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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
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 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)
[NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
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
Magnetic moment of \[\ce{[MnCl4]^{2-}}\] is 5.92 BM. Explain giving reason.
