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प्रश्न
[Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2− has square planar, yet both exhibit diamagnetism. Explain.
[Atomic number: Ni = 28]
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उत्तर
In [Ni(CN)4]2−, nickel is in a +2 oxidation state and the ion has the electronic configuration 3d8. The hybridisation scheme is shown in the diagram.
Whereas in [Ni(CO)4], Ni is in a +2 oxidation state and shows sp2 hybridisation due to which its geometry is tetrahedral.
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संबंधित प्रश्न
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Write the hybridisation and number of unpaired electrons in the complex `[CoF_6]^(3-)`. (Atomic No. of Co = 27)
[NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? (Atomic no. Ni = 28)
Using valence bond theory, explain the following in relation to the complexes given below:
\[\ce{[FeCl6]^{4-}}\]
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
In a coordination entity, the electronic configuration of the central metal ion is t2g3 eg1
Is the coordination compound a high spin or low spin complex?
Which of the following methods is used for measuring bond length?
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[Given: Atomic number of Co = 27]
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:
Given below are two statements:
Statement I: Both [Co(NH3)6]3+ and [CoF6]3− complexes are octahedral but differ in their magnetic behavior.
Statement II: [Co(NH3)6]3+ is diamagnetic whereas [CoF6]3− is paramagnetic.
In the light of the above statements, Choose the correct answer form the options given below:
