Important Questions for Class 12 Chemistry Chapter 9

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Important Questions for Class 12 Chemistry Chapter 9 Coordination Compounds Class 12 Important Questions

Question 1. Give an example of linkage isomerism. (Delhi) 2010

Answer: Linkage isomerism : When more than one atom in an ambidentate ligand is linked with central metal ion to form two types of complexes, then the formed isomers are called linkage isomers and the phenomenon is called linkage isomerism.

[Cr(H₂O)₅(NCS)]²⁺ Pentaaquathiocyanate chromium (III) ion

[Cr(H₂O)₅(NCS)]²⁺

Pentaaquaisothiocyanate chromium (III) ion

Question 2. Give an example of coordination isomerism. (Delhi 2010)

Answer:

Example : [Co(NH₃)₆] [Cr(CN)₆] and

[Cr(NH₃)₆] [CO(CN)₆]

Question 3. Give an example of ionization isomerism. (Delhi 2010)

Answer:

Example : [Pt (NH₃)₅ (Br)₃] SO₄ and

[Co (NH₃)₅ (SO₄)] Br

Question 4. Give IUPAC name of ionization isomer of [Ni(NH₃)₃NO₃]Cl. (Comptt. All India 2012)

Answer: IUPAC name : Triammine nitrato nickel (III) chloride

Question 5. Write down the formula of : Tetraamineaquachloridocobalt(III) chloride. (Comptt. All India 2012)

Answer: [Co(NH₃)₄(H₂O)Cl]Cl₂

Question 6. Indicate the types of isomerisms exhibited by the complex [Co(NH₃)₅ (NO₂)] (NO₃)₂. (At. no. Co = 27) (Comptt. All India 2012)

Answer: It shows ionisation isomerism and linkage isomerism.

Question 7. What type of bonding helps in stabilishing the a-helix structure of proteins? (Delhi 2013)

Answer: a-helix formation -» Intramolecular hydrogen bonding.

Question 8. Which complex ion is formed when undecomposed AgBr is washed with hypo solution in photography? (Comptt. All India 2013)

Answer: Sodium dithiosulphato argentate (I) complex is formed

Question 9. Give IUPAC name of the ionization isomer of [Ni(NH₃)₃NO₃]Cl. (Comptt. All India 2013)

Answer: IUPAC name : Triammine chlorido nickel (II) nitrate [Ni(NH₃)₃NO₃]Cl

Question 10. Give two examples of ligands which form coordination compounds useful in analytical chemistry. (Comptt. All India 2013)

Answer:

Examples :

(i) EDTA (Ethylene diamine tetra-acetic acid)

(ii) Dimethyl glyoxime (DMG)

Question 11. Which of the following is more stable complex and why?

[Co(NH₃)₆]³⁺ and [Co(en)₃]³⁺ (Delhi 2014)

Answer: [Co(en)₃]³⁺ is more stable complex than [CO(NH₃)₆]³⁺ because of chelate effect.

Question 12. What is the IUPAC name of the complex [Ni(NH₃)₆]Cl₂? (Comptt. Delhi 2015)

Answer: [Ni(NH₃)₆]Cl₂

IUPAC name : Hexaamminenickel (II) chloride.

Question 13. What is meant by chelate effect? (Comptt. All India 2015)

Answer: Chelate effect : When a bidentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six membered ring is formed. This effect is called Chelate effect. As a result, the stability of the complex increases.
Example: the complex of Ni²⁺ with ‘+ion’ is more stable than NH₃.

Question 14. Write the IUPAC name of the following complex : [CO(NH₃)₆]³⁺ (Comptt. Delhi 2016)

Answer: Hexaamminecobalt (III) ion.

Question 15. Write the IUPAC name of the following coordination compound [NiCl₄]^2-. (Comptt. All India 2016)

Answer: Tetrachloridonickelate (II) ion.

Question 16. Why are low spin tetrahedral complexes not formed? (Comptt. Delhi 2017)

Answer: Law spin tetrahedral complexes are rarely observed because orbital splitting energies for tetrahedral complexes are sufficiently large for forcing pairing.

Question 17. Write IUPAC name of the complex: [CoCl₂(en)₂]⁺ (Comptt. All India 2017)

Answer: [CoCl₂(en)₂]⁺: Dichloridobis(ethylenediamine)Cobalt(III)ion.

Question 18. Write IUPAC name of the complex [Co(NH₃)₄Cl(NO₂)]⁺. (Comptt. All India 2017)

Answer: Tetra amminechloridonitro cobalt (III) ion.

Question 19. Write IUPAC name of the complex: [CoCl₂(en²)]⁺ (Comptt. All India 2017)

Answer: Dichloridobis ethylenediamine cobalt (III) ion.

Coordination Compounds Class 12 Important Questions Short Answer Type -I [SA – I]

Question 20. Name the following coordination compounds according to IUPAC system of nomenclature :

(i) [Co(NH₃)₄ (H₂O)Cl]Cl₂

(ii) [CrCl₂(en)₂]Cl,

(en = ethane – 1, 2 – diamine) (Delhi 2010)

Answer:

(i) [CO(NH₃)₄ (H₂O)Cl]Cl₂

Tetraammine aquachlorido cobalt (III) chloride

(ii) [CrCl₂(en)₂]Cl

Dichlorobis (ethane-1, 2-diamine) chromium (III) chloride

Question 21. Describe the shape and magnetic behaviour of following complexes :

(i) [CO(NH₃)₆]³⁺

(ii) [Ni(CN)₄]^2- (At. No. Co = 27, Ni = 28) (Delhi 2010)

Answer:

(i) [CO(NH₃)₆]³⁺ :

Orbitals of CO³⁺ ion :

Hybridization : d²sp³ Shape : Octahedral Magnetic behaviour : Diamagnetic (absence of unpaired electrons)

(ii) [Ni(CN)₄]^2-

Question 22. How is the stability of a co-ordination compound in solution decided ? How is the dissociation constant of a complex defined? (Comptt. All India 2012)

Answer: Stability of a complex in solution means the measure of resistance to the replacement of a ligand by some other ligand. This stablility can be expressed in terms of equilibrium constant. Let the reaction between metal and ligand be represented as M^a+ + nL^x- ⇌ ML_n^b+

Stability or Dissociation constant (K)

= \(\frac{\left[\mathrm{ML}_{\mathrm{n}}^{b+}\right]}{\left[\mathrm{M}^{a+}\right]\left[\mathrm{L}^{x-}\right]^{\mathrm{n}}}\)

The reciprocal of the stability constant K is known as instability constant or dissociation constant

\(\mathrm{K}_{\mathrm{i}}=\frac{1}{\mathrm{K}}=\frac{\left[\mathrm{M}^{a+}\right]\left[\mathrm{L}^{x-}\right]^{\mathrm{n}}}{\left[\mathrm{ML}_{\mathrm{n}}^{b+}\right]}\)

Factors affecting the stability of a complex ion

(i) Nature of metal ion : Greater the charge and smaller the size of the ion, more is its charge density and greater will be stability of the complex.

(ii) Nature of ligand : More the basicity of ligand, more is its tendency to donate electron pair and therefore, more is the stability of the complex.

Question 23. [Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain. (At. no. Fe = 26) (Comptt. All India 2012)

Answer: In both the cases, Fe is in oxidation state +3. Outer electronic configuration of Fe+3 is :

In the presence of CN^–, the 3d electrons pair up leaving only one unpaired electron. The hybridisation involved is d²sp³ forming inner orbital complex which is weakly paramagnetic. In the presence of H₂O (a weak ligand), 3d electrons do not pair up. The hybridisation involved is sp³d² forming an outer orbital complex. As it contains five unpaired electrons so it is strongly paramagnetic. .

Question 24. Explain why [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex. (At. no. Co = 27, Ni = 28) (Comptt. All India 2013)

Answer: In [Co(NH₃)₆]³⁺, the d-electrons of Co³⁺ ([Ar]3d⁶ 45°) get paired leaving behind two empty d-orbital and undergo d²sp³ hybridization and hence inner orbital complex, while in [Ni(NH₃)₆]²⁺ the d-electrons of Ni²⁺ ([Ar]3d⁸ 45°) do not pair up and use outer 4d subshell hence outer orbital complex.

Question 25. Write the IUPAC name of the complex [Cr(NH₃)₄Cl₂]⁺. What type of isomerism does it exhibit? (Delhi 2014)

Answer: IUPAC name : Tetraamine dichlorido chromium (III) ion. It exhibits geometrical isomerism.

