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[Ni(CN)4]2- and [NiCl4]2- are
- a)both diamagnetic
- b)both paramagnetic
- c)diamagnetic and paramagnetic respectively
- d)antiferromagnetic and diamagnetic respectively
Correct answer is option 'C'. Can you explain this answer?
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[Ni(CN)4]2- and [NiCl4]2- area)both diamagneticb)both paramagneticc)di...
[Ni(CN)4]2- in this CN is strong field ligand so pairing of electrons takes place and there are no unpaired electrons. So it is Diamagnetic. where as in [NiCl4]2- , Cl is weak filed ligand so pairing of electrons not possible. So there are 2 unpaired electrons in [NiCl4]2- . So it is paramagnetic.
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[Ni(CN)4]2- and [NiCl4]2- area)both diamagneticb)both paramagneticc)di...
Introduction:
In this question, we are given two compounds, [Ni(CN)4]2- and [NiCl4]2-, and we need to determine their magnetic properties. Specifically, we need to identify if they are diamagnetic or paramagnetic.
Explanation:
To determine the magnetic properties of a compound, we need to consider the electronic configuration of the central metal ion and the ligands surrounding it. Let's analyze each compound separately.
[Ni(CN)4]2-:
To determine the electronic configuration of the central Ni2+ ion, we need to know the atomic number of Nickel (Ni), which is 28. The 2+ charge indicates that two electrons have been removed from the neutral atom. Therefore, the electronic configuration of Ni2+ is 1s22s22p63s23p63d8.
The CN- ligands are strong-field ligands, which means they pair electrons in the d orbitals. Thus, the electronic configuration of [Ni(CN)4]2- is predicted to be d8 with all the electrons paired.
[NiCl4]2-:
The electronic configuration of the central Ni2+ ion remains the same as in the previous compound, which is 1s22s22p63s23p63d8.
The Cl- ligands are weak-field ligands, which means they do not pair electrons in the d orbitals. Thus, the electronic configuration of [NiCl4]2- is also predicted to be d8, but with unpaired electrons.
Conclusion:
- [Ni(CN)4]2- has all the electrons paired in its d orbitals, making it diamagnetic.
- [NiCl4]2- has unpaired electrons in its d orbitals, making it paramagnetic.
Therefore, the correct answer is option 'C': [Ni(CN)4]2- is diamagnetic, and [NiCl4]2- is paramagnetic.
In this question, we are given two compounds, [Ni(CN)4]2- and [NiCl4]2-, and we need to determine their magnetic properties. Specifically, we need to identify if they are diamagnetic or paramagnetic.
Explanation:
To determine the magnetic properties of a compound, we need to consider the electronic configuration of the central metal ion and the ligands surrounding it. Let's analyze each compound separately.
[Ni(CN)4]2-:
To determine the electronic configuration of the central Ni2+ ion, we need to know the atomic number of Nickel (Ni), which is 28. The 2+ charge indicates that two electrons have been removed from the neutral atom. Therefore, the electronic configuration of Ni2+ is 1s22s22p63s23p63d8.
The CN- ligands are strong-field ligands, which means they pair electrons in the d orbitals. Thus, the electronic configuration of [Ni(CN)4]2- is predicted to be d8 with all the electrons paired.
[NiCl4]2-:
The electronic configuration of the central Ni2+ ion remains the same as in the previous compound, which is 1s22s22p63s23p63d8.
The Cl- ligands are weak-field ligands, which means they do not pair electrons in the d orbitals. Thus, the electronic configuration of [NiCl4]2- is also predicted to be d8, but with unpaired electrons.
Conclusion:
- [Ni(CN)4]2- has all the electrons paired in its d orbitals, making it diamagnetic.
- [NiCl4]2- has unpaired electrons in its d orbitals, making it paramagnetic.
Therefore, the correct answer is option 'C': [Ni(CN)4]2- is diamagnetic, and [NiCl4]2- is paramagnetic.
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[Ni(CN)4]2- and [NiCl4]2- area)both diamagneticb)both paramagneticc)diamagnetic and paramagnetic respectivelyd)antiferromagnetic and diamagnetic respectivelyCorrect answer is option 'C'. Can you explain this answer?
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