CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. How many ions are produced from the complex, [Co(NH₃)₆]Cl₂ in solution? [CBSE SQP 2016]
Answer:
Three ions. [Co(NH₃)₆]²⁺, 2Cl⁻.

Teacher's Note:
1. The whole bracket stays together as one ion; count it once and then add the counter ions.

Question 2. Low spin configuration are rarely observed in tetrahedral coordination entity formation. Explain. [CBSE SQP 2018-19]
Answer:
The orbital splitting energies, Δ_t are not sufficiently large for forcing pairing of electrons in the tetrahedral coordination entity formation.

Teacher's Note:
1. Tetrahedral splitting is small because there are only four ligands and none of them point directly at the d orbitals.

Question 3. Explain the following—
[Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. [CBSE Foreign 2012]
Answer:
[Fe(CN)₆]⁴⁻ does not have unpaired electrons (CN is a strong field ligand) whereas [Fe(H₂O)₆]²⁺ has unpaired electrons and it absorbs light from visible region and radiates complementary colour.

Teacher's Note:
1. The oxidation state of Fe is the same (+2) in both; the different colours come only from the different field strengths of CN⁻ and H₂O.

Question 4. 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 Δ₀ < P. [CBSE SQP 2018-19]
Answer:
t_2g⁴ e_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 5. What is ambidentate ligand? Give one example.
Answer:
Monodentate ligands which contain more than one coordinating atom and can ligate through either of them are called ambidentate ligands. Example, CN⁻.

Teacher's Note:
1. Other common examples are NO₂⁻ (N or O) and SCN⁻ (S or N); they give rise to linkage isomerism.

Question 6. Write the IUPAC name of [PtCl₂(en)₂](NO₃)₂.
Answer:
Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

Teacher's Note:
1. The nitrate ions are outside the bracket, so they are named last as the counter ion; Pt: x − 2 + 0 − 2 = 0 gives +4.

Question 7. What is a chelate ligand? Give one example.
Answer:
The ligand which binds through two or more donor sites simultaneously to form cyclic complexes. Example, ethane-1,2-diamine.

Teacher's Note:
1. A chelate ring makes the complex more stable than one with similar unidentate ligands (chelate effect).

Question 8. How many geometrical isomers are possible for [Ni(NH₃)₄]²⁺?
Answer:
None. Geometrical isomerism is not possible because all 4 ligands are same.

Teacher's Note:
1. Geometrical isomerism needs at least two different kinds of ligands, as in [MA₂B₂].

Question 9. Define coordination polyhedron.
Answer:
The spatial arrangement of the ligand atoms which are directly attached to the central metal atom/ion is called coordination polyhedron.

Teacher's Note:
1. Common coordination polyhedra are octahedral, square planar and tetrahedral.

Question 10. Give the chemical formula of potassium hexacyanidoferrate(II).
Answer:
K₄[Fe(CN)₆]

Teacher's Note:
1. Fe(II) with six CN⁻ gives a 4− complex ion, so four K⁺ ions are needed.

Question 11. Name the metal present in: (i) Chlorophyll (ii) Haemoglobin (iii) Vitamin B₁₂ (iv) cisplatin
Answer:
(i) Mg   (ii) Fe   (iii) Co   (iv) Pt

Teacher's Note:
1. Vitamin B₁₂ (cyanocobalamin) is a cobalt complex; cisplatin, cis-[PtCl₂(NH₃)₂], is an anticancer drug.

Short Answer Questions (2 Marks)

Question 12. Explain why [Fe(H₂O)₆]³⁺ has magnetic moment value of 5.92 BM whereas [Fe(CN)₆]³⁻ has a value of only 1.74 BM. [NCERT Exemplar]
Answer:
[Fe(CN)₆]³⁻ involves d²sp³ hybridisation with one unpaired electron and [Fe(H₂O)₆]³⁺ involves sp³d² hybridisation with five unpaired electrons. This difference is due to the presence of strong CN⁻ and weak ligand H₂O in these complexes.

Teacher's Note:
1. Check with μ = √[n(n + 2)]: n = 5 gives 5.92 BM and n = 1 gives 1.73 BM.

Question 13. 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) Δ₀ > P    (ii) Δ₀ < P [CBSE (AI) 2013]
Answer:
The difference of energy between two sets of degenerate orbitals after crystal field splitting is known as crystal field splitting energy.
(i) t_2g⁴e_g⁰     (ii) t_2g³e_g¹

Teacher's Note:
1. The fourth electron decides: if pairing costs less than Δ₀ it pairs in t_2g, otherwise it goes up to e_g.

Question 14. The π-complexes are known for transition metals only. Why?
Answer:
Transition metals have vacant d-orbitals in their atoms or ions into which the electron pairs can be donated by ligands containing π-electrons, e.g., C₆H₆, CH₂=CH₂, etc. Thus, dπ–pπ bonding is possible.

Teacher's Note:
1. The key word is the availability of suitable d-orbitals for both accepting π-electrons and back bonding.

Question 15. On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands. [NCERT Exemplar]
Answer:
With weak field ligands, Δ₀ < P, the electronic configuration of Co(III) will be t_2g⁴e_g² and it has 4 unpaired electrons and is paramagnetic. With strong field ligands, Δ₀ > P the electronic configuration will be t_2g⁶e_g⁰. It has no unpaired electrons and is diamagnetic.

Teacher's Note:
1. Examples to quote: [CoF₆]³⁻ (paramagnetic) and [Co(NH₃)₆]³⁺ (diamagnetic).

Question 16. What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex? [NCERT Exemplar]
Answer:
When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over.

Teacher's Note:
1. The colour seen is the complementary colour of the light absorbed – e.g., a complex absorbing green-yellow appears purple.

Question 17. (i) Write the IUPAC name of the isomer of the following complex:
[Co(NH₃)₅Cl]SO₄
(ii) Write the formula for the following:
Diamminechloridonitrito-N-platinum(II) [CBSE Foreign 2017]
Answer:
(i) Pentaamminesulphatocobalt (III) chloride
(ii) [Pt(NH₃)₂Cl(NO₂)]

Teacher's Note:
1. The isomer here is the ionisation isomer [Co(NH₃)₅SO₄]Cl, in which sulphate moves inside and chloride becomes the counter ion.

Question 18. Write the IUPAC name of the complex [Cr(NH₃)₄Cl₂]⁺. What type of isomerism does it exhibit? [CBSE Delhi 2014]
Answer:
(i) Tetraamminedichlorido chromium (III) ion.
(ii) Geometrical isomerism / cis-trans.

Teacher's Note:
1. MA₄B₂ octahedral complexes show cis and trans forms; neither form is optically active.

Question 19. Indicate the types of isomerisms exhibited by the complex [Co(NH₃)₅(NO₂)](NO₃)₂. (At. No. Co = 27). [CBSE Comptt. OD 2012]
Answer:
Linkage and ionisation isomerism.
It can show linkage isomerism:
[Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(ONO)](NO₃)₂
It can show ionization isomerism:
[Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(NO₃)](NO₃)(NO₂)

Teacher's Note:
1. Look for two clues: an ambidentate ligand (NO₂) means linkage isomerism, and an anion that could swap with the counter ion means ionisation isomerism.

Question 20. Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers. [NCERT]
Answer:
When ionisation isomers are dissolved in water, they ionise to give different ions. These ions then react differently with different reagents to give different products.
[Co(NH₃)₅Cl]SO₄ + Ba²⁺ → BaSO₄↓ (White precipitate)
[Co(NH₃)₅Cl]SO₄ + Ag⁺ → No reaction
[Co(NH₃)₅SO₄]Cl + Ba²⁺ → No reaction
[Co(NH₃)₅SO₄]Cl + Ag⁺ → AgCl↓ (White precipitate)

Teacher's Note:
1. Only ions outside the bracket are free in solution, so only they give the usual tests (BaCl₂ for sulphate, AgNO₃ for chloride).

Teacher's Note:
1. There is confusion in electronic configuration of ions: remove the 4s electrons first, so Co³⁺ is 3d⁶ (not 3d⁴4s²).

Teacher's Note:
1. Inner or outer depends on the number of 3d orbitals that can be emptied: two are needed for d²sp³, and d⁸ can never provide two.

Teacher's Note:
1. While describing the reason, state both the cause (oxidation state, pairing) and the consequence (hybridisation, shape).

Question 24. Calculate the magnetic moments of the following complexes:
(i) [Fe(CN)₆]⁴⁻   (ii) [FeF₆]³⁻
Answer:
(i) Fe²⁺ → 3d⁶. CN⁻ is a strong ligand so electrons pair up; there is no unpaired electron (n = 0), so μ = \( \sqrt{n(n+2)} \) = 0, i.e., magnetic moment is zero.
(ii) Fe³⁺ → 3d⁵. F⁻ is a weak ligand, so n = 5.
μ = \( \sqrt{n(n+2)} \) = \( \sqrt{5(5+2)} \) = \( \sqrt{35} \) = 5.92 BM

Teacher's Note:
1. Spin-only formula: μ = √[n(n+2)] BM; memorise 1.73, 2.83, 3.87, 4.90 and 5.92 BM for n = 1 to 5.

Question 25. Explain the following:
(i) NH₃ acts as a ligand but NH₄⁺ does not.
(ii) CN⁻ is an ambidentate ligand.
Answer:
(i) NH₃ has one lone pair while NH₄⁺ does not.
(ii) Because it has two donor atoms (C and N) in a monodentate ligand; it can bind through either of them.

Teacher's Note:
1. A ligand must be a Lewis base – no free lone pair means no coordinate bond.

Question 26. Mention the main postulates of Werner’s theory.
Answer:
(i) Metal ion has two types of valency – primary valency and secondary valency.
(ii) Primary valencies are ionisable and are satisfied by negative ions; they correspond to the oxidation state.
(iii) Secondary valency is equal to coordination number; it is satisfied by neutral molecules or negative ions and is non-ionisable.
(iv) Secondary valencies are directional, so they decide the geometry of the complex.

Teacher's Note:
1. In [Co(NH₃)₆]Cl₃, primary valency = 3 (the Cl⁻ ions) and secondary valency = 6 (the NH₃ ligands).

Short Answer Questions (3 Marks)

Question 27. Write the correct formulae for the following coordination compounds:
(i) CrCl₃.6H₂O (violet with 3 chloride ions precipitated as AgCl)
(ii) CrCl₃.6H₂O (light green colour, with 2 chloride ions precipitated as AgCl)
(iii) CrCl₃.6H₂O (dark green colour, with 1 chloride ion precipitated as AgCl) [CBSE Sample Paper 2017]
Answer:
(i) [Cr(H₂O)₆]Cl₃
(ii) [Cr(H₂O)₅Cl]Cl₂.H₂O
(iii) [Cr(H₂O)₄Cl₂]Cl.2H₂O

Teacher's Note:
1. These three are hydrate isomers: each Cl that enters the coordination sphere pushes one H₂O out as water of crystallisation.

Question 28. Using valence bond theory, predict the hybridisation and magnetic character of the following:
(i) [CoF₆]³⁻    (ii) [Ni(CN)₄]²⁻
[Atomic number: Co = 27, Ni = 28] [CBSE 2019 (56/5/2)]
Answer:
(i) sp³d², paramagnetic
(ii) dsp², diamagnetic

Teacher's Note:
1. Weak F⁻ keeps the 4 unpaired 3d electrons of Co³⁺; strong CN⁻ pairs the two unpaired electrons of Ni²⁺.

Teacher's Note:
1. The trans isomer has a mirror plane through Pt and the two en rings, so it is optically inactive; the cis isomer is chiral.

Question 30. Name the following coordination entities and describe their structures:
(i) [Fe(CN)₆]⁴⁻    (ii) [Cr(NH₃)₄Cl₂]⁺    (iii) [Ni(CN)₄]²⁻
(Atomic Numbers: Fe = 26, Cr = 24, Ni = 28) [CBSE (AI) 2012]
Answer:
(i) [Fe(CN)₆]⁴⁻ = hexacyanidoferrate(II) ion; Fe²⁺(3d⁶)

d²sp³ hybridisation in [Fe(CN)₆]⁴⁻ leads to octahedral structure.
(ii) [Cr(NH₃)₄Cl₂]⁺ = tetraamminedichloridochromium(III) ion; Cr³⁺(3d³)

d²sp³ hybridisation in [Cr(NH₃)₄Cl₂]⁺ leads to octahedral structure.
(iii) [Ni(CN)₄]²⁻ = tetracyanidonickelate(II) ion; Ni²⁺(3d⁸)

dsp² hybridisation in [Ni(CN)₄]²⁻ leads to square planar structure.

Teacher's Note:
1. A d³ ion like Cr³⁺ always has two empty inner d-orbitals, so its octahedral complexes are inner orbital whatever the ligand.

Question 31. Describe for any two of the following complex ions, the type of hybridisation, shape and magnetic property:
(i) [Fe(H₂O)₆]²⁺    (ii) [Co(NH₃)₆]³⁺    (iii) [NiCl₄]²⁻
(At. Nos. Fe = 26, Co = 27, Ni = 28)
Answer:
(i)

Octahedral because of sp³d² hybridisation. Paramagnetic, as there are four unpaired electrons present in the complex ion.
(ii)

d²sp³ hybridisation leads to octahedral geometry. Diamagnetic, as there is no unpaired electron present in the complex.
(iii)

sp³ hybridisation leads to tetrahedral geometry. Paramagnetic, as there are two unpaired electrons.
(any two)

Teacher's Note:
1. In VBT answers, show three steps: free ion, hybridised orbitals and filling by ligand electron pairs – then state shape and magnetism.

Question 32. Fe³⁺ \( \xrightarrow[\text{excess}]{\text{SCN}^-} \) A \( \xrightarrow[\text{excess}]{\text{F}^-} \) B
What are A and B? Give IUPAC names of A and B. Find spin only magnetic moment of B. [HOTS]
Answer:
A = [Fe(SCN)₃], trithiocyanatoiron(III)
B = [FeF₆]³⁻, hexafluoridoferrate(III)
EC of Fe(III) = d⁵, Unpaired electrons (n) = 5
Spin only magnetic moment of B = \( \sqrt{n(n+2)} \) = \( \sqrt{5(5+2)} \) = 5.916 BM

Teacher's Note:
1. F⁻ is weak field, so all five d-electrons of Fe³⁺ stay unpaired; √35 ≈ 5.92 BM.

Teacher's Note:
1. Mn²⁺ is d⁵: even with strong CN⁻ one electron stays unpaired, so it is paramagnetic.
2. Current IUPAC names use “cyanido”, e.g., potassium hexacyanidomanganate(II); either form is accepted.

Question 34. 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 [Fe(H₂O)₆]²⁺. [NCERT Exemplar]
Answer:

[Fe(H₂O)₆]²⁺: Fe²⁺ = 3d⁶, i.e., t_2g⁴e_g²
Number of unpaired electrons (n) = 4
Magnetic moment (μ) = \( \sqrt{4(4+2)} = \sqrt{24} \) = 4.9 BM

Teacher's Note:
1. H₂O is a weak field ligand for Fe²⁺ (Δ₀ < P), so the sixth electron is the only one that pairs.

Question 35. Explain the following: [HOTS]
(i) Low spin octahedral complexes of nickel are not known.
(ii) Co²⁺ is easily oxidised to Co³⁺ in the presence of a strong ligand.
(iii) CO is a stronger complexing reagent than NH₃.
Answer:
(i)

For the formation of low spin complex, electrons present in 3d electrons of Ni²⁺ should pair up. This will produce only one empty d orbital. Hence, d²sp³ hybridisation in nickel is not possible to form low spin octahedral complex.
(ii) With the electronic configuration 3d⁷4s⁰, Co²⁺ has three unpaired electrons. H₂O being a weak ligand, the unpaired electrons present in 3d orbitals of Co(II) do not pair up. In the presence of strong ligands, two unpaired electrons in 3d orbitals pair up and the third unpaired electron shifts to higher energy orbital from where it can be easily lost and hence shows an oxidation state of III.
(iii) CO has empty π-orbitals which overlap with filled d-orbitals (t_2g orbitals) of transition metals and form π-bonds by back bonding. These π-interactions increase the value of crystal field stabilisation energy (Δ₀).
As NH₃ cannot form π bonds by back bonding, therefore, CO is stronger ligand than NH₃.

Teacher's Note:
1. Ni²⁺ (d⁸) can free at most one 3d orbital, so its octahedral complexes are always sp³d² with two unpaired electrons.
2. Strong ligands also give a large CFSE to low spin d⁶ Co³⁺, which is why Co(III) complexes are so stable.

Question 36. Give the electronic configuration of the [HOTS]
(i) d-orbitals of Ti in [Ti(H₂O)₆]³⁺ ion in an octahedral crystal field.
(ii) Why is this complex coloured? Explain on the basis of distribution of electrons in the d-orbitals.
(iii) How does the colour change on heating [Ti(H₂O)₆]³⁺ ion?
Answer:
(i) In [Ti(H₂O)₆]³⁺ ion, Ti is in +3 oxidation state. There is one electron in d-orbital and the electronic configuration is t_2g¹e_g⁰.
(ii) Due to d-d transition, the electron present in t_2g absorbs green and yellow radiation of white light for excitation to e_g and the configuration becomes e_g¹. The complementary colour is purple.
(iii) On heating, H₂O is lost. In the absence of ligand, crystal field splitting does not occur hence the substance becomes colourless.

Teacher's Note:
1. A d-d transition needs at least one d-electron and a vacant higher d-level; d⁰ and d¹⁰ ions give colourless complexes.

Question 37. Indicate the types of isomerism exhibited by the following complexes :
(i) [Co(NH₃)₅(NO₂)]²⁺
(ii) [Co(en)₃]Cl₃ (en = ethylenediamine)
(iii) [Pt(NH₃)₂Cl₂] [CBSE Comptt. Delhi 2015]
Answer:
(i) Linkage isomerism.
(ii) Optical isomerism.
(iii) Cis - trans / Geometrical isomerism.

Teacher's Note:
1. Write "linkage isomerism", not "ligand isomerism" – a common slip in exams.

Long Answer Questions (5 Marks)

Question 38. A metal ion Mⁿ⁺ having d⁴ valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming Δ₀ > P: [HOTS]
(i) Explain orbital splitting during this complex formation.
(ii) Write the electronic configuration of the valence electrons of the metal Mⁿ⁺ ion in terms of t_2g and e_g.
(iii) What type of hybridisation will Mⁿ⁺ ion have?
(iv) Name the type of isomerism exhibited by this complex.
Answer:
(i) As Δ₀ > P pairing will occur in the t_2g orbitals and e_g orbitals will remain vacant.
(ii) t_2g⁴e_g⁰
(iii) As there are three bidentate ligands to combine therefore hybridisation will be d²sp³.
(iv) Optical isomerism.

Teacher's Note:
1. With Δ₀ > P and the e_g set empty, two inner d-orbitals are free for d²sp³.
2. [M(AA)₃] complexes (e.g., [Co(en)₃]³⁺) exist as non-superimposable mirror images.

Question 39. Specify the oxidation numbers of the metals in the following coordination entities :
(a) [Co(H₂O)(CN)(en)₂]²⁺
(b) [PtCl₄]²⁻
(c) [CoBr₂(en)₂]⁺
(d) K₃[Fe(CN)₆]
(e) [Cr(NH₃)₃Cl₃] [NCERT]
Answer:
(a) [Co(H₂O)(CN)(en)₂]²⁺
Let the oxidation number of Co be x. The charge on the complex is +2.
x + 0 + (−1) + 2(0) = +2
x − 1 = +2
x = +3
(b) [PtCl₄]²⁻
Let the oxidation number of Pt be x. The charge on the complex is −2.
x + 4(−1) = −2
x − 4 = −2
x = +2
(c) [CoBr₂(en)₂]⁺
x + 2(−1) + 2(0) = +1
x − 2 = +1
x = +3
(d) K₃[Fe(CN)₆], i.e., [Fe(CN)₆]³⁻
x + 6(−1) = −3
x = +3
(e) [Cr(NH₃)₃Cl₃]
x + 3(0) + 3(−1) = 0
x − 3 = 0
x = +3

Teacher's Note:
1. Neutral ligands (H₂O, NH₃, en) contribute 0; only the anionic ligands and the overall charge decide the oxidation number.

Question 40. A metal ion Mⁿ⁺ having d⁴ valence electronic configuration combines with three didentate ligands to form a complex compound. Assuming Δ_o > P:
(i) Draw the diagram showing d-orbital splitting during this complex formation.
(ii) What type of hybridisation will Mⁿ⁺ have?
(iii) Name the type of isomerism exhibited by this complex.
(iv) Write the electronic configuration of metal Mⁿ⁺.
Answer:
(i) If Δ_o > P, the fourth electron pairs up in the t_2g set (diagram below).

(ii) d²sp³
(iii) [M(AA)₃] type complex shows optical isomerism.
(iv) t_2g⁴e_g⁰

Teacher's Note:
1. Δ_o > P means pairing is cheaper than promotion, so all four electrons stay in t_2g and two inner d orbitals stay free for d²sp³.