Page No 254: - Chapter 9 Coordination Compounds Intext Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 9.5:

Explain on the basis of valence bond theory that [Ni(CN)₄]²⁻ ion with square

planar structure is diamagnetic and the [NiCl₄]²⁻ ion with tetrahedral geometry is paramagnetic.

Answer:

Ni is in the +2 oxidation state i.e., in d⁸ configuration.

There are 4 CN⁻ ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN⁻ ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.

It now undergoes dsp² hybridization. Since all electrons are paired, it is diamagnetic.

In case of [NiCl₄]²⁻, Cl⁻ ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp³ hybridization.

Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.

Question 9.6:

[NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?

Answer:

Though both [NiCl₄]²⁻ and [Ni(CO)₄] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl⁻ is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl₄]²⁻ is paramagnetic.

In Ni(CO)₄, Ni is in the zero oxidation state i.e., it has a configuration of 3d⁸ 4s^2.

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp³ hybridization. Since no unpaired electrons are present in this case, [Ni(CO)₄] is diamagnetic.

Question 9.7:

[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]³⁻ is weakly paramagnetic. Explain.

Answer:

In both  and  , Fe exists in the +3 oxidation state i.e., in d⁵ configuration.

Since CN⁻ is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.

Therefore,

On the other hand, H₂O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.

Therefore,

Thus, it is evident that  is strongly paramagnetic, while  is weakly paramagnetic.

Question 9.8:

Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.

Answer:

Predict the number of unpaired electrons in the square planar [Pt(CN)₄]²⁻ ion.Question 9.9:

Answer:

In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp² hybridization. Now, the electronic configuration of Pd(+2) is 5d⁸.

CN⁻ being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in

Question 9.10:

The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.

Answer:

Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron