CBSE · NCERT · Class 12 Chemistry · Chapter 5

NCERT Solutions: Class 12 Chemistry Chapter 5 - Coordination Compounds

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Chapter-wise NCERT intext questions and exercise answers for Coordination Compounds, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 24
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1Exercises24 questions
Q.5.1Explain the bonding in coordination compounds in terms of Werner’s postulates.v
Solution

According to Werner, metals in coordination compounds show primary and secondary valences. Primary valences are ionisable and are satisfied by negative ions. Secondary valences are non-ionisable, equal to coordination number and are satisfied by neutral molecules or negative ions directly attached to the metal. The groups satisfying secondary valences occupy definite spatial positions around the metal, giving characteristic geometries.

Answer:

Werner proposed primary ionisable valences and secondary non-ionisable valences in coordination compounds.

Q.5.2FeSO4 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?v
Solution

FeSO4 and (NH4)2SO4 form Mohr's salt, which dissociates in water into simple ions including Fe2+, so Fe2+ tests are obtained. CuSO4 with excess NH3 forms the coordination entity [Cu(NH3)4]2+. The concentration of free Cu2+ becomes very small, so ordinary Cu2+ tests are not obtained.

Answer:

Mohr's salt is a double salt and dissociates to give Fe2+, whereas Cu2+ forms the stable complex [Cu(NH3)4]2+.

Q.5.3Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.v
Solution

Coordination entity examples: [Co(NH3)6]3+ and [Fe(CN)6]4-. Ligand examples: NH3 and Cl-. Coordination number examples: [Co(NH3)6]3+ has CN 6 and [PtCl4]2- has CN 4. Coordination polyhedron examples: [Co(NH3)6]3+ is octahedral and [PtCl4]2- is square planar. Homoleptic examples: [Ni(CO)4] and [Co(NH3)6]3+. Heteroleptic examples: [Co(NH3)4Cl2]+ and [PtCl2(NH3)2].

Answer:

A coordination entity is the central atom/ion with attached ligands; ligands are donor species; coordination number is number of donor atoms; coordination polyhedron is ligand arrangement; homoleptic has one ligand type; heteroleptic has more than one ligand type.

Q.5.4What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.v
Solution

Unidentate examples: Cl- and NH3. Didentate examples: ethane-1,2-diamine (en) and oxalate, C2O4^2-. Ambidentate examples: NO2- which can bind through N or O, and SCN- which can bind through S or N.

Answer:

Unidentate ligands bind through one donor atom; didentate through two donor atoms; ambidentate through either of two different donor atoms.

Q.5.5Specify the oxidation numbers of the metals in the following coordination entities: (i) [Co(H2O)(CN)(en)2]2+ (iii) [PtCl4]2– (v) [Cr(NH3)3Cl3] (ii) [CoBr2(en)2]+ (iv) K3[Fe(CN)6]v
Solution

Use ligand charges: H2O, NH3 and en are neutral; CN-, Br- and Cl- are -1. For [Co(H2O)(CN)(en)2]2+, x - 1 = +2 so Co = +3. For [CoBr2(en)2]+, x - 2 = +1 so Co = +3. For [PtCl4]2-, x - 4 = -2 so Pt = +2. In K3[Fe(CN)6], the complex is 3-, so x - 6 = -3 and Fe = +3. In [Cr(NH3)3Cl3], x - 3 = 0 so Cr = +3.

Answer:

(i) Co = +3; (ii) Co = +3; (iii) Pt = +2; (iv) Fe = +3; (v) Cr = +3.

Q.5.7Using IUPAC norms write the systematic names of the following: (i) [Co(NH3)6]Cl3 (iv) [Co(NH3)4Cl(NO2)]Cl (vii) [Ni(NH3)6]Cl2 (ii) [Pt(NH3)2Cl(NH2CH3)]Cl (v) [Mn(H2O)6]2+ (viii) [Co(en)3]3+ (iii) [Ti(H2O)6]3+ (vi) [NiCl4]2– (ix) [Ni(CO)4]v
Solution

Ligands are named alphabetically before the metal. Anionic ligands take -ido endings, neutral NH3 is ammine, H2O is aqua and CO is carbonyl. The oxidation state of the metal is written in Roman numerals.

Answer:

(i) Hexaamminecobalt(III) chloride; (ii) diamminechlorido(methanamine)platinum(II) chloride; (iii) hexaaquatitanium(III) ion; (iv) tetraamminechloridonitrito-N-cobalt(III) chloride; (v) hexaaquamanganese(II) ion; (vi) tetrachloridonickelate(II) ion; (vii) hexaamminenickel(II) chloride; (viii) tris(ethane-1,2-diamine)cobalt(III) ion; (ix) tetracarbonylnickel(0).

Q.5.8List various types of isomerism possible for coordination compounds, giving an example of each.v
Solution

Structural isomerism includes ionisation isomerism, e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br; linkage isomerism, e.g. [Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2; coordination isomerism, e.g. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]; and solvate/hydrate isomerism, e.g. [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2.H2O. Stereoisomerism includes geometrical isomerism, e.g. cis/trans-[Pt(NH3)2Cl2], and optical isomerism, e.g. Δ/Λ-[Co(en)3]3+.

Answer:

Coordination compounds show structural isomerism and stereoisomerism.

Q.5.9How many geometrical isomers are possible in the following coordination entities? (i) [Cr(C2O4)3]3– (ii) [Co(NH3)3Cl3]v
Solution

[Cr(C2O4)3]3- has three identical didentate ligands in an octahedral arrangement, so it has optical isomers but no geometrical isomers. [Co(NH3)3Cl3] has octahedral MA3B3 type geometry, giving fac and mer geometrical isomers.

Answer:

(i) No geometrical isomer; (ii) two geometrical isomers, facial and meridional.

Q.5.12Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?v
Solution

The complex is square planar with four different ligands. The distinct geometrical isomers are obtained by choosing which ligand is trans to NH3: Br, Cl or py. Square planar complexes are planar and have a plane of symmetry, so these isomers are not optically active.

Answer:

Three geometrical isomers are possible; none exhibits optical isomerism.

Q.5.13Aqueous copper sulphate solution (blue in colour) gives: (i) a green precipitate with aqueous potassium fluoride and (ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.v
Solution

Aqueous CuSO4 contains blue [Cu(H2O)6]2+. With F-, copper(II) fluoride precipitates, giving a green precipitate. With excess Cl-, ligand substitution forms the soluble complex [CuCl4]2-, which gives a bright green solution.

Answer:

Fluoride gives sparingly soluble copper(II) fluoride, while chloride forms a soluble tetrachloridocuprate(II) complex.

Q.5.15Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)6]4– (ii) [FeF6]3– (iii) [Co(C2O4)3]3– (iv) [CoF6]3–v
Solution

In [Fe(CN)6]4-, Fe2+ is d6 and CN- is strong-field, causing pairing and d2sp3 octahedral hybridisation. In [FeF6]3-, Fe3+ is d5 and F- is weak-field, so electrons remain unpaired and sp3d2 octahedral hybridisation is used. In [Co(C2O4)3]3-, Co3+ is d6 and the octahedral complex is inner orbital. In [CoF6]3-, Co3+ is d6 but F- is weak-field, so an outer orbital high-spin octahedral complex forms.

Answer:

(i) inner orbital d2sp3, diamagnetic; (ii) outer orbital sp3d2, paramagnetic; (iii) inner orbital d2sp3, diamagnetic; (iv) outer orbital sp3d2, paramagnetic.

Q.5.17What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.v
Solution

A typical order is I- < Br- < S2- < SCN- < Cl- < N3- < F- < OH- < C2O4^2- < H2O < NCS- < edta4- < NH3 < en < CN- < CO. Weak-field ligands generally do not pair d electrons and produce high-spin complexes. Strong-field ligands favour electron pairing and often produce low-spin complexes.

Answer:

The spectrochemical series orders ligands by increasing crystal field splitting power. Weak field ligands cause small splitting; strong field ligands cause large splitting.

Q.5.18What is crystal field splitting energy? How does the magnitude of Do decide the actual configuration of d orbitals in a coordination entity?v
Solution

If Δo is smaller than the pairing energy, electrons occupy higher eg orbitals before pairing, giving high-spin complexes. If Δo is larger than pairing energy, electrons pair in lower t2g orbitals, giving low-spin complexes. Thus ligand field strength controls the d electron arrangement.

Answer:

Crystal field splitting energy Δo is the energy gap between t2g and eg levels in an octahedral field; its size relative to pairing energy decides high-spin or low-spin configurations.

Q.5.19[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why?v
Solution

Cr3+ is d3, so in octahedral [Cr(NH3)6]3+ it has t2g3 configuration with three unpaired electrons and is paramagnetic. Ni2+ is d8, and CN- is a strong field ligand. In square planar [Ni(CN)4]2-, d electrons pair, leaving no unpaired electrons, so it is diamagnetic.

Answer:

[Cr(NH3)6]3+ has three unpaired d electrons; [Ni(CN)4]2- is square planar with all electrons paired.

Q.5.20A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.v
Solution

[Ni(H2O)6]2+ is an octahedral d8 complex with weak-field water ligands and absorbs visible light by d-d transitions, appearing green. CN- is a strong-field ligand and forms square planar [Ni(CN)4]2- with paired electrons and large splitting; it does not show the same visible d-d absorption, so the solution is colourless.

Answer:

The aqua complex has d-d transitions in the visible region; the cyanido complex is square planar low-spin with a different, larger splitting so visible absorption responsible for colour is absent or very weak.

Q.5.21[Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?v
Solution

Both complexes contain Fe2+, but CN- is a strong-field ligand whereas H2O is a weak-field ligand. The magnitude of Δo is therefore different, so the energy and wavelength of absorbed light are different. As a result, the complexes show different colours.

Answer:

Because CN- and H2O produce different crystal field splittings around Fe2+.

Q.5.22Discuss the nature of bonding in metal carbonyls.v
Solution

The M-C sigma bond is formed by donation of the lone pair on carbon of CO into a vacant metal orbital. The M-C pi bond is formed by back donation from filled metal d orbitals into empty π* orbitals of CO. These two interactions reinforce each other and stabilise metal carbonyls.

Answer:

Metal carbonyl bonding is synergic: CO donates a lone pair to the metal and the metal back-donates d electrons into CO antibonding orbitals.

Q.5.23Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes: (i) K3[Co(C2O4)3] (iii) (NH4)2[CoF4] (ii) cis-[CrCl2(en)2]Cl (iv) [Mn(H2O)6]SO4v
Solution

Oxalate is 2-, Cl- and F- are 1-, en and H2O are neutral. Charge balance gives Co3+ in K3[Co(C2O4)3], Cr3+ in cis-[CrCl2(en)2]Cl, Co2+ in (NH4)2[CoF4], and Mn2+ in [Mn(H2O)6]SO4. The d counts follow from the metal ions. Coordination numbers count donor atoms: oxalate and en are didentate, while F- and H2O are unidentate.

Answer:

(i) Co(III), d6, CN 6; (ii) Cr(III), d3, CN 6; (iii) Co(II), d7, CN 4; (iv) Mn(II), d5, CN 6.

Q.5.25Explain the violet colour of the complex [Ti(H2O)6]3+ on the basis of crystal field theory.v
Solution

In octahedral [Ti(H2O)6]3+, the d orbitals split into lower t2g and higher eg sets. Ti3+ has one d electron in t2g. Absorption of light with energy equal to Δo promotes it to eg. The transmitted/reflected complementary light gives the complex its violet colour.

Answer:

Ti3+ is d1; absorption of visible light promotes the electron from t2g to eg, and the complementary colour observed is violet.

Q.5.26What is meant by the chelate effect? Give an example.v
Solution

When a ligand forms a ring by binding through two or more donor atoms, the complex is called a chelate. Chelated complexes are usually more stable because dissociation would require breaking more than one metal-ligand bond and because of favourable entropy. Example: [Co(en)3]3+ is more stable than analogous ammine complexes.

Answer:

Chelate effect is the extra stability of complexes containing didentate or polydentate ligands compared with similar complexes of unidentate ligands.

Q.5.27Discuss briefly giving an example in each case the role of coordination compounds in: (i) biological systems (iii) analytical chemistry (ii) medicinal chemistry and (iv) extraction/metallurgy of metals.v
Solution

In biological systems, haemoglobin is an iron coordination compound and chlorophyll is a magnesium coordination compound. In medicinal chemistry, cisplatin is an anticancer drug and EDTA is used in lead poisoning. In analytical chemistry, EDTA titrations estimate hardness of water and dimethylglyoxime detects Ni2+. In metallurgy, silver and gold are extracted using cyanide complexes such as [Ag(CN)2]- and [Au(CN)2]-.

Answer:

Coordination compounds are important in biological systems, medicine, analysis and metallurgy.

Q.5.29Amongst the following ions which one has the highest magnetic moment value? (i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6]2+ (iii) [Zn(H2O)6]2+v
  1. i. [Cr(H2O)6]3+
  2. ii. [Fe(H2O)6]2+
  3. iii. [Zn(H2O)6]2+
Solution

Cr3+ is d3 with three unpaired electrons. Fe2+ in weak-field water is high-spin d6 with four unpaired electrons. Zn2+ is d10 with no unpaired electrons. Therefore [Fe(H2O)6]2+ has the highest magnetic moment.

Answer:

(ii) [Fe(H2O)6]2+.

Q.5.30Amongst the following, the most stable complex is (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+ (iii) [Fe(C2O4)3]3– (iv) [FeCl6]3–v
  1. i. [Fe(H2O)6]3+
  2. ii. [Fe(NH3)6]3+
  3. iii. [Fe(C2O4)3]3-
  4. iv. [FeCl6]3-
Solution

Oxalate is a didentate chelating ligand. Chelate complexes are more stable than comparable complexes with unidentate ligands such as H2O, NH3 or Cl-. Therefore [Fe(C2O4)3]3- is the most stable among the given complexes.

Answer:

(iii) [Fe(C2O4)3]3-.

Q.5.31What will be the correct order for the wavelengths of absorption in the visible region for the following: [Ni(NO2)6]4–, [Ni(NH3)6]2+, [Ni(H2O)6]2+ ?v
  1. i. [Ni(NO2)6]4- > [Ni(NH3)6]2+ > [Ni(H2O)6]2+
  2. ii. [Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4-
  3. iii. [Ni(NH3)6]2+ > [Ni(H2O)6]2+ > [Ni(NO2)6]4-
Solution

In the spectrochemical series, field strength increases as H2O < NH3 < NO2-. Larger crystal field splitting means higher energy absorption and therefore shorter wavelength. Hence wavelength of absorption decreases in the order [Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4-.

Answer:

[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4-.