CBSE Class 12 Chemistry Coordination Compunds Questions Answers

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These Class 12 Chemistry questions and answers on Coordination Compounds cover 1, 2, 3 and 5 mark questions on ligands, coordination number, nomenclature, isomerism, valence bond theory, crystal field theory, magnetic behaviour, colour and stability of complexes, each with a verified answer and a Teacher's Note.

COORDINATION COMPOUNDS

1 mark questions

1. Explain coordination entity with example. [1 Mark]

Answer: A coordination entity consists of a central metal atom or ion bonded to a fixed number of ions or molecules called ligands, for example [Co(NH₃)₃Cl₃] or [Fe(CN)₆]^4-.

Teacher's Note:
a) Remember both parts: the central metal and the fixed number of ligands bonded to it.

2. What do you understand by coordination compounds? [1 Mark]

Answer: Coordination compounds are compounds that contain complex ions or neutral coordination entities, in which a central metal atom or ion is bonded to a fixed number of anions or molecules (ligands) through coordinate bonds, for example [Co(NH₃)₆]Cl₃ and K₄[Fe(CN)₆].

Teacher's Note:
a) A coordination compound may be a neutral complex, or a complex ion together with its counter ions.

3. What is coordination number? [1 Mark]

Answer: The coordination number of a metal ion in a complex is the total number of ligand donor atoms to which the metal ion is directly bonded; for example, it is 6 in [Co(NH₃)₆]³⁺.

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

4. Name the different types of isomerisms in coordination compounds. [1 Mark]

Answer: The two main types are structural isomerism (ionisation, linkage, coordination and solvate isomerism) and stereoisomerism (geometrical and optical isomerism).

Teacher's Note:
a) Stereoisomers have the same bonds but a different arrangement in space; structural isomers have different bonds.

5. Draw the structure of xenon difluoride. [1 Mark]

Answer: In XeF₂ the central Xe atom is sp³d hybridised, so its electron pair geometry is trigonal bipyramidal. The three lone pairs occupy the equatorial positions and the two F atoms occupy the axial positions, so the shape of the molecule is linear (F-Xe-F, 180°).

Teacher's Note:
a) The lone pairs sit in the equatorial plane because that arrangement has the least repulsion; the molecular shape is linear, not trigonal bipyramidal.

6. What is spectrochemical series? [1 Mark]

Answer: The series in which ligands are arranged in the order of increasing field strength (increasing crystal field splitting) is called the spectrochemical series. The order is: I^- < Br^- < SCN^- < Cl^- < S^2- < F^- < OH^- < C₂O₄^2- < H₂O < NCS^- < EDTA^4- < NH₃ < en < CN^- < CO.

Teacher's Note:
a) Weak field ligands such as I^- and Br^- give high spin complexes, and strong field ligands such as CN^- and CO give low spin complexes.

7. What do you understand by denticity of a ligand? [1 Mark]

Answer: The number of coordinating (donor) groups present in a ligand is called its denticity. For example, ethane-1,2-diamine is bidentate because it has 2 donor nitrogen atoms which can link to the central metal atom.

Teacher's Note:
a) Denticity is the number of donor atoms that actually bind the metal, so EDTA is hexadentate.

8. Why is CO a stronger ligand than Cl-? [1 Mark]

Answer: CO forms a sigma bond by donating a lone pair of carbon and also accepts electron density from filled metal d orbitals into its empty pi* orbitals (back bonding), which produces a large splitting, whereas Cl^- is only a weak field sigma and pi donor ligand.

Teacher's Note:
a) Synergic bonding (sigma donation plus pi back donation) is the key phrase to write.

9. Why are low spin tetrahedral complexes not formed? [1 Mark]

Answer: In a tetrahedral complex the crystal field splitting is small, Δ_t = \( \frac{4}{9} \) Δ_o, so it is always smaller than the pairing energy and electrons do not pair up.

Teacher's Note:
a) Because Δ_t is only about 44% of Δ_o, tetrahedral complexes are always high spin.

10. Square planar complexes with coordination number 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why? [1 Mark]

Answer: In a tetrahedral complex all four positions are equivalent, as each ligand has the same relative position with respect to every other ligand, so cis and trans arrangements are not possible. In a square planar complex the ligands can be cis or trans to each other.

Teacher's Note:
a) Geometrical isomerism needs positions that are not equivalent, such as cis and trans positions.

11. What are crystal fields? [1 Mark]

Answer: The electrostatic field produced by the ligands (treated as negative point charges or dipoles) around the central metal ion is called the crystal field.

Teacher's Note:
a) Crystal field theory treats the metal-ligand bond as purely electrostatic.

12. What is meant by chelate effect? Give an example . [1 Mark]

Answer: When a didentate or polydentate ligand binds the central metal atom through two or more donor atoms and forms a five or six membered ring, the complex is more stable than a similar complex with monodentate ligands; this is called the chelate effect. For example, [PtCl₂(en)] or [Co(en)₃]³⁺.

Teacher's Note:
a) The ring must be 5 or 6 membered for the effect to be large.

13. What do you understand by ambidentate ligand? [1 Mark]

Answer: An ambidentate ligand has two donor atoms, but only one of them forms a coordinate bond with the central metal atom or ion at a time, for example NO₂^- (nitrito-N or nitrito-O) and SCN^- (thiocyanato-S or isothiocyanato-N).

Teacher's Note:
a) Ambidentate ligands lead to linkage isomerism.

14. What is the difference between homoleptic and heteroleptic complexes? [1 Mark]

Answer: In homoleptic complexes the central metal atom is bound to only one kind of donor group, for example [Co(NH₃)₆]³⁺, whereas in heteroleptic complexes it is bound to more than one kind of donor group, for example [Co(NH₃)₄Cl₂]⁺.

Teacher's Note:
a) Compare the kinds of ligand, not the number of ligands.

15. Give one limitation for crystal field theory. [1 Mark]

Answer: The ligands are treated as point charges, so anionic ligands should exert a greater splitting effect than neutral ligands; however, anionic ligands lie at the low end of the spectrochemical series.

Teacher's Note:
a) Another limitation that you could use is that the theory ignores the covalent character of the metal-ligand bond; give only one when asked for one.

16. How many ions are produced from the complex: [Co(NH₃)₆]Cl₂ [1 Mark]

Answer: 3 ions, one [Co(NH₃)₆]²⁺ ion and two Cl^- ions.

Teacher's Note:
a) Ligands inside the square bracket do not ionise; only the counter ions outside do.

17. The oxidation number of cobalt in K[Co(CO)₄] [1 Mark]

Answer: -1

Teacher's Note:
a) K is +1 and CO is neutral, so the oxidation number of cobalt balances the charge: x + 0 = -1.

18. Which compound is used to estimate the hardness of water volumetrically? [1 Mark]

Answer: EDTA (ethylenediaminetetraacetic acid, as its sodium salt), which forms stable chelates with Ca²⁺ and Mg²⁺ ions.

Teacher's Note:
a) EDTA titration is the standard method for total hardness.

19. Magnetic moment of [MnCl₄]^2- is 5.92B.M explain with reason. [1 Mark]

Answer: In [MnCl₄]^2- the oxidation state of Mn is +2, so Mn²⁺ is d⁵. Cl^- is a weak field ligand, so the 5 electrons stay unpaired and the magnetic moment is \( \sqrt{5(5+2)} \) = 5.92 B.M. The hybridisation is sp³ (tetrahedral), not dsp².

Teacher's Note:
a) 5.92 B.M corresponds to five unpaired electrons, the maximum for d⁵.

20. How many donor atoms are present in EDTA ligand? [1 Mark]

Answer: 6 (two nitrogen atoms and four oxygen atoms of the carboxylate groups).

Teacher's Note:
a) EDTA is a hexadentate ligand.

