:
Polydentate ligands.

Teacher's Note:
1. A chelate needs a ligand with two or more donor atoms that bind the same metal ion and close a ring, e.g. en, oxalate, EDTA.

Question 4. Why complexes are preferred in the electrolytic bath for electroplating ?
Answer:
They dissociate slowly and hence give a smooth and even deposit.

Teacher's Note:
1. Example: [Ag(CN)₂]⁻ is used for silver plating and [Au(CN)₂]⁻ for gold plating.

Question 5. Complete the equation :
[Fe(H₂O)₆]²⁺ + NO → (CBSE Foreign 2017)
Answer:
[Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅NO]²⁺ (dark brown) + H₂O

Teacher's Note:
1. This is the brown ring complex of the nitrate test; NO replaces one water molecule.

Question 6. Define coordination number. (Hr. Board 2012)
Answer:
Coordination number is number of coordinate bonds formed by the ligands with the central metal atom.

Teacher's Note:
1. Count donor atoms, not ligands: in [Co(en)₃]³⁺ there are 3 ligands but the coordination number is 6.

Question 7. What is the coordination number of Fe in [Fe(EDTA)] ?
Answer:
Six.

Teacher's Note:
1. EDTA⁴⁻ is hexadentate (2 N + 4 O donor atoms), so one EDTA alone gives a coordination number of 6.

Question 8. What is oxidation state and coordination number of central metal ion in
(a) (NH₄)₂[CoF₄] ?
(b) [Pt(O₂)(en)₂Br]⁺ (Hr. Board 2011)
Answer:
(a) [CoF₄]²⁻ : x − 4 = −2 or x = +2, Coordination no. = 4.
(b) O₂²⁻ (peroxo) is monodentate, en is didentate and Br⁻ is monodentate. Hence, coordination no. = 1 + 4 + 1 = 6.
[Pt(O₂)²⁻(en)₂Br⁻]⁺ : x − 2 − 1 = +1 or x = +4

Teacher's Note:
1. Neutral ligands like en add nothing to the charge sum but each en adds 2 to the coordination number.

Question 9. What are ambidentate ligands ? Give an example. (Hr. Board 2012)
Answer:
Ligands which have two donor atoms but can coordinate with the central metal atom/ion through either of the two donor atoms (only one at a time) are called ambidentate ligands.
e.g., NO₂⁻ can coordinate through N (nitrito-N, –NO₂) or through O (nitrito-O, –ONO); similarly SCN⁻ can coordinate through S (thiocyanato) or N (isothiocyanato).

Teacher's Note:
1. Do not confuse ambidentate with bidentate: an ambidentate ligand uses only one of its two donor atoms at a time; this gives rise to linkage isomerism.

Question 10. Give one example of a bidentate ligand. (HP Board 2011, Manipur Board 2011)
Answer:
ethylene diamine, CH₂NH₂–CH₂NH₂ (en).

Teacher's Note:
1. Oxalate, C₂O₄²⁻ (ox), is the other common bidentate example; both bind through two donor atoms forming a five-membered chelate ring.

Short Answer Questions (2 Marks)

Question 11. Draw the structure of:
(i) [Ni(CO)₄]   (ii) [Fe(H₂O)₆]³⁺

Answer:
(i) [Ni(CO)₄] is tetrahedral (sp³). (ii) [Fe(H₂O)₆]³⁺ is octahedral (sp³d²). The structures are shown in the figure below.

Teacher's Note:
1. In [Ni(CO)₄] the CO ligands bond through carbon, so draw Ni–C bonds, not Ni–O.

Question 12. How does EDTA help as a cure for lead poisoning?
Answer:
Calcium in Ca-EDTA complex is replaced by lead in the body. The more soluble complex lead-EDTA is eliminated in urine.

Teacher's Note:
1. Pb²⁺ forms a more stable chelate with EDTA than Ca²⁺ does, which drives the replacement.

Question 13. Define homoleptic and heteroleptic complexes.
Answer:
Homoleptic: When metal atom/ion is linked with only one type of ligands. Example, [Ni(CO)₄].
Heteroleptic: When metal atom/ion is linked with more than one kind of ligands. Example, [Co(NH₃)₄Cl₂]⁺.

Teacher's Note:
1. Only heteroleptic complexes can show geometrical isomerism.

