The correct order of [FeF_6]^{3-}, [CoF_6]^{3 | vedclass

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

Question

(A) To determine the number of unpaired electrons,we analyze the electronic configuration of the metal ion in each complex:$1$. $[FeF_6]^{3-}$: $Fe^{3

Vedclass · Accepted Answer

$[FeF_6]^{3-} > [CoF_6]^{3-} > [Ni(CN)_4]^{2-} = [Ni(CO)_4]$

Vedclass · Answer

(A) To determine the number of unpaired electrons,we analyze the electronic configuration of the metal ion in each complex:$1$. $[FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. Since $F^-$ is a weak field ligand,electrons remain unpaired. Number of unpaired electrons = $5$.$2$. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. Since $F^-$ is a weak field ligand,it forms a high-spin complex. Number of unpaired electrons = $4$.$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. Number of unpaired electrons = $0$.$4$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of all electrons in $3d$ orbital. Number of unpaired electrons = $0$.Thus,the order is $[FeF_6]^{3-} (5) > [CoF_6]^{3-} (4) > [Ni(CN)_4]^{2-} (0) = [Ni(CO)_4] (0)$.

The correct order of $[FeF_6]^{3-}, [CoF_6]^{3-}, [Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ complex species based on the number of unpaired electrons present is $:$

Similar Questions

Which of the following statements is correct?

$A$ complex with the composition $[MA_3B]^{n \pm}$ is found to have no geometrical isomers. The possible structure$(s)$ of the complex is/are (where $A$ and $B$ are monodentate ligands):

In the complex $[SbF_5]^{2-}$,$sp^3d$ hybridization is present. The geometry of the complex is

Which from the following statements about $[Co(NH_3)_6]^{3+}$ complex is $NOT$ correct?

Which of the following statements are true about $[NiCl_4]^{2-}$?
$(a)$ The complex has tetrahedral geometry
$(b)$ Coordination number of $Ni$ is $2$ and oxidation state is $+4$
$(c)$ The complex is $sp^3$ hybridised
$(d)$ It is a high spin complex
$(e)$ The complex is paramagnetic

Vedclass Test Series

Exam Paper Generator

Online Exam Module