Advertisements
Advertisements
प्रश्न
Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
(At.no. of Ni = 28)
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
For the complex [Fe(CN)6]3–, write the hybridization type, magnetic character and spin nature of the complex. (At. number : Fe = 26).
[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]
Write down the IUPAC name of the following complex and indicate the oxidation state, electronic configuration and coordination number. Also, give the stereochemistry and magnetic moment of the complex:
K[Cr(H2O)2(C2O4)2].3H2O
Why is [NiCl4]2− paramagnetic while [Ni(CN)4]2− is diamagnetic? (Atomic number of 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
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.
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.
