The number of unpaired electrons in [NiCl_{4} | vedclass

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

Question

(B) $1$. For $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand, so no pairing occurs. It has $2$ unpaired ele

Vedclass · Accepted Answer

$2, 0, 1$

Vedclass · Answer

(B) $1$. For $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand, so no pairing occurs. It has $2$ unpaired electrons.$2$. For $Ni(CO)_{4}$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand, causing pairing of $4s$ electrons into $3d$ orbitals, resulting in $0$ unpaired electrons.$3$. For $[Cu(NH_{3})_{4}]^{2+}$: $Cu$ is in $+2$ oxidation state $(3d^9)$. It has $1$ unpaired electron.Thus, the number of unpaired electrons are $2, 0, 1$ respectively. The correct option is $B$.

The number of unpaired electrons in $[NiCl_{4}]^{2-}$, $Ni(CO)_{4}$ and $[Cu(NH_{3})_{4}]^{2+}$ respectively are

Similar Questions

In an octahedral geometry,which pair of $d$-orbitals is involved in $d^2sp^3$ hybridization?

If an iron $(III)$ complex with the formula $[Fe(NH_3)_{x}(CN)_{y}]^{-}$ has no electron in its $e_g$ orbital, then the value of $x+y$ is

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

Discuss the complex species $[Co(NH_3)_6]^{3+}$ based on Valence Bond Theory.

The hybridization of the complex $[CrCl_2(NO_2)_2(NH_3)_2]^-$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module