CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 01

Very Short Answer Questions (1 Mark)

Teacher's Note:
1. F⁻ is a weak field ligand, so the 3d electrons of Co³⁺ do not pair and the outer 4d orbitals are used.

Question 2. What do you understand by ‘denticity of a ligand’?
Answer:
Denticity : The number of coordinating groups present in a ligand is called the denticity of ligand.
For example, bidentate ligand ethane-1, 2-diamine has two donor nitrogen atoms which can link to central metal atom.
H₂N̈–CH₂–CH₂–N̈H₂ (Ethane-1, 2-diamine)

Teacher's Note:
1. Always quote an example with its donor atoms – e.g. en (2 N), C₂O₄²⁻ (2 O), EDTA⁴⁻ (2 N + 4 O).

Question 3. Explain the term crystal field splitting in an octahedral field.
Answer:
The splitting of the degenerate d-orbitals into three orbitals of lower energy, t_2g set and two orbitals of higher energy e_g set due to the interaction of ligand in an octahedral crystal field is known as crystal field splitting in an octahedral field.

d-orbital splitting in an octahedral crystal field

Teacher's Note:
1. In an octahedral field the e_g orbitals (d_x²−y², d_z²) point at the ligands, so they are raised in energy.

Question 4. Write the formula of the following coordination compound :
Iron (III) hexacyanoferrate(II)
Answer:
Fe₄[Fe(CN)₆]₃

Teacher's Note:
1. Balance charges: four Fe³⁺ (+12) need three [Fe(CN)₆]⁴⁻ (−12). This is Prussian blue.

Question 5. When a coordination compound CrCl₃·6H₂O is mixed with AgNO₃, 2 moles of AgCl are precipitated per mole of the compound. Write structural formula of the complex.
Answer:
For one mole of the compound, two moles of AgCl are precipitated which indicates that two ionisable chloride ions in the complex. Hence, its structural formula is [CrCl(H₂O)₅]Cl₂·H₂O

Teacher's Note:
1. Coordination number 6 is completed by 1 Cl + 5 H₂O; the sixth water stays outside as water of crystallisation.

Question 6. 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)
Answer:
[Ni(H₂O)₆]²⁺ is a high spin complex (Δ_o small) while [Ni(CN)₄]²⁻ is a low spin square planar complex.
In [Ni(H₂O)₆]²⁺ complex, d-d transitions are taking place on absorbing low energy radiation (red component of spectrum) from visible region showing green as the complementary colour.
In [Ni(CN)₄]²⁻ complex, d-d transitions do not take place in the visible region of spectrum, d-d transitions take place in the UV region and hence, complex is colourless.

Teacher's Note:
1. Colour seen = complementary colour of the light absorbed; if absorption moves into the UV the complex looks colourless.

Question 7. Write the IUPAC name of [Cr(NH₃)₆][Co(CN)₆].
Answer:
Hexaamminechromium(III) hexacyanidocobaltate(III)

Teacher's Note:
1. Both ions are 3±; name the cation first with the plain metal name and the anion with ‘-ate’.

Question 8. What is the difference between a complex and a double salt?
Answer:
Double salts dissociate into ions completely when dissolved in water. On the other hand, in complexes, the complex ion does not dissociate.

Teacher's Note:
1. Example: Mohr’s salt gives the test for Fe²⁺, but K₄[Fe(CN)₆] does not, since [Fe(CN)₆]⁴⁻ stays intact.

Question 9. Chelates are generally more stable than the complexes of unidentate ligands. Explain.
Answer:
Chelates are cyclic compounds so they are more stable than normal complexes. In chelates ligands are held by two or more bonds with the transition metals. e.g., [Co(en)₃]³⁺, in which each en is bonded to Co through both N atoms forming three five-membered rings.

Teacher's Note:
1. To dissociate a chelating ligand, two (or more) metal–ligand bonds must break at once – this is the chelate effect.

Question 10. Write the coordination number and oxidation state of platinum in the complex [Pt(en)₂Cl₂].
Answer:
Coordination number and oxidation state of Pt in the complex [Pt(en)₂Cl₂] are 6 and +2 because en is a bidentate and neutral ligand.

Teacher's Note:
1. Coordination number counts donor atoms, not ligands: 2 en × 2 + 2 Cl = 6.

Short Answer Questions (2 Marks)

Teacher's Note:
1. Both are tetrahedral sp³; the difference is the oxidation state of Ni (+2 vs 0) and the ligand strength.

Teacher's Note:
1. When Δ_o < P it costs less energy to put an electron in e_g than to pair it in t_2g.