2 marks questions

1. Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.
i) [CoF₆]^3-
ii) [Fe(CN)₆]^4- [2 Marks]

Answer:
i) Co³⁺ is d⁶ and F^- is a weak field ligand, so the configuration is high spin: t_2g⁴e_g² (four unpaired electrons).
ii) Fe²⁺ is d⁶ and CN^- is a strong field ligand, so the configuration is low spin: t_2g⁶e_g⁰ (no unpaired electrons).

Teacher's Note:
a) Compare Δ_o with the pairing energy: a weak ligand gives high spin and a strong ligand gives low spin.
b) In [CoF₆]^3- the charge on the complex is 3-, so cobalt is +3.

2. Explain the following with examples:
i) Linkage isomerism
ii) Outer orbital complex [2 Marks]

Answer:
i) Linkage isomerism arises when an ambidentate ligand is bonded through different donor atoms in a coordination compound. For example, [Co(NH₃)₅NO₂]Cl₂ (nitro, bonded through N) and [Co(NH₃)₅ONO]Cl₂ (nitrito, bonded through O).
ii) When the outer ns, np and nd orbitals are involved in hybridisation, an outer orbital complex is formed. For example, [CoF₆]^3- in which cobalt is sp³d² hybridised (high spin).

Teacher's Note:
a) The printed example [CoF₆]^2- should be [CoF₆]^3-, since cobalt is in the +3 state.
b) Outer orbital complexes use 4d orbitals and are high spin; inner orbital complexes use 3d orbitals.

3. i)Low spin octahedral complexes of nickel are not found . Explain why?
ii)the π complexes are known for transition elements only.explain. [2 Marks]

Answer:
i) Ni²⁺ is d⁸, so even after pairing only one 3d orbital is vacant, and d²sp³ hybridisation needs two vacant 3d orbitals. Hence nickel forms only outer orbital (sp³d²) octahedral complexes.
ii) Transition metals have vacant d orbitals that accept the pi electrons of ligands such as benzene and ethene, and filled d orbitals that can back donate, so d pi - p pi bonding is possible.

Teacher's Note:
a) Count the vacant 3d orbitals after pairing to decide the hybridisation.
b) Pi complexes involve metal d orbitals; main group elements lack suitable d orbitals.

4. How would you account for the following:
i) [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless.
ii) [Ni(CO)₄] possess tetrahedral geometry whereas [Ni(CN)₄]^2- is square planar. [2 Marks]

Answer:
i) Ti³⁺ has one electron in the 3d subshell, so d-d transition takes place by absorption of visible light and the complex appears coloured. Sc³⁺ is d⁰, so no d-d transition is possible and it is colourless.
ii) In [Ni(CO)₄], nickel is in the zero oxidation state (3d⁸4s² becomes 3d¹⁰), so it is sp³ hybridised and tetrahedral. In [Ni(CN)₄]^2-, Ni²⁺ is d⁸ and CN^- pairs the electrons, so it is dsp² hybridised and square planar.

Teacher's Note:
a) Colour of a transition metal complex needs at least one d electron and an empty d orbital to jump to.
b) Both nickel complexes are diamagnetic even though their shapes differ.

5. State reasons for each of the following:
i) All the P-Cl bonds in PCl₅ molecule are not equivalent.
ii) S has greater tendency for catenation than O. [2 Marks]

Answer:
i) PCl₅ is trigonal bipyramidal, with 3 equatorial and 2 axial P-Cl bonds. The axial bond pairs suffer more repulsion than the equatorial bond pairs, so the 2 axial bonds are longer than the 3 equatorial bonds.
ii) Catenation depends on bond strength. The S-S bond (213 kJ/mol) is much stronger than the O-O bond (138 kJ/mol), so S has a greater tendency for catenation than O.

Teacher's Note:
a) Axial bonds make 90° angles with three equatorial bonds, whereas equatorial bonds make 90° with only two axial bonds.
b) The O-O bond is weak because of repulsion between the lone pairs on the small oxygen atoms.

6. Give the stereochemistry and the magnetic behaviour of the following complexes:
i) [Co(NH₃)₅Cl]Cl₂
ii) K₂[Ni(CN)₄] [2 Marks]

Answer:
i) Co³⁺ is d⁶ and NH₃ is a strong field ligand, so the hybridisation is d²sp³ and the shape is octahedral. All electrons are paired, so it is diamagnetic.
ii) Ni²⁺ is d⁸ and CN^- is a strong field ligand, so the hybridisation is dsp² and the shape is square planar. All electrons are paired, so it is diamagnetic.

Teacher's Note:
a) Write the hybridisation first and then deduce the shape and magnetic behaviour from it.
b) Magnetic behaviour depends on the number of unpaired electrons: none means diamagnetic.

7. Draw the structures of isomers if any and write the names of the following complexes:
i) [Cr(NH₃)₄Cl₂]⁺
ii) [Co(en)₃]³⁺ [2 Marks]

Answer:
i) tetraamminedichloridochromium(III) ion. It has two geometrical isomers, cis and trans (shown below).
ii) tris(ethane-1,2-diamine)cobalt(III) ion. It has two optical isomers (non-superimposable mirror images, d and l), shown below.

Teacher's Note:
a) The printed name "tris(ethane-1,2-diammine)" should be written tris(ethane-1,2-diamine).
b) The cis form of [Cr(NH₃)₄Cl₂]⁺ is not optically active because it has a plane of symmetry.

8. State reasons for each of the following:
i) The N-O bond in NO₂^- is shorter than the N-O bond in NO₃^-
ii) SF₆ is kinetically an inert substance. [2 Marks]

Answer:
i) In NO₂^- the N-O bond is an average of a single and a double bond (bond order 1.5), whereas in NO₃^- it is an average of two single bonds and one double bond (bond order 1.33). The higher bond order makes the N-O bond in NO₂^- shorter.
ii) In SF₆ the S atom is sterically protected by six F atoms, which do not allow water molecules to attack the S atom, and the S-F bonds are strong. Hence SF₆ is kinetically inert.

Teacher's Note:
a) Bond length decreases as bond order increases.
b) Kinetic inertness (slow reaction) is different from thermodynamic stability.

9. Hydrated copper sulphate is blue in colour whereas anhydrous copper sulphate is colourless. Why? [2 Marks]

Answer:
In hydrated copper sulphate, water molecules act as ligands and split the d orbitals of Cu²⁺ (d⁹). The d-d transition t_2g⁶e_g³ to t_2g⁵e_g⁴ absorbs light in the red-orange region, so the crystal appears blue.
In anhydrous copper sulphate there are no ligands to split the d orbitals, so no d-d transition occurs and it is colourless.

Teacher's Note:
a) The colour seen is the complementary colour of the light absorbed.
b) Heating the blue crystals removes the water ligands, which is why the colour disappears.

10. Calculate the magnetic moment of the metal ions present in the following complexes:
i) [Cu(NH₃)₄]SO₄
ii) [Ni(CN)₄]^2- [2 Marks]

Answer:
i) In [Cu(NH₃)₄]SO₄, Cu²⁺ is d⁹ (t_2g⁶e_g³), so there is n = 1 unpaired electron. μ = \( \sqrt{n(n+2)} \) = \( \sqrt{3} \) = 1.73 B.M.
ii) In [Ni(CN)₄]^2-, Ni²⁺ is d⁸ and CN^- is a strong field ligand, so the electrons pair up (square planar, dsp²) and n = 0. μ = 0 B.M (diamagnetic).

Teacher's Note:
a) The printed key gives n = 2 and 2.828 B.M for [Ni(CN)₄]^2-; that is wrong because the complex is square planar and diamagnetic, as the same key states elsewhere.
b) Use μ = \( \sqrt{n(n+2)} \) B.M with n as the number of unpaired electrons.

3 marks questions

1. (a) What is a ligand? Give an example of a bidentate ligand.
(b) explain as to how the 2 complexes of nickel,[Ni(CN)₄]^2- and Ni(CO)₄ have different structures but donot differ in their magnetic behaviour.( Ni=28) [3 Marks]

Answer:
(a) The ion, atom or molecule bound to the central atom or ion in a coordination entity is called a ligand. It has a lone pair of electrons in its valence orbital which it donates to the central metal. An example of a bidentate ligand is ethane-1,2-diamine (en) or the oxalate ion.
(b) In [Ni(CN)₄]^2-, Ni²⁺ is d⁸ and CN^- pairs the electrons: dsp² hybridisation, square planar, n = 0, diamagnetic.
In Ni(CO)₄, nickel is Ni(0) with 3d⁸4s² rearranged to 3d¹⁰ by the strong field CO: sp³ hybridisation, tetrahedral, n = 0, diamagnetic.

Teacher's Note:
a) A ligand must have at least one lone pair to donate.
b) Both complexes have no unpaired electrons, so the magnetic behaviour is the same although the hybridisation and shape differ.
c) Atomic number 28 gives the configuration 3d⁸4s² for Ni.

2. Nomenclate the following complexes:
i) [Co(NH₃)₅(CO₃)]Cl
ii)[CoCl₂(en)₂]Cl
iii) Fe₄[Fe(CN)₆]₃ [3 Marks]

Answer:
i) pentaamminecarbonatocobalt(III) chloride
ii) dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
iii) iron(III) hexacyanidoferrate(II)

Teacher's Note:
a) Ligands are named in alphabetical order, ignoring prefixes such as di and bis.
b) In an anionic complex the metal ends in -ate, so the iron inside the bracket is ferrate(II).
c) The printed formula Fe₄[Fe(CN)₆] lacks the subscript 3; four Fe³⁺ ions balance three [Fe(CN)₆]^4- ions.

3. (a)why do compounds with similar geometry have different magnetic moment?
(b)what is the relationship between the observed colour and wavelength of light absorbed by the complex? [3 Marks]

Answer:
(a) It is due to the presence of weak and strong field ligands. A strong ligand gives large CFSE (low spin, few unpaired electrons, low magnetic moment) and a weak ligand gives small CFSE (high spin, high magnetic moment). For example, [CoF₆]^3- is paramagnetic and [Co(NH₃)₆]³⁺ is diamagnetic, though both are octahedral.
(b) The higher the CFSE, the shorter is the wavelength of the light absorbed. The colour of the complex is the complementary colour of the light absorbed, that is the colour of the light left over.

Teacher's Note:
a) Magnetic moment depends on the number of unpaired electrons, which depends on Δ_o versus pairing energy.
b) A large splitting means absorption at a shorter wavelength.
c) The observed colour is complementary to the absorbed colour.

4. Explain the following terms giving a suitable example.
(a) ambident ligand
(b) denticity of a ligand
(c) crystal field splitting in an octahedral field [3 Marks]

Answer:
(a) An ambidentate ligand has two donor atoms, but only one of them forms a coordinate bond at a time with the central metal atom or ion. Example: nitrito-N and nitrito-O (NO₂^-).
(b) The number of coordinating groups present in a ligand is called its denticity. Example: the bidentate ligand ethane-1,2-diamine has 2 donor nitrogen atoms which can link to the central metal atom.
(c) In an octahedral crystal field the five degenerate d orbitals split into three orbitals of lower energy (t_2g) and two orbitals of higher energy (e_g); the energy gap is Δ_o.

Teacher's Note:
a) The spelling is ambidentate; the question prints ambident.
b) Splitting is measured by Δ_o, and the t_2g orbitals lie at -0.4 Δ_o and the e_g orbitals at +0.6 Δ_o.
c) Give an example with each term to earn the full marks.

5. (a) Copper sulphate pentahydrate is blue in colour while anhydrous copper sulphate is colourless. Why?
(b) Sulphur has greater tendency for catenation than oxygen.Why? [3 Marks]

Answer:
(a) In the pentahydrate, water molecules act as ligands and split the d orbitals of Cu²⁺. A d-d transition (t_2g⁶e_g³ to t_2g⁵e_g⁴) then absorbs part of the visible light and imparts the blue colour. Anhydrous copper sulphate has no ligands, so there is no splitting and no colour.
(b) Catenation depends on bond strength. The S-S bond (213 kJ/mol) is much stronger than the O-O bond (138 kJ/mol), so sulphur has a greater tendency for catenation.

Teacher's Note:
a) No ligand means no d orbital splitting, so no d-d transition.
b) The O-O single bond is weak due to lone pair repulsion between small oxygen atoms.
c) Use the bond enthalpy values to support the reason.

6. draw structures of geometrical isomers of the following complexes:
(a) [Fe(NH₃)₂(CN)₄]^- (b)[CrCl₂(ox)₂]^3- (c)[Co(en)₃]Cl₃ [3 Marks]

Answer:
(a) [Fe(NH₃)₂(CN)₄]^- has two geometrical isomers: trans (the two NH₃ opposite to each other) and cis (the two NH₃ adjacent).
(b) [CrCl₂(ox)₂]^3- has two geometrical isomers: trans (the two Cl opposite) and cis (the two Cl adjacent; the cis form also has a mirror-image pair).
(c) [Co(en)₃]Cl₃ has no geometrical isomers, because all three bidentate ligands are identical; it shows only optical isomerism (two enantiomers).

Teacher's Note:
a) Octahedral M(AA)₃ complexes show optical isomerism but no geometrical isomerism.
b) The cis isomer of [CrCl₂(ox)₂]^3- is chiral, but the trans isomer is not.
c) Draw cis and trans with the ligand positions clearly marked.

7. write the state of hybridisation, the shape and the magnetic behaviour of the following complexes:
(i) [Co(en)₃]Cl₃
(II) K₂[Ni(CN)₄]
(III)[Fe(CN)₆]^3- [3 Marks]

Answer:
(i) [Co(en)₃]Cl₃: Co³⁺ is d⁶ and en is a strong ligand, so d²sp³ hybridisation, octahedral shape, diamagnetic.
(ii) K₂[Ni(CN)₄]: Ni²⁺ is d⁸ and CN^- is strong, so dsp² hybridisation, square planar shape, diamagnetic.
(iii) [Fe(CN)₆]^3-: Fe³⁺ is d⁵ and CN^- is strong, giving t_2g⁵, d²sp³ hybridisation, octahedral shape, paramagnetic with one unpaired electron (1.73 B.M).

Teacher's Note:
a) Find the oxidation state first, then the d configuration, then the field strength of the ligand.
b) Low spin d⁵ has one unpaired electron, so it is weakly paramagnetic.
c) Use the table of strong and weak ligands from the spectrochemical series.

8. how would you account for the following:
(i) [Ti(H₂O)₆]³⁺is coloured while [Sc(H₂O)₆]³⁺ is colourless .
(II) [ Fe(CN)₆]^3- is weakly paramagnetic while [ Fe(CN)₆]^4- is diamagnetic.
(III) Ni(CO)₄ possess tetrahedral geometry while [Ni (CN)₄]^2- is square planar. [3 Marks]

Answer:
(i) Ti³⁺ has one electron in the 3d subshell, so d-d transition takes place by absorption of visible light and the complex is coloured. Sc³⁺ has no d electron, so no d-d transition is possible and it is colourless.
(ii) Paramagnetism is due to unpaired electrons. In [Fe(CN)₆]^3-, Fe³⁺ is d⁵ (low spin t_2g⁵) with one unpaired electron, so it is weakly paramagnetic. In [Fe(CN)₆]^4-, Fe²⁺ is d⁶ (t_2g⁶) and all electrons are paired, so it is diamagnetic.
(iii) In Ni(CO)₄, Ni is sp³ hybridised and tetrahedral. In [Ni(CN)₄]^2-, Ni is dsp² hybridised and square planar.

