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 $Q < R < P$.

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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