Coordination Compounds

Multiple Choice Questions (MCQs)

1. Which of the following complexes formed by Cu2+ ions is most stable?

(a) Cu2++4NH3[Cu(NH3)4]2+,logK=11.6

(b) Cu2++4CN[Cu(CN)4]2,logK=27.3

(c) Cu2++2en[Cu(en)2]2+,logK=15.4

(d) Cu2++4H2O[Cu(H2O)4]2+,logK=8.9

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Answer

(b) Greater the value of logK, greater will be stability of complex compound formed. For reaction,

Cu2++4CN[Cu(CN)4]2K=[(Cu(CN)4)2][Cu2+][CN]4 and logK=27.3

For this reaction, logK has highest value among the given four reactions. Hence, K will also be higher among these four. i.e., stability of the complexes will be highest among these four complexes.

  • (a) Cu2++4NH3[Cu(NH3)4]2+,logK=11.6

    The value of logK for this complex is 11.6, which is lower than the logK value of 27.3 for the [Cu(CN)4]2 complex. A lower logK value indicates lower stability of the complex.

  • (c) Cu2++2en[Cu(en)2]2+,logK=15.4

    The value of logK for this complex is 15.4, which is lower than the logK value of 27.3 for the [Cu(CN)4]2 complex. A lower logK value indicates lower stability of the complex.

  • (d) Cu2++4H2O[Cu(H2O)4]2+,logK=8.9

    The value of logK for this complex is 8.9, which is the lowest among the given options. A lower logK value indicates lower stability of the complex.

2. The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+,[Co(CN)6]3, [Co(H2O)6]3+

(a) [Co(CN)6]3>[Co(NH3)6]3+>[Co(H2O)6]3+

(b) [Co(NH3)6]3+>[Co(H2O)6]3+>[Co(CN)6]3

(c) [Co(H2O)6]3+>[Co(NH3)6]3+>[Co(CN)6]3.

(d) [Co(CN)6]3>[Co(NH3)6]3+>[Co(H2O)6]3+

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Thinking Process

This problem is based on the concept of crystal field splitting and colour of coordination compounds. Follow the steps to answer this question

(i) Arrange the given complexes in increasing order of their crystal field splitting energy.

(ii) Now arrange them in decreasing order of the wavelength of light.

(iii) As energy and wavelength are related as ΔE=hv=hcλ

Answer

(c) As we know that, strong field ligand split the five degenerate energy levels with more energy separation than weak field ligand, i.e., as strength of ligand increases crystal field splitting energy increases.

Hence, ΔE=hcλ

ΔE1λλ1ΔE

As energy separation increases, the wavelength decreases.

Thus, the correct order is

[Co(H2O)6]3+>[Co(NH3)6]3+>[Co(CN)6]3

Here, strength of ligand increases, ΔE increases, CFSE increases and λ absored decreases.

Hence, correct choice is (c).

  • Option (a) [Co(CN)6]3>[Co(NH3)6]3+>[Co(H2O)6]3+: This option is incorrect because it suggests that [Co(CN)6]3 absorbs the longest wavelength of light, which implies it has the smallest crystal field splitting energy (ΔE). However, CN⁻ is a strong field ligand and should have the largest ΔE and thus absorb the shortest wavelength of light.

  • Option (b) [Co(NH3)6]3+>[Co(H2O)6]3+>[Co(CN)6]3: This option is incorrect because it suggests that [Co(NH3)6]3+ absorbs a longer wavelength than [Co(H2O)6]3+. However, NH₃ is a stronger field ligand than H₂O, meaning [Co(NH3)6]3+ should have a larger ΔE and thus absorb a shorter wavelength of light compared to [Co(H2O)6]3+.

  • Option (d) [Co(CN)6]3>[Co(NH3)6]3+>[Co(H2O)6]3+: This option is incorrect for the same reason as option (a). It suggests that [Co(CN)6]3 absorbs the longest wavelength of light, which is not true because CN⁻ is a strong field ligand and should have the largest ΔE and thus absorb the shortest wavelength of light.

3. When 0.1molCoCl(NH3)5 is treated with excess of AgNO3,0.2mol of AgCl are obtained. The conductivity of solution will correspond to

(a) 1:3 electrolyte

(b) 1:2 electrolyte

(c) 1:1 electrolyte

(d) 3:1 electrolyte

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Answer

(b) One mole of AgNO3 precipitates one mole of chloride ion. In the above reaction, when 0.1 mole CoCl3(NH3)5 is treated with excess of AgNO3,0.2 mole of AgCl are obtained thus, there must be two free chloride ions in the solution of electrolyte.

So, molecular formula of complex will be [Co(NH3)5Cl]Cl2 and electrolytic solution must contain [Co(NH3)5Cl]2+ and two Clas constituent ions. Thus, it is 1:2 electrolyte.

[Co(NH3)5Cl]Cl2[Co(NH3)5Cl]2(aq)+2Cl(aq)

Hence, option (b) is the correct.

  • Option (a) 1:3 electrolyte: This option is incorrect because a 1:3 electrolyte would dissociate into four ions in total, with one cation and three anions. In the given problem, the complex [Co(NH3)5Cl]Cl2 dissociates into one cation [Co(NH3)5Cl]2+ and two anions Cl, making it a 1:2 electrolyte, not 1:3.

  • Option (c) 1:1 electrolyte: This option is incorrect because a 1:1 electrolyte would dissociate into two ions in total, with one cation and one anion. In the given problem, the complex [Co(NH3)5Cl]Cl2 dissociates into one cation [Co(NH3)5Cl]2+ and two anions Cl, making it a 1:2 electrolyte, not 1:1.

  • Option (d) 3:1 electrolyte: This option is incorrect because a 3:1 electrolyte would dissociate into four ions in total, with three cations and one anion. In the given problem, the complex [Co(NH3)5Cl]Cl2 dissociates into one cation [Co(NH3)5Cl]2+ and two anions Cl, making it a 1:2 electrolyte, not 3:1.

