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प्रश्न
Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
(At.no. of Ni = 28)
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उत्तर
Carbonyl, CO being a strong field ligand causes the pairing of up valence electrons in the Ni atom against the Hund's Rule of Maximum Multiplicity. This results in the formation of an inner orbital complex, [Ni(CO)4] having diamagnetic character. [Ni(CO)4] has sp3 hybridization.
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संबंधित प्रश्न
For the complex [Fe(H2O)6]+3, write the hybridisation, magnetic character and spin of the complex. (At, number : Fe = 26)
For the complex [Fe(CN)6]3–, write the hybridization type, magnetic character and spin nature of the complex. (At. number : Fe = 26).
Write the hybridization and shape of the following complexe : [CoF6]3–
(Atomic number : Co = 27, Ni = 28)
[NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
Explain [Co(NH3)6]3+ is an inner orbital complex, whereas [Ni(NH3)6]2+ is an outer orbital complex.
[At. No.: Co = 27, Ni = 28]
Which of the following options are correct for \[\ce{[Fe(CN)6]^{3-}}\] complex?
(i) d2sp3 hybridisation
(ii) sp3d2 hybridisation
(iii) paramagnetic
(iv) diamagnetic
Why do compounds having similar geometry have different magnetic moment?
Assertion: \[\ce{[Fe(CN)6]^{3-}}\] ion shows magnetic moment corresponding to two unpaired electrons.
Reason: Because it has d2sp3 type hybridisation.
Explain [Fe(CN)6]3− is an inner orbital complex, whereas [FeF6]3− is an outer orbital complex.
[Atomic number: Fe = 26]
Given below are two statements:
Statement I: Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II: The number of unpaired electrons in a Cr2+ ion (Z = 24) is the same as that of a Nd3+ ion (Z = 60).
In the light of the above statements, choose the correct answer from the options given below:
