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प्रश्न
[Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2− has square planar, yet both exhibit diamagnetism. Explain.
[Atomic number: Ni = 28]
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उत्तर
In [Ni(CN)4]2−, nickel is in a +2 oxidation state and the ion has the electronic configuration 3d8. The hybridisation scheme is shown in the diagram.
Whereas in [Ni(CO)4], Ni is in a +2 oxidation state and shows sp2 hybridisation due to which its geometry is tetrahedral.
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संबंधित प्रश्न
Discuss the nature of bonding in the following coordination entity on the basis of valence bond theory:
[Fe(CN)6]4−
Discuss the nature of bonding in the following coordination entity on the basis of valence bond theory:
[Co(C2O4)3]3−
Discuss the nature of bonding in the following coordination entity on the basis of valence bond theory:
[CoF6]3−
Write the hybridisation and number of unpaired electrons in the complex `[CoF_6]^(3-)`. (Atomic No. of Co = 27)
Using valence bond theory, explain the following in relation to the complexes given below:
\[\ce{[Mn(CN)6]^{3-}}\]
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
If orbital quantum number (l) has values 0, 1, 2 and 3, deduce the corresponding value of principal quantum number, n.
How many radial nodes for 3p orbital?
As the s-character of hybridised orbital increases, the bond angle
Which of the following has square planar structures?
During chemistry class, a teacher wrote \[\ce{[Ni(CN)4]^2-}\] as a coordination complex ion on the board. The students were asked to find out the magnetic behaviour and shape of the complex. Pari, a student, wrote the answer paramagnetic and tetrahedral whereas another student Suhail wrote diamagnetic and square planer.
Evaluate Pari’s and Suhail’s responses.