4. When 1 mole of CrCl36H2O is treated with excess of AgNO3,3 moles of AgCl are obtained. The formula of the complex is

(a) [CrCl3(H2O)3]3H2O

(b) [CrCl2(H2O)4]Cl2H2O

(c) [CrCl(H2O)5]Cl2H2O

(d) [Cr(H2O)6]Cl3

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Answer

(d) 1 mole of AgNO3 precipitates one free chloride ion (Cl).

Here, 3 moles of AgCl are precipitated by excess of AgNO3. Hence, there must be three free Clions.

So, the formula of the complex can be [Cr(H2O)6]Cl3 and correct choice is (d).

  • (a) [CrCl3(H2O)3]3H2O: This formula suggests that all three chloride ions are coordinated to the chromium ion, leaving no free chloride ions to react with AgNO3. Therefore, it would not result in the precipitation of 3 moles of AgCl.

  • (b) [CrCl2(H2O)4]Cl2H2O: This formula indicates that only one chloride ion is free (outside the coordination sphere) and can react with AgNO3. This would result in the precipitation of only 1 mole of AgCl, not 3 moles.

  • (c) [CrCl(H2O)5]Cl2H2O: This formula implies that two chloride ions are free and can react with AgNO3. This would result in the precipitation of only 2 moles of AgCl, not 3 moles.

5. The correct IUPAC name of [Pt(NH3)2Cl2] is

(a) Diamminedichloridoplatinum (II)

(b) Diamminedichloridoplatinum (IV)

(c) Diamminedichloridoplatinum (0)

(d) Dichloridodiammineplatinum (IV)

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Thinking Process

This problem is based on IUPAC nomenclature of coordination compound. IUPAC nomenclature of any coordination compound can be done as follows

(i) Positively charged ions are named first.

(ii) Negatively charged ions are named in alphabetical order of ligands including their numbers followed by metal ending with -ium and oxidation state in the bracket.

Answer

(a) The complex compound is [Pt(NH3)2Cl2].

The ligands present in the compound are

(i) NH3 - neutral ligand represented as amine.

(ii) Cl - anion ligand (ending with-o-) represented as chlorido di prefixed to represent two ligands.

The oxidation number of platinum in the compound is 2 . Hence, correct IUPAC name of [Pt(NH3)2Cl2] is

Diammine dichloridoplatinum (II)

So, (a) option is correct.

  • Option (b): Diamminedichloridoplatinum (IV) is incorrect because the oxidation state of platinum in the given complex [Pt(NH3)2Cl2] is +2, not +4. The correct oxidation state is determined by the charges of the ligands and the overall neutrality of the complex.

  • Option (c): Diamminedichloridoplatinum (0) is incorrect because the oxidation state of platinum in the given complex [Pt(NH3)2Cl2] is +2, not 0. The correct oxidation state is determined by the charges of the ligands and the overall neutrality of the complex.

  • Option (d): Dichloridodiammineplatinum (IV) is incorrect because the oxidation state of platinum in the given complex [Pt(NH3)2Cl2] is +2, not +4. Additionally, the correct naming convention places the ammine ligands before the chlorido ligands, making “diammine” come before “dichlorido” in the name.

6. The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?

(a) [Fe(CO)5]

(b) [Fe(CN)6]3

(c) [Fe(C2O4)3]3

(d) [Fe(H2O)6]3+

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Answer

(c) Chelation (formation of cycle by linkage between metal ion and ligand) stabilises the coordination compound. The ligand which chelates the metal ion are known as chelating ligand.

Here, only [Fe(C2O4)3]3 is a coordination compound which contains oxalate ion as a chelating ligand. Hence, it stabilises coordination compound by chelating Fe3+ ion.

  • (a) [Fe(CO)5]: This complex does not involve chelation. The carbonyl (CO) ligands are monodentate, meaning they bind to the metal ion through a single donor atom, which does not provide the stabilizing chelate effect.

  • (b) [Fe(CN)6]3: This complex also does not involve chelation. The cyanide (CN) ligands are monodentate, binding to the metal ion through a single donor atom, and thus do not provide the stabilizing chelate effect.

  • (d) [Fe(H2O)6]3+: This complex does not involve chelation either. The water (H₂O) molecules are monodentate ligands, binding to the metal ion through a single donor atom, and do not provide the stabilizing chelate effect.

7. Indicate the complex ion which shows geometrical isomerism.

(a) [Cr(H2O)4Cl2]+

(b) [Pt(NH3)3Cl]

(c) [Co(NH3)6]3+

(d) [Co(CN)5(NC)]3

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Thinking Process

This problem includes concept of isomerism in coordination compound. Complex of MA4B2 type show geometrical isomerism.

Answer

(a) [Cr(H2O)4Cl2]+shows geometrical isomerism because it is a MA4B2 type coordination compound which contains two set of equivalent ligands, four H2O and 2Cl.

Hence, the possible geometrical isomers are

Hence, correct choice is (a).

  • (b) [Pt(NH3)3Cl]: This complex does not show geometrical isomerism because it is a MA3B type coordination compound, which does not have the necessary arrangement of ligands to form different geometrical isomers.

  • (c) [Co(NH3)6]3+: This complex does not show geometrical isomerism because it is a MA6 type coordination compound, where all six ligands are identical, making it impossible to form different geometrical isomers.

  • (d) [Co(CN)5(NC)]3: This complex does not show geometrical isomerism because it is a MA5B type coordination compound, where the arrangement of ligands does not allow for the formation of different geometrical isomers.

8. The CFSE for octahedral [CoCl6]4 is 18,000cm1. The CFSE for tetrahedral [CoCl4]2 will be

(a) 18,000cm1

(b) 16,000cm1

(c) 8,000cm1

(d) 20,000cm1

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Answer

(c) CFSE for octahedral and tetrahedral complexes are closely related to each other by formula Δt=49Δ0

where, Δ0= CFSE for octahedral complex, Δt= CFSE for tetrahedral complex According to question, Δ0=18,000cm1

Δt=49Δ0=49×18,000cm1=4×2,000cm1=8,000cm1

Hence, correct choice is (c).

  • Option (a) 18,000cm1: This option is incorrect because it assumes that the CFSE for the tetrahedral complex is the same as that for the octahedral complex. However, the CFSE for a tetrahedral complex is generally less than that for an octahedral complex and is related by the formula Δt=49Δ0.

  • Option (b) 16,000cm1: This option is incorrect because it does not follow the correct relationship between the CFSE of octahedral and tetrahedral complexes. According to the formula Δt=49Δ0, the CFSE for the tetrahedral complex should be significantly less than 18,000cm1, and 16,000cm1 is too high.

  • Option (d) 20,000cm1: This option is incorrect because it exceeds the CFSE of the octahedral complex, which is 18,000cm1. The CFSE for a tetrahedral complex should be less than that of an octahedral complex, as given by the formula Δt=49Δ0.

9. Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are

(a) linkage isomers

(b) coordination isomers

(c) ionisation isomers

(d) geometrical isomers

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Answer

(a) The ligand(s) which has two different bonding sites are known as ambident ligands e.g., NCS,NO2 etc.

Here, NCS has two binding sites at N and S.

Hence, NCS (thiocyanate) can bind to the metal ion in two ways

MNCS or MSNC

Thus, coordination compounds containing NCS as a ligand can show linkage isomerism i.e., [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are linkage isomers.

Hence, correct choice is (a).

  • Coordination isomers: Coordination isomers occur when there is an interchange of ligands between the cationic and anionic parts of a coordination compound. In this case, both compounds have the same ligands and the same metal center, so they cannot be coordination isomers.

  • Ionisation isomers: Ionisation isomers occur when a compound can produce different ions in solution. Since both compounds have the same ligands and metal center, they will produce the same ions in solution, so they cannot be ionisation isomers.

  • Geometrical isomers: Geometrical isomers occur due to different spatial arrangements of ligands around the central metal atom. In this case, the isomerism is due to the different binding sites of the ambidentate ligand (NCS), not due to different spatial arrangements of the same ligands. Therefore, they cannot be geometrical isomers.

10. The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl represent

(a) linkage isomerism

(b) ionisation isomerism

(c) coordination isomerism

(d) no isomerism

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Answer

(d) Compounds having same molecular formula but different structural formula are known as isomers. [Co(SO4)2(NH3)5]Br and [Co(SO4)(NH3)5]Cl have not same molecular formula. Hence, they are not isomers.

  • (a) Linkage isomerism: This type of isomerism occurs when a ligand can coordinate to the metal through two different atoms. In the given compounds, the sulfate ligand (SO4) cannot coordinate through different atoms, so linkage isomerism is not possible.

  • (b) Ionisation isomerism: This type of isomerism occurs when two compounds have the same composition but yield different ions in solution. The given compounds do not produce different ions in solution; they simply have different counterions (Br and Cl), so ionisation isomerism is not applicable.

  • (c) Coordination isomerism: This type of isomerism occurs when there is an interchange of ligands between the cationic and anionic entities of different coordination compounds. The given compounds do not involve such an interchange of ligands, so coordination isomerism is not relevant.

11. A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?

(a) Thiosulphato

(b) Oxalato

(c) Glycinato

(d) Ethane-1, 2-diamine

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Answer

(a) A chelating ligand has two or more binding donor atoms to a single metal ion e.g.,

Here () denotes binding site.

thiosulphato (S2O32) is not a chelating ligand because geometrically it is not favourable for S2O32 to chelate a metal ion.

  • Oxalato (b): Oxalato (C₂O₄²⁻) is a chelating agent because it has two oxygen atoms that can donate electron pairs to bind to a single metal ion, forming a five-membered ring.

  • Glycinato (c): Glycinato (NH₂CH₂COO⁻) is a chelating agent because it has two donor atoms: the nitrogen atom in the amine group and the oxygen atom in the carboxylate group, which can bind to a single metal ion.

  • Ethane-1,2-diamine (d): Ethane-1,2-diamine (NH₂CH₂CH₂NH₂) is a chelating agent because it has two nitrogen atoms that can donate electron pairs to bind to a single metal ion, forming a five-membered ring.

12. Which of the following species is not expected to be a ligand?

(a) NO

(b) NH4+

(c) NH2CH2CH2NH2

(d) CO

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Answer

(b) Ligand must donate a pair of electron or loosely held electron pair to metal and form a ML bond.

e.g.

Among N+H4 does not have any pair of electron.

Hence N+H4 is not a ligand.

  • (a) NO: Nitric oxide (NO) can act as a ligand because it has an unpaired electron and can donate this electron to form a bond with a metal center.

  • (c) NH₂CH₂CH₂NH₂: Ethylenediamine (NHCHCHNH) is a bidentate ligand, meaning it has two nitrogen atoms each with a lone pair of electrons that can be donated to a metal center to form coordinate bonds.

  • (d) CO: Carbon monoxide (CO) is a strong ligand because the carbon atom has a lone pair of electrons that can be donated to a metal center, forming a coordinate bond.

13. What kind of isomerism exists between [Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2H2O (greyish-green)?

(a) Linkage isomerism

(b) Solvate isomerism

(c) lonisation isomerism

(d) Coordination isomerism

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Answer

(c) Soluate isomerism to shown when two compounds having same molecular formula differ by whether or solvent molecule is directly bonded to metal ion or is present as free solvent molecules in the crystal lattice.

When water is present as solvent and show this type of isomerism then it is known as hydrate isomerism.

Coordination compound [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl1H2OCl2. are solvate isomers, because water is exchanged by chloride ion. This is why both of them show different colour on exposure to sunlight.

  • Linkage isomerism: This type of isomerism occurs when a ligand can coordinate to the metal ion through two different atoms. For example, the ligand NO₂⁻ can bind through the nitrogen or the oxygen atom. In the given compounds, there is no such ligand that can bind through different atoms, so linkage isomerism is not applicable.

  • Ionisation isomerism: This type of isomerism occurs when two compounds have the same composition but yield different ions in solution. For example, [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are ionisation isomers. In the given compounds, the ions produced in solution are not different, so ionisation isomerism is not applicable.

  • Coordination isomerism: This type of isomerism occurs when there is an interchange of ligands between the cationic and anionic entities of different metal ions in a complex. For example, [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are coordination isomers. In the given compounds, there is no such interchange of ligands between different metal ions, so coordination isomerism is not applicable.

14. IUPAC name of [Pt(NH3)2Cl(NO2)] is

(a) Platinum diaminechloronitrite

(b) Chloronitrito-N-ammineplatinum (II)

(c) Diamminechloridonitrito-N-platinum (II)

(d) Diamminechloronitrito-N-platinate (II)

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Answer

(c) Correct IUPAC name can be written as The ligands present in the given coordination compound are

(i) (NH3) represented as amine

(ii) Cl3 represented as chlorido

(iii) NO2s represented as nitrito- N

According to IUPAC rule, ligands are named in an alphabetical order before central atom. Prefex di-will be used to indicate the number of NH3 ligands present.

Oxidation state of metal is indicated by Roman numeral in parenthesis.

So, IUPAC name will be

diamminechloronitrito-N-platinum (II)

Hence, option (c) is correct.

  • Option (a): “Platinum diaminechloronitrite” is incorrect because:

    • The term “diamine” should be “diammine” to correctly represent the two ammonia (NH3) ligands.
    • The term “nitrite” should be “nitrito-N” to specify the binding through the nitrogen atom.
    • The oxidation state of the metal (II) is not indicated.
  • Option (b): “Chloronitrito-N-ammineplatinum (II)” is incorrect because:

    • The ligands are not listed in alphabetical order. “Ammine” should come before “chloronitrito-N”.
    • The prefix “di-” is missing to indicate the presence of two ammonia (NH3) ligands.
  • Option (d): “Diamminechloronitrito-N-platinate (II)” is incorrect because:

    • The suffix “platinate” is used for anionic complexes, but the given complex is neutral. The correct term should be “platinum”.

Multiple Choice Questions (More Than One Options)

15. Atomic number of Mn. Fe and Co are 25,26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

(a) [Co(NH3)6]3+

(b) [Mn(CN)6]3

(c) [Fe(CN)6]4

(d) [Fe(CN)6]3

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Thinking Process

This problem is based on magnetic property of coordination compound. Coordination compound containing at least one unpaired electron(s) are paramagnetic and coordination compounds all containing all paired electrons are diamagnetic in nature.

Answer

(a,c)

Molecular orbital electronic configuration of Co3+ in [Co(NH3)6]3+ is

[18Ar]

Number of unpaired electron =0

Magnetic property = Diamagnetic

Molecular orbital electronic configuration of Mn3+ in [Mn(CN)6]3

Number of unpaired electrons =2

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of Fe2+ in [Fe(CN)6]4 is

Number of unpaired electron =0

Magnetic property = Diamagnetic

Molecular orbital electronic configuration of Fe3+ in [Fe(CN)6]3

Number of unpaired electron =1

Magnetic property = Paramagnetic

Thus, [Co(NH3)6]3+ and [Fe(CN)6]4 are diamagnetic.

Hence, correct choices are options (a) and (c).

  • Option (b) [Mn(CN)6]3:

    • The molecular orbital electronic configuration of Mn3+ in [Mn(CN)6]3 shows that it has 2 unpaired electrons.
    • Magnetic property: Paramagnetic (due to the presence of unpaired electrons).
  • Option (d) [Fe(CN)6]3:

    • The molecular orbital electronic configuration of Fe3+ in [Fe(CN)6]3 shows that it has 1 unpaired electron.
    • Magnetic property: Paramagnetic (due to the presence of unpaired electrons).

16. Atomic number of Mn,Fe,Co and Ni are 25,26,27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons?

(a) [MnCl6]3

(b) [FeF6]3

(c) [CoF6]3

(d) [Ni(NH3)6]2+

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Answer

(a, c)

Molecular orbital electronic configuration of Mn3+ in [MnCl6]3 is

Number of unpaired electrons =4

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of Co3+ in [CoF6]3 is

Number of unpaired electrons =4

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of Fe3+ in [FeF6]3 is

Number of unpaired electrons =5

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of Ni2+ in [Ni2(NH3)6]2+ is

Number of unpaired electrons =2

Magnetic property = Paramagnetic

Thus, [MnCl6]3 and [CoF6]3 are paramagnetic having four electrons each.

Hence, correct choices are (a) and (c).

  • Option (b) [FeF6]3:

    • The molecular orbital electronic configuration of Fe3+ in [FeF6]3 shows that it has 5 unpaired electrons. This is different from the 4 unpaired electrons found in [MnCl6]3 and [CoF6]3.
  • Option (d) [Ni(NH3)6]2+:

    • The molecular orbital electronic configuration of Ni2+ in [Ni(NH3)6]2+ shows that it has 2 unpaired electrons. This is different from the 4 unpaired electrons found in [MnCl6]3 and [CoF6]3.

17. Which of the following options are correct for [Fe(CN)6]3 complex?

(a) d2sp3 hybridisation

(b) sp3d2 hybridisation

(c) Paramagnetic

(d) Diamagnetic

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Answer

(a,c)

According to VBT, the molecular orbital electronic configuration of Fe3 in [Fe(CN)6]3 is

Hybridisation =d2sp3

Number of unpaired electron =1

Magnetic property = Paramagnetic

Hence, correct choices are options (a) and (c).

  • Option (b) sp3d2 hybridisation is incorrect because the correct hybridisation for [Fe(CN)6]3 is d2sp3, not sp3d2. This is due to the involvement of inner d-orbitals in the hybridisation process, which leads to the formation of an octahedral complex.

  • Option (d) Diamagnetic is incorrect because [Fe(CN)6]3 has one unpaired electron, making it paramagnetic, not diamagnetic. A diamagnetic substance would have all electrons paired, resulting in no net magnetic moment.

18. An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because…… .

(a) [Co(H2O)6]2+ is transformed into [CoCl6]4

(b) [Co(H2O)6]2+ is transformed into [CoCl4]2

(c) tetrahedral complexes have smaller crystal field splitting than octahedral complexes

(d) tetrahedral complexes have larger crystal field splitting than octahedral complex

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Answer

(b,c)

Aqueous pink solution of cobalt (II) chloride is due to electronic transition of electron from t2g to eg energy level of [Co(H2O)6]2+ complex. When excess of HCl is added to this solution

(i) [Co(H2O)6]2+ is transformed into [CoCl4]2.

(ii) Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because Δt=49Δ0

Hence, options (b) and (c) are correct choices.

  • Option (a) is incorrect because the transformation that occurs is from [Co(H2O)6]2+ to [CoCl4]2, not to [CoCl6]4.
  • Option (d) is incorrect because tetrahedral complexes have smaller crystal field splitting than octahedral complexes, not larger.

19. Which of the following complexes are homoleptic?

(a) [Co(NH3)6]3+

(b) [Co(NH3)4Cl2]+

(c) [Ni(CN)4]2

(d) [Ni(NH3Cl2].

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Answer

(a, c)

Homoleptic complex The complex containing only one species or group as ligand is known as homoleptic ligand.

e.g.,

[Co(NH3)6]3+,[Ni(CN)4]2

Here, [Co(NH3)6]3+ contain only NH3 as a ligand and [Ni(CN)4]2 contain CN as a ligand.

While other two complexes [Co(NH3)4Cl2]+and [Ni(NH3)4Cl2] contain NH3 and Cl as ligands.

Hence, options (a) and (c) are correct choices.

  • Option (b) [Co(NH3)4Cl2]+: This complex is not homoleptic because it contains two different types of ligands, NH3 and Cl.

  • Option (d) [Ni(NH3)4Cl2]: This complex is not homoleptic because it also contains two different types of ligands, NH3 and Cl.

20. Which of the following complexes are heteroleptic?

(a) [Cr(NH3)6]3+

(b) [Fe(NH3)4Cl2]+

(c) .[Mn(CN)6)]4

(d) [Co(NH3)4Cl2]

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Answer

( b,d)

Heteroleptic complexes Coordination complexes which contain more than one type of ligands are known as heteroleptic complexes.

e.g., [Fe(NH3)4Cl2]+contain NH3 and Cl as a ligand is as heteroleptic complex. Similarly, [Co(NH3)4Cl2] contain NH3 and Cl as ligand is also a heteroleptic complex.

Hence, optons (b) and (d) are correct choices.

  • (a) [Cr(NH3)6]3+: This complex is homoleptic because it contains only one type of ligand, NH3.

  • (c) [Mn(CN)6]4: This complex is homoleptic because it contains only one type of ligand, CN.

21. Identify the optically active compounds from the following

(a) [Co(en)3]3+

(b) trans [Co(en)2Cl2]+

(c) cis [Co(en)2Cl2]+

(d) [Cr(NH3)5Cl]

Show Answer

Answer

(a, c)

[Co(en)3]3+ and cis [Co(en)2Cl2]+are optically active compounds because their mirror images are non-superimposable isomer.

Hence, (a) and (c) are correct choices.

  • (b) trans [Co(en)2Cl2]+: This compound is not optically active because it has a plane of symmetry. The trans configuration allows for the mirror image to be superimposable on the original molecule, thus making it optically inactive.

  • (d) [Cr(NH3)5Cl]: This compound is not optically active because it does not have any chiral centers. The arrangement of the ligands around the central metal ion is such that the mirror image can be superimposed on the original molecule, making it optically inactive.

22. Identify the correct statements for the behaviour of ethane-1, 2diamine as a ligand.

(a) It is a neutral ligand

(b) It is a didentate ligand

(c) It is a chelating ligand

(d) It is a unidentate ligand

Show Answer

Answer

(a,b,c)

Molecular formula of ethane-1, 2-diamine is

(a) Ethane-1, 2-diamine is a neutral ligand due to absence of any charge.

(b) It is a didentate ligand due to presence of two donor sites one at each nitrogen atom of amino group.

(c) It is a chelating, ligand due to its ability to chelate with the metal.

Hence, options (a), (b) and (c) are correct choices.

  • Option (d) is incorrect because ethane-1, 2-diamine is not a unidentate ligand. A unidentate ligand has only one donor site that can bind to a metal atom, whereas ethane-1, 2-diamine has two donor sites (one at each nitrogen atom of the amino groups), making it a didentate ligand.

23. Which of the following complexes show linkage isomerism?

(a) [Co(NH3)5(NO2)]2+

(b) [Co(H2O)5CO]3+.

(c) [Cr(NH3)5]SCN2+

(d) [Fe(en)2Cl2]+

Show Answer

Answer

(a,c)

Coordination compounds containing a ligand with more than one non-equivalent binding position (known as ambident ligand) show linkage isomerism.

e.g., [Co(NH3)5(NO2)+.contains NO2 which have two donor sites N and O can be shown by arrow () as

[Cr(NH3)5SCN]2+ contains SCN which have two different donor sites S and N can be shown by arrow () as

SCN

Hence, [Co(NH3)5(NO2)]2+ and [Cr(NH3)5SCN]2+ show linkage isomerism. While [Co(H2O)5CO]3+ and [Fe(en)2Cl2]+has no ambident ligand. So, these two will not show linkage isomerism.

Hence, options (a) and (c) are correct choices.

  • Option (b): [Co(H2O)5CO]3+ does not show linkage isomerism because it does not contain an ambident ligand. The ligands present (water and carbon monoxide) do not have multiple non-equivalent binding positions.

  • Option (d): [Fe(en)2Cl2]+ does not show linkage isomerism because it does not contain an ambident ligand. The ligands present (ethylenediamine and chloride) do not have multiple non-equivalent binding positions.

Short Answer Type Questions

24. Arrange the following complexes in the increasing order of conductivity of their solution

[Co(NH3)3Cl3],[Co(NH3)4Cl2]Cl,[Co(NH3)6]Cl,[Cr(NH3)5Cl]Cl2

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Thinking Process

This problem is based on the concept of conductivity of coordination compound. Greater the number of ions, greater the conductivity of coordination compound.

Answer

Ions or molecules present outside the coordination sphere are ionisable. A complex which gives more ions on dissolution, is more conducting.

[Co(NH3)3Cl3](1 ion )<[Co(NH3)4Cl2]Cl(2 ions )<[Cr(NH3)5Cl]Cl2(3 ions )<[Co(NH3)6]Cl3(4 ions )

Here, number of ions increases and conductivity increases.

25. A coordination compound CrCl34H20 precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write structural formula of the compound and name it.

Show Answer

Answer

Formation of white precipitate with AgNO3 shows that atleast one Cl ion is present outside the coordination sphere. Moreover only two ions are obtained in solution, so only one Clis present outside the sphere.

Thus, the formula of the complex is [Co(H2O)4Cl2]Cl and its IUPAC name is Tetraaquadichloridocobalt (III) chloride.

26. A complex of the type [M(AA)2X2]n+ is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.

Show Answer

Answer

An optically active complex of the type [M(AA)2X2]n+ indicates cis-octahedral structure, e.g., cis- [Pt(en)2Cl2]2+ or cis- [Cr(en)2Cl2]+because its mirror image isomers are non-superimposable.

Non-superimposable isomers of [Pt(en)2Cl2]2+.

27. Magnetic moment of [MnCl4]2 is 5.92BM. Explain giving reason present.

Show Answer

Answer

The magnetic moment 5.92BM shows that there are five unpaired electrons present 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 5.92BM magnetic moment value.

28. 0n the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.

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Answer

With weak field ligands; ΔO<P, (pairing energy) so, the electronic configuration of Co (III) will be t2g4eg2 i.e., it has 4 unpaired electrons and is paramagnetic.

With strong field ligands, Δ0>P (pairing energy), so pairing occurs thus, the electronic configuration will be t2g6eg0. It has no unpaired electrons and is diamagnetic.

29. Why are low spin tetrahedral complexes not formed?

Show Answer

Answer

In tetrahedral complex, the d-orbital is splitting to small as compared to octahedral. For same metal and same ligand Δt=49Δ0.

Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.

30. Give the electronic configuration of the following complexes on the basis of crystal field splitting theory. [CoF6]3,[Fe(CN)6]4 and [Cu(NH3)6]2+.

Show Answer

Answer

According to spectrochemical series, ligands can be arranged in a series in the order of increasing field strength i.e., F<NH3<CN.

Hence, CNand NH3 being strong field ligand pair up the t2g electrons before filling eg set. [CoF6]3;Co3+=(d6)t2g4eg2

[Fe(CN)64,Fe2+=(d6)t2g6eg0.

[Cu(NH3)c6]2+,Cu2+=(d9)t3g6eg3

31. Explain why [Fe(H2O)6]3+ has magnetic moment value of 5.92BM whereas [Fe(CN)6]3 has a value of only 1.74BM ?

Show Answer

Answer

As we know, where,

μm=n(n+2BMμm= magnetic moment μn= number of unpaired electrons Itμm=1.74 i.e., n=1andμm=5.92 i.e., n=5

[Fe(CN)6]3 involves d2sp3 hybridisation with one unpaired electron (as shown by its magnetic moment 1.74BM ) and [Fe(H2O)6]3+ involves sp3d2 hybridisation with five unpaired electrons (because magnetic moment equal to 5.92BM ).

CNis stronger ligand than H2O according to spectrochemical series. Δ0>P for CNhence, fourth electron will pair itself. Whereas for water pairing will not happen for [Fe(CN)6]3 the electronic configuration of Fe3+ is

[Ar]

One unpaired electron

For [Fe(H2O)6]3+ the electronic configuration of Fe3+ is

Five unpaired electron

Hence, [Fe(CN)6]3 and [Fe(H2O)6]3+ are inner orbital and outer orbital complex respectively.

32. Arrange following complex ions in increasing order of crystal field splitting energy (Δ0).

[Cr(Cl)6]3,[Cr(CN)6]3,[Cr(NH3)6]3+.

Show Answer

Answer

CFSE is higher when the complex contains strong field ligand. Thus, crystal field splitting energy increases in the order

[Cr(Cl)6]3<[Cr(NH3)6]3+<[Cr(CN)6]3.

Because according to spectrochemical series the order of field strength is

Cl<NH3<CN

33. Why do compounds having similar geometry have different magnetic moment?

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Answer

It is due to the presence of weak and strong field ligands in complexes. If CFSE is high, the complex will show low value of magnetic moment and vice-versa, e.g. [CoF6]3 and [Co(NH3)6]3+, the former is paramagnetic, and the latter is diamagnetic because Fis a weak field ligand and NH3 is a strong field ligand while both have similar geometry.

Q.34 CuSO45H2O is blue in colour while CuSO4 is colourless. Why?

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Answer

In CuSO45H2O, water acts as ligand and causes crystal field splitting. Hence, dd transition is possible thus CuSO45H2O is coloured. In the anhydrous CuSO4 due to the absence of water (ligand), crystal field splitting is not possible and hence, it is colourless.

35. Name the type of isomerism when ambidentate ligands are attached to central metal ion. Give two examples of ambidentate ligands.

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Answer

Ligand having more than one different binding position are known as ambidentate ligand. e.g., SCN has two different binding positions S and N. Coordination compound containing ambidentate ligands are considered to show linkage isomerism due to presence of two different binding positions.

e.g., (i) [Co(NH3)5SCN]3+ and

(ii) [Fe(NH3)5(NO2)]3+

Matching The Columns

36. Match the complex ions given in Column I with the colours given in Column II and assign the correct code.

Column I
(Complex ion)
Column II
(Colour)
A. [Co(NH3)6]3+ 1. Violet
B. [Ti(H2O)6]3+ 2. Green
C. [Ni(H2O)6]2+ 3. Pale blue
D. [Ni(H2O)4(en)]2+( aq ) 4. Yellowish orange

Codes

A B C D
(a) 1 2 4 5
(b) 4 3 2 1
(c) 3 2 4 1
(d) 4 1 2 3
Show Answer

Answer

A. (4)

B. (3)

C. (2)

D. (1)

Colour of coordination compound is closely related to CFSE of coordination compound. Depending upon the CFSE of given coordination compounds. Correct matching will be as follows

Column I
(Complex ion)
Column II
(Colour)
A. [Co(NH3)6]3+ 4. Yellowish orange
B. [Ti(H2O)6]3+ 3. Pale blue
C. [Ni(H2O)6]2+ 2. Green
D. [Ni(H2O)4(en)]2+(. aq ) 1. Violet

Hence, correct choice is (b).

37. Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code.

Column I
(Coordination compound)
Column II
(Central metal atom)
A. Chlorophyll 1. Rhodium
B. Blood pigment 2. Cobalt
C. Wilkinson catalyst 3. Magnesium
D. Vitamin B 12 4. Iron

Codes

A B C D
(a) 3 4 1 2
(b) 3 4 5 1
(c) 4 3 2 1
(d) 3 4 1 2
Show Answer

Answer

A. (3)

B. (4)

C. (1)

D. (2)

Central metal ions present on coordination compounds determine the properties of coordination compound and their biological role.

Column I
(Coordination compound)
Column II
(Central metal atom)
A. Chlorophyll 3. Magnesium
B. Blood pigment 4. Iron
C. Wilkinson catalyst 1. Rhodium
D. Vitamin B 12 2. Cobalt

Hence, correct choice is (a).

38. Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code.

Column I
(Complex ion)
Column II
(Hybridisation, number
of unpaired electrons)
A. [Cr(H2O)6]3+ 1. dsp2,1
B. [Co(CN)4]2 2. sp3d2,5
C. [Ni(NH3)6]2+ 3. d2sp3,3
D. [MnF6]4 4. sp3d2,2

Codes

A B C D A B C D
(a) 3 1 4 2 (b) 4 3 2 1
(c) 3 2 4 1 (d) 4 1 2 3
Show Answer

Answer

A. (3)

B. (1)

C. (4)

D. (2)

Formation of inner orbital complex and outer orbital complex determines hybridisation of molecule which inturn depends upon field strength of ligand and number of vacant d orbitals.

(i) Strong field ligand forms inner orbital complex with hybridisation d2sp3.

(ii) Weak field ligand forms outer orbital complex with hybridisation sp3d2.

According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as A. [Cr(H2O)6]3+

MOEC (Molecular orbital electronic configuration) of Cr3+ in [Cr(H2O)6]3+ is

Hybridisation =d2sp3

n (number of unrpaired electrons) =3

B. [Co(CN)4]2

MOEC of Co2+ in [Co(CN)4]2 is

Hybridisation =dsp2

n=1

C. [Ni(NH3)6]2+

MOEC of Ni2+ in [Ni(NH3)6]2+ is

Hybridisation =sp3d2

n=2

D. [MnF6]4

MOEC of Mn2+ in [MnF6]4 is

Hybridisation =sp3d2

n=5

Hence, correct choice can be represented by (a).

39. Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code.

Column I
(Complex species)
Column II
(Isomerism)
A. [Co(NH3)4Cl2]+ 1. Optical
B. cis [Co(en)2Cl2]+ 2. Ionisation
C. [Co(NH3)5(NO2)]Cl2 3. Coordination
D. [Co(NH3)6][Cr(CN)6] 4. Geometrical

Codes

A B C D
(a) 1 2 4 3
(b) 4 3 2 1
(c) 4 2 1 3
(d) 4 1 2 3
Show Answer

Answer

A. (4)

B. (1)

C. (2)

D. (3)

Isomerism in coordination compound is decided by type of ligands and geometry of coordination and arrangement of ligands.

A. [Co(NH3)4Cl2]+shows geometrical isomerism due to presence of two types of ligand whose [Co(NH3)4Cl2]+arrangement around central metal ion.

B. cis [Co(en)2Cl2]+shows optical isomer due to its non-superimposable mirror image relationship.

C. [Co(NH3)5(NO2)]Cl2 shows ionisation isomer due to its interchanging ligand from outside the ionisation sphere.

D. [Co(NH3)6][Cr(CN)6] shows coordination isomer due to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.

Hence, correct choice is (d).

40. Match the compounds given in Column I with the oxidation state of cobalt present in it (given in column II) and assign the correct code.

Column I
( Compound )
Column II
(Oxidation state of Co)
A. [Co(NCS)(NH3)5](SO3) 1. +4
B. [Co(NH3)4Cl2]SO4 2. 0
C. Na4[Co(S2O3)3] 3. +2
D. [CO2(CO)8] 4. +3

Code

A B C D
(a) 1 2 4 3
(b) 4 3 2 1
(c) 3 1 4 2
(d) 4 1 3 2
Show Answer

Answer

A. (4)

B. (1)

C. (3)

D. (2)

Oxidation state of CMI (central metal ion) can be calculated by considering the oxidation state of whole molecule is equal to charge present on coordination sphere.

A. [Co(NCS)(NH3)5]SO3.

Let oxidation state of Co is x.

x1+5×0=+2x=+2+1=+3

B. [Co(NH3)4Cl2]SO4

Let oxidation state of Co=x

x+4×0+2×(1)=+2x2=+2x=4

C. Na4[Co(S2O3)3]

Let oxidation state of Co=x

x+3×(2)=4x6=4x=4+6=+2

D. [Co(CO)8]

Let oxidation state of Co=x

x8×0=0x=0

Hence, correct choice is (d).

Assertion and Reason

In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. choose the correct answer out of the following choices.

(a) Assertion and reason both are true, reason is correct explanation of assertion.

(b) Assertion and reason both are true but reason is not the correct explanation of assertion.

(c) Assertion is true, reason is false.

(d) Assertion is false, reason is true.

41. Assertion (A) Toxic metal ions are removed by the chelating ligands.

Reason ( R ) Chelate complexes tend to be more stable.

Show Answer

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

Toxic metal ions are removed by chelating ligands. When a solution of chelating ligand is added to solution containing toxic metals ligands chelates the metal ions by formation of stable complex.

42. Assertion (A) [Cr(H2O6)]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature. Reason ( R ) Unpaired electrons are present in their d-orbitals.

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Answer

(b) Assertion and reason both are true but reason is not correct explanation of assertion. Correct reason is

[Cr(H2O6)]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature due to formation of more stable complex ion after gaining of electron.

43. Assertion (A) Linkage isomerism arises in coordination compounds containing ambidentate ligand.

Reason (R) Ambidentate ligand has two different donor atoms.

Show Answer

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Linkage isomerism arises in coordination compounds containing ambidentate ligands because ambidentate ligand has two different donor atoms.

e.g., SCN,NO2 etc.

44. Assertion (A) Complexes of MX6 and MX5L type ( X and L are unidentate) do not show geometrical isomerism.

Reason (R) Geometrical isomerism is not shown by complexes of coordination number 6.

Show Answer

Answer

(b) Assertion and reason both are correct and reason is not correct explanation of assertion.

Complexes of MX6 and MX5L type ( X and L are unidentate) do not show geometrical isomerism due to presence of plane of symmetry and necessary condition for showing geometrical isomerism is that complex is must of MA4B2 type or [M(AB)2X2] type

45. Assertion (A) [Fe(CN)6]3 ion shows magnetic moment corresponding to two unpaired electrons.

Reason ( R ) Because it has d2sp3 type hybridisation.

Show Answer

Answer

(d) Assertion is false but reason is true.

According to VBT, MOEC of Fe3+ in [Fe(CN)6]3 is

Hybridisation =d2sp3

n=1

Hence, correct assertion is

[Fe(CN)6]3 ion shows magnetic moment corresponding to one unpaired electron.

i.e.,

μ=n(n+2)=1(1+2)=3=1.73BM

Long Answer Type Questions

46. Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following

(a) [CoF6]3,[Co(H2O)6]2+,[Co(CN)6]3

(b) FeF63,[Fe(H2O)6]2+,[Fe(CN)6]4

Show Answer

Thinking Process

This problem includes conceptual mixing of crystal field theory and magnetic moment (μ) determination.

μ=n(n+2)BM where, n= number of unpaired electrons

Answer

(a) [CoF6]3.

Fis a weak field ligand.

Configuration of Co3+=3d6(. or .t2g4eg2)

Number of unpaired electrons (n)=4

Magnetic moment (μ)=n(n+2)=4(4+2)=24=4.9BM

[Co(H2O)6]2+,

H2O is a weak field ligand.

Configuration of Co2+=3d7(. or .t2g5eg2)

Number of unpaired electrons (n)=3

μ=3(3+2)=15=3.87BM

[Co(CN)6]3 i.e., Co3+

CN is strong field ligand.

Co3+=3d6(. or .t2g6eg0)

There is no unpaired electron, so it is diamagnetic.

μ=0

(b) [FeF6]3,

Number of unpaired electrons, n=5

μ=5(5+2)=35=5.92BM

[Fe(H2O)6]2+

Fe2+=3d6( or t2g4eg2)

Number of unpaired electrons, n=4

μ=4(4+2)Z=24=4.98BM

[Fe(CN)6]4

Since, CNis a strong field ligand, all the electrons get paired.

Fe2+=3d6(. or .t2g6eg0)

Because there is no unpaired electron, so it is diamagnetic in nature.

47. Using valence bond theory, explain the following in relation to the complexes given below

[Mn(CN)6]3,[Co(NH3)6]3+,[Cr(H2O)6]3+,[FeCl6]4

(a) Type of hybridisation

(b) Inner or outer orbital complex

(c) Magnetic behaviour

(d) Spin only magnetic moment value.

Show Answer

Answer

(a) [Mn(CN)6]3

(i) d2sp3 hybridisation

(ii) Inner orbital complex because (n1)d-orbitals are used.

(iii) Paramagnetic, as two unpaired electrons are present.

(iv) Spin only magnetic moment (μ)=2(2+2)=8=2.82BM

(b) [Co(NH3)6]3+

Co3+=3d64s0

(NH3. pair up the unpaired 3d electrons.)

(i) d2sp3 hybridisation

(ii) Inner orbital complex because of the involvement of (n1)d-orbital in bonding.

(iii) Diamagnetic, as no unpaired electron is present.

(iv) μ=n(n+2)=0(0+2)=0 (Zero)

(c) [Cr(H2O)6]3+

(i) d2sp3 hybridisation

(ii) Inner orbital complex (as (n1)d-orbital take part.)

(iii) Paramagnetic (as three unpaired electrons are present.)

(iv) μ=n(n+2)=3(3+2)=15=3.87BM

(d) [Fe(Cl)6]4

Fe2+=3d6

(i) sp3d2 hybridisation

(ii) Outer orbital complex because nd-orbitals are involved in hybridisation.

(iii) Paramagnetic (because of the presence of four unpaired electrons).

(iv) μ=n(n+2)=4(4+2)=24=4.9BM

Q.48 CoSO4Cl5NH3 exists in two isomeric forms ’ A ’ and ’ B ‘. Isomer ’ A ’ reacts with AgNO3 to give white precipitate, but does not react with BaCl2. Isomer ’ B ’ gives white precipitate with BaCl2 but does not react with AgNO3. Answer the following questions.

(a) Identify ’ A ’ and ’ B ’ and write their structural formulae.

(b) Name the type of isomerism involved.

(c) Give the IUPAC name of ’ A ’ and ’ B ‘.

Show Answer

Thinking Process

This problem is based on chemical properties of coordination compounds, ionisation isomerism, and nomenclature of coordination compounds.

Answer

A ’ gives precipitate with AgNO3, so in it Cl is present outside the coordination sphere.

B ’ gives precipitate with BaCl2, so in it SO42 is present outside the coordination sphere.

(a) So,A - [Co(NH3)5SO4]Cl B - [Co(NH3)5Cl]SO4

(b) Ionisation isomerism (as give different ions when subjected to ionisation.)

(c) [A], Pentaamminesulphatocobalt (III) chloride.

[B], Pentaamminechloridocobalt (III) sulphate.

49. What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?

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Answer

When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting energy, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over.

e.g., if green light is absorbed, the complex appears red.

In terms of crystal field theory, suppose there is an octahedral complex with empty eg level and unpaired electrons in the t2g level in ground level. If the unpaired electron absorbs light corresponding is blue-green region, it will excite to eg level and the complex will appear violet in colour.

In absence of ligand, crystal field splitting does not occur and the substance is colourless.

e.g., anhydrous CuSO4 is while, but CuSO45H2O is blue in colour.

50. Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?

Show Answer

Answer

Extent of splitting of d-orbitals is different in octahedral and tetrahedral field. CFSE in octahedral and tetra federal field are closely related as.

Δt=(49)ΔO

where,Δt= crystal field splitting energy in tetrahedral field

Δ0= crystal field splitting energy in octahedral field

Wavelength of light and CFSE are related to each other by formula

Δ0=E=hcλE1λ

So, higher wavelength of light is absorbed in octahedral complexes than tetrahedral complexes for same metal and ligands. Thus, different colours are observed.