The correct order of the spin only magnetic m | vedclass

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(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.

Vedclass · Accepted Answer

$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$

Vedclass · Answer

(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.						$Complex$			$Configuration 	ext{ and } n$							$[V(CN)_6]^{4-}$ ($V^{2+}$,$d^3$)			$t_{2g}^3 e_g^0$,$n = 3$							$[Cr(NH_3)_6]^{2+}$ ($Cr^{2+}$,$d^4$)			$t_{2g}^4 e_g^0$,$n = 2$							$[Ru(NH_3)_6]^{3+}$ ($Ru^{3+}$,$d^5$)			$t_{2g}^5 e_g^0$,$n = 1$							$[Fe(CN)_6]^{4-}$ ($Fe^{2+}$,$d^6$)			$t_{2g}^6 e_g^0$,$n = 0$			Since the magnetic moment is directly proportional to the number of unpaired electrons $(n)$,the order is $n=3 > n=2 > n=1 > n=0$.Therefore,the order is $[V(CN)_6]^{4-} > [Cr(NH_3)_6]^{2+} > [Ru(NH_3)_6]^{3+} > [Fe(CN)_6]^{4-}$.This corresponds to the order of metal ions: $V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$.

The correct order of the spin only magnetic moment of metal ions in the following low spin complexes,$[V(CN)_6]^{4-}$,$[Fe(CN)_6]^{4-}$,$[Ru(NH_3)_6]^{3+}$,and $[Cr(NH_3)_6]^{2+}$,is

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