CO (carbonyl) is a strong ligand so it causes pairing of electrons. Was this answer helpful? The hybridization of N i in [N i(CO)4] is ?

The empty 4s and three 4p orbitals undergo sp 3 hybridization and form bonds with CO ligands to give Ni (CO) 4. Thus Ni (CO) 4 is diamagnetic. KEY POINTS: Hybridization of Ni (CO) 4 : sp 3. Shape & Structure (geometry) of Ni (CO) 4: Tetrahedral. Magnetic nature: Diamagnetic (low spin) [Ni (CN) 4] 2- = Ni 2+ + 4CN -

The valence shell electronic configuration of the ground state of N i is 3 d 8 4 s 2, but all the 10 electrons are pushed into 3 d orbital due to the strong field C O ligands approaching the N i atom, thereby forming s p 3 hybridization.

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 sp 3 hybridization.

In traditional hybridisation theory, tetrahedral and square planar 4-coordinate complexes are considered to be $\mathrm{sp^3}$ and $\mathrm{dsp^2}$ respectively. Since $\ce{[Ni(CO)4]}$ is tetrahedral and $\ce{[Ni(CN)4]^2-}$ is square planar, one can assign the hybridisations accordingly. If you have a textbook that still mentions this outdated ...

⇒ CO as strong filled ligand paired all the electrons of Ni including 4s electrons in 3d orbitals. ⇒ Now one 4s and three 3d orbitals are available for four CO ligands participating in hybridization. ⇒ So, the hybridization of [Ni(CO) 4] is sp 3. ⇒ ∴ the possible geometry for [Ni(CO) 4] is tetrahedral.

2018年11月5日 · Ni(CO) 4: The outermost electronic configuration will be 3d 8 4s 2 4p o. CO is the strong ligand, causes pairing up of the 4s electrons into the 3d electrons. Thus the hybridization will be sp 3 (tetrahedral) Ni(CO) 4 will be diamagnetic in nature.