Consider the following complex ions P = [FeF_ | vedclass

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(C) $[FeF_6]^{3-}$: $Fe^{3+}$ is $[Ar] 3d^5$. $F^-$ is a weak field ligand,so the number of unpaired electrons is $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35

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$Q < R < P$

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(C) $[FeF_6]^{3-}$: $Fe^{3+}$ is $[Ar] 3d^5$. $F^-$ is a weak field ligand,so the number of unpaired electrons is $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$$[V(H_2O)_6]^{2+}$: $V^{2+}$ is $3d^3$. The number of unpaired electrons is $3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$$[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,so the number of unpaired electrons is $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$Comparing the values: $\sqrt{15} Therefore,the correct order is $Q < R < P$.

List-$I$	List-$II$
$I. [FeF_6]^{3-}$	$A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$	$B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$	$C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$	$D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$	$E. 3.88 \ BM$

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

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Match List-$I$ with List-$II$ and select the correct answer using the codes given below:
List-$I$ List-$II$
$I. [FeF_6]^{3-}$ $A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$ $B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$ $C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$ $D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$ $E. 3.88 \ BM$

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