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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.
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.
[NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
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 4s2.
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.
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
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.
Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
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| Oxidation state of cobalt = +3 | Oxidation state of Ni = +2 |
| Electronic configuration of cobalt = d6 | Electronic configuration of nickel = d8 |
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NH3 being a strong field ligand causes the pairing. Therefore, Cobalt can undergo d2sp3 hybridization.
Hence, it is an inner orbital complex. |
If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo d2sp3hybridization. Therefore, it undergoes sp3d2 hybridization.
Hence, it forms an outer orbital complex. |
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion.
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
The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
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| Mn is in the +2 oxidation state. | Mn is in the +2 oxidation state. |
| The electronic configuration is d5. | The electronic configuration is d5. |
| The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in | The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in T2g5eg0. |
Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron.
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