According to the valence bond theory,the hybr | vedclass

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

Question

(C) To determine the hybridization of the central metal atom,we analyze the coordination number and the nature of the ligands:$1$. In $Na_{2}[NiCl_{4}

Vedclass · Accepted Answer

$K_{2}[Ni(CN)_{4}]$

Vedclass · Answer

(C) To determine the hybridization of the central metal atom,we analyze the coordination number and the nature of the ligands:$1$. In $Na_{2}[NiCl_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $Cl^-$ is a weak field ligand,leading to $sp^{3}$ hybridization (tetrahedral geometry).$2$. In $NiCl_{2} \cdot 6 H_{2}O$,the complex is $[Ni(H_{2}O)_{6}]^{2+}$. $Ni^{2+}$ is in an octahedral environment with $sp^{3}d^{2}$ hybridization.$3$. In $K_{2}[Ni(CN)_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This leaves one $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals available for hybridization,resulting in $dsp^{2}$ hybridization (square planar geometry).$4$. In $[Ni(CO)_{4}]$,$Ni$ is in the $0$ oxidation state ($3d^{8} 4s^{2}$ configuration). $CO$ is a strong field ligand,causing the $4s$ electrons to pair into the $3d$ orbitals,resulting in $sp^{3}$ hybridization (tetrahedral geometry).Therefore,the correct compound is $K_{2}[Ni(CN)_{4}]$.

According to the valence bond theory,the hybridization of the central metal atom is $dsp^{2}$ for which one of the following compounds?

Similar Questions

Which of the following statements is incorrect?

An octahedral complex of $Ni(II)$ must be:

Which complex has a square planar structure?

Which of the following complexes is paramagnetic,an inner orbital complex,and coloured?

Which of the following is an outer orbital complex? (Atomic numbers: $Mn = 25, Fe = 26, Co = 27, Ni = 28$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module