Which of the following compounds is diamagnet | vedclass

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

Question

(C) The complex $[Ni(CO)_4]$ involves $Ni$ in the $0$ oxidation state. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In the presence of st

Vedclass · Accepted Answer

$[Ni(CO)_4]$

Vedclass · Answer

(C) The complex $[Ni(CO)_4]$ involves $Ni$ in the $0$ oxidation state. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In the presence of strong field ligand $CO$,the $4s$ electrons pair up into the $3d$ orbitals,resulting in a $3d^{10}$ configuration. This leads to $sp^3$ hybridization,which corresponds to a tetrahedral geometry. Since all electrons are paired,it is diamagnetic. The other options like $[NiCl_4]^{2-}$ are tetrahedral but paramagnetic due to the presence of unpaired electrons in the $3d$ orbitals.

Which of the following compounds is diamagnetic and tetrahedral?

Similar Questions

The correct order of hybridization of the central atom in the following species $NH_3$,$[PtCl_4]^{2-}$,$PCl_5$,and $BCl_3$ is

Number of identical $Cr-O$ bonds in dichromate ion $Cr_2O_7^{2-}$ is

$NiCl_2\{P(C_2H_5)_2(C_6H_5)\}_2$ exhibits temperature-dependent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of $Ni^{2+}$ in the paramagnetic and diamagnetic states are respectively

$A$ coordination compound is made of $Co^{3+}$,$NH_3$,and $Cl^{-}$. $A$ $0.1 \ M$ solution of this complex when treated with excess silver nitrate gave no precipitate. The formula of the complex and secondary valency of metal are respectively.

Which of the following configurations will show an outer orbital octahedral complex?

Vedclass Test Series

Exam Paper Generator

Online Exam Module