Advertisements
Advertisements
प्रश्न
Write the hybridization type and magnetic behaviour of the complex [Ni(CN)4]2−. (Atomic number of Ni = 28)
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.
APPEARS IN
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)
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)
Magnetic moment of \[\ce{[MnCl4]^{2-}}\] is 5.92 BM. Explain giving reason.
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.
Assertion: \[\ce{[Fe(CN)6]^{3-}}\] ion shows magnetic moment corresponding to two unpaired electrons.
Reason: Because it has d2sp3 type hybridisation.
The correct order of magnetic moment (spin only value in B.m.) is
Explain [Fe(CN)6]3− is an inner orbital complex, whereas [FeF6]3− is an outer orbital complex.
[Atomic number: Fe = 26]
