CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. Using IUPAC norms, name the complex : [Co(en)₂(ONO)Cl]Cl
Answer:
chloridobis(ethane-1,2-diamine)nitrito-O-cobalt (III) chloride.

Teacher's Note:
1. –ONO means NO₂⁻ is bonded through oxygen, so the ligand is nitrito-O; bonded through N it would be nitrito-N.

Question 2. Write the name of the coordination isomer of the complex [Co(en)₃][Cr(CN)₆].
Answer:
[Cr(en)₃][Co(CN)₆] = tris(ethane-1,2-diamine)chromium (III) hexacyanocobaltate (III).

Teacher's Note:
1. In coordination isomerism the ligands are exchanged between the cationic and anionic complex entities; the metals stay in the same oxidation states.

Question 3. What type of isomerism is shown by the following complexes ?
(i) [Cr(en)₃]³⁺ (ii) [Pt(NH₃)₄] [PtCl₄]
Answer:
(i) Optical isomerism
(ii) Coordination isomerism (because some Cl and NH₃ ligands may interchange, e.g., [Pt(NH₃)₃Cl] [Pt(NH₃)Cl₃].

Teacher's Note:
1. [M(AA)₃] octahedral complexes have no plane of symmetry, so they always exist as a pair of enantiomers (d and l).

Question 4. Define hydrate isomerism or ionization isomerism or geometrical isomerism or linkage isomerism with suitable example.
Answer:
Hydrate isomerism : Isomers which differ in the number of water molecules present as ligands inside the coordination sphere and as water of crystallisation outside it, e.g., [Cr(H₂O)₆]Cl₃ (violet) and [Cr(H₂O)₅Cl]Cl₂.H₂O (grey-green).
Ionization isomerism : Isomers which give different ions in solution because a counter ion is itself a potential ligand and exchanges place with a ligand, e.g., [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄.
Geometrical isomerism : Isomers which have the same ligands but differ in their spatial arrangement around the central atom, e.g., cis- and trans-[Pt(NH₃)₂Cl₂].
Linkage isomerism : Isomers which differ in the donor atom of an ambidentate ligand attached to the metal, e.g., [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂.

Teacher's Note:
1. Hydrate, ionisation, linkage and coordination isomerism are structural; geometrical and optical are stereoisomerism (same bonds, different spatial arrangement).

Question 5. Give an example of linkage isomerism. (CBSE 2010)
Answer:
[Co(NH₃)₅NO₂]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂

Teacher's Note:
1. Linkage isomerism needs an ambidentate ligand such as NO₂⁻ or SCN⁻.

Question 6. Give an example of coordination isomerism. (CBSE 2010)
Answer:
[Cr(NH₃)₆][CoF₆] and [Co(NH₃)₆][CrF₆]

Teacher's Note:
1. Coordination isomerism is possible only when both cation and anion are complex ions.

Question 7. Give an example of ionization isomerism. (CBSE 2010)
or Name the isomerism shown by the following pair of coordination compounds : [Co(NH₃)₅Br] SO₄ and [Co(NH₃)₅SO₄] Br (Assam Board 2012)
Answer:
[Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl
or Ionisation isomerism.

Teacher's Note:
1. Ionisation isomers can be told apart chemically: [Co(NH₃)₅Br]SO₄ gives a white precipitate with BaCl₂, while [Co(NH₃)₅SO₄]Br gives a pale yellow precipitate with AgNO₃.

Question 8. Illustrate the geometrical isomers of [Pt(NH₃)₄Cl₂]²⁺.

Answer:
The complex is octahedral (Pt(IV), coordination number 6). It exists as trans- (two Cl opposite to each other) and cis- (two Cl adjacent) isomers, as shown.

Teacher's Note:
1. In an octahedral [MA₄B₂] complex, B–M–B angle is 180° in the trans isomer and 90° in the cis isomer.

Question 9. Draw the structure and write the hybridisation
(i) [Fe(CO)₅] (ii) trans-[Co(NH₃)₄ Cl₂]⁺
(iii) Ni(CO)₄ (iv) cis-[Co(NH₃)₄ Cl₂]⁺
(v) [NiCl₄]²⁻

Answer:
(i) dsp³ (trigonal bipyramidal: three CO in the equatorial plane at 120° and two CO axial)
(ii) d²sp³ (octahedral) (iii) sp³ (tetrahedral)
(iv) d²sp³ (octahedral) (v) sp³ (tetrahedral)
Structures of the two octahedral isomers (ii) and (iv) :

Teacher's Note:
1. Hybridisation decides geometry: sp³ tetrahedral, dsp² square planar, dsp³ trigonal bipyramidal, d²sp³/sp³d² octahedral.

Question 10. What is crystal field splitting energy (CFSE) ?
Answer:
The difference of energy between two sets of d-orbitals after splitting is called crystal field splitting energy or crystal field stabilization energy (CFSE).

Teacher's Note:
1. For octahedral complexes it is written Δ_o and for tetrahedral complexes Δ_t.

Short Answer Questions (2 Marks)

Question 11. On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ_o > P. (CBSE Foreign 2015, 2017)
Answer:
In this case, the 4th electron will enter into t_2g, i.e., pairing in one of the t_2g orbitals will take place. Hence, electronic configuration will be t_2g⁴ (t_2g⁴e_g⁰).

Teacher's Note:
1. When pairing costs less energy than promotion to e_g (P < Δ_o), the electron pairs up in t_2g.

Question 12. With the help of crystal field theory, predict the number of unpaired electrons in [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ complexes.
Answer:
In both complex, Fe is present as Fe (II). It has d⁶ configuration, i.e., 4 unpaired electrons are present. As CN⁻ is a strong ligand, these electrons will pair up. Hence, unpaired electrons = 0 (t_2g⁶).
But H₂O is a weak ligand, electrons will not pair up. Hence, unpaired electrons = 4 (t_2g⁴e_g²).

Teacher's Note:
1. Same metal ion, same oxidation state – only the ligand strength changes the number of unpaired electrons.

Question 13. What happens to the colour of coordination compound [Ti(H₂O)₆]Cl₃ when heated gradually ?
Answer:
On heating, its colour becomes lighter as H₂O is gradually lost. When all H₂O molecules are lost, it becomes colourless because in the absence of ligands, crystal field splitting does not occur.

Teacher's Note:
1. No ligands → no splitting → no d–d transition → no colour.

Question 14. The values of dissociation constants of [Cu(NH₃)₄]²⁺ and [Co(NH₃)₆]³⁺ are 1·0 × 10⁻¹² and 6·2 × 10⁻³⁶ respectively. Which complex would be more stable and why ?
Answer:
Smaller the value of the dissociation constant, more stable is the complex in the solution. Hence, [Co(NH₃)₆]³⁺ is more stable than [Cu(NH₃)₄]²⁺ ion.

Teacher's Note:
1. Stability constant is the reciprocal of the dissociation (instability) constant: small dissociation constant = large stability constant.

Question 15. Which of the two compounds is more stable and why ? K₄[Fe(CN)₆], K₃[Fe(CN)₆]
Answer:
K₄[Fe(CN)₆] is more stable than K₃[Fe(CN)₆]. This follows from the Effective Atomic Number Rule (Sidgwick theory).
EAN of Fe in [Fe(CN)₆]⁴⁻ = 26 − 2 + 6 × 2 = 36 (equal to the atomic number of krypton), whereas in [Fe(CN)₆]³⁻, EAN = 26 − 3 + 6 × 2 = 35.

Teacher's Note:
1. EAN = atomic number − electrons lost in forming the ion + 2 × number of ligand donor atoms.

Question 16. [Ni(CN)₄]²⁻ is colourless whereas [Ni(H₂O)₆]²⁺ is green. Why ?
Answer:
H₂O is a weak ligand. 3d electrons of Ni²⁺ do not pair up. The unpaired electrons absorb red wavelength of white light for d-d transition and transmitted colour (complementary colour) is green. Hence, the solution of [Ni(H₂O)₆]²⁺ is green.
CN⁻ is a strong ligand. The 3d electrons of Ni²⁺ pair up. There is no unpaired electron left to absorb energy. Hence, solution of [Ni(CN)₄]²⁻ is colourless.

Teacher's Note:
1. The colour we see is the complementary colour of the light absorbed, not the absorbed colour itself.

Question 17. Do we call metal carbonyls as organometallics ? Why or why not ?
Answer:
Yes, metal carbonyls are called organometallics because C atom of CO is linked to the metal atom. In fact, the metal-carbon bonds have both σ and π character.

Teacher's Note:
1. An organometallic compound is defined by at least one metal–carbon bond.

Question 18. In the formation of metal carbonyls, which orbitals are involved in the overlap ?
Answer:
(i) Filled bonding π_2p orbital of CO with empty metal d-orbital resulting in σ-bond between metal atom and carbon atom of CO.
(ii) Filled metal d-orbital with empty antibonding π_2p* orbital of CO resulting in additional π-bond between the metal and same CO molecule (back bonding).

Teacher's Note:
1. The σ and π parts strengthen each other (synergic bonding), which makes the M–CO bond strong.

Question 19. How is nickel detected and estimated in a salt ? Give the reaction involved.
Answer:
This is done by adding dimethylglyoxime solution in presence of NH₄OH when a red ppt. is obtained:
Ni²⁺ + 2 dmgH₂ + 2 NH₄OH → [Ni(dmgH)₂] (red ppt.) + 2 NH₄⁺ + 2 H₂O
The red precipitate of nickel dimethylglyoximate is filtered, dried and weighed to estimate nickel.

Teacher's Note:
1. dmgH₂ = dimethylglyoxime; each dmgH⁻ acts as a bidentate ligand through two N atoms.

Teacher's Note:
1. A quick test: Mohr's salt gives the test for Fe²⁺, but K₄[Fe(CN)₆] does not, because Fe²⁺ is locked inside [Fe(CN)₆]⁴⁻.

Question 21. What is meant by unidentate, bidentate and ambidentate ligands ? Illustrate with examples.
Answer:
Unidentate ligand : A ligand bound to the metal through a single donor atom, e.g., Cl⁻, H₂O, NH₃.
Bidentate ligand : A ligand which binds through two donor atoms, e.g., H₂NCH₂CH₂NH₂ (ethane-1,2-diamine) and C₂O₄²⁻ (oxalate).
Ambidentate ligand : A ligand which can bind through two different donor atoms but only one at a time, e.g., NO₂⁻ (through N or O) and SCN⁻ (through S or N).

Teacher's Note:
1. Bidentate = two atoms bonded at the same time; ambidentate = either of two atoms, one at a time.

Question 22. What are chelates? Giving one example write the importance of chelate. (MP Board 2013)
Answer:
When a bidentate or a polydentate ligand binds the same metal ion through two or more donor atoms, a ring is formed. Such complexes are called chelates and the ligand is called a chelating ligand, e.g., [Co(en)₃]³⁺, [Ni(dmg)₂], [Ca(EDTA)]²⁻.
Importance : Chelates are more stable than similar complexes containing unidentate ligands (chelate effect). Hence, chelating agents are used in the estimation of hardness of water (EDTA), in the detection of metal ions (dmg for Ni²⁺) and in removing toxic metals from the body (EDTA in lead poisoning).

Teacher's Note:
1. Five- and six-membered chelate rings are the most stable.

Question 23. Define cis and trans isomerism. Draw the cis and trans isomers of [Co (NH₃)₄Cl₂]⁺ ion.

Answer:
When two identical ligands occupy adjacent positions (at 90°), the isomer is called the cis isomer, and when they occupy opposite positions (at 180°), it is called the trans isomer. This type of geometrical isomerism is called cis-trans isomerism.
The isomers of [Co(NH₃)₄Cl₂]⁺ are :

Teacher's Note:
1. The cis isomer of [Co(NH₃)₄Cl₂]⁺ is violet and the trans isomer is green - a classic Werner result.

Question 24. The aqueous solution of a complex compound having the formula CoBrCl₂.4 NH₃ conducts electricity and also gives white precipitate with AgNO₃. Write the IUPAC notation and IUPAC name of the compound. (Manipur Board 2011)
Answer:
As the solution conducts electricity and gives a white precipitate (AgCl) with AgNO₃, one Cl⁻ is present outside the coordination sphere. Coordination number of Co being 6, the formula is [Co(NH₃)₄BrCl]Cl.
IUPAC name : Tetraamminebromidochloridocobalt (III) chloride.

Teacher's Note:
1. White precipitate means AgCl, not AgBr (pale yellow) - this tells you which halide is outside the bracket.

Question 25. How would you account for the following ? [Ni(CO)₄] has tetrahedral geometry while [Pt(NH₃)₂Cl₂] is square planar.
Answer:
In [Ni(CO)₄], Ni is in zero oxidation state (3d⁸4s²). CO is a strong field ligand; the 4s electrons pair up in 3d (3d¹⁰). No 3d orbital is free, so one 4s and three 4p orbitals hybridise (sp³) and the geometry is tetrahedral.
In [Pt(NH₃)₂Cl₂], Pt is in +2 state (5d⁸). For 5d elements the splitting is large, so the 8 electrons pair up in four 5d orbitals, leaving one 5d orbital empty. One 5d, one 6s and two 6p orbitals hybridise (dsp²) and the geometry is square planar.

Teacher's Note:
1. An empty inner d orbital available for hybridisation is what decides dsp² (square planar) versus sp³ (tetrahedral) for 4-coordinate complexes.

Short Answer Questions (3 Marks)

Question 26. A complex is prepared by mixing CoCl₃ and NH₃ in the molar ratio of 1 : 4. 0·1 M solution of this complex was found to freeze at −0·372°C. What is the formula of the complex ? Given that molal depression constant of water (K_f) = 1·86°C/m.
Answer:
Theoretical ΔT_f = K_f × m = 1·86 × 0·1 = 0·186°
Observed ΔT_f = 0·372° (Given)
As observed ΔT_f is double of the theoretical value, this shows that each molecule of the complex dissociates to form two ions. This can be so only if the formula of the complex is [Co(NH₃)₄Cl₂] Cl.

Teacher's Note:
1. van’t Hoff factor i = observed ΔT_f / calculated ΔT_f gives the number of ions per formula unit (here i = 2).

Teacher's Note:
1. For (ii), the low spin Co³⁺ (t_2g⁶) formed has a large CFSE, which is the driving force for the oxidation.

Question 28. When dilute ferrous sulphate solution is added to an aqueous solution containing nitrate ion followed by careful addition of concentrated sulphuric acid along the sides of the test tube, a brown ring is formed at the interface between the solution and sulphuric acid layers. Which anion is confirmed by the appearence of brown ring ? What is the composition of the brown ring ? (CBSE 2019)
Answer:
The appearance of brown ring at the interface between the solution and sulphuric acid layers indicates the presence of nitrate ion in the solution
NO₃⁻ + 3Fe²⁺ + 4H⁺ → NO + 3 Fe³⁺ + 2 H₂O
[Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅NO]²⁺ (Brown) + H₂O
The brown ring is [Fe(H₂O)₅NO]²⁺ (as [Fe(H₂O)₅NO]SO₄).

Teacher's Note:
1. Add the H₂SO₄ slowly along the side so that it forms a separate layer; the ring forms only at the junction.

Question 29. Explain with examples the following :
(i) coordination number (ii) oxidation state (iii) ligand (iv) coordination sphere (v) chelate. (Hr. Board 2012, Assam Board 2013)
Answer:
(i) Coordination number : The number of ligand donor atoms directly bonded to the central metal atom/ion is called its coordination number, e.g., in [Co(NH₃)₆]³⁺ and [Co(en)₃]³⁺ it is 6; in [Ni(CO)₄] it is 4.
(ii) Oxidation state : The charge that the central atom would carry if all the ligands were removed along with the electron pairs shared with it, e.g., in [Cu(CN)₄]³⁻, x − 4 = −3, so oxidation state of Cu = +1.
(iii) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity through donor atoms, e.g., Cl⁻, H₂O, NH₃, CN⁻, en.
(iv) Coordination sphere : The central atom/ion and the ligands attached to it, enclosed in square brackets, together form the coordination sphere, e.g., in K₄[Fe(CN)₆], [Fe(CN)₆]⁴⁻ is the coordination sphere and K⁺ are counter ions.
(v) Chelate : When a bidentate or polydentate ligand uses its two or more donor atoms to bind the same metal ion, a ring is formed. The complex so formed is called a chelate, e.g., [Co(en)₃]³⁺, [Cu(en)₂]²⁺.

Teacher's Note:
1. Counter ions outside the square bracket are not part of the coordination sphere and are the only ions that ionise in water.

Question 30. Determine the oxidation states of central metal ions in the following complex compounds :
(i) (a) K₄[Fe(CN)₆] (b) [Co(NH₃)₆]Cl₃ (Jharkhand Board 2011)
(ii) (a) K₃[Fe(CN)₆] (b) [PtCl₄]²⁻ (c) [CoBr₂(CN)₂]⁺ (Jharkhand Board 2012)
(iii) (a) [Pt(NH₃)₃Cl₃]⁻ ion (b) Na₄[Ni(CN)₄] (c) [Zn(H₂O)₃OH]⁺ (Chhatisgarh Board 2012)
Answer:
(i) (a) [Fe(CN)₆]⁴⁻ : x − 6 = −4, x = +2 (b) [Co(NH₃)₆]³⁺ : x + 0 = +3, x = +3
(ii) (a) [Fe(CN)₆]³⁻ : x − 6 = −3, x = +3 (b) x − 4 = −2, x = +2 (c) x − 2 − 2 = +1, x = +5
(iii) (a) x + 0 − 3 = −1, x = +2 (b) [Ni(CN)₄]⁴⁻ : x − 4 = −4, x = 0 (c) x + 0 − 1 = +1, x = +2

Teacher's Note:
1. Find the charge on the complex ion first from the counter ions, then subtract the ligand charges; neutral ligands (NH₃, H₂O) count as zero.

Question 31. Write IUPAC names of the following coordination compounds :
(a) (i) [Cu(NH₃)₄]SO₄ (ii) [Cr(H₂O)₆]Cl₃ (iii) [Ni(CO)₄] (iv) K₂[HgI₄] (MP Board 2011)
(b) (i) K₃[Fe(CN)₆] (ii) [CoCl₂(NH₃)₄]⁺ (iii) [Cr(NH₃)₄Cl₂]⁺ (iv) [CoCl₂(en)₂]Cl. (J & K Board 2011, Raj. Board 2012)
Answer:
(a) (i) Tetraamminecopper (II) sulphate (ii) Hexaaquachromium (III) chloride (iii) Tetracarbonylnickel (0) (iv) Potassium tetraiodidomercurate (II)
(b) (i) Potassium hexacyanoferrate (III) (ii) Tetraamminedichloridocobalt (III) ion (iii) Tetraamminedichloridochromium (III) ion (iv) Dichloridobis(ethane-1,2-diamine)cobalt (III) chloride

Teacher's Note:
1. Name the cation first; in an anionic complex the metal name ends in '-ate' (mercurate, ferrate), and the oxidation state goes in Roman numerals.

Question 32. List various types of isomerism possible for coordination compounds, giving an example of each. (Hr. Board 2011)
Answer:
A. Structural isomerism
(i) Ionisation isomerism : [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄
(ii) Hydrate (solvate) isomerism : [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂.H₂O
(iii) Linkage isomerism : [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂
(iv) Coordination isomerism : [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
B. Stereoisomerism
(v) Geometrical isomerism : cis- and trans-[Pt(NH₃)₂Cl₂]; fac- and mer-[Co(NH₃)₃(NO₂)₃]
(vi) Optical isomerism : d- and l-[Co(en)₃]³⁺; d- and l-cis-[PtCl₂(en)₂]²⁺

Teacher's Note:
1. Structural isomers differ in which atoms are bonded; stereoisomers have the same bonds but a different arrangement in space.

Question 33. Write what you know about any two of the following with respect to coordination compounds. Illustrate each with one example.
(i) ionisation isomerism (ii) linkage isomerism (iii) geometrical isomerism (iv) optical isomerism (v) coordination isomerism (vi) polymerisation isomerism (vii) hydrate isomerism. (MP Board 2011, Chhatisgarh Board 2011)
Answer:
(i) Ionisation isomerism : Compounds having the same composition but giving different ions in solution, because a ligand inside the coordination sphere and an ion outside exchange places, e.g., [Co(NH₃)₅SO₄]Br (gives Br⁻) and [Co(NH₃)₅Br]SO₄ (gives SO₄²⁻).
(ii) Linkage isomerism : Shown by complexes containing an ambidentate ligand, which differ in the donor atom bonded to the metal, e.g., [Co(NH₃)₅(NO₂)]Cl₂ (yellow) and [Co(NH₃)₅(ONO)]Cl₂ (red).
(iii) Geometrical isomerism : Arises in heteroleptic complexes due to different possible geometric arrangements of ligands. In the cis isomer, identical ligands are adjacent; in the trans isomer they are opposite, e.g., cis- and trans-[Pt(NH₃)₂Cl₂].
(iv) Optical isomerism : Isomers which are non-superimposable mirror images of each other (enantiomers) and rotate the plane of polarised light in opposite directions (d and l), e.g., [Co(en)₃]³⁺.
(v) Coordination isomerism : Arises from interchange of ligands between cationic and anionic complex entities, e.g., [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆].
(vi) Polymerisation isomerism : Compounds having the same empirical formula but molecular formulae which are multiples of it, e.g., [Pt(NH₃)₂Cl₂] and [Pt(NH₃)₄][PtCl₄].
(vii) Hydrate isomerism : Isomers differing in the number of water molecules inside the coordination sphere as ligands, e.g., [Cr(H₂O)₆]Cl₃ (violet), [Cr(H₂O)₅Cl]Cl₂.H₂O (grey-green) and [Cr(H₂O)₄Cl₂]Cl.2H₂O (dark green).