Teacher's Note:
1. Oxalate is a weaker ligand than NH₃ in general, but with Co³⁺ (high charge) it still pairs the 3d⁶ electrons.

Question 14. Using IUPAC norms write the formulae for the following :
(i) Pentaamminenitrito–O–cobalt(III) chloride
(ii) Potassium tetracyanonickelate(II)
Answer:
(i) [Co(NH₃)₅(ONO)]Cl₂
(ii) K₂[Ni(CN)₄]

Teacher's Note:
1. Write ONO (not NO₂) for nitrito-O, so the formula shows bonding through oxygen.

Question 15. Explain the following term giving a suitable example : Ambidentate ligand
Answer:
Ambidentate ligand : A unidentate ligand which can coordinate to central metal atom through two different atoms is called ambidentate ligand.
For example, NO₂⁻ ion can coordinate either through nitrogen or through oxygen to the central metal atom/ion.

Teacher's Note:
1. SCN⁻ (through S or N) is the other standard example; ambidentate ligands give rise to linkage isomerism.

Teacher's Note:
1. NH₃ pairs the 3d⁶ electrons of Co³⁺ to free two inner 3d orbitals – compare with [CoF₆]³⁻, which is outer orbital and paramagnetic.

Question 17. Why Co²⁺ is easily oxidised to Co³⁺ in presence of a strong ligand?
Answer:
In presence of strong field ligand, Co(II) has electronic configuration t_2g⁶e_g¹ (Δ_o > P).
It can easily lose one electron present in e_g orbital to give stable t_2g⁶ configuration. This is why Co²⁺ is easily oxidised to Co³⁺ in the presence of strong field ligand.

Teacher's Note:
1. The lone high-energy e_g electron is the key point: removing it leaves the very stable low spin t_2g⁶ arrangement.

Question 18. Write down the IUPAC name of the following complex : [Cr(NH₃)₂Cl₂(en)]Cl
Answer:
Diamminedichlorido(ethane-1,2-diamine)chromium(III) chloride

Teacher's Note:
1. Order: ammine, chlorido, ethane-1,2-diamine (alphabetical); Cr + 2(−1) = +1 gives Cr(III).

Question 19. A student prepared a complex compound of cobalt with NH₃ and NO₂ as donor ligands. He got a red precipitate. Another student also prepared the same complex using same metal salt solution and same ligands. He obtained yellow crystals. The second student complained to the teacher that his chemicals were different so he got different product. But their teacher is satisfied with both the results. Now answer the following questions :
(i) What type of ligand is present in the given compounds which is responsible for changing colour?
(ii) Write IUPAC name of both the compounds.
Answer:
(i) Ambidentate ligands (NO₂, ONO) present in the given compounds.
(ii) [Co(NH₃)₅(NO₂)]Cl₂ (Yellow) : Pentaamminenitrito-N-cobalt(III) chloride
[Co(NH₃)₅(ONO)]Cl₂ (Red) : Pentaamminenitrito-O-cobalt(III) chloride

Teacher's Note:
1. The two products are linkage isomers: N-bonded (nitro) is yellow and O-bonded (nitrito) is red.

Question 20. Give the formula of each of the following coordination entities :
(i) Co³⁺ ion is bound to one Cl⁻, one NH₃ molecule and two bidentate ethylene diamine (en) molecules.
(ii) Ni²⁺ ion is bound to two water molecules and two oxalate ions.
Write the name and magnetic behaviour of each of the above coordination entities.
(At. nos. Co = 27, Ni = 28)
Answer:
(i) [Co(en)₂Cl(NH₃)]²⁺
Amminechloridobis(ethane-1,2-diamine)cobalt(III) ion
In presence of strong NH₃ and en ligand, Co³⁺ (3d⁶) forms low spin complex. Hence, complex is diamagnetic.
(ii) [Ni(ox)₂(H₂O)₂]²⁻ : Diaquadioxalatonickelate(II) ion
In the presence of weak ox and H₂O ligand, Ni(II) forms high spin complex (sp³d² hybridisation). It is paramagnetic.

Teacher's Note:
1. Find the charge first: Co(+3) + Cl(−1) = +2; Ni(+2) + 2 ox(−2 each) = −2, so the nickel entity is an anion (‘nickelate’).

Question 21. (i) What is the IUPAC name of the complex [Ni(NH₃)₆]Cl₂?
(ii) What is meant by chelate effect?
Answer:
(i) Hexaamminenickel (II) chloride.
(ii) Formation of stable complex with a polydentate ligand due to stronger bonding is known as chelate effect.
OR
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. Write "hexaammine" with a double a and double m: hexa + ammine. Ammine (NH₃ ligand) is different from amine.

Question 22. (i) Which of the following is more stable complex and why?
[Co(NH₃)₆]³⁺ and [Co(en)₃]³⁺.
(ii) How many ions are produced from the complex, [Co(NH₃)₆]Cl₂ in solution?
Answer:
(i) [Co(en)₃]³⁺: Because (en) is a chelating ligand/bidentate ligand.
(ii) Three ions: [Co(NH₃)₆]²⁺, 2Cl⁻

Teacher's Note:
1. When counting ions, the whole complex ion counts as one ion; then add the counter ions.

Question 23. (i) Write the IUPAC name of the following coordination compound [NiCl₄]²⁻.
(ii) Write IUPAC name of the complex: [CoCl₂(en)₂]⁺.
Answer:
(i) Tetrachloridonickelate (II) ion.
(ii) Dichloridobis (ethane-1, 2-diamine) cobalt (III) ion.

Teacher's Note:
1. Calculate the oxidation state only within the coordination sphere, and use the suffix "-ate" for the metal only when the complex ion is an anion.

Question 24. (i) Write IUPAC name of the complex: [Co(NH₃)₄Cl(NO₂)]⁺.
(ii) Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂].
Answer:
(i) Tetraamminechloridonitrocobalt (III) ion.
(ii) Coordination Number = 6, Oxidation State = +2

Teacher's Note:
1. en is bidentate, so two en ligands give 4 donor atoms; with 2 Cl⁻ the coordination number is 6, and x + 0 + 2(−1) = 0 gives Pt = +2.

Question 25. (i) Why are low spin tetrahedral complexes not formed? [CBSE Comptt. Delhi Set-1, 2, 3 2017]
(ii) Low spin configuration are rarely observed in tetrahedral coordination entity formation. Explain.
Answer:
(i) Orbital splitting energies are not sufficiently large for forcing pairing of electrons.
(ii) The orbital splitting energies, Δ are not sufficiently large for forcing pairing of electrons in the tetrahedral coordination entity formation.

Teacher's Note:
1. Support the answer with the relation Δ_t = (4/9)Δ_o, which shows Δ_t is usually less than the pairing energy P.

Short Answer Questions (3 Marks)

Question 26. Write the IUPAC names of the following coordination compounds :
(i) [Cr(NH₃)₃Cl₃]   (ii) K₃[Fe(CN)₆]   (iii) [CoBr₂(en)₂]⁺
Answer:
(i) Triamminetrichloridochromium(III)
(ii) Potassium hexacyanoferrate(III)
(iii) Dibromidobis(ethane-1,2-diamine)cobalt(III) ion

Teacher's Note:
1. A neutral complex like (i) is written as one word with no ‘ion’ or counter-ion name.

Question 27. The splitting pattern of d-orbitals in octahedral and tetrahedral geometry are reverse of each other. Why?
Answer:
In octahedral complex ligands approach along the axes. So axial orbitals (d_x²−y², d_z²) lie directly in the path of ligand and experience greater repulsion than the non-axial orbitals. Whereas in tetrahedral complex ligands are closer to non-axial orbitals (i.e. d_xy, d_xz and d_yz). So, non-axial orbitals experience greater force of repulsion than the axial orbitals. i.e. approach of ligands in octahedral and tetrahedral fields is opposite of each other. This is why splitting pattern of d-orbitals in octahedral and tetrahedral geometry is reverse of each other.

Teacher's Note:
1. In tetrahedral complexes the labels lose the ‘g’ (t₂ and e) because a tetrahedron has no centre of symmetry.

Question 28. (a) What is d-d transition?
(b) Tetrahedral complexes are always of high spin. Explain.
Answer:
(a) When ligands approach the central metal, atom or ion of complex its d-orbital breaks into two parts t_2g and e_g levels. When light falls on the complex the complex absorbs light of suitable frequency for transfer of electron from lower level to higher level. This jump of electron from one d-level to another is called d-d transition.

(b) For tetrahedral complexes crystal field splitting energy Δ_t is always less than pairing energy. So, tetrahedral complexes are always high spin.

Teacher's Note:
1. Δ_t ≈ (4/9)Δ_o, which is why pairing is never favoured in tetrahedral complexes.

Teacher's Note:
1. An anionic complex takes ‘-ate’: nickel becomes nickelate.