Question 14. [NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:
In [NiCl₄]²⁻, Ni²⁺ has 3d⁸4s⁰ configuration; Cl⁻ is a weak ligand and can’t pair up the electrons, so two unpaired electrons remain (paramagnetic).
In [Ni(CO)₄], Ni has 3d⁸4s² configuration; CO is a strong ligand and pairs up the electrons (3d¹⁰), so there is no unpaired electron (diamagnetic).

Teacher's Note:
1. Note the oxidation states: Ni is +2 in [NiCl₄]²⁻ but 0 in [Ni(CO)₄] – both use sp³ hybridisation.

Question 15. Write the oxidation number of cobalt in the complex:
(i) K[Co(CO)₄]   (ii) [Co(C₂O₄)₃]³⁻
Answer:
(i) −1    (ii) +3

Teacher's Note:
1. (i) +1 + x + 0 = 0 gives x = −1; (ii) x + 3(−2) = −3 gives x = +3.

Question 16. What are t_2g and e_g orbitals?
Answer:
In a free transition metal ion, the d-orbitals are degenerate. When it forms a complex, the degeneracy is split and d-orbitals split into t_2g (d_xy, d_yz, d_xz) and e_g (d_x²−y², d_z²) orbitals.

Teacher's Note:
1. In an octahedral field e_g lies higher (pointing at the ligands); in a tetrahedral field the order is reversed.

Question 17. What is the solution in which photographic film is washed? What reaction takes place?
Answer:
Hypo solution (sodium thiosulphate).
AgBr + 2Na₂S₂O₃ → Na₃[Ag(S₂O₃)₂] + NaBr

Teacher's Note:
1. Unexposed AgBr is removed as the soluble complex sodium dithiosulphatoargentate(I).

Question 18. What are the assumptions of Crystal Field Theory?
Answer:
(i) Ligands act as point charges (anions) or point dipoles (neutral molecules).
(ii) Metal ion has electrostatic attraction force with the ligands; the bonding is purely ionic.

Teacher's Note:
1. CFT ignores covalent character, which is why it cannot explain why neutral CO is a stronger ligand than charged ions.

Question 19. CuSO₄ is colourless while CuSO₄·5H₂O is coloured. Why?
Answer:
CuSO₄ does not have any ligand, so splitting of d-orbitals does not take place, while in CuSO₄·5H₂O the water ligands split the d-orbitals and d-d transition gives colour.

Teacher's Note:
1. Cu²⁺ is d⁹; colour needs both an incomplete d set and ligands to split it.

Question 20. What is the difference between inner and outer orbital complexes?
Answer:
Inner orbital complex: When d-orbitals of the inner shell, (n−1)d, take part in hybridisation (d²sp³).
Outer orbital complex: When d-orbitals of the outermost shell, nd, take part in hybridisation (sp³d²).

Teacher's Note:
1. Inner orbital complexes are usually low spin; outer orbital complexes are high spin.

Question 21. How is stability of coordination compound determined in aqueous solution?
Answer:
By using stability constant (formation constant).
More the stability constant, more the stability.

Teacher's Note:
1. For M + 4L ⇄ ML₄, β₄ = [ML₄]/[M][L]⁴; the reciprocal is the instability constant.

Question 22. In the complex [Co(NH₃)₅NO₂]Cl₂,
(i) Identify the ligands.
(ii) Write the oxidation number of metal ion.
Answer:
(i) NH₃, NO₂⁻
(ii) +3

Teacher's Note:
1. x + 5(0) + (−1) = +2 for the complex cation, so x = +3.

Question 23. Explain how the nature of ligand affects the stability of complex ion.
Answer:
Strong ligand: More stability
Weak ligand: Less stability
Chelating (polydentate) ligands form more stable complexes than unidentate ligands.

Teacher's Note:
1. More basic ligands and chelating ligands give larger stability constants.

Question 24. What is meant by denticity of a ligand? Find out denticity of: (i) C₂O₄²⁻   (ii) EDTA
Answer:
The number of coordinating (donor) groups in a ligand is called its denticity.
(i) 2    (ii) 6

Teacher's Note:
1. EDTA⁴⁻ binds through 2 N and 4 O atoms, so it is hexadentate.

Question 25. Why only transition metals are known to form π complexes ? (CBSE 2009, 2010)
Answer:
Transition metals/ions have empty d-orbitals into which the electron-pairs can be donated by ligands containing π electrons, i.e., electrons in their π molecular orbitals, e.g., CH₂ = CH₂, C₅H₅, C₆H₆, etc.

Teacher's Note:
1. Examples to quote: Zeise’s salt K[PtCl₃(η²-C₂H₄)] and ferrocene [Fe(η⁵-C₅H₅)₂].

Short Answer Questions (3 Marks)

Question 26. (i) What type of isomerism is shown by the complex [Co(NH₃)₆][Cr(CN)₆]?
(ii) Why a solution of [Ni(H₂O)₆]²⁺ is green while a solution of [Ni(CN)₄]²⁻ is colourless? (At. No. of Ni = 28)
(iii) Write the IUPAC name of the following complex: [Co(NH₃)₅(CO₃)]Cl. [CBSE Delhi 2017]
Answer:
(i) Coordination isomerism
(ii) [Ni(CN)₄]²⁻ has no unpaired electron in its d-subshell therefore d-d transition is not possible whereas [Ni(H₂O)₆]²⁺ has unpaired electron in its d-subshell resulting in d-d transition imparting colour.
(iii) Pentaamminecarbonatocobalt(III) Chloride

Teacher's Note:
1. Colour needs a d–d transition: strong-field CN⁻ pairs up the two 3d electrons of Ni²⁺, so no unpaired electron is left to be excited.

Question 27. (i) What type of isomerism is shown by the complex [Co(en)₃]Cl₃?
(ii) Write the hybridization and magnetic character of [Co(C₂O₄)₃]³⁻. (At. no. of Co = 27)
(iii) Write IUPAC name of the following Complex [Cr(NH₃)₃Cl₃] [CBSE Delhi 2017]
Answer:
(i) Optical isomerism
(ii) d²sp³, diamagnetic
(iii) Triamminetrichloridochromium (III)

Teacher's Note:
1. Many students write sp³d² and paramagnetic here; oxalate pairs the six 3d electrons of Co³⁺, so the complex is inner orbital d²sp³ and diamagnetic.

Question 28. For the complex ion [Fe(en)₂Cl₂]⁺ write the hybridization type and magnetic behaviour. Draw one of the geometrical isomer of the complex ion which is optically active. [Atomic No.: Fe = 26] [CBSE Comptt. OD 2017]
Answer:
Hybridisation: d²sp³
Magnetic character: Paramagnetic
The cis-isomer is optically active:

Teacher's Note:
1. Fe³⁺ is d⁵; even after pairing by en one electron stays unpaired, so the complex remains paramagnetic.

Teacher's Note:
1. A quick test: Mohr's salt gives the tests of Fe²⁺, NH₄⁺ and SO₄²⁻, but K₄[Fe(CN)₆] does not give the test of Fe²⁺.

Teacher's Note:
1. Use μ = √[n(n + 2)] BM with n = number of unpaired electrons; remember the unit BM, or marks are deducted.

Teacher's Note:
1. Square planar MA₂B₂ gives exactly two isomers; a tetrahedral MA₂B₂ would give none, which is one proof that [Pt(NH₃)₂Cl₂] is square planar.

Question 32. (i) Write the IUPAC name of the complex [Cr(NH₃)₄Cl₂] Cl.
(ii) What type of isomerism is exhibited by the complex [Co(en)₃]³⁺? (en = ethane-1, 2-diamine)
(iii) Why is [NiCl₄]²⁻ paramagnetic but [Ni(CO)₄] is diamagnetic? (At. nos. : Cr = 24, Co = 27, Ni = 28) [CBSE OD 2014]
Answer:
(i) Tetraamminedichloridochromium (III) chloride.
(ii) Optical isomerism.
(iii) In [NiCl₄]²⁻, Cl⁻ act as weak ligand therefore does not cause forced pairing, thus electrons will remain unpaired hence paramagnetic.
In [Ni(CO)₄], CO act as strong ligand therefore causes forced pairing, thus electrons will become paired hence diamagnetic.

Teacher's Note:
1. Students get confused between the magnetic terms: unpaired electrons present means paramagnetic, all electrons paired means diamagnetic.

