Advertisements
Advertisements
Question
[Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2− has square planar, yet both exhibit diamagnetism. Explain.
[Atomic number: Ni = 28]
Advertisements
Solution
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.
APPEARS IN
RELATED QUESTIONS
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion.
[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why?
Discuss the nature of bonding in the following coordination entity on the basis of valence bond theory:
[CoF6]3−
Using valence bond theory, explain the following in relation to the complexes given below:
\[\ce{[Co(NH3)6]^{3+}}\]
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
Write the hybridization and shape of the following complexes:
[Ni(CN)4]2−
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?
Which of the statement given below is incorrect about H2O2?
Using Valence bond theory, explain the following in relation to the paramagnetic complex [Mn(CN)6]3−
- type of hybridization
- magnetic moment value
- type of complex – inner, outer orbital complex
The geometry and magnetic behaviour of the complex [Ni(CO)4] are ______.
