Consider the following complex ions,P,Q and R | vedclass

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(B) $P = [FeF_6]^{3-}$,oxidation number of $Fe = +3$,configuration: $3d^5$.Since $F^-$ is a weak field ligand,no pairing occurs,resulting in $5$ unpai

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$Q < R < P$

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(B) $P = [FeF_6]^{3-}$,oxidation number of $Fe = +3$,configuration: $3d^5$.Since $F^-$ is a weak field ligand,no pairing occurs,resulting in $5$ unpaired electrons.$Q = [V(H_2O)_6]^{2+}$,oxidation number of $V = +2$,configuration: $3d^3$.It has $3$ unpaired electrons.$R = [Fe(H_2O)_6]^{2+}$,oxidation number of $Fe = +2$,configuration: $3d^6$.Since $H_2O$ is a weak field ligand,no pairing occurs,resulting in $4$ unpaired electrons.The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Number of unpaired electrons: $Q(3) Therefore,the order of magnetic moments is $Q Thus,the correct option is $B$.

Consider the following complex ions,$P$,$Q$ and $R$.
$P = [FeF_6]^{3-}$,$Q = [V(H_2O)_6]^{2+}$ and $R = [Fe(H_2O)_6]^{2+}$.
The correct order of the complex ions,according to their spin-only magnetic moment values (in $B.M.$) is

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Consider the following complex ions,$P$,$Q$ and $R$.$P = [FeF_6]^{3-}$,$Q = [V(H_2O)_6]^{2+}$ and $R = [Fe(H_2O)_6]^{2+}$.The correct order of the complex ions,according to their spin-only magnetic moment values (in $B.M.$) is