Chemistry · Volume 1 · Chapter 5

Samacheer Class 12 Chemistry - Coordination Chemistry

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Q.1 The sum of primary valence and secondary valence of the metal M in the complex [M(en)3]Cl3 is:
Answer: d

In [M(en)3]Cl3 the three chloride ions are outside → metal oxidation state (primary valence) = +3. en is bidentate and there are three en → coordination number (secondary valence) = 3×2 = 6. Sum = 3 + 6 = 9.

Q.2 An excess of silver nitrate is added to 100 mL of a 0.01 M solution of pentaaquachloridochromium(III) chloride [Cr(H2O)5Cl]Cl2. The number of moles of AgCl precipitated would be:
Answer: b

Moles of complex = 0.100 L × 0.01 M = 0.001 mol. In [Cr(H2O)5Cl]Cl2 there are two free Cl− per complex (counter ions) → number of moles AgCl = 2 × 0.001 = 0.002 mol.

Q.3 A complex has molecular formula M SO4 Cl · 6H2O. The aqueous solution gives white precipitate with BaCl2 (indicating free sulfate) and no precipitate with AgNO3 (no free chloride). If the secondary valence (coordination number) of the metal is six, which correctly represents the complex?
Answer: c

Free sulfate (gives BaSO4) means SO4^2− is outside the coordination sphere; absence of AgCl ppt means Cl− is coordinated. Coordination number = 6. [M(H2O)5Cl]+ (five H2O + one Cl = 6 donors) with one crystal water gives overall formula M SO4 Cl · 6H2O → [M(H2O)5Cl]SO4 · H2O.

Q.4 Oxidation state of iron and the charge on the ligand NO in [Fe(H2O)5NO]SO4 are:
Answer: a

Sulfate is 2− so the complex cation is 2+. Water is neutral. If NO is neutral (NO·), oxidation state of Fe = +2 to give overall +2. Thus Fe = +2 and NO is neutral (charge 0).

Q.5 As per IUPAC guidelines, the name of the complex Co en ONO C C ( ) ( )  l l is a) Chlorobisethylenediaminenitritocobalt(III) chloride b) Chloridobis(ethane-1,2-diamine)nitro -Ocobaltate(III) chloride c) Chloridobis(ethane-1,2-diammine)nitrito -Ocobalt(II) chloride d) Chloridobis(ethane-1,2-diammine)nitrito κ -Ocobalt(III)chloride
Answer: d

IUPAC: neutral or anionic donor named before metal; bidentate en = ethane-1,2-diamine. Chlorido for Cl− as ligand; nitrito bound through O is 'nitrito-κO'. Metal oxidation = +3 (one outer Cl−). Correct full name: chloridobis(ethane-1,2-diamine)(nitrito-κO)cobalt(III) chloride (option d).

Q.6 IUPAC name of the complex K Al C O ( )   is a) Potassiumtrioxalatoaluminium(III) b) Potassiumtrioxalatoaluminate(II) c) Potassiumtrisoxalatoaluminate(III) d) Potassiumtrioxalatoaluminate(III)
Answer: d

Anionic complex is trioxalatoaluminate(III): Al is +3, oxalato is bidentate (oxalato). Counterion = potassium. IUPAC: potassium trioxalatoaluminate(III).

Q.7 A magnetic moment of 1.73 BM corresponds to one unpaired electron. Which of the following shows this value?
Answer: c

μ( spin-only ) = √[n(n+2)] BM. μ = 1.73 BM → n = 1 unpaired electron. Cu(II) (d9) species such as [Cu(NH3)4]+ have one unpaired electron → μ ≈ 1.73 BM. Other choices are diamagnetic or have more unpaired electrons.

Q.8 Crystal field stabilization energy (CFSE) for a high-spin d5 octahedral complex is:
Answer: b

For high-spin d5 in octahedral: t2g^3 eg^2 → CFSE = 3(−0.4Δo) + 2(+0.6Δo) = −1.2Δo + 1.2Δo = 0.

Q.9 In which of the following coordination entities will the magnitude of Δ0 be maximum?
Answer: a

Ligand field strength order (spectrochemical series): CN− and CO are strong field; CN− typically gives the largest Δ0 in common complexes. Among the options, [Co(CN)6]3− will have the largest Δ0.

Q.10 Which one of the following will give a pair of enantiomorphs (optical isomers)?
Answer: b

cis‑[Co(en)2Cl2]+ (an MA2B2 type with two bidentate en ligands and two identical monodentate ligands cis) is chiral and exists as a pair of enantiomers. Hence it gives enantiomorphs.

Q.11 Which type of isomerism is exhibited by Pt(NH3)2Cl2 ?
Answer: d

Pt(NH3)2Cl2 is a square planar complex and exhibits cis–trans (geometrical) isomerism (cis and trans forms).

Q.12 How many geometrical isomers are possible for a square planar Pt complex with four different monodentate ligands (e.g. Pt(Py)(NH3)(Br)(Cl))?
Answer: a

A square planar complex with four different monodentate ligands (ML1L2L3L4) has three distinct geometrical isomers (different relative positions of ligands about the plane).

Q.13 Which one of the following pairs represents linkage isomers?
Answer: d

Linkage isomerism arises when an ambidentate ligand binds through different atoms. (b) nitro (–NO2 through N) vs nitrito (–ONO through O) and (c) NCS vs SCN are examples. So both (b) and (c) are linkage isomers.

Q.14 Which kinds of isomerism are possible for [Co(NH3)4BrCl]Cl ?
Answer: b

For an MA4BC octahedral type ([Co(NH3)4BrCl]+) there are two geometrical isomers (Br and Cl cis or trans); the cis form is chiral (optical) giving enantiomers. Thus both geometrical and optical isomerism are possible.

Q.15 Which one of the following complexes is not expected to exhibit isomerism?
Answer: d

[FeCl6]3− is an octahedral complex with six identical ligands (all Cl−) → no possible isomerism. Complexes with different ligand types can show isomerism.

Q.16 A complex in which the oxidation number of the metal is zero is:
Answer: c

Metal in Fe(CO)5 has oxidation state 0 (carbonyl complexes often have metal in zero oxidation state). The cyanide complexes and ammine complexes typically have positive oxidation states.

Q.17 Formula of tris(ethane-1,2-diamine)iron(II) phosphate (salt formed between [Fe(en)3]2+ and PO4 3−) is:
Answer: c

[Fe(en)3]2+ is a 2+ cation; PO4^3− is 3− anion. To get a neutral compound use 3 cations and 2 anions: [Fe(en)3]3(PO4)2 (i.e. 3[Fe(en)3]2+ with 2 PO4^3−).

Q.18 Which of the following is paramagnetic in nature?
Answer: c

Zn2+ (d10) and Ni(CN)4^2− (square planar Ni2+, d8, diamagnetic) are diamagnetic. [Co(NH3)6]3+ is Co3+ (d6) with NH3 as a strong/medium field ligand often low-spin (diamagnetic). [Ni(H2O)6]2+ (Ni2+, d8 in octahedral) is paramagnetic (has unpaired electrons).

Q.19 Fac–mer isomerism is shown by which of the following?
Answer: c

Fac–mer isomerism occurs in octahedral complexes of type MA3B3 (three identical ligands A and three identical ligands B). [Co(NH3)3Cl3] (MA3B3 type) exhibits fac and mer isomers.

Q.20 Choose the correct statement. a) Square planar complexes are more stable than octahedral complexes b) The spin only magnetic moment of Cu Cl ( )   - is 1.732 BM and it has square planar structure. c) Crystal field splitting energy ∆ ( ) 0 of FeF6 [ ] - 4 is higher than the ∆ ( ) 0 of Fe CN 6 ( )   - d) crystal field stabilization energy of V H O ( )   + is higher than the crystal field stabilization of Ti H O ( )   +
Answer: d

Compare CFSE: V2+ is d3 → CFSE = 3(−0.4Δo) = −1.2Δo; Ti3+ is d1 → CFSE = −0.4Δo. Thus CFSE magnitude for V2+ (d3) is greater than for Ti3+ (d1). Statements a–c are incorrect (a is a generalization, b is false for chloride complex geometry, c is opposite of spectrochemical order).

