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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)
Write the hybridization and shape of the following complexe : [Ni(CN)4]2–
(Atomic number : Co = 27, Ni = 28)
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
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]
Write the hybridization and magnetic character of the following complexes:
[Fe(CO)5]
(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
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]
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:
