The magnetic moments of the following,arrange | vedclass

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(C) The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$(i)$ $Co^{3+}$ $(3d^6)$: Stro

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$ii > iii > iv > i$

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(C) The magnetic moment is calculated as $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$(i)$ $Co^{3+}$ $(3d^6)$: Strong field ligand causes pairing,$t_{2g}^6 e_g^0$,$n = 0$,$\mu = 0 	ext{ BM}$.$(ii)$