Teacher's Note:
a) Colour needs a partly filled d subshell; d⁰ and d¹⁰ ions are colourless.
b) The number of unpaired electrons decides the extent of paramagnetism.
c) Ni(0) in the carbonyl has no vacant 3d orbital, hence sp³.

9. Explain the following ::
(i) low spin octahedral complexes of Ni are not known.
(ii) The pi - complexes are known for the transition elements only.
(iii) CO is a stronger ligand than NH₃ for many metals [3 Marks]

Answer:
(i) Ni²⁺ (d⁸) cannot provide two vacant 3d orbitals, so d²sp³ hybridisation is not possible.
(ii) Transition metals have vacant d orbitals in their atoms or ions into which the pi electrons of ligands such as benzene and ethene can be donated, so d pi - p pi bonding is possible.
(iii) In CO, back bonding takes place: the central atom uses its filled d orbitals to donate electrons into the empty antibonding pi* molecular orbital of CO, which strengthens the metal-ligand bond.

Teacher's Note:
a) Three parts, three points: one reason for each statement.
b) Back bonding is also called synergic bonding because sigma donation and pi back donation strengthen each other.
c) NH₃ is only a sigma donor.

10. What is meant by stability of a coordination compounds in solutions? State the factors which govern the stability of complexes. [3 Marks]

Answer:
The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. For M + 4L ⇌ ML₄, the stability constant is K = [ML₄]/([M][L]⁴); the larger the K, the more stable the complex.
Factors on which stability depends: (i) charge on the central metal ion (more charge, more stable), (ii) nature of the metal ion, (iii) basic nature of the ligand (more basic, more stable).
(iv) presence of a chelate ring, and (v) the effect of multidentate cyclic (macrocyclic) ligands, which give extra stability.

Teacher's Note:
a) The stability constant is the equilibrium constant for the formation of the complex.
b) Chelation increases stability markedly compared with monodentate ligands.
c) Do not confuse thermodynamic stability with kinetic inertness.

5 marks questions

1. Draw the structures of the following molecules:
(a) [Fe(NH₃)₂(CN)₄]^- (b)[CrCl₂(ox)₂]^3- (c)[Co(en)₃]Cl₃(d) [Co(en)₃]Cl₃(e)[Fe(CN)₆]^3- [5 Marks]

Answer:
(a) [Fe(NH₃)₂(CN)₄]^- is octahedral, with Fe³⁺ at the centre; the two NH₃ can be trans (opposite) or cis (adjacent), the other four positions being CN^-.
(b) [CrCl₂(ox)₂]^3- is octahedral, with two bidentate oxalate ions forming five membered rings and two Cl^- either trans or cis.
(c) and (d) [Co(en)₃]³⁺ is octahedral, with three bidentate en ligands each forming a five membered ring; (d) is the same formula as (c) in the printed question, so the figure also shows its two optical isomers, d and l.
(e) [Fe(CN)₆]^3- is a regular octahedron with Fe³⁺ at the centre and six CN^- ligands bonded through carbon.

Teacher's Note:
a) Show the central metal, every ligand position and the chelate rings clearly.
b) Mark chelating ligands with an arc joining their two donor atoms.
c) The question repeats [Co(en)₃]Cl₃ as parts (c) and (d); one structure covers both, with the mirror image shown for optical isomerism.
d) The coordination number is 6 in all these structures.

2. What is crystal field theory for octahedral complexes? Also write the limitations of this theory. [5 Marks]

Answer:
Crystal field theory (CFT) assumes that the metal-ligand bond is purely electrostatic: the ligands are point charges (or dipoles) and the central metal ion has five degenerate d orbitals.
In an octahedral field the ligands approach along the x, y and z axes. The d_x2-y2 and d_z2 orbitals point towards the ligands and are raised in energy (e_g, +0.6 Δ_o); the d_xy, d_yz and d_zx orbitals point between the ligands and are lowered (t_2g, -0.4 Δ_o). The gap between the two sets is Δ_o.
If Δ_o is less than the pairing energy, electrons occupy e_g before pairing (weak field, high spin); if Δ_o is greater, they pair in t_2g first (strong field, low spin). The d-d transitions between t_2g and e_g give the colour.
Limitations: (i) ligands are treated as point charges, so anionic ligands should split more than neutral ones, but they lie at the low end of the spectrochemical series; (ii) the covalent character of the metal-ligand bond is ignored; (iii) pi bonding is not explained.

Teacher's Note:
a) Include the splitting diagram and the CFSE values to earn the diagram marks.
b) State clearly when a complex is high spin (Δ_o less than P) and low spin (Δ_o greater than P).
c) Colour arises from d-d transitions, absorbing light of energy equal to Δ_o.
d) Do not forget at least two limitations of the theory.

3. Write the state of hybridisation the shape and the magnetic behaviour of the following complex entities:
(i) [Cr(NH₃)₄Cl₂]Cl
(ii) [Co(en)₃]Cl₃
(iii) K₂[NiCl₄]
(iv) [Fe(H₂O)₆]²⁺
(v) [NiCl₄]^2- [5 Marks]

Answer:
(i) [Cr(NH₃)₄Cl₂]Cl: Cr³⁺ is d³, d²sp³ hybridisation, octahedral, paramagnetic (3 unpaired electrons).
(ii) [Co(en)₃]Cl₃: Co³⁺ is d⁶, en is strong, d²sp³ hybridisation, octahedral, diamagnetic.
(iii) K₂[NiCl₄]: Ni²⁺ is d⁸, Cl^- is weak, sp³ hybridisation, tetrahedral, paramagnetic (2 unpaired electrons).
(iv) [Fe(H₂O)₆]²⁺: Fe²⁺ is d⁶, H₂O is weak, sp³d² hybridisation, octahedral, paramagnetic (4 unpaired electrons).
(v) [NiCl₄]^2-: the same as (iii): sp³ hybridisation, tetrahedral, paramagnetic (2 unpaired electrons).

Teacher's Note:
a) Find the oxidation state, then the d configuration, then decide using the ligand field strength.
b) With three d electrons, d²sp³ leaves them unpaired in t_2g.
c) Parts (iii) and (v) are the same complex ion; K₂[NiCl₄] contains [NiCl₄]^2-.
d) Tetrahedral nickel(II) with Cl^- has two unpaired electrons, giving μ = 2.83 B.M.

4. Using valence bond theory explain the following questions in relation to [Co(NH₃)₆]³⁺.
(i) Nomenclature
(ii) Type of hybridisation
(iii) Inner or outer orbital complex
(iv) Magnetic behaviour
(v) Spin only magnetic moment [5 Marks]

Answer:
(i) Nomenclature: hexaamminecobalt(III) ion.
(ii) Co³⁺ is 3d⁶; NH₃ is a strong field ligand and pairs the electrons, leaving two 3d orbitals vacant, so the hybridisation is d²sp³ (octahedral).
(iii) It uses the inner 3d orbitals, so it is an inner orbital (low spin) complex.
(iv) All electrons are paired, so it is diamagnetic.
(v) n = 0, so μ = \( \sqrt{n(n+2)} \) = 0 B.M.

Teacher's Note:
a) The d² in d²sp³ comes from the 3d orbitals, which is why the complex is inner orbital.
b) Zero unpaired electrons means diamagnetic and a spin only moment of zero.
c) Check the oxidation state: 3 charges on the ion with neutral NH₃ gives Co(III).
d) Write the electronic configuration of Co³⁺ before the hybridisation.

5. Compare the following complexes with respect to structural shape of units, magnetic behaviour and hybrid orbitals involved in units:
[Co(NH₃)₆]³⁺, [Cr(NH₃)₆]³⁺,[ Ni(CO)₄] [5 Marks]

Answer:
[Co(NH₃)₆]³⁺: Co³⁺ is d⁶ (low spin), hybrid orbitals d²sp³, shape octahedral, diamagnetic (μ = 0).
[Cr(NH₃)₆]³⁺: Cr³⁺ is d³, hybrid orbitals d²sp³, shape octahedral, paramagnetic (3 unpaired electrons, μ = 3.87 B.M).
[Ni(CO)₄]: Ni(0) is 3d¹⁰, hybrid orbitals sp³, shape tetrahedral, diamagnetic (μ = 0).
The first two are octahedral inner orbital complexes, but they differ in magnetic behaviour; the third is tetrahedral with a different hybridisation.

Teacher's Note:
a) Present the comparison as a table of shape, hybridisation and magnetic behaviour.
b) Both octahedral complexes use d²sp³, but only the chromium complex has unpaired electrons.
c) Ni(CO)₄ is diamagnetic because Ni(0) has a completely filled 3d subshell.
d) Spin only moment \( \sqrt{3(3+2)} \) = 3.87 B.M for Cr³⁺.

ONE MARK QUESTIONS:

Q.1 Write the formula forTetraamineaquachloridocobalt(III) chloride [1 Mark]

Answer: [Co(NH₃)₄(H₂O)Cl]Cl₂

Teacher's Note:
a) Cobalt is +3, the complex ion has charge +2, so two Cl^- ions lie outside the bracket.

Q.2 Write the IUPAC name of [Co (NH₃)₄ Br₂]₂ [Zn Cl₄] [1 Mark]

Answer: bis[tetraamminedibromidocobalt(III)] tetrachloridozincate(II)

Teacher's Note:
a) The cation has charge +1 (Co³⁺ with two Br^-), so two cations balance the [ZnCl₄]^2- anion; use bis for the complex cation.

Q.3 Which of these cannot act as ligand and why: NH₃, H₂O, CO, CH₄. Give reason? [1 Mark]

Answer: CH₄ cannot act as a ligand because it has no lone pair of electrons to donate to the metal.

Teacher's Note:
a) A ligand needs at least one lone pair (or pi electrons) to donate.

Q.4 How many EDTA (lethylendiamine tetra acetic acid) molecules are required to make an octahedral complex with a Ca²⁺ ion. [1 Mark]

Answer: One. EDTA is a hexadentate ligand, so a single EDTA molecule fills all 6 octahedral positions.

Teacher's Note:
a) Divide the coordination number 6 by the denticity 6.

Q.5 Why tetrahedral complexes do not exhibit geometrical isomerism ? [1 Mark]

Answer: Because the relative positions of the ligands attached to the central atom are the same with respect to one another, all four positions being equivalent.

Teacher's Note:
a) Cis and trans positions do not exist in a tetrahedron.

Q.6 What is the hybridisation of central metal ion and shape of Wilkinson's catalyst ? [1 Mark]

Answer: Wilkinson's catalyst is [RhCl(PPh₃)₃]. Rh(I) is d⁸ and the hybridisation is dsp², so the shape is square planar.

Teacher's Note:
a) Rh(I) is a d⁸ ion with strong field phosphine ligands, like Ni²⁺ in [Ni(CN)₄]^2-.

Q.7 Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H₂O)(CN)(en)₂]²⁺ and (ii) [CoBr₂(en)₂]⁺ [1 Mark]

Answer: (i) +3 and (ii) +3

Teacher's Note:
a) Solve x + (charges of the ligands) = charge of the complex; H₂O and en are neutral.

Q.8 Draw the structure of optical isomers of [Co(en)₃]³⁺. [1 Mark]

Answer: [Co(en)₃]³⁺ has two optical isomers, which are non-superimposable mirror images of each other (d and l forms), as shown in the figure.

Teacher's Note:
a) The two forms rotate plane polarised light in opposite directions.

Q.9 Name the types of isomerism exhibited by [Co(NH₃)₅(NO₂)](NO₃)₂ [1 Mark]

Answer: It shows linkage isomerism, [Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(ONO)](NO₃)₂, and also ionisation isomerism, [Co(NH₃)₅(NO₃)](NO₃)(NO₂).

Teacher's Note:
a) NO₂^- is ambidentate and NO₃^- can be inside or outside the bracket.

Q.10 Write the formula of Amminebromidochloridonitrito-N-platinate(II) ion [1 Mark]

Answer: [Pt(NH₃)BrCl(NO₂)]^-

Teacher's Note:
a) Pt(II) with 3 anionic ligands gives a charge of 2 - 3 = -1.

Q.11 which one is more stable complex among(i) [Fe(H₂O)₆]³⁺ (ii) [Fe(NH₃)₆]³⁺(iii) [Fe(C₂O₄)₃]^3- [1 Mark]

Answer: (iii) [Fe(C₂O₄)₃]^3- is the most stable, because the oxalate ligand forms chelate rings.

Teacher's Note:
a) Chelate rings increase the stability of a complex.

Q.12 How many ions are produced from the complex Co(NH₃)₆Cl₂ in solution? [1 Mark]

Answer: 3 ions: the complex can be written as [Co(NH₃)₆]Cl₂, which gives one [Co(NH₃)₆]²⁺ ion and two Cl^- ions.

Teacher's Note:
a) Only the species outside the square bracket ionise separately.

Q.13 What do you meant by degenerate d-orbitals? [1 Mark]

Answer: Degenerate d orbitals are the d orbitals having the same energy, as the five d orbitals of a free ion (or in a spherical field).

Teacher's Note:
a) In an octahedral field the degeneracy is removed to give t_2g and e_g sets.

Q.14 Out of the following two coordination entities which is chiral (optically active)? (a) cis-[CrCl₂ (ox)₂]^3- (b) trans-[CrCl₂ (ox)₂]^3- [1 Mark]

Answer: (a) cis-[CrCl₂(ox)₂]^3- is chiral, because it has no plane of symmetry.

Teacher's Note:
a) The trans isomer has a plane of symmetry, so it is optically inactive.

Q.15 The spin only magnetic moment of [MnBr₄]^2- is 5.9 BM. Predict the geometry of the complex ion? [1 Mark]

Answer: Tetrahedral (sp³ hybridisation): 5.9 B.M means five unpaired electrons on Mn²⁺ (d⁵), the weak field Br^- does not pair them.

Teacher's Note:
a) Four ligands with weak field ligand and no pairing means tetrahedral, not square planar.

Q.16 What is the oxidation state of Ni in [Ni(CO)₄]. [1 Mark]

Answer: Zero (0).

Teacher's Note:
a) CO is a neutral ligand and the complex is neutral.

Q.17 What is the magnetic behavior of [Ni(CN)₄]^2- . [1 Mark]

Answer: Diamagnetic, since Ni²⁺ (dsp², square planar) has no unpaired electrons.

Teacher's Note:
a) Strong field CN^- pairs the d⁸ electrons.

Q.18 Name the metal ion present in vitamin B₁₂ [1 Mark]

Answer: Cobalt (Co³⁺).

Teacher's Note:
a) Vitamin B₁₂ is also called cobalamin.

Q.19 Name the isomerism shown by complex K[Cr(H₂O)₂(C₂O₄)₂] [1 Mark]

Answer: It shows geometrical isomerism (cis and trans) and the cis isomer also shows optical isomerism.

Teacher's Note:
a) The cis form has no plane of symmetry, so it exists as an enantiomeric pair.

Q.20 Name the compound used for inhibiting the growth of tumours. (cancer treatment) [1 Mark]

Answer: cis-platin, cis-[PtCl₂(NH₃)₂], a compound of platinum.

Teacher's Note:
a) Only the cis isomer is active; the trans isomer is not.

TWO MARK QUESTIONS:

Q.1 Write the IUPAC names of the following coordination compounds:
(i) [Pt(NH₃)₂Cl(NO₂ )] (ii) K₃ [Cr(C₂O₄)₃ ] [2 Marks]

Answer:
(i) diamminechloridonitrito-N-platinum(II)
(ii) potassium trioxalatochromate(III)

Teacher's Note:
a) Ligands are named alphabetically: ammine, chlorido, nitrito.
b) The printed key reads "Diaminechloridonitro(o)platinum(II)"; the correct name has ammine for NH₃.

Q.2 A cationic complex has two isomers A & B. Each has one Co³⁺, five NH₃, one Br and one SO₄^2-. A gives a white precipitate with BaCl₂ solution while B gives a yellow precipitate with AgNO₃ solution.
(a) What are the possible structures of the complexes A and B?
(a) Write the name of structural isomerism shown by A and B. [2 Marks]

Answer:
(a) A gives a white precipitate with BaCl₂, so SO₄^2- is outside the bracket: A = [Co(NH₃)₅Br]SO₄. B gives a yellow precipitate with AgNO₃, so Br^- is outside: B = [Co(NH₃)₅SO₄]Br.
(a) (second part) The two complexes show ionisation isomerism.

Teacher's Note:
a) BaSO₄ is a white precipitate and AgBr is pale yellow.
b) Ionisation isomers give different ions in solution.

Q.3 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 ? [2 Marks]

Answer:
FeSO₄ and (NH₄)₂SO₄ in 1 : 1 ratio form a double salt, FeSO₄.(NH₄)₂SO₄.6H₂O, which dissociates completely in water into its constituent ions, so it gives the test of Fe²⁺.
CuSO₄ and NH₃ in 1 : 4 ratio form the complex [Cu(NH₃)₄]SO₄, in which Cu²⁺ is held in the stable complex ion [Cu(NH₃)₄]²⁺ and does not dissociate to give the test of Cu²⁺.

Teacher's Note:
a) A double salt loses its identity in solution; a complex keeps it.
b) The deep blue colour is due to [Cu(NH₃)₄]²⁺.

Q.4 What is meant by ambidentate ligands? Give two examples. [2 Marks]

Answer:
Ligands that can bind to the central metal atom through two different donor atoms (only one at a time) are called ambidentate ligands.
Examples: NO₂^- (donor atom N in nitrito-N, or O in nitrito-O) and SCN^- (donor atom S in thiocyanato, or N in isothiocyanato).

Teacher's Note:
a) Two donor atoms but only one is used at a time.
b) Such ligands give linkage isomers.

Q.5 [Co(NH₃)₆]³⁺ is diamagnetic whereas [Co(F₆)]^3- is paramagnetic. Give reasons. [2 Marks]

Answer:
In [Co(NH₃)₆]³⁺, Co³⁺ is 3d⁶ and NH₃ is a strong field ligand, so the electrons pair up, d²sp³ hybridisation takes place and there are no unpaired electrons: diamagnetic.
In [CoF₆]^3-, Co³⁺ is 3d⁶ and F^- is a weak field ligand, so the electrons do not pair, sp³d² hybridisation takes place and four electrons stay unpaired: paramagnetic.

Teacher's Note:
a) The printed key writes [CoF₆]³⁺; the charge on the complex ion is 3-.
b) Weak field ligand means high spin.

Q.6 [Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why? [2 Marks]

Answer:
In [Cr(NH₃)₆]³⁺, Cr³⁺ is d³; the three electrons occupy separate t_2g orbitals, so d²sp³ hybridisation leaves three unpaired electrons: paramagnetic.
In [Ni(CN)₄]^2-, Ni²⁺ is d⁸ and CN^- is a strong field ligand that pairs all the electrons, so dsp² hybridisation occurs with no unpaired electron: diamagnetic.

Teacher's Note:
a) d³ always has three unpaired electrons in an octahedral field.
b) Strong field means pairing.

Q.7 Draw figure to show the splitting of d orbitals in an octahedral crystal field. How is the magnitude of ∆₀ affected by
(i) Nature of ligand.
(ii) Oxidation State of metal ion. [2 Marks]

Answer:
In an octahedral field the d_x2-y2 and d_z2 orbitals (e_g) rise in energy by 0.6 Δ_o, while d_xy, d_yz and d_zx orbitals (t_2g) fall by 0.4 Δ_o, as shown in the figure.
(i) The stronger the ligand (higher in the spectrochemical series), the greater is the splitting Δ_o.
(ii) The higher the oxidation state of the metal ion, the greater is the magnitude of Δ_o.

Teacher's Note:
a) Label e_g above and t_2g below the average energy level.
b) Δ_o for a given ligand increases with the charge on the metal ion.

Q.8 Metal carbonyl are much more stable than normal complexes, why? [2 Marks]

Answer:
The metal-carbon bond in metal carbonyls has both sigma and pi character. A sigma bond is formed when the carbonyl carbon donates a lone pair of electrons to a vacant orbital of the metal.
A pi bond is formed by donation of a pair of electrons from a filled metal d orbital into the vacant antibonding pi* orbital of CO (back bonding). The sigma bond strengthens the pi bond and vice versa, a synergic effect, which makes the metal-CO bond strong.

Teacher's Note:
a) Synergic bonding is the single key word the examiner looks for.
b) Back bonding makes the metal-carbon bond stronger than a simple sigma bond.

Q.9 Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionization isomers. [2 Marks]

Answer:
The two isomers give different ions in water, so they react differently with the same reagents.
[Co(NH₃)₅Cl]SO₄ + Ba²⁺ gives a white precipitate of BaSO₄ and no reaction with Ag⁺.
[Co(NH₃)₅SO₄]Cl + Ag⁺ gives a white precipitate of AgCl and no reaction with Ba²⁺.

Teacher's Note:
a) The ion outside the bracket is the one that precipitates.
b) Same molecular formula but different ions in solution means ionisation isomers.

Q.10 Write the formulas for the following coordination compounds and name the isomerism shown by them:
i. potassium tetracyanonickelate(II)
ii. tris(ethane-1,2-diamine) chromium(III) chloride [2 Marks]

Answer:
i. K₂[Ni(CN)₄]. It shows no isomerism, being square planar with four identical ligands.
ii. [Cr(en)₃]Cl₃. It shows optical isomerism (two enantiomers).

Teacher's Note:
a) Write the oxidation state and charge to get the number of K⁺ ions: Ni(II) with four CN^- gives 2-.
b) Tris chelates have no geometrical isomers but are optically active.

THREE MARK QUESTIONS:

Q.1 How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C₂O₄)₃]^3- (ii) [Co(NH₃)₃Cl₃] [3 Marks]

Answer:
(i) [Cr(C₂O₄)₃]^3- has no geometrical isomers, because all three bidentate ligands are identical; it shows only optical isomerism (two enantiomers).
(ii) [Co(NH₃)₃Cl₃] has two geometrical isomers: facial (fac), in which the three Cl are on one triangular face, and meridional (mer), in which the three Cl lie in one plane around the metal.
Hence 0 geometrical isomers in (i) and 2 in (ii).

Teacher's Note:
a) The printed key says no isomer is possible "as it is bidentate"; the real reason is that the three identical chelate rings are symmetric.
b) fac and mer isomers are the pair seen in M(A)₃(B)₃.
c) Both isomers of (ii) are optically inactive.

Q.2 (a) Write the IUPAC name of [Ti(H₂O)₆]⁺³.
(b) [Ti(H₂O)₆]⁺³ is coloured why?
(c) A Complex having scandium in +3 oxidation-state was found colorless why? [3 Marks]

Answer:
(a) hexaaquatitanium(III) ion
(b) Ti³⁺ has one d electron (3d¹), so a d-d transition (t_2g¹ to e_g¹) is possible by absorbing visible light and the complex is coloured.
(c) Sc³⁺ has the configuration 3d⁰, with no electron in the d orbitals, so no d-d transition is possible and the complex is colourless.

Teacher's Note:
a) The printed key writes "hexaaquo"; the IUPAC prefix is aqua.
b) d⁰ and d¹⁰ ions give colourless complexes.
c) Colour needs an electron that can be excited from t_2g to e_g.

Q.3 (a) Write IUPAC name of [Co(en)₃]³⁺
(b) [NiCl₄]^2- is paramagnetic while [Ni(CO)]₄ is diamagnetic though both are tetrahedral. Why?
(c) Explain why K₃[Fe(CN)₆] is more stable than K₄[Fe(CN)₆]. [3 Marks]

Answer:
(a) tris(ethane-1,2-diamine)cobalt(III) ion
(b) In [NiCl₄]^2-, Ni²⁺ is d⁸ and Cl^- is a weak field ligand that does not pair the 3d electrons, so there are two unpaired electrons (sp³): paramagnetic. In Ni(CO)₄, Ni is in the zero oxidation state (3d⁸4s²) and CO, a strong field ligand, pairs the electrons and shifts the 4s electrons into 3d (3d¹⁰, sp³), so there is no unpaired electron: diamagnetic.
(c) The stability of a complex depends on the charge density (charge/radius ratio) on the central ion. Fe³⁺ is smaller and has a higher charge than Fe²⁺, so it binds CN^- more strongly and K₃[Fe(CN)₆] is more stable.

Teacher's Note:
a) The same shape need not mean the same magnetic behaviour; ligand field strength decides the number of unpaired electrons.
b) The formula [Ni(CO)]₄ in the question means [Ni(CO)₄].
c) Higher charge density on the metal ion means a more stable complex.

Q.4 (a) What is the hybridization state of nickel in [Ni(CN)₄]^2-
(b) Draw the structure of [Ni(CN)₄]^2-
(c) A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain. [3 Marks]

Answer:
(a) dsp² hybridisation.
(b) The complex is square planar, with Ni²⁺ at the centre and four CN^- ligands at the corners of the square, as in the figure.
(c) In [Ni(H₂O)₆]²⁺, H₂O is a weak field ligand and Ni²⁺ has unpaired electrons, so d-d transitions are possible by absorbing visible light and the solution is coloured (green). In [Ni(CN)₄]^2-, CN^- is a strong field ligand that pairs all the electrons, so d-d transition is not possible and the solution is colourless.

Teacher's Note:
a) Square planar means dsp² with one vacant 3d orbital used.
b) No unpaired electrons gives a colourless d⁸ square planar complex.
c) Draw bond angles of 90° between neighbouring CN^-.

Q.5 Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex. [3 Marks]

Answer:
In [Co(NH₃)₆]³⁺, cobalt is +3 and 3d⁶. NH₃ is a strong field ligand and pairs the electrons, leaving two 3d orbitals vacant, so Co can undergo d²sp³ hybridisation using inner 3d orbitals: an inner orbital complex.
In [Ni(NH₃)₆]²⁺, nickel is +2 and 3d⁸. Even if NH₃ pairs the electrons, only one 3d orbital is empty, so d²sp³ hybridisation is not possible.
Hence Ni²⁺ uses its outer 4s, 4p and 4d orbitals (sp³d²) and forms an outer orbital complex.

Teacher's Note:
a) Count the vacant 3d orbitals: two are needed for an inner orbital complex.
b) The printed key puts the d²sp³ remark under the wrong complex; it belongs to cobalt.
c) Outer orbital complexes are high spin.

Q.6 For the complexes (i) [Ni(CN)₄]^2- (ii) [Ni(Cl)₄]^2- (iii) [Ni(CO)₄] Identify:
(a) The oxidation No. of nickel
(b) The hybrid orbitals and the shape of the complexes [3 Marks]

Answer:
(i) [Ni(CN)₄]^2-: oxidation number of Ni = +2; hybrid orbitals dsp²; shape square planar.
(ii) [NiCl₄]^2-: oxidation number of Ni = +2; hybrid orbitals sp³; shape tetrahedral.
(iii) [Ni(CO)₄]: oxidation number of Ni = 0; hybrid orbitals sp³; shape tetrahedral.

Teacher's Note:
a) Strong field CN^- makes nickel(II) dsp²; weak field Cl^- gives sp³.
b) Ni(0) in the carbonyl is d¹⁰, so sp³.
c) Present your answer as a small table of complex, oxidation number, hybridisation and shape.

Q.7 Specify the (i) oxidation numbers (ii) coordination numbers and (iii) IUPAC name of the following coordination entities:
(a) [Co(H₂O)(CN)(en)₂]²⁺
(b) [PtCl₄]^2- [3 Marks]

Answer:
(a) oxidation number = +3; coordination number = 6; IUPAC name: aquacyanidobis(ethane-1,2-diamine)cobalt(III) ion.
(b) oxidation number = +2; coordination number = 4; IUPAC name: tetrachloridoplatinate(II) ion.

Teacher's Note:
a) Each en counts for two donor atoms, so (a) has 1 + 1 + 4 = 6.
b) An anionic complex ends in -ate, so platinum becomes platinate.
c) The printed key spells it "aquacynobis" and "tetachlorido"; the correct spellings are used here.

Q.8 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(a) [Fe(CN)₆]^4-
(b) [FeF₆]^3- [3 Marks]

Answer:
(a) In [Fe(CN)₆]^4-, iron is +2 and 3d⁶. CN^- is a strong field ligand and pairs the 3d electrons, leaving two 3d orbitals vacant. The six CN^- donate electron pairs into the six d²sp³ hybrid orbitals, so the complex is octahedral and diamagnetic (inner orbital, low spin).
(b) In [FeF₆]^3-, iron is +3 and 3d⁵. F^- is a weak field ligand and does not pair the 3d electrons, so the hybridisation is sp³d² using the 4s, 4p and 4d orbitals. The complex is octahedral and paramagnetic with five unpaired electrons (outer orbital, high spin).

Teacher's Note:
a) Name the hybridisation, shape and magnetic property for each, as the examiner expects.
b) d²sp³ means inner orbital; sp³d² means outer orbital.
c) Spin only moment of [FeF₆]^3- is 5.92 B.M.

Q.9 Dimethyl glyoxime is added to alcoholic solution of NiCl₂. When ammonium hydroxide is slowly added to it, a rosy red precipitate of a complex appears.
(a) Give the str. of the complex showing hydrogen bond.
(b) Give oxidation state and hybridization of central metal ion.
(c) Identify whether it is paramagnetic or diamagnetic. [3 Marks]

[Figure: Square planar complex with Ni at the centre, bonded to four nitrogen atoms of two dimethylglyoximate ligands; each ligand has a CH₃-C=N-OH group and a CH₃-C=N-O^- group, and the two ligands are joined by two O-H...O hydrogen bonds between the neighbouring oxygen atoms]

Answer:
(a) The complex is bis(dimethylglyoximato)nickel(II), [Ni(dmgH)₂]. Ni²⁺ is at the centre, bonded through the four nitrogen atoms of two dimethylglyoxime ligands in a square planar arrangement, and two intramolecular O-H...O hydrogen bonds link the two ligands.
(b) The oxidation state of nickel is +2 and the hybridisation is dsp².
(c) There is no unpaired electron, so the complex is diamagnetic.

Teacher's Note:
a) Bidentate dimethylglyoxime forms two five membered chelate rings with Ni²⁺.
b) The rosy red precipitate is the standard test for Ni²⁺.
c) dsp² means square planar, as in [Ni(CN)₄]^2-.

Q.10 Draw all the isomers (geometrical and optical) of:
(a) [CoCl₂(en)₂]⁺
(b) [Co(NH₃)Cl(en)₂]²⁺ [3 Marks]

Answer:
(a) [CoCl₂(en)₂]⁺ has three isomers in total: trans (optically inactive) and cis, which exists as a pair of optical isomers (d and l).
(b) [Co(NH₃)Cl(en)₂]²⁺ also has three isomers in total: trans (optically inactive) and cis, which exists as a pair of optical isomers (d and l).
The trans isomers have a plane of symmetry; the cis isomers are optically active.