Question 33. (i) Describe the type of hybridisation for the complex ion [Fe(H₂O)₆]²⁺.
(ii) Write the IUPAC name of the ionisation isomer of the coordination compound [Co(NH₃)₅Br]SO₄. Give one chemical test to distinguish between the two compounds. [CBSE SQP 2016]
Answer:
(i) Fe exists as Fe²⁺. Fe (II) in [Fe(H₂O)₆]²⁺ = 3d⁶ 4s⁰ 4p⁰ 4d⁰
As water is a weak ligand, pairing does not occur and the 6 lone pairs available from each water molecule moves to one 4s, three 4p and two 4d orbitals. Thus, the hybridisation involved is sp³d².
(ii) The ionisation isomer is [Co(NH₃)₅SO₄]Br. The IUPAC name is Pentaamminesulphatocobalt (III) bromide.
The isomer [Co(NH₃)₅Br]SO₄ gives a white precipitate of BaSO₄ with BaCl₂ solution whereas the isomer [Co(NH₃)₅SO₄]Br does not form this precipitate. (or any other relevant test)

Teacher's Note:
1. Another valid test: [Co(NH₃)₅SO₄]Br gives a pale yellow precipitate of AgBr with AgNO₃, while the other isomer does not.

Teacher's Note:
1. Three unpaired electrons give μ = √15 ≈ 3.87 BM, a value worth quoting when asked for magnetic behaviour.

Question 35. A coordination compound has the formula CoCl₃·4NH₃. It does not liberate NH₃ but forms a precipitate with AgNO₃. Write the structure and IUPAC name of the complex compound. Does it show geometrical isomerism?
Answer:
Formula: [Co(NH₃)₄Cl₂]Cl
Name: Tetraamminedichloridocobalt(III) chloride
Yes, it shows geometrical isomerism (cis and trans).

Teacher's Note:
1. Coordination number 6 is filled by 4 NH₃ + 2 Cl; only the third Cl, outside the bracket, is precipitated by AgNO₃.

Question 36. Why does a tetrahedral complex of the type [MA₂B₂] not show geometrical isomerism?
Answer:
Because all positions of tetrahedron are same in orientation (every position is adjacent to the other three), so cis and trans arrangements are not possible.

Teacher's Note:
1. Geometrical isomerism is shown by square planar and octahedral complexes, never by tetrahedral ones.

Long Answer Questions (5 Marks)

Question 37. (i) Discuss the nature of bonding in metal carbonyls.
(ii) Draw figure to show the splitting of d-orbitals in an octahedral crystal field and write electronic configuration of M²⁺ ion (having d⁵ configuration) when: (a) P > Δ_o   (b) Δ_o > P
Answer:
(i) The metal–carbon bond in metal carbonyls possesses both σ and π character. The M–C σ bond is formed by the donation of the lone pair of electrons on carbon of CO into a vacant orbital of the metal (M ← C≡O), while the M–C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of CO.
The synergic (σ + π) bonding strengthens the metal–ligand bond.
(ii) Splitting of d-orbitals in an octahedral crystal field:

(a) t_2g³e_g²    (b) t_2g⁵e_g⁰

Teacher's Note:
1. In an octahedral field e_g is raised by (3/5)Δ_o and t_2g is lowered by (2/5)Δ_o – label Δ_o on the diagram for full marks.

Question 38. (i) [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ are of different colours in dilute solution. Why?
(ii) A complex is prepared by mixing CoCl₃ and NH₃ in the molar ratio of 1 : 4. 0.1 M solution of this complex was found to freeze at −0.372 °C. What is the formula of the complex? (K_f = 1.86 °C/m)
Answer:
(i) In both the cases Fe is in +2 state (3d⁶), but CN⁻ and H₂O have different crystal field splitting energy. CN⁻ is a strong field ligand (t_2g⁶, no unpaired electron) while H₂O is a weak field ligand (t_2g⁴e_g², 4 unpaired electrons). Hence they absorb light of different wavelengths and show different colours.
(ii) ΔT_f = i K_f m
0.372 = i × 1.86 × 0.1  ⇒  i = \( \frac{0.372}{0.186} \) = 2
i = 2 means complex dissociates into two ions.
Hence the formula is [Co(NH₃)₄Cl₂]Cl.

Teacher's Note:
1. The van’t Hoff factor i tells you the number of ions per formula unit, which fixes how many Cl⁻ are outside the bracket.