Teacher's Note:
1. In the exam, write any two only, each with a definition and a correct pair of formulae.

Question 34. Name the following coordination entities and draw the structures of their stereoisomers :
(i) [Co(en)₂Cl₂]⁺ (en = ethan-1, 2-diamine)
(ii) [Cr(C₂O₄)₃]³⁻ (iii) [Co(NH₃)₃Cl₃]
(Atomic numbers Cr = 24, Co = 27) (CBSE 2012)
Answer:
(i) Dichloridobis(ethane-1,2-diamine)cobalt (III) ion. It shows geometrical isomerism (cis and trans) and the cis form is optically active (d and l). Thus, there are three stereoisomers :
(ii) Trioxalatochromate (III) ion. It shows optical isomerism (d and l forms) :
(iii) Triamminetrichloridocobalt (III). It shows geometrical isomerism (fac and mer forms) :

Teacher's Note:
1. [M(AA)₂B₂]: trans has a plane of symmetry (optically inactive); cis has none, so it exists as d and l forms.

Question 35. Draw all the possible isomers (structural and stereoisomeric) having the composition CrBr₂Cl (NH₃)₄.

Answer:
Cr is in +3 state with coordination number 6. The possible structural (ionisation) isomers are [Cr(NH₃)₄Br₂]Cl and [Cr(NH₃)₄BrCl]Br.
Each of these shows cis-trans isomerism (neither form is optically active). Thus, there are four isomers :
(i) cis-[Cr(NH₃)₄Br₂]Cl (ii) trans-[Cr(NH₃)₄Br₂]Cl (iii) cis-[Cr(NH₃)₄BrCl]Br (iv) trans-[Cr(NH₃)₄BrCl]Br

Teacher's Note:
1. First find the structural isomers by swapping counter ion and ligand, then look for cis/trans forms of each.

Teacher's Note:
1. The fewer NH₃ molecules, the more Cl⁻ must enter the sphere to keep CN = 6, so precipitable chloride drops 3, 2, 1, 0.

Long Answer Questions (5 Marks)

Teacher's Note:
1. In orbital box diagrams, show the ligand electron pairs (××) only in the hybridised orbitals and keep the metal's own electrons in the remaining d-orbitals.

Question 38. What are the shortcomings of valence bond theory for bonding in complexes ? Briefly describe crystal field theory.
Answer:
Shortcomings of VBT :
(i) It involves a number of assumptions.
(ii) It does not give quantitative interpretation of magnetic data.
(iii) It does not explain the colour exhibited by coordination compounds.
(iv) It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.
(v) It does not make exact predictions regarding tetrahedral and square planar structures of 4-coordinate complexes.
(vi) It does not distinguish between weak and strong ligands.
Crystal field theory : It is an electrostatic model which considers the metal-ligand bond to be ionic, arising purely from electrostatic interaction between the metal ion and the ligands. Ligands are treated as point charges (anions) or point dipoles (neutral molecules).
The five d-orbitals of an isolated metal ion are degenerate. When ligands approach, the negative field causes splitting of the d-orbitals (crystal field splitting). In an octahedral field, the e_g orbitals (d_x²−y², d_z²), pointing towards the ligands, are raised in energy and the t_2g orbitals (d_xy, d_yz, d_xz) are lowered; the difference is Δ_o. In a tetrahedral field the order is reversed and Δ_t = 4/9 Δ_o.
The magnitude of Δ_o depends on the field produced by the ligand and the charge on the metal ion (spectrochemical series). If Δ_o < P, high spin complexes are formed; if Δ_o > P, low spin complexes are formed. CFT explains the magnetic properties and colour (d-d transitions) of complexes.

Teacher's Note:
1. CFT treats ligands as point charges, so it cannot explain why neutral CO is a stronger ligand than anionic halides - a limitation worth remembering.