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प्रश्न
Write the hybridization type and magnetic behaviour of the complex [Ni(CN)4]2−. (Atomic number of Ni = 28)
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उत्तर
Cyanide, CN− being a strong-field ligand causes the pairing up of valence electrons in the Ni2+ ion against the Hund's rule of maximum multiplicity. This results in the formation of an inner orbital complex, [Ni(CN)4]2− having diamagnetic character and dsp2 hybridisation.
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संबंधित प्रश्न
Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
(At.no. of Ni = 28)
Write the hybridization and shape of the following complexe : [Ni(CN)4]2–
(Atomic number : Co = 27, Ni = 28)
Write the hybridisation and magnetic character of [Co(C2O4)3]3–.
(At. no. of Co = 27)
Why is [NiCl4]2− paramagnetic while [Ni(CN)4]2− is diamagnetic? (Atomic number of Ni = 28)
Write the hybridization and magnetic character of the following complexes:
[Fe(H2O)6]2+
(Atomic no. of Fe = 26)
Which of the following options are correct for \[\ce{[Fe(CN)6]^{3-}}\] complex?
(i) d2sp3 hybridisation
(ii) sp3d2 hybridisation
(iii) paramagnetic
(iv) diamagnetic
Explain why \[\ce{[Fe(H2O)6]^{3+}}\] has magnetic moment value of 5.92 BM whereas \[\ce{[Fe(CN)6]^{3-}}\] – has a value of only 1.74 BM.
The correct order of magnetic moment (spin only value in B.m.) is:
Low oxidation state of metals in their complexes are common when ligands ______.
Explain [Fe(CN)6]3− is an inner orbital complex, whereas [FeF6]3− is an outer orbital complex.
[Atomic number: Fe = 26]