Question 30. Write the name, the structure and the magnetic behaviour of each one of the following complexes :
(i) [Pt(NH₃)₂Cl(NO₂)]   (ii) [Co(NH₃)₄Cl₂]Cl   (iii) Ni(CO)₄
(At. nos. Co = 27, Ni = 28, Pt = 78)
Answer:
(i) [Pt(NH₃)₂Cl(NO₂)] : Diamminechloridonitrito-N-platinum(II)
It is square planar and diamagnetic.
(ii) [Co(NH₃)₄Cl₂]Cl : Tetraamminedichloridocobalt(III) chloride
It is octahedral and diamagnetic.
(iii) Ni(CO)₄ : Tetracarbonylnickel(0)
It is tetrahedral and diamagnetic.

Teacher's Note:
1. Pt²⁺ (5d⁸) is dsp² square planar; Co³⁺ with NH₃ is d²sp³; Ni(0) with CO is sp³ – all three have paired electrons.

Question 31. [Mn(CN)₆]³⁻ has two unpaired electrons whereas [MnCl₆]³⁻ has four unpaired electrons. Why?
Answer:
In [Mn(CN)₆]³⁻, Mn is in +3 state so, it has configuration of 3d⁴.
Since CN⁻ is a strong field ligand hence pairing of electrons in 3d-orbital takes place: t_2g⁴e_g⁰ (two unpaired electrons).
So, [Mn(CN)₆]³⁻ has two unpaired electrons. But in [MnCl₆]³⁻, Cl⁻ is a weak field ligand, so no pairing takes place (t_2g³e_g¹) and it has 4 unpaired electrons.

Teacher's Note:
1. Same metal, same oxidation state – any difference in unpaired electrons must come from ligand strength.

Teacher's Note:
1. Mn + 6(−1) = −4 gives Mn(II), 3d⁵; strong field CN⁻ squeezes the five electrons into three orbitals, leaving one unpaired.

Question 33. For the complex [Fe(en)₂Cl₂]Cl, identify the following :
(i) Oxidation number of iron
(ii) Hybrid orbitals and shape of the complex
(iii) Magnetic behaviour of the complex
(iv) Name of the complex.
Answer:
(i) [Fe(en)₂Cl₂]Cl
x + 0 × 2 + (−1) × 2 + (−1) × 1 = 0   ∴ x = +3
Oxidation number of iron = +3
(ii) d²sp³ hybridisation and octahedral shape.
(iii) Paramagnetic due to presence of one unpaired electron.
(iv) Dichloridobis(ethane-1,2-diamine)iron(III) chloride

Teacher's Note:
1. Fe³⁺ is 3d⁵; with the strong field ligand en it is low spin t_2g⁵, so one electron stays unpaired.

Question 34. Name the following coordination entities and describe their structures.
(i) [Fe(CN)₆]⁴⁻   (ii) [Cr(NH₃)₄Cl₂]⁺   (iii) [Ni(CN)₄]²⁻
Answer:
(i) [Fe(CN)₆]⁴⁻ : Hexacyanidoferrate(II) ion
Hybridisation – d²sp³
Structure : Inner orbital octahedral complex

(ii) [Cr(NH₃)₄Cl₂]⁺ : Tetraamminedichloridochromium(III) ion
Hybridisation – d²sp³
Structure : Inner orbital octahedral complex

(iii) [Ni(CN)₄]²⁻ : Tetracyanidonickelate(II) ion
Hybridisation – dsp²
Structure – Square planar

Teacher's Note:
1. Describe structure by naming hybridisation and geometry together: d²sp³ → octahedral, dsp² → square planar.

Question 35. Explain the following :
(i) Anhydrous CuSO₄ is white while hydrated CuSO₄ is blue in colour.
(ii) [Ti(H₂O)₆]Cl₃ is violet in colour but becomes colourless on heating.
Answer:
(i) Anhydrous CuSO₄ has no ligand. So, crystal field splitting does not occur so it does not show any colour but in hydrated form it is linked with H₂O ligand so, it shows colour due to d-d transition.
(ii) [Ti(H₂O)₆]Cl₃ is a complex compound. In presence of 6 H₂O molecules the d-orbitals of Ti³⁺ undergo splitting. The compound is coloured (violet) due to d-d transition. On heating water molecules escape, d-orbitals become degenerate. There is no d-d transition. Hence compound becomes colourless.

Teacher's Note:
1. Both parts test the same idea: no ligand field → no splitting → no d–d transition → no colour.

Teacher's Note:
1. Spin-only μ = √3 ≈ 1.73 B.M. for this low spin d⁵ complex, compared with 5.92 B.M. for high spin [FeF₆]³⁻.

Long Answer Questions (5 Marks)

Question 37. (i) Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following :
(a) [CoF₆]³⁻, (b) [FeF₆]³⁻, (c) [Fe(CN)₆]⁴⁻
(ii) FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
(i) (a) [CoF₆]³⁻ :

Electronic configuration : t_2g⁴e_g²
No. of unpaired electrons = 4, μ = \( \sqrt{4(4+2)} \) = 4.9 B.M.
(b) [FeF₆]³⁻ :

Electronic configuration : t_2g³e_g²
No. of unpaired electrons = 5, μ = \( \sqrt{5(5+2)} \) = 5.92 B.M.
(c) [Fe(CN)₆]⁴⁻ :

Electronic configuration : t_2g⁶e_g⁰
No. of unpaired electrons = 0, μ = 0
(ii) When FeSO₄ and (NH₄)₂SO₄ solutions are mixed in 1 : 1 molar ratio, Mohr’s salt (a double salt) is formed.
FeSO_4(aq) + (NH₄)₂SO_4(aq) → FeSO₄·(NH₄)₂SO₄·6H₂O
FeSO₄·(NH₄)₂SO₄·6H₂O ⇄ Fe²⁺_(aq) + 2NH₄⁺_(aq) + 2SO₄²⁻_(aq) + 6H₂O
Because Fe²⁺ ions are formed on dissolution of Mohr’s salt, its aqueous solution gives the test of Fe²⁺ ions.
When CuSO₄ is mixed with ammonia, following reaction occurs :
CuSO_4(aq) + 4NH_3(aq) → [Cu(NH₃)₄]SO₄
This complex does not produce Cu²⁺ ion, so the solution of CuSO₄ and NH₃ does not give the test of Cu²⁺ ion.
[Cu(NH₃)₄]SO₄ ⇄ [Cu(NH₃)₄]²⁺_(aq) + SO₄²⁻_(aq)

Teacher's Note:
1. Decide weak or strong field first (F⁻ weak, CN⁻ strong), then fill t_2g/e_g and count unpaired electrons for μ.
2. Part (ii) is the classic double salt vs complex distinction: a double salt gives tests for all its ions, a complex ion does not.

Question 38. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H₂O)₂(C₂O₄)₂].3H₂O
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) [CrCl₃(py)₃]
(iv) Cs[FeCl₄]
(v) K₄[Mn(CN)₆]
Answer:
(i) K[Cr(H₂O)₂(C₂O₄)₂].3H₂O
Potassium diaquadioxalatochromate(III) trihydrate
Oxidation state of Cr = +3; coordination number = 6
Stereochemistry : octahedral
E.C. of Cr³⁺, 3d³ = (t_2g)³, (e_g)⁰
Magnetic moment (μ) = \( \sqrt{n(n+2)} \) B.M. = \( \sqrt{3(3+2)} \) = \( \sqrt{15} \) B.M. = 3.87 B.M.
(ii) [Co(NH₃)₅Cl]Cl₂
Pentaamminechloridocobalt(III) chloride
Oxidation state of Co = +3, coordination number = 6
Stereochemistry = octahedral
E.C. of Co³⁺, 3d⁶ = (t_2g)⁶, (e_g)⁰
Magnetic moment (μ) = 0
(iii) [CrCl₃(py)₃]
Trichloridotripyridinechromium(III)
Oxidation state of Cr = +3, coordination number = 6
Stereochemistry : octahedral
E.C. of Cr³⁺, 3d³ = (t_2g)³, (e_g)⁰
Magnetic moment = 3.87 B.M.
(iv) Cs[FeCl₄]
Caesium tetrachloridoferrate(III)
Oxidation state of Fe = +3; coordination number = 4
Stereochemistry = Tetrahedral
E.C. of Fe³⁺, 3d⁵ = (e)², (t₂)³
Magnetic moment = \( \sqrt{5(5+2)} \) B.M. = \( \sqrt{35} \) B.M. = 5.92 B.M.
(v) K₄[Mn(CN)₆]
Potassium hexacyanomanganate(II)
Oxidation state of Mn = +2, coordination number = 6
Stereochemistry = octahedral
E.C. of Mn²⁺, 3d⁵ = (t_2g)⁵ (e_g)⁰
Magnetic moment
(μ) = \( \sqrt{1(1+2)} \) B.M. = \( \sqrt{3} \) B.M. = 1.73 B.M.

Teacher's Note:
1. Water of crystallisation (.3H₂O) is named at the end as ‘trihydrate’ and is not part of the coordination sphere.
2. For tetrahedral Cs[FeCl₄] use e and t₂ (no ‘g’), with e lower in energy.