Both [Ni(CO)_4] and [Ni(CN)_4]^{2-} are diama | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pa

Vedclass · Accepted Answer

$sp^3, dsp^2$

Vedclass · Answer

(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,leading to $sp^3$ hybridization and tetrahedral geometry.In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The electronic configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the $3d$ electrons to pair up,resulting in $dsp^2$ hybridization and square planar geometry.

Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes,respectively,are

Similar Questions

Which statement is incorrect regarding the complex $[Cr(NH_3)_6]Cl_3$?

Which of the following is an incorrect pair?

What is the type of hybridization of $Ag$ in the linear complex $[Ag(NH_3)_2]^+$?

According to Valence Bond Theory,the number of unpaired electrons present in $[MnCl_6]^{3-}$,$[Fe(CN)_6]^{3-}$ and $[Co(C_2O_4)_3]^{3-}$,respectively,are

What is the number of unpaired electrons in $[Co(NH_3)_6]^{3+}$ complex?

Vedclass Test Series

Exam Paper Generator

Online Exam Module