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The tetrachloro complexes of Ni(ll) and Pd(ll) respectively are (atomic number of Ni and Pd are 28 and 46 respectively).
- a)diamagnetic and diamagnetic
- b)diamagnetic and paramagnetic
- c)paramagnetic and diamagnetic
- d)paramagnetic and paramagnetic
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The tetrachloro complexes of Ni(ll) and Pd(ll) respectively are (atomi...
ln [NiCI4]2- , Ni has sp3 hybridisation with 2 unpaired electrons, i.e. paramagnetic.
In [PdCI4]2- Pd has dsp2 hybridisation and it has no unpaired electrons, i.e. diamagnetic.
In [PdCI4]2- Pd has dsp2 hybridisation and it has no unpaired electrons, i.e. diamagnetic.
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The tetrachloro complexes of Ni(ll) and Pd(ll) respectively are (atomi...
The tetrachloro complexes of Ni(ll) and Pd(ll) have Ni and Pd ions in the +2 oxidation state, which means they have two fewer electrons than their neutral atoms. To determine the magnetic properties of these complexes, we need to consider the electronic configuration and the presence of unpaired electrons.
Electronic configuration of Ni(ll): [Ar] 3d8 4s2
Electronic configuration of Pd(ll): [Kr] 4d8 5s2
Diamagnetic vs. Paramagnetic:
- Diamagnetic substances have all their electrons paired up and do not have any net magnetic moment.
- Paramagnetic substances have unpaired electrons, which result in a net magnetic moment.
Analyzing the electronic configurations:
- Both Ni(ll) and Pd(ll) have partially filled d-orbitals (d8 configuration), which means they can exhibit paramagnetic behavior due to the presence of unpaired electrons.
- However, in the case of tetrachloro complexes, the chloride ions (Cl-) act as strong ligands, which cause a significant amount of pairing of electrons in the d-orbitals. This effect is known as the "ligand field stabilization energy" (LFSE).
- The LFSE is greater for Ni(ll) compared to Pd(ll) because Ni has a smaller atomic size and higher charge density, leading to stronger interaction between the metal ion and the ligands.
- As a result, the tetrachloro complex of Ni(ll) has a greater tendency to pair up its electrons and becomes diamagnetic.
- On the other hand, the tetrachloro complex of Pd(ll) has a weaker LFSE and retains some unpaired electrons, making it paramagnetic.
Summary:
- The tetrachloro complex of Ni(ll) is diamagnetic because of the strong ligand field stabilization energy, which causes complete pairing of electrons.
- The tetrachloro complex of Pd(ll) is paramagnetic because it has a weaker ligand field stabilization energy and retains some unpaired electrons.
Electronic configuration of Ni(ll): [Ar] 3d8 4s2
Electronic configuration of Pd(ll): [Kr] 4d8 5s2
Diamagnetic vs. Paramagnetic:
- Diamagnetic substances have all their electrons paired up and do not have any net magnetic moment.
- Paramagnetic substances have unpaired electrons, which result in a net magnetic moment.
Analyzing the electronic configurations:
- Both Ni(ll) and Pd(ll) have partially filled d-orbitals (d8 configuration), which means they can exhibit paramagnetic behavior due to the presence of unpaired electrons.
- However, in the case of tetrachloro complexes, the chloride ions (Cl-) act as strong ligands, which cause a significant amount of pairing of electrons in the d-orbitals. This effect is known as the "ligand field stabilization energy" (LFSE).
- The LFSE is greater for Ni(ll) compared to Pd(ll) because Ni has a smaller atomic size and higher charge density, leading to stronger interaction between the metal ion and the ligands.
- As a result, the tetrachloro complex of Ni(ll) has a greater tendency to pair up its electrons and becomes diamagnetic.
- On the other hand, the tetrachloro complex of Pd(ll) has a weaker LFSE and retains some unpaired electrons, making it paramagnetic.
Summary:
- The tetrachloro complex of Ni(ll) is diamagnetic because of the strong ligand field stabilization energy, which causes complete pairing of electrons.
- The tetrachloro complex of Pd(ll) is paramagnetic because it has a weaker ligand field stabilization energy and retains some unpaired electrons.
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The tetrachloro complexes of Ni(ll) and Pd(ll) respectively are (atomic number of Ni and Pd are 28 and 46 respectively).a)diamagnetic and diamagneticb)diamagnetic and paramagneticc)paramagnetic and diamagneticd)paramagnetic and paramagneticCorrect answer is option 'C'. Can you explain this answer?
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