Compounds with spin-only magnetic moment equi | vedclass

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

Question

(B, C, D) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} 	ext{ BM}$, where $n$ is the number of unpaired electrons. For $n=5$, $\mu =

Vedclass · Accepted Answer

$K_{4}[MnF_{6}]$

Vedclass · Answer

(B, C, D) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} 	ext{ BM}$, where $n$ is the number of unpaired electrons. For $n=5$, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 	ext{ BM}$.$(A)$ In $K_{4}[Mn(CN)_{6}]$, $Mn$ is in $+2$ oxidation state $(3d^{5})$. $CN^{-}$ is a strong field ligand, causing pairing of electrons. Thus, $n=1$.$(B)$ In $[Fe(H_{2}O)_{6}]Cl_{3}$, $Fe$ is in $+3$ oxidation state $(3d^{5})$. $H_{2}O$ is a weak field ligand, so no pairing occurs. Thus, $n=5$.$(C)$ In $K_{3}[FeF_{6}]$, $Fe$ is in $+3$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand, so no pairing occurs. Thus, $n=5$.$(D)$ In $K_{4}[MnF_{6}]$, $Mn$ is in $+2$ oxidation state $(3d^{5})$. $F^{-}$ is a weak field ligand, so no pairing occurs. Thus, $n=5$.Therefore, the compounds with five unpaired electrons are $(B)$, $(C)$, and $(D)$.

Compounds with spin-only magnetic moment equivalent to five unpaired electrons are

Similar Questions

The compound which does not show paramagnetism is

The diamagnetic species is

The correct order of the spin-only magnetic moments of the following complexes is
$(I) [Cr(H_2O)_6]Br_2$
$(II) Na_4[Fe(CN)_6]$
$(III) Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$
$(IV) (Et_4N)_2[CoCl_4]$

An octahedral complex of $M^{+3}$ with $NH_3$ and $NO_2^{-}$ ligands has four $M-N$ linkages and two $M-O$ linkages. Which type of isomerism will such a complex exhibit?

What is the magnetic moment (in $B.M.$) of ${K_3}[FeF_6]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module