Teacher's Note:
a) Trans forms are inactive; cis forms exist as enantiomers.
b) Two different monodentate ligands with two en rings follow the same pattern as two identical ones.
c) Draw the en rings as arcs joining two adjacent positions.

FIVE MARK QUESTIONS:

Q.1 For the complex [Fe(en)₂Cl₂]Cl identify : (2009)
(a) the oxidation No. of Iron.
(b) the hybrid orbitals and the shape of the complex.
(c) the magnetic behavior of the complex.
(d) No. of geometrical isomers.
(e) whether there is an optical isomer also
(f) name of the complex. [5 Marks]

Answer:
(a) The oxidation number of iron = +3.
(b) The hybrid orbitals are d²sp³ and the shape is octahedral.
(c) The complex is paramagnetic (one unpaired electron).
(d) There are two geometrical isomers, cis and trans.
(e) The cis isomer shows optical isomerism (a pair of enantiomers), while the trans isomer does not.
(f) dichloridobis(ethane-1,2-diamine)iron(III) chloride

Teacher's Note:
a) The printed key names the complex as a platinum compound; the metal is iron, so the name is iron(III).
b) Only the cis form lacks a plane of symmetry.
c) Fe³⁺ is d⁵; with the strong field en ligands the electrons pair up as far as possible, leaving one unpaired.
d) Total stereoisomers: trans, cis-d and cis-l, that is three.

Q.2 State a reason for each of the following situations
(a) Co²⁺ ion is easily oxidized to Co³⁺ in the presence of strong ligand. (2010,12)
(b) CO is a stronger complexing reagent than NH₃ . (2009,12)
(c) The molecular shape of [Ni(CO)₄ ] is not the same as that of [Ni(CN)₄]^2-. (2012)
(d) Ni does not form low spin octahedral complexes. (2009,10)
(e) The π complexes are known for the transition metals only. (2009,10) [5 Marks]

Answer:
(a) With a strong field ligand, Co³⁺ has the stable t_2g⁶e_g⁰ configuration, while Co²⁺ has t_2g⁶e_g¹ with an electron in the higher e_g level, which is easily lost.
(b) CO can form a synergic bond: a sigma bond by donating a lone pair to the metal and a pi bond by back donation from filled metal d orbitals into the pi* orbitals of CO, so it is a stronger complexing agent than NH₃, which is only a sigma donor.
(c) In [Ni(CO)₄], nickel is sp³ hybridised, so it is tetrahedral. In [Ni(CN)₄]^2-, Ni²⁺ has a vacant 3d orbital after pairing, so it is dsp² hybridised and square planar.
(d) Ni²⁺ (d⁸) does not have two vacant inner 3d orbitals for d²sp³ hybridisation even after pairing the electrons.
(e) Transition metals have vacant d orbitals that accept pi electrons of the ligand and filled d orbitals that back donate, so pi complexes form.

Teacher's Note:
a) Give one clear reason for each lettered part.
b) A strong ligand raises Δ_o, which stabilises the low spin Co³⁺ ion.
c) Use the keywords synergic bonding for (b) and vacant d orbitals for (e).
d) The year tags printed with the parts are exam references, not part of the answer.

Q.3 Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronicconfiguration 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 [5 Marks]

Answer:
(i) Potassium diaquabis(oxalato)chromate(III) trihydrate: oxidation state +3, configuration 3d³ (t_2g³), coordination number 6, octahedral. It has two geometrical isomers (cis and trans), and the cis isomer is optically active, so there are three stereoisomers. μ = \( \sqrt{3(3+2)} \) = 3.87 B.M (about 4 B.M).
(ii) Pentaamminechloridocobalt(III) chloride: oxidation state +3, configuration 3d⁶ (t_2g⁶), coordination number 6, octahedral with no geometrical or optical isomers. μ = 0 (diamagnetic).
(iii) Trichloridotris(pyridine)chromium(III): oxidation state +3, configuration 3d³ (t_2g³), coordination number 6, octahedral. It has two geometrical isomers, fac and mer, and both are optically inactive. μ = 3.87 B.M (about 4 B.M).

Teacher's Note:
a) The printed key says CrCl₃(py)₃ has four optical isomers; fac and mer are both achiral, so it has only two isomers.
b) d³ has three unpaired electrons, giving 3.87 B.M, and d⁶ low spin has none.
c) The water of crystallisation is named as trihydrate at the end.
d) Use the spin only formula \( \sqrt{n(n+2)} \) B.M.

Q.4 (i) Write the IUPAC name and type of isomerism shown by following complex compounds
a) [Co(NH₃)₅(NO₂)](NO₃)₂
b) [Pt(NH₃)(H₂O)Cl₂]
(ii) Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] and how many of these will exhibit optical isomers? [5 Marks]

Answer:
(i) a) pentaamminenitrito-N-cobalt(III) nitrate. It shows linkage isomerism, [Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(ONO)](NO₃)₂, and ionisation isomerism, [Co(NH₃)₅(NO₃)](NO₃)(NO₂).
(i) b) ammineaquadichloridoplatinum(II). It shows geometrical isomerism: cis and trans.
(ii) [Pt(NH₃)(Br)(Cl)(py)] is square planar, with three geometrical isomers: the NH₃ can be trans to Br, trans to Cl or trans to py (the other two ligands being trans to each other).
None of these isomers shows optical isomerism, since a square planar complex has a plane of symmetry (its own plane).

Teacher's Note:
a) The printed key writes ammineaquochloridoplatinum(II); the full name is ammineaquadichloridoplatinum(II).
b) The key explains the lack of optical isomerism by tetrahedral complexes; the correct reason is that the square planar complex is planar and so achiral.
c) For M(ABCD) square planar there are three geometrical isomers.
d) Linkage isomerism arises from the ambidentate NO₂^- ligand.

Q.5 (i) If to an aqueous solution of CuSO₄ in two tubes, we add ammonia solution in one tube and HCl(aq) to the other tube, how the colour of the solutions will change? Explain with the help of reaction.
(ii) Write the IUPAC name of complex compounds formed during the process [5 Marks]

Answer:
(i) With ammonia the colour changes from blue to deep blue: [Cu(H₂O)₄]²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ + 4H₂O (deep blue).
With HCl the colour changes to yellow (green): [Cu(H₂O)₄]²⁺ + 4Cl^- → [CuCl₄]^2- + 4H₂O (yellow).
(ii) [Cu(NH₃)₄]²⁺ is tetraamminecopper(II) ion.
[CuCl₄]^2- is tetrachloridocuprate(II) ion.

Teacher's Note:
a) The printed key shows [CuCl₄]²⁺; the chloro complex is an anion, [CuCl₄]^2-, so it is named cuprate.
b) Ammonia and chloride replace the water ligands and change Δ_o, which shifts the colour.
c) NH₃ is a stronger field ligand than H₂O and Cl^- is weaker, so the colours differ.
d) The change in colour is evidence of ligand substitution.

Ans: structure :trigonalbipyramidal

Shape: linear

Ans : because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.

Ans: the ligands has around them negatively charged field because of which they are called crystal fields.

Ans: when a didentate or polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal atom, a 5 or 6 membered ring is formed , the effect is called chelate effect. Eg. [PtCl₂(en)]

ii) it does not take into account the covalent character of metal ligand bond. ( any one )

Ans: the magnetic moment of 5.9 B.M. corresponds to the presence of 5 unpaired electrons in the d- orbitals of Mn2+ ion. As a result the hybridisation involved is sp3 rather than dsp2. Thus tetrahedral structure of [MnCl4]2- complex will show 5.92 B.M magnetic moment value.

Ans: i) due to the presence of 1 electron in 3d subshell in [Ti(H2O)6]3+ complex d-d transition takes place by the absorption of visible light. Hence the complex appears coloured. On the other hand, [Sc(H2O)6]3+ does not possess any unpaired electron .Hence d-d transition is not possible (which is responsible for colour) in this complex is not possible, therefore it is colourless.

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