From the aboveisomers, none will exhibit optical isomers. Tetrahedral complexes rarely showoptical isomerization. They do so only in the presence of unsymmetricalchelating agents.

Question - 23 : -
Aqueous copper sulphate solution (blue incolour) gives:

Answer - 23 : -

(i) a green precipitate with aqueous potassium fluoride, and

(ii) a bright green solution with aqueous potassium chloride

Explain these experimental results.

Answer

AqueousCuSO₄ exists as [Cu(H₂O)₄]SO₄.It is blue in colour due to the presence of

[Cu[H₂O)₄]²⁺ ions.

(i) When KF is added:

(ii) When KCl is added:

Inboth these cases, the weak field ligand water is replaced by the F⁻ andCl⁻ ions.

Question - 24 : -
Whatis the coordination entity formed when excess of aqueous KCN is added to anaqueous solution of copper sulphate?

Answer - 24 : - Why is it that no precipitate of copper sulphide is obtained when H_2S(g) is passed through this solution?

Answer

i.e.,

Thus, the coordination entity formed in the process is K₂[Cu(CN)₄].

is a very stable complex, which does not ionize to give Cu²⁺ ions when added to water. Hence, Cu²⁺ ions are not precipitated when H₂S₍g₎is passed through the solution.

Question - 25 : -
Discuss the nature of bonding in thefollowing coordination entities on the basis of valence bond theory:

Answer - 25 : -

(i) [Fe(CN)_6]_4−

(ii) [FeF_6]_3−

(iii) [Co(C_2O_4)3]_3−

(iv) [CoF_6]_3−

_Answer

(i) [Fe(CN)₆]⁴⁻

In the above coordination complex, ironexists in the +II oxidation state.

Fe²⁺ :Electronic configuration is 3d⁶

Orbitalsof Fe²⁺ ion:

As CN⁻ isa strong field ligand, it causes the pairing of the unpaired 3d electrons.

Sincethere are six ligands around the central metal ion, the most feasiblehybridization is d²sp³.

d²sp³hybridizedorbitals of Fe²⁺ are:

6electron pairs from CN⁻ ions occupy thesix hybrid d²sp³orbitals.

Then

Hence, the geometry of the complex isoctahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF₆]³⁻

In this complex, the oxidation state of Fe is+3.

Orbitalsof Fe⁺³ ion:

Thereare 6 F⁻ ions. Thus, it will undergo d²sp³or sp³d² hybridization.As F⁻is a weak field ligand, it does not cause thepairing of the electrons in the 3d orbital. Hence, the mostfeasible hybridization is sp³d².

sp³d² hybridizedorbitals of Fe are:

Hence, the geometry of the complex is foundto be octahedral.

(iii) [Co(C₂O₄)₃]³⁻

Cobalt exists in the +3 oxidation state inthe given complex.

Orbitals of Co³⁺ ion:

Oxalateis a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbitalelectrons. As there are 6 ligands, hybridization has to be either sp³d² or d²sp³ hybridization.

sp³d² hybridization of Co^3+:

The 6electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand)occupy these sp³d² orbitals.

Hence, the geometry of the complex is foundto be octahedral.

(iv) [CoF₆]³⁻

Cobalt exists in the +3 oxidation state.

Orbitalsof Co³⁺ ion:

Again,fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons.As a result, the Co³⁺ ion will undergo sp³d² hybridization.

sp³d² hybridizedorbitals of Co³⁺ ion are:

Hence, the geometryof the complex is octahedral and paramagnetic.

Question - 26 : -
Drawfigure to show the splitting of d orbitals in an octahedralcrystal field.

Answer - 26 : - The splitting of the d orbitals in an octahedral fieldtakes palce in such a way that ,

experience a rise in energy and form the eglevel, while dxy, dyzand dzx experience a fall in energy and form the t₂g level.

Question - 27 : -
What isspectrochemical series? Explain the difference between a weak field ligand anda strong field ligand.

Answer - 27 : -

Aspectrochemical series is the arrangement of common ligands in the increasingorder of their crystal-field splitting energy (CFSE) values. The ligandspresent on the R.H.S of the series are strong field ligands while that on theL.H.S are weak field ligands. Also, strong field ligands cause higher splittingin the d orbitals than weak field ligands.

I−< Br⁻ < S²⁻ ⁻ < Cl⁻₃ < F⁻ ⁻ < C₂O₄²⁻ ∼ H₂O< NCS⁻ ∼ H⁻ ⁻< NH₃ < en ∼SO₃²⁻ < NO₂⁻ < phen < CO

Question - 28 : -
Whatis crystal field splitting energy? How does the magnitude of Δ_o decidethe actual configuration of d-orbitals in a coordination entity?

Answer - 28 : -

Thedegenerate d-orbitals (in a spherical field environment) split intotwo levels i.e., eg and t_2g inthe presence of ligands. The splitting of the degenerate levels due to thepresence of ligands is called the crystal-field splitting while the energydifference between the two levels (e_g and t_2g)is called the crystal-field splitting energy. It is denoted by Δ_o.

After the orbitalshave split, the filling of the electrons takes place. After 1 electron (each)has been filled in the three t_2g orbitals, thefilling of the fourth electron takes place in two ways. It can enter the e_g orbital (giving rise to t_2g³ e_g¹ like electronic configuration) or thepairing of the electrons can take place in the t_2gorbitals (giving rise to t_2g⁴ e_g⁰ like electronic configuration). If the Δ_o value of a ligand is less than the pairing energy (P), then theelectrons enter the e_g orbital. On theother hand, if the Δ_o value of a ligand is more than the pairingenergy (P), then the electrons enter the t_2gorbital.

Question - 29 : -
[Cr(NH₃)₆]³⁺ isparamagnetic while [Ni(CN)₄]²⁻ isdiamagnetic. Explain why?

Answer - 29 : -

Cr isin the +3 oxidation state i.e., d³ configuration.Also, NH₃ is a weak field ligand that does notcause the pairing of the electrons in the 3d orbital.

Cr³⁺

Therefore,it undergoes d²sp³ hybridizationand the electrons in the 3d orbitals remain unpaired. Hence, it isparamagnetic in nature.

In[Ni(CN)₄]²⁻, Ni exists in the +2oxidation state i.e., d⁸ configuration.

Ni^2+:

CN⁻ isa strong field ligand. It causes the pairing of the 3d orbitalelectrons. Then, Ni²⁺ undergoes dsp² hybridization.

As there are nounpaired electrons, it is diamagnetic.

Question - 30 : -
Asolution of [Ni(H₂O)₆]²⁺ isgreen but a solution of [Ni(CN)₄]²⁻ iscolourless. Explain.

Answer - 30 : - In [Ni(H₂O)₆]²⁺, isa weak field ligand. Therefore, there are unpaired electrons in Ni²⁺.In this complex, the d electrons from the lower energy levelcan be excited to the higher energy level i.e., the possibility of d−d transitionis present. Hence, Ni(H₂O)₆]²⁺ iscoloured.
In [Ni(CN)₄]²⁻, the electrons are all paired as CN^– is a strong field ligand. Therefore, d-d transitionis not possible in [Ni(CN)₄]²⁻. Hence, it iscolourless.

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