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प्रश्न
[NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? (Atomic no. Ni = 28)
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उत्तर
[NiCl4]2-
Ni is in + 2 oxidation state.
3d8 configuration 
Cl- is weak field ligand. So, pairing doesnot occur.
sp3 hybridised orbitals of 2 Ni2+ 
[ ] NiCl is paramagnetic as n = 2
4 [ ( ) ] Ni CO Ni is in ‘0’ oxidation state
In presence of ‘CO’ pairing of e– takes place ‘CO’ is strong field ligand. So, with ‘CO
[Ni(CO)4] → 
[Ni(CO)4] → 
So, [Ni(CO)4] is diamagnetic as n = 0.
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संबंधित प्रश्न
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\[\ce{[Cr(H2O)6]^{3+}}\]
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(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
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(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
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Which of the following has square planar structures?
[Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2− has square planar, yet both exhibit diamagnetism. Explain.
[Atomic number: Ni = 28]
Which of the following are paramagnetic?
- [NiCl4]2−
- Ni(CO)4
- [Ni(CN)4]2−
- [Ni(H2O)6]2+
- Ni(PPh3)4
Choose the correct answer from the options given below:
