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Chemistry Class 12 NCERT Solutions: Chapter 9 Coordination Compounds Part 13
Q: 4. Give evidence that are ionization isomers.
Answer:
When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products.
Q: 5. Explain on the basis of valence bond theory that with square planar structure is diamagnetic and the with tetrahedral geometry is paramagnetic.
Answer:
Ni is in the oxidation state i.e.. , in configuration.
There are ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN β ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.
It now undergoes hybridization. Since all electrons are paired, it is diamagnetic. In case of is a weak field ligand. Therefore, it does not lead to the pairing of unpaired electrons. Therefore, it undergoes sp3 hybridization.
Since there are unpaired electrons in this case, it is paramagnetic in nature.
Q: 6. is paramagnetic while is diamagnetic though both are tetrahedral. Why?
Answer:
Though both and are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. is a weak field ligand and it does not cause the pairing of unpaired electrons. Hence, is paramagnetic.
In , Ni is in the zero oxidation state i.e.. , it has a configuration of .
But is a strong field ligand. Therefore, it causes the pairing of unpaired electrons. Also, it causes the electrons to shift to the orbital, thereby giving rise to hybridization. Since no unpaired electrons are present in this case, is diamagnetic.
Q: 7. is strongly paramagnetic whereas is weakly paramagnetic. Explain.
In both , exists in the oxidation state i.e.. , in configuration.
Since 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, is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is .
Therefore,
Thus, it is evident that is strongly paramagnetic, while is weakly paramagnetic.