Answer the following questionsAnswer the following questions25 questions
Q.1Write the IUPAC names for the following complexes. i) Na[Ni(EDTA)] ii) [Ag(CN)2]- iii) [Co(en)2(SO4)?] (?) iv) [Co(ONO)(NH3)5]2+ v) [Pt(NH3)Cl(NO2)] (forms shown in source OCR)v
Solution

i) EDTA is the ethylenediaminetetraacetato ligand (EDTA4−); Na[Ni(EDTA)] is named sodium nickel(II) ethylenediaminetetraacetate. ii) [Ag(CN)2]−: CN− is named cyanido (or cyanido), so [Ag(CN)2]− is dicyanidoargentate(I). iii) Interpreted as a complex in which sulfate coordinates to Co together with two en ligands: name = sulfato-bis(ethylenediamine)cobalt(III) (coordination-site indicated as sulfato- if O‑bound). iv) ONO denotes nitrito bound through O; [Co(NH3)5(ONO)]2+ is named pentaamminenitrito-Ocobalt(III) ion. v) Interpreted as a platinum complex containing NH3, Cl and NO2; when NO2 is N‑bound it is named nitro; an appropriate IUPAC style name is amminechloridonitroplatinum(II).

Answer:

i) sodium nickel(II) ethylenediaminetetraacetate (Na[Ni(EDTA)]) ii) dicyanidoargentate(I) ([Ag(CN)2]−) iii) sulfato-bis(ethylenediamine)cobalt(III) (sulfato-bis(ethylenediamine)cobalt(III) complex) iv) pentaamminenitrito-Ocobalt(III) ion ([Co(NH3)5(ONO)]2+) v) amminechloridonitroplatinum(II) (e.g. [Pt(NH3)Cl(NO2)])

Q.2Write the formula for the following coordination compounds. a) Potassiumhexacyanidoferrate(II) b) Pentacarbonyliron(0) c) Pentaamminenitrito-κNcobalt(III)ion d) Hexaamminecobalt(III)sulphate e) Sodiumtetrafluoridodihydroxidochromate(III)v
Solution

a) Fe(II) + 6 CN− → complex [Fe(CN)6]4−; potassium salt K4[Fe(CN)6]. b) Pentacarbonyliron(0) is neutral Fe(CO)5. c) Nitrito-κN (NO2− bound through N) is an anionic ligand; Co(III) with five neutral NH3 and one NO2− gives [Co(NH3)5(NO2)]2+. d) [Co(NH3)6]3+ cation; to balance sulfate (SO42−) need overall formula [Co(NH3)6]2(SO4)3. e) Cr(III) with 4 F− and 2 OH−: complex ion [CrF4(OH)2]3−; sodium salt Na3[CrF4(OH)2].

Answer:

a) K4[Fe(CN)6] b) Fe(CO)5 c) [Co(NH3)5(NO2)]2+ d) [Co(NH3)6]2(SO4)3 e) Na3[CrF4(OH)2]

Q.3Arrange the following in order of increasing molar conductivity i) Mg Cr NH Cl ( )( )   5 ii) Cr NH Cl CoF ( )  [ ] iii) Cr NH Cl ( )  v
Solution

The molar conductivity of a coordination compound in solution depends on how many ions it dissociates into (i.e. the total number of ions per formula unit). Arrange from least to greatest dissociation (fewest → most ions). (Because the OCRed formulas are unclear, apply this rule to the given complexes: the species that gives the smallest ionic count in solution has the lowest molar conductivity and the one giving the largest ionic count has the highest.)

Answer:

Molar conductivity increases with the number of ions produced on dissolution; the compound that yields the fewest ions has the lowest molar conductivity and vice versa.

Q.4Give an example of coordination compound used in medicine and two examples of biologically important coordination compounds.v
Solution

Cisplatin (cis‑Pt(NH3)2Cl2) is an anticancer drug. Important biological coordination compounds include haemoglobin (Fe2+/Fe3+ in a porphyrin ring), chlorophyll (Mg2+ in a porphyrin ring) and vitamin B12 (corrin ring with Co).

Answer:

Medicine: cisplatin, Pt(NH3)2Cl2. Biologically important: heme (iron porphyrin in haemoglobin) and chlorophyll (magnesium porphyrin); also vitamin B12 (cobalamin).

Q.5Based on VB theory explain why [Cr(NH3)6]3+ is paramagnetic, while [Ni(CN)4]2- is diamagnetic.v
Solution

VB theory: Cr3+ (d3) in [Cr(NH3)6]3+ with weak‑field NH3 does not pair electrons; three unpaired electrons give paramagnetism. Ni2+ (d8) in presence of strong‑field CN− pairs electrons and forms a square‑planar dsp2 hybrid (as in [Ni(CN)4]2−), resulting in all electrons paired and diamagnetism.

Answer:

[Cr(NH3)6]3+ (Cr3+, d3) is paramagnetic with three unpaired electrons (NH3 is a weak field ligand so electrons remain unpaired). [Ni(CN)4]2− (Ni2+, d8) with strong field CN− undergoes pairing and adopts square‑planar dsp2 hybridisation; all electrons are paired → diamagnetic.

Q.6Draw all possible geometrical isomers of the complex [Co(en)2Cl2]+ and identify the optically active isomer.v
Solution

In the octahedral [Co(en)2Cl2]+ (two bidentate en and two Cl−): - trans: the two Cl− are opposite → achiral. - cis: the two Cl− adjacent → the arrangement of the two bidentate ligands gives non‑superposable mirror images (Δ and Λ) → cis isomer is optically active (enantiomeric pair).

Answer:

For [Co(en)2Cl2]+ (octahedral, en = bidentate) there are two geometric types: cis and trans. The trans isomer is achiral. The cis isomer exists as a pair of enantiomers (Δ and Λ) and is optically active.

Q.7Explain why [Ti(H2O)6]3+ is coloured, while [Sc(H2O)6]3+ is colourless.v
Solution

Colour in many transition complexes arises from d–d electronic transitions. Ti3+ (3d1) has one d electron that can be excited (giving colour). Sc3+ (3d0) has no d electrons, so there are no d–d transitions and the complex is colourless (except for possible weak charge‑transfer bands).

Answer:

[Ti(H2O)6]3+ is coloured because Ti3+ is d1 (one d electron) — d–d transitions are possible. [Sc(H2O)6]3+ is colourless because Sc3+ is d0 (no d electrons) — no d–d transitions.

Q.8Give an example for complex of the type Ma b c [ ] where a, b, c are monodentate ligands and give the possible isomers.v
Solution

For an octahedral complex of type MA2B2C2 (three different pairs of monodentate ligands) there are several distinct geometric isomers. One important classification: the facial (fac) isomer (each pair of identical ligands adjacent on a face) and meridional (mer) isomer (identical ligands lie in a plane through the centre). Specific counts and chirality depend on which ligands are trans/cis; e.g. [Co(NH3)2Cl2Br2] shows these distinct arrangements.

Answer:

Example: [Co(NH3)2Cl2Br2] (type MA2B2C2). Possible geometric isomers (octahedral): fac and mer isomers; several arrangements exist depending on positions of identical ligands — fac (three identical ligands occupy one face) and mer (three lie in one meridian).

Q.9Give one test to differentiate [Co(NH3)4Cl]SO4 and [Co(NH3)4SO4]Cl.v

Add aqueous AgNO3 to the solutions. If Cl− is present as a free (ionic) counter‑ion (as in [Co(NH3)4]Cl with external Cl−) a white precipitate of AgCl forms. If Cl− is coordinated to Co (inside the coordination sphere) it is not available as a free ion and no AgCl precipitate is obtained. (Conversely, BaCl2 can be used to test for free sulfate: BaSO4 precipitates if SO4 2− is outside the coordination sphere.)

Q.10In an octahedral crystal field, show the splitting of the five d orbitals.v

In an octahedral field the five d orbitals split into two sets: the lower‑energy t2g set (dxy, dxz, dyz) and the higher‑energy eg set (dz2, dx2−y2). The t2g orbitals are stabilized by −0.4Δo each and the eg orbitals are destabilized by +0.6Δo each; the energy gap between the sets is Δo.

Q.11What is linkage isomerism? Explain with an example.v
Solution

Ambidentate ligands (e.g. NO2−, SCN−) can coordinate through alternative atoms, giving linkage isomers that differ in which atom is bonded to the metal. The chemical and spectroscopic properties of the isomers are different.

Answer:

Linkage isomerism arises when an ambidentate ligand can bind through two different donor atoms. Example: nitrite ion NO2− binds via N (nitro) or via O (nitrito): [Co(NH3)5(NO2)]2+ (nitro, N‑bound) vs [Co(NH3)5(ONO)]2+ (nitrito, O‑bound).

Q.12 Classify the following ligands based on the number of donor atoms. a) NH3 b) en c) ox2- d) pyridine
Answer: a) NH3 — monodentate b) en (ethylenediamine) — bidentate c) ox2− (oxalate, C2O4 2−) — bidentate (can act as a chelating ligand) d) pyridine — monodentate

NH3 and pyridine each donate a single lone pair → monodentate. Ethylenediamine (en) has two donor N atoms → bidentate. Oxalate (C2O4 2−) has two oxygen donor atoms and commonly binds through both oxygens → bidentate.

Q.13Give the difference between double salts and coordination compounds.v
Solution

Key differences: (1) Bonding — double salts are held by ionic interactions between simple ions; coordination compounds have coordinate (dative covalent) bonds between metal and ligands. (2) Dissociation — double salts dissociate into original ions in solution; coordination compounds retain the complex ion. (3) Properties — coordination compounds often show distinct stereochemistry and characteristic colours, whereas double salts behave as mixtures of their component salts in solution.

Answer:

Double salts: ionic compounds formed by combining two simple salts in crystalline form; they dissociate into their constituent ions completely in solution. Coordination compounds: contain complex ions with coordinate covalent bonds between metal and ligands; on dissolution they usually give complex ions and counter‑ions, not the original separate salts.

Q.14Write the postulates of Werner's theory.v
Solution

Werner explained coordination compounds by introducing primary valence (oxidation state) and secondary valence (coordination number/geometry). He used these to rationalize the composition and isomerism of coordination complexes, predicting octahedral and tetrahedral arrangements corresponding to different coordination numbers.

Answer:

Werner's postulates (concise): 1) Metals have two kinds of valences — primary (ionisable, oxidation state) and secondary (non‑ionisable, coordination number). 2) Primary valences are satisfied by negative ions; secondary valences are directed in space and fixed in number giving definite geometry (e.g. 4 → tetrahedral/square planar, 6 → octahedral). 3) Secondary valences are responsible for stereochemistry (isomerism) of complexes. 4) In solution, complexes retain the arrangement of ligands (secondary valences) while primary valence ions may be ionisable.

Q.15Why tetrahedral complexes do not exhibit geometrical isomerism.v
Solution

Because a tetrahedron has no pair of positions that are uniquely trans or cis relative to others (every ligand sees three equivalent neighbours), arrangements that would be distinct in an octahedron are equivalent in a tetrahedron. Thus geometrical isomers (cis/trans) are not observed for most tetrahedral ML4 complexes.

Answer:

In a tetrahedral ML4 complex all four ligand positions are equivalent; swapping positions produces superposable arrangements, so cis/trans type geometrical isomerism (as in octahedral complexes) does not occur for simple tetrahedral complexes.

Q.16Explain optical isomerism in coordination compounds with an example.v
Solution

Chiral coordination complexes lack mirror planes or inversion centres; their mirror images are non‑superposable (enantiomers). In tris(bidentate) complexes like [Co(en)3]3+, the arrangement of three bidentate en ligands can be right‑ or left‑handed (Δ and Λ), giving optical activity.

Answer:

Optical isomerism occurs when a complex is chiral (non‑superposable on its mirror image). Example: [Co(en)3]3+ exists as two enantiomers denoted Δ and Λ which rotate plane‑polarized light in opposite directions.

Q.17What are hydrate isomers? Explain with an example.v
Solution

Hydrate (or solvate) isomerism arises when water molecules interchange between being coordinated to the metal and being present as crystal‑water. For example, Cr(III) chloride hydrates include [Cr(H2O)6]Cl3 (all six waters coordinated) and [Cr(H2O)5Cl]Cl2·H2O (one Cl− coordinated, one water uncoordinated as lattice water).

Answer:

Hydrate isomers are coordination isomers differing in the number of water molecules inside the coordination sphere versus outside (as waters of crystallisation). Example: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2·H2O are hydrate isomers.

Q.18What is crystal field splitting energy?v
Solution

Ligand electric fields remove the degeneracy of metal d orbitals; the resulting energy difference between the higher‑energy and lower‑energy sets of d orbitals is the crystal field splitting energy (Δ). In an octahedral field Δo separates eg (higher) and t2g (lower) levels.

Answer:

Crystal field splitting energy (Δ or Δo) is the energy gap between sets of d orbitals (eg and t2g) generated when ligands approach a metal ion in a particular geometry (e.g. octahedral Δo).

Q.19What is crystal field stabilization energy (CFSE) ?v
Solution

When d electrons occupy split orbital levels (t2g and eg) the lower‑lying electrons lower the energy of the complex. CFSE quantifies this stabilization (in units of Δ). For example, a d3 ion in octahedral field has CFSE = 3(−0.4Δo) = −1.2Δo.

Answer:

CFSE is the net stabilization gained by placing the metal d electrons in the lower‑energy crystal‑field split orbitals relative to the energy they would have in an unsplit (spherical) field. Calculated as (number of t2g electrons × −0.4Δo) + (number of eg electrons × +0.6Δo) plus pairing corrections if needed.

Q.20A solution of [Ni(H2O)6]2+ is green, whereas a solution of [Ni(CN)4]2− is colourless. Explain.v

[Ni(H2O)6]2+ is an octahedral, weak‑field (water) complex of Ni2+ (d8). Weak‑field ligands give moderate d–d splitting so d–d transitions absorb visible light, producing a green colour. [Ni(CN)4]2− has CN−, a strong‑field ligand, and the complex adopts square‑planar geometry for d8 Ni2+; electrons are paired and d–d transitions are suppressed or shifted out of the visible region, so the complex appears colourless.

Q.21Discuss briefly the nature of bonding in metal carbonyls.v
Solution

Bonding in metal carbonyls is described by synergic (two-way) interaction: (1) σ-donation — the filled lone pair on carbon (CO HOMO, 5σ) donates electron density into an empty metal orbital; (2) π-backbonding — filled metal d orbitals donate electron density into the empty π* antibonding orbitals of CO (LUMO, 2π*). The net result strengthens M–C bonding and weakens the C≡O bond (lower C–O stretching frequency). This synergic bonding often leads to complexes that obey the 18‑electron rule, explains stability of terminal vs. bridging CO, and accounts qualitatively for spectroscopic (IR) shifts and observed geometries. MO descriptions combine these interactions into bonding and antibonding molecular orbitals delocalized over metal and CO.

Answer:

Synergic σ-donation and π-backbonding between metal and CO (18‑electron rule, CO acts as a neutral ligand).

Q.22What is the coordination entity formed when excess of liquid ammonia is added to an aqueous solution of copper sulphate?v
Solution

When excess NH3 is added to CuSO4·5H2O, ammonia replaces water ligands to give the tetraammine complex: Cu^{2+} + 4 NH3 → [Cu(NH3)4]^{2+}. In aqueous medium two water molecules may remain axially: [Cu(NH3)4(H2O)2]^{2+}. The salt is commonly written as [Cu(NH3)4]SO4 (or with waters coordinated); the complex is deep blue.

Answer:

[Cu(NH3)4(H2O)2]^{2+} (tetraamminecopper(II) complex, deep blue).

Q.23On the basis of valence bond theory explain the nature of bonding in [Co(CN)6]3−.v

Co in [Co(CN)6]3− is in the +3 oxidation state (Co3+, d6). CN− is a strong‑field ligand and pairs the 3d electrons. The paired d electrons occupy three 3d orbitals leaving two 3d orbitals (along with 4s and 4p) available for hybridization to form six d2sp3 hybrid orbitals directed octahedrally. Six CN− ligands coordinate by donating lone pairs into these hybrids to form coordinate covalent bonds. All 3d electrons are paired (low‑spin d6), so the complex is diamagnetic.

Q.24What are the limitations of VB theory?v
Solution

Main limitations of valence‑bond (VB) theory for coordination compounds: (1) It gives only a qualitative picture and cannot predict energy levels or spectral lines (electronic spectra). (2) It fails to account for the variable magnetic moments and the existence of both high‑spin and low‑spin states in some cases quantitatively. (3) It does not treat metal–ligand π‑bonding (π‑backbonding) adequately — important in carbonyls and π‑acceptor ligands. (4) It cannot explain the bonding in complexes with extensive delocalization or metal–metal bonding. (5) It sometimes gives incorrect predictions of geometry (e.g. complexes of d10 metals) and does not account well for thermodynamic stabilities and reaction kinetics. (6) It lacks the molecular orbital treatment needed for quantitative description.

Answer:

Fails to explain spectra, variable magnetic moments, metal–ligand π-bonding (e.g. metal carbonyls), delocalization, and some geometries/stabilities; no quantitative energy predictions.

Q.25For the complex K3[Mn(CN)6], write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in an octahedral crystal field.v

Oxidation state of Mn: +3 (Mn3+, d4). Coordination number: 6. Nature of ligand: CN− (monodentate, strong‑field). In an octahedral crystal field CN− causes pairing → low‑spin d4 with electronic configuration t2g4 eg0. Magnetic property: paramagnetic with two unpaired electrons.