CBSE Class 12 Chemistry Coordination Compounds Worksheet Set 06

Coordination Compounds Worksheet

Question 1. Distinguish between a double salt and a coordination compound. Give one example of each.
Answer:
Double salt: an addition compound that loses its identity in water and gives all the ions of the simple salts that formed it. Example: Mohr's salt, FeSO₄.(NH₄)₂SO₄.6H₂O, which gives Fe²⁺, NH₄⁺ and SO₄²⁻ ions.
Coordination compound: an addition compound that keeps its identity in solution because the complex ion does not split into simple ions. Example: K₄[Fe(CN)₆], which gives K⁺ and [Fe(CN)₆]⁴⁻ ions, and no free Fe²⁺ or CN⁻ ions.
In simple words: A double salt breaks up completely in water; a complex keeps its complex ion.

Teacher's Note:
1. Test the solution: a double salt answers the tests of every simple ion, a complex does not.
2. Alum and Mohr's salt are the usual double salt examples.

Question 2. For the compound [Co(NH₃)₆]Cl₃ name the central metal ion, ligands, coordination number, coordination sphere, counter ions and the oxidation number of the metal.
Answer:
Central metal ion: Co³⁺.
Ligands: six NH₃ molecules (neutral, Lewis bases).
Coordination number: 6.
Coordination sphere (coordination entity): [Co(NH₃)₆]³⁺.
Counter ions (ionisation sphere): three Cl⁻.
Oxidation number of Co: x + 6(0) = +3, so +3.
In simple words: The square bracket holds the metal with its ligands; whatever is outside is a counter ion.

Teacher's Note:
1. The charge on the entity equals the total charge of the counter ions with the opposite sign.
2. Neutral ligands add nothing to the charge.

Question 3. State the main postulates of Werner's theory. How does it explain why only some of the chloride ions in CoCl₃.5NH₃ are precipitated by AgNO₃?
Answer:
1. A metal shows primary valency (oxidation state, ionisable, non-directional) and secondary valency (coordination number, non-ionisable, directional).
2. Every metal tries to satisfy both valencies. Negative ions can satisfy primary valency; negative ions or neutral molecules can satisfy secondary valency.
3. The secondary valencies are directed in space, which fixes the geometry (six: octahedral; four: tetrahedral or square planar).
For CoCl₃.5NH₃ the primary valency of Co is 3 and the secondary valency is 6. Five NH₃ and one Cl⁻ satisfy the secondary valency, so the formula is [Co(NH₃)₅Cl]Cl₂. The Cl⁻ in the sphere is held by a coordinate bond and is not precipitated, while the two outside ions give 2 mol AgCl.
In simple words: Chlorides inside the bracket are bonded to cobalt and do not precipitate; those outside do.

Teacher's Note:
1. Count the chlorides outside the bracket to get the moles of AgCl.
2. The coordination number of Co(III) is always 6 in this series.

Question 4. Write the IUPAC names of (a) [Cu(NH₃)₄]SO₄, (b) K₂[Zn(OH)₄], (c) [Co(NH₃)₄(H₂O)Cl]Cl₂ and (d) Na₃[Co(NO₂)₆].
Answer:
(a) tetraamminecopper(II) sulphate (Cu is +2 because sulphate is −2).
(b) potassium tetrahydroxidozincate(II) (2(+1) + x + 4(−1) = 0, so Zn is +2; anionic complex, so -ate).
(c) tetraammineaquachloridocobalt(III) chloride (ligands in alphabetical order: ammine, aqua, chlorido; x − 1 = +2, so Co is +3).
(d) sodium hexanitrito-N-cobaltate(III) (3(+1) + x + 6(−1) = 0, so Co is +3; NO₂⁻ bound through N).
In simple words: List the ligands alphabetically, then the metal with its Roman numeral; add -ate for an anionic complex.

Teacher's Note:
1. Ignore di, tri, tetra when alphabetising.
2. Find the metal's oxidation number from the charge before writing the name.

Question 5. Write the formulae of (a) diamminechloridonitrito-N-platinum(II), (b) tris(ethane-1,2-diamine)cobalt(III) sulphate, (c) sodium tetrachloridoaurate(III) and (d) hexaaquairon(II) ion.
Answer:
(a) Pt²⁺ + 2 NH₃ + Cl⁻ + NO₂⁻ gives a neutral complex: [Pt(NH₃)₂Cl(NO₂)].
(b) [Co(en)₃]³⁺ and SO₄²⁻ need ratio 2 : 3: [Co(en)₃]2(SO₄)₃.
(c) Au³⁺ + 4 Cl⁻ gives [AuCl₄]⁻, so one Na⁺ is needed: Na[AuCl₄].
(d) [Fe(H₂O)₆]²⁺.
In simple words: Write the metal first, then ligands alphabetically, and balance the charge with counter ions.

Teacher's Note:
1. In formulae, the abbreviation en is put in parentheses.
2. Check that the total charge is zero for the compound.

Question 6. Explain the terms unidentate, didentate, polydentate, chelating and ambidentate ligands with one example of each.
Answer:
Unidentate: one donor atom, e.g. NH₃.
Didentate: two donor atoms, e.g. ethane-1,2-diamine (en) or oxalate.
Polydentate: several donor atoms, e.g. EDTA⁴⁻ (hexadentate).
Chelating ligand: a didentate or polydentate ligand that forms a ring with the metal, e.g. en; the complex is a chelate.
Ambidentate: a ligand with two different donor atoms that can bind through either, e.g. NO₂⁻ (N or O) or SCN⁻ (S or N).
In simple words: The denticity is the number of donor atoms one ligand uses; an ambidentate ligand has a choice of donor atom.

Teacher's Note:
1. Chelating ligands must have at least two donor atoms.
2. Ambidentate ligands cause linkage isomerism.

Question 7. Using valence bond theory, explain why [Co(NH₃)₆]³⁺ is diamagnetic whereas [CoF₆]³⁻ is paramagnetic. State the hybridisation of each.
Study the orbital diagrams below and answer.
Answer:
Co³⁺ is 3d⁶ (four unpaired electrons).
[Co(NH₃)₆]³⁺: NH₃ is a strong field ligand, so the six electrons pair in three 3d orbitals and two 3d orbitals become empty. With 4s and 4p, this gives d²sp³ hybridisation (inner orbital, low spin). No unpaired electrons, so it is diamagnetic.
[CoF₆]³⁻: F⁻ is a weak field ligand and cannot pair the electrons, so outer 4d orbitals are used: sp³d² (outer orbital, high spin). Four unpaired electrons remain, so it is paramagnetic (μ = 4.90 BM).
In simple words: The strong field ligand pairs up the d electrons, but the weak field ligand does not.

Teacher's Note:
1. Inner orbital = (n−1)d orbitals used; outer orbital = nd orbitals used.
2. Count the unpaired electrons to decide the magnetic behaviour.

Question 8. Predict the shape, hybridisation and magnetic behaviour of [Ni(CN)₄]²⁻ and [NiCl₄]²⁻ from the diagrams.
The diagrams show the orbital filling in the two complexes.
Answer:
[Ni(CN)₄]²⁻: CN⁻ is strong field, so the 3d⁸ electrons of Ni²⁺ pair up and one 3d orbital is freed. dsp², square planar, diamagnetic.
[NiCl₄]²⁻: Cl⁻ is weak field and leaves the two unpaired electrons as they are. sp³, tetrahedral, paramagnetic (n = 2, μ = 2.83 BM).
In simple words: Strong field ligand: square planar and no unpaired electron; weak field ligand: tetrahedral with two unpaired electrons.

Teacher's Note:
1. Both complexes have the same metal ion, Ni²⁺; only the ligand differs.
2. Compare the number of empty 3d orbitals.

Question 9. Calculate the spin-only magnetic moments of (a) [Fe(H₂O)₆]²⁺ (high spin), (b) [Fe(CN)₆]⁴⁻, (c) [Cr(H₂O)₆]³⁺ and (d) [Cu(NH₃)₄]²⁺.
Answer:
Use μ = \( \sqrt{n(n+2)} \) BM.
(a) Fe²⁺ is d⁶, high spin t₂g⁴ e_g²: n = 4, μ = \( \sqrt{24} \) = 4.90 BM.
(b) Fe²⁺ is d⁶, low spin t₂g⁶: n = 0, μ = 0 BM (diamagnetic).
(c) Cr³⁺ is d³: n = 3, μ = \( \sqrt{15} \) = 3.87 BM.
(d) Cu²⁺ is d⁹: n = 1, μ = \( \sqrt{3} \) = 1.73 BM.
In simple words: Count the unpaired electrons, then substitute into the square-root formula.

Teacher's Note:
1. Decide high or low spin from the ligand first.
2. d³ and d⁹ ions have a fixed number of unpaired electrons.

Question 10. Explain crystal field splitting in an octahedral complex. How does the strength of the ligand decide whether a d⁴ ion is high spin or low spin?
The two diagrams show a d⁴ ion in a weak field and in a strong field.
Answer:
The ligands approach along the axes. The orbitals d(x²−y²) and d(z²) point at them and are raised (e_g); dxy, dyz and dzx point between the ligands and are lowered (t₂g). The gap is Δ₀: e_g is raised by 0.6 Δ₀ and t₂g is lowered by 0.4 Δ₀.
For d⁴ the fourth electron has a choice. If Δ₀ < P (weak field ligand) it enters e_g: t₂g³ e_g¹, high spin, four unpaired electrons.
If Δ₀ > P (strong field ligand) it pairs in t₂g: t₂g⁴ e_g⁰, low spin, two unpaired electrons.
In simple words: A weak field ligand gives a small gap, so electrons spread out; a strong field ligand gives a large gap, so they pair.

Teacher's Note:
1. P is the energy needed to pair two electrons in one orbital.
2. The two diagrams show weak and strong field arrangements.

Question 11. Why are tetrahedral complexes mostly high spin? How does Δt compare with Δ₀?
Refer to the splitting diagram below.
Answer:
In a tetrahedral field only four ligands are present, and none points directly at the d orbitals. So the splitting is small: Δt = 4/9 Δ₀ for the same metal and ligands.
The e set (lower) and t₂ set (higher) are separated by Δt, which is usually smaller than the pairing energy P. The electrons therefore occupy the orbitals singly, giving high spin complexes.
In simple words: A small splitting never beats the cost of pairing electrons.

Teacher's Note:
1. Low spin tetrahedral complexes are very rare.
2. Note the inverted order compared with the octahedral case.

Question 12. (a) Why is [Ti(H₂O)₆]³⁺ purple? (b) [Co(NH₃)₆]³⁺ absorbs light of wavelength 475 nm (blue). What colour does it appear?
Answer:
(a) Ti³⁺ is d¹. Its one electron is in t₂g. The complex absorbs blue-green light (about 498 nm) to promote the electron to e_g, and the transmitted light is purple.
(b) The complex absorbs blue light, so the colour seen is the complement, yellow-orange.
A different ligand changes Δ₀, so the wavelength absorbed and the colour change.
In simple words: The colour you see is what remains after the complex absorbs part of white light.

Teacher's Note:
1. Complementary pairs: blue and orange, red and green, yellow and violet.
2. d⁰ and d¹⁰ ions are colourless.

Question 13. Name and explain, with one example each, the four kinds of structural isomerism in coordination compounds.
Answer:
Ionisation isomerism: a ligand and the counter ion exchange places. [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄.
Solvate (hydrate) isomerism: water is inside or outside the sphere. [Cr(H₂O)₆]Cl₃ and [CrCl(H₂O)₅]Cl₂.H₂O.
Linkage isomerism: an ambidentate ligand binds through different atoms. [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂.
Coordination isomerism: ligands are exchanged between a cationic and an anionic complex. [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆].
In simple words: Same formula, different ligands inside the coordination sphere.

Teacher's Note:
1. Ionisation isomers give different ions in solution.
2. Linkage isomerism needs an ambidentate ligand.

Question 14. Draw the stereoisomers of [Co(en)₂Cl₂]⁺. Which of them are optically active, and why?
The figure shows the two cis enantiomers.
Answer:
Three stereoisomers exist: trans and the two cis forms. In the figures, the two cis structures are the enantiomers drawn with each curved line joining two N atoms of one en.
The trans isomer has a plane of symmetry, so it is optically inactive.
The cis isomer has no plane or centre of symmetry, so its mirror image cannot be superimposed on it. The two cis forms are optically active (d and l).
In simple words: Only the cis form is chiral, so it comes in a left-handed and a right-handed version.

Teacher's Note:
1. A complex with a plane of symmetry is optically inactive.
2. Count d and l of cis as two separate isomers.