Question 26. (i) Write down the IUPAC name of the following complex :

[Cr(NH₃)₂CI₃(en)]Cl (en = ethylenediamine)

(ii) Write the formula for the following complex : Pentaamminenitrito-o-Cobalt (III) (Delhi 2015)

Answer:

(I) [Cr(NH₃)₂Cl₃(en)]Cl

IUPAC name : Diammine dichlorido ethylenediamine chromium (III) chloride.

(ii) [Co(NH₃)₅(ONO)]²⁺

Question 27.

(i) Write down the IUPAC name of the following complex : [CO(NH₃)₅Cl]²⁺

(ii) Write the formula for the following complex : Potassium tetrachloridonickelate (II) (All India 2015)

Answer:

(i) [CO(NH₃)₅Cl]²⁺

IUPAC name : Pentaammine chlorido cobalt (III) ion

(ii) Formula of the complex potassium tetrachloridonickelate (II) K₂[NiCl₄]

Question 28. When a co-ordination compound CrCl₃.6H₂O is mixed with AgNO₃, 2 moles of AgCl are precipitated per mole of the compound. Write

(i) Structural formula of the complex.

(ii) IUPAC name of the complex. (Delhi 2016)

Answer:

(i) The complex formed on mixing a coordination compound CrCl₃.6H₂O with AgNO₃ is as follows

CrCl₃.6H₂O + AgNO₃ → [Cr(H₂O₅)Cl]Cl₂. H₂O

(ii) Pentaaquachloridochromium (III) chloride monohydrate

Question 29. When a coordination compound CoCl₃.6NH₃ is mixed with AgNO₃, 3 moles of AgCl are precipitated per mole of the compound. Write

(i) Structural formula of the complex

(ii) IUPAC name of the complex (All India 2016)

Answer:

(i) Complex so formed is: CoCl₃.6NH₃ + AgNO₃ → [Co(NH₃)₆]Cl₃

(ii) IUPAC name of complex is: Hexaamminecobalt (III) chloride

Question 30. Using IUPAC norms write the formulae for the following:

(i) Sodium dicyanidoaurate (I)

(it) Tetraamminechloridonitrito-N-platinum (IV) sulphate (Delhi 2017)

Answer:

(i) Na[Au(CN)₂]

(ii) [Pt(NH₃)₄ Cl(NO₂)] (SO₄)

Question 31. Using IUPAC norms write the formulae for the following:

(a) Tris(ethane-1,2-diamine) chromium (III) chloride

(b) Potassium tetrahydroxozincate(II) (All India 2017)

Answer:

(a) [Cr(en)₃] Cl₃

(b) K₂[Zn(OH)₄]

Question 32. Using IUPAC norms write the formulae for the following:

(a) Potassium trioxalatoaluminate (III)

(b) Dichloridobis(ethane-l, 2-diamine) cobalt (III) (All India 2017)

Answer:

(a) K₃[Al(C₂O₄)₃]

(b) [Co(Cl)₂(en)₂]⁺

Coordination Compounds Class 12 Important Questions Short Answer Type -II [SA – II]

Question 33. For the complex [Fe(en)₂Cl₂], Cl, (en = ethylene diamine), identify

(i) the oxidation number of iron,

(ii) the hybrid orbitals and the shape of the complex,

(iii) the magnetic behaviour of the complex,

(iv) the number of geometrical isomers,

(v) whether there is an optical isomer also, and

(vi) name of the complex. (At. no. of Fe = 26) (Delhi 2009)

Answer:

(i) [Fe(en)₂Cl₂] Cl or x + 0 + 2 (-1) + (-1) = 0

x + (- 3) = 0 or x = + 3

∴ Oxidation number of iron, x = + 3

(ii) The complex has two bidentate ligands and two monodentate ligands. Therefore, the coordination number is 6 and hybridization will be d²sp³ and shape will be octahedral.

(iii) In the complex ₂₆Fe³⁺ = 3d⁵ 4s⁰ 4p⁰

Due to presence of one unpaired electrons in d orbitals the complex is paramagnetic.

The number of geometrical isomers are two.

(v) In coordination complex of [Fe(en)₂Cl₂] Cl, only cis-isomer shows optical isomerism.

(vi) Name of complex: Dichloridobis (ethane-1, 2- diamine) Iron (III) chloride.

Question 34. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved :

(i) [CoF₄]^2-

(ii) [Cr(H₂O)₂(C₂O₄)₂]^–

(iii) [Ni(CO)₄] (Atomic number : Co = 27, Cr = 24, Ni = 28) (Delhi 2009)

Answer:

(i) [COF4]2_ : Tetrafluorido cobalt (III) ion

Coordination number = 4 Shape = Tetrahedral Hybridisation = sp³

Question 35. Giving a suitable example for each, explain the following :

(i) Crystal field splitting

(ii) Linkage isomerism

(iii) Ambidentate ligand (All India 2009)

Answer:

(i) Crystal field splitting: It is the splitting of the degenerate energy levels due to the presence of ligands. When ligand approaches a transition metal ion, the degenerate d-orbitals split into two sets, one with lower energy and the other with higher energy. This is known as crystal field splitting and the difference between the lower energy set and higher energy set is known as crystal field splitting energy (CFSE)

Example : 3d⁵ of Mn²⁺

(ii) Linkage isomerism: When more than one atom in an ambidentate ligand is linked with central metal ion to form two types of complexes, then the formed isomers are called linkage isomers and the phenomenon is called linkage isomerism.

[Cr(H₂O)₅(NCS)]²⁺ Pentaaquathiocyanate chromium (III) ion

[Cr(H₂O)₅(NCS)]²⁺

Pentaaquaisothiocyanate chromium (III) ion

(iii) Ambidentate ligand: The monodentate ligands with more than one coordinating atoms is known as ambidentate ligand. Monodentate ligands have only one atom capable of binding to a central metal atom or ion. For example, the nitrate ion NO₂^– can bind to the central metal atom/ion at either the nitrogen atom or one of the oxygen atoms.

Example : — SCN thiocyanate, — NCS isothiocyanate

Question 36. Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units :

[Co(NH₃)₆]⁺³, [Cr(NH₃)₆]³⁺, Ni(CO)₄

(At. nos. : Co = 27, Cr = 24, Ni = 28) (All India 2009)

Answer:

(i) [Co(NH₃)₆]⁺³ → Octahedral shape, d²sp³ hybridisation, diamagnetic

Formation of [Co(NH₂)₆]⁺³ → oxidation state of Co is +3.

Question 37. Explain the following:

(i) Low spin octahedral complexes of nickel are not known.

(ii) The π-complexes are known for transition elements only.

(iii) CO is a stronger ligand than NTL, for many metals. (All India 2009)

Answer:

(i) The electronic configuration of Ni is [Ar] 3d⁸ 4s² which shows that it can only form two types of complexes i.e. square planar (dsp²) in presence of strong ligand and tetrahedral (sp3) in presence of weak ligand. There are four empty orbitals in Ni while octahedral complexes require six empty orbitals.

(ii) Due to presence of empty d-orbitals in transition metals, they can accept electron pairs from ligands containing π electrons and hence can form ic-bonding complexes.
Example : ligands like C₅H₅, C₆H₆ etc.

(iii) Due to greater magnitude of Δ₀, CO produces strong fields which cause more splitting of d-orbitals and moreover it is also able to form π bond due to back bonding.

Question 38. Compare the following complexes’ with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units:

(i) [Ni(CN)₄]^2- (ii) [NiCl₄]^2- (iii) [CoF₆]^3-

[At. Nos. : Ni = 28; Co = 27] (All India 2009)

Answer:

(i) [Ni(CN)₄]^2-

Shape : Octahedral outer orbital complex

Hybridisation : sp³d²

Magnetic behaviour : Paramagnetic (4 unpaired electrons)

Question 39. Explain the following cases giving appropriate reasons :

(i) Nickel does not form low spin octahedral complexes.

(ii) The n-complexes are known for the transition metals only. (All India 2010)

Answer:

(i) The electronic configuration of Ni is [Ar] 3d⁸ 4s² which shows that it can only form two types of complexes i.e. square planar (dsp²) in presence of strong ligand and tetrahedral (sp³) in presence of weak ligand. There are four empty orbitals in Ni while octahedral complexes require six empty orbitals.

(ii) Due to presence of empty d-orbitals in transition metals, they can accept electron pairs from ligands containing π electrons and hence can form ic-bonding complexes.

Example : ligands like C₅H₅, C₆H₆ etc.

(iii) Due to greater magnitude of Δ₀, CO produces strong fields which cause more splitting of d-orbitals and moreover it is also able to form π bond due to back bonding.

Question 40. Write the name, the state of hybridization, the shape and the magnetic behaviour of the following complexes :

[CoCl₄]^2-, [Ni(CN)₄]^2-, [Cr(H₂O)₂(C₂O₄)₂]^–

(At No. : Co = 27, Ni = 28, Cr = 240 (All India 2010)

Answer:

(i) [CoCl₂]^– :

Name – Tetra chlorido Cobalt (II) ion

Shape = Tetrahedral

Hybridization = sp³

Magnetic property = Paramagnetic

(ii) [Ni(CN)₄]^2-

(iii) [Cr(H₂O)₂(C₂O₄)₂]^– :

Name = Diaquabis (oxalato) chromium (III) ion
Shape = Octahedral

Hybridization = d²sp³

Magnetic property = Paramagnetic

Question 41. Write the name, stereochemistry and magnetic behaviour of the following : (At. nos. Mn = 25, Co = 27, Ni = 28) (Delhi 2011)

(i) K₄[Mn(CN)₆]

(ii) [CO(NH₃)₅ Cl]Cl₂

(iii) K₂ [Ni(CN)₂]

Answer:

(i) K₄[Mn(CN)₂] : IUPAC name : Potassium Hexacyano manganate (II)

Geometry : Octahedral

Magnetic behaviour: Paramagnetic (one unpaired electron)

(ii) [CO(NH₃)₅ Cl]Cl₂ :

Name : Pentaammine chlorido cobalt (III) chloride

Shape : Octahedral (∵ Coordination number = 6)

Hybridization : d²sp³ Magnetic behaviour : Diamagnetic (no unpaired electrons)

(iii) K₂ [Ni(CN)₄] :

Name : Potassium tetracyanonickelate (II)

Shape : Square planar Hybridization : dsp² (∵ Coordination number = 4)

Hybridization : dsp² Magnetic behaviour : Diamagnetic

Question 42. Explain the following terms giving a suitable example in each case :

(i) Ambident ligand

(ii) Denticity of a ligand

(iii) Crystal field splitting in an octahedral field (All India 2011)

Answer: (i) Ambidentate ligand : The monodentate ligands with more than one coordinating atoms is known as ambidentate ligand. Monodentate ligands have only one atom capable of binding to a central metal atom or ion. For example, the nitrate ion NO₂^– can bind to the central metal atom/ion at either the nitrogen atom or one of the oxygen atoms.
Example : — SCN thiocyanate, — NCS isothiocyanate

(ii) Denticity of a ligand: The number of donor atoms in a ligand which forms coordinate bond with the central metal atom are called denticity of a ligand.

Example : If donor atom is one then it is called Monodentate ligand, if it is two, then it is called Bidendentate and so on.

(iii) Crystal field splitting: It is the splitting of the degenerate energy levels due to the presence of ligands. When ligand approaches a transition metal ion, the degenerate d-orbitals split into two sets, one with lower energy and the other with higher energy. This is known as crystal field splitting and the difference between the lower energy set and higher energy set is known as crystal field splitting energy (CFSE)

Example : 3d⁵ of Mn²⁺

Question 43. Write the structures and names of all the stereoisomers of the following compounds :

(i) [Co(en)₃]Cl₃

(ii) [Pt(NH₃)₂Cl₂]

(iii) [Fe(NH₃)₄Cl₂]Cl (All India 2011)

Answer:

(i) [Co(en)₃]Cl₃

Name ; Tris (ethane -1,2-diamine cobalt (III) chloride)

Hybridization : d²sp² (∵ Coordination number = 6)

Shape : Octahedral

Magnetic behaviour : Diamagnetic

(ii) [Pt(NH₃)₂Cl₂]

Name : Diammine dichlorido platinum (II) ion

Hybridization : dsp²( ∵ Coordination number = 4)

Shape : Square planar

Magnetic behaviour : Diamagnetic

(iii) [Fe(NH₃)₄Cl₂]Cl

Name : Tetraammine dichlorido Iron (III) chloride

Hybridization : d²sp³ (∵ Coordination number = 6)

Shape : Octahedral

Magnetic behaviour diamagnetic :

Question 44. Write the state of hybridization, the shape and the magnetic behaviour of the following complex entities :

(i) [Cr(NH₃)₄Cl₂]Cl

(ii) [Co(en)₃]Cl₃

(iii) K₂[Ni(CN)₄] (All India 2011)

Answer:

(i) [Cr(NH₃)_s4Cl₂]Cl :

Hybridization : d²sp³

Shape : Octahedral

Magnetic behaviour: Paramagnetic

(ii) [Co(en)₃] Cl₃ :

Hybridization : d2sp³

Shape : Octahedral

Magnetic behaviour : Diamagnetic

(iii) K₃[Ni(CN)₄] :

Hybridization ; dsp²

Shape : Square planar

Magnetic behaviour: Diamagnetic

Question 45. Give the formula of each of the following coordination entities :

(i) Co³⁺ ion is bound to one Cl^–, one NH₃ molecule and two bidentate ethylene diamine (en) molecules.

(ii) Ni²⁺ ion is bound to two water molecules and two oxalate ions.

Write the name and magnetic behaviour of each of the above coordination entities.

(At nos. Co = 27, Ni = 28) (Delhi 2011)

Answer:

(i) [Co (en)₂ (NH₃)Cl]²⁺ : amminechloridobis = (ethane-1, 2-diamine) cobalt (ID), “diamagnetic”

(ii) [Ni(C₂O₄)₂ (H₂O)₂]^-2 : diaquadioxalatonickelate (II), “paramagnetic”

Question 46. State a reason for each of the following situations:

(i) Co²⁺ is easily oxidized to Co³⁺ in presence of a strong ligand.

(ii) CO is a stronger complexing reagent than NH₃.

(iii) The molecular shape of [Ni(CO)₄] is not the same as that of [Ni(CN)₄]^2- (Delhi 2011)

Answer:

(i) Because in the presence of strong ligands, the crystal field splitting energy is more than the energy required to oxidise Co²⁺.

(ii) This is due to the formation of π – bond by back donation of electrons from metal to carbon of CO or due to synergic bonding.

(iii) CO is a stronger field ligand than CN. Ni is in zero oxidation state in Ni(CO)₄ and has tetrahedral geometry. But, Ni is in +2 oxidation state in [Ni(CN)₄]^2- and has dsp² hybridization (different geometry than tetrahedral sp³).

Question 47. Write the name, the structure and the magnetic behaviour of each one of the following complexes:

(i) [Pt(NH₃)₂Cl(NO₂)]

(ii) [Co(NH₃)₄Cl₂]Cl

(iii) Ni(CO)₄ (Atmos. Co = 27, Ni = 28, Pt = 78) (Delhi 2011)

Answer:

(i) [Pt(NH₃)₂Cl(NO₂)]

Name : Diamine chloridonitroplatinum II

Structure :

Magnetic behaviour: paramagnetic

(ii) [Co(NH₃)₄Cl₂]Cl

Name : Tetraamminedichloridocobalt (III) chloride

Structure : octahedral

Magnetic behaviour : diamagnetic

(iii) Ni(CO)₄

Name : Tetracarbonylnickel (O)

Structure : tetrahedral

Magnetic behaviour : diamagnetic.

More Resources for Class 12

Question 48. 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)₃ Cl₃] (Atomic numbers Cr = 24, Co = 27) (All India 2011)

Answer:

(i) [Co(en)₂Cl₂]⁺

Name : Dichlorido bis (en = ethan-1, 2-diamine) Cobalt (III)

The geometrical isomers of [CoCl₂(en)₂]⁺ (2 isomers)

(ii) [Cr(C₂O₄)₃]^3-

Name : Trioxalatochromate (III) ion

(iii) [Co(NH₃)₃ Cl₃]

Name : Triamminetrichlorido cobalt (III)

Structure : The geometrical isomers of [Co(NH₃)₃Cl₃] (2 isomers) :

Question 49. Name the following coordination entities and describe their structures :

(i) [Fe(CN)₆]^4-

(ii) [Cr(NH₃)₄Cl₂]⁺

(iii) [Ni(CN)₄]^2-

(Atomic numbers Fe = 26, Cr = 24, Ni = 28) (All India 2011)

Answer:

(i) [Fe(CN)₆]^4-

Name : Hexacyanoferate (II) ion

Question 50.

(a) Give two examples of coordination compounds used in industries.

(b) Using valence bond theory, explain the geometry and magnetic behaviour of [Co(NH₃)₆]³⁺

(At. no. of Co = 27) (Comptt. Delhi 2011)

Answer:

(a) Examples:

(i) Pure Ni can be obtained from Ni(CO)₄

(ii) Gold and Ag are extracted by the use of complex formation like Na[Ag(CN)₂].

It involves d²sp³ hybridisation, Octahedral shape and diamagnetic due to absence of unpaired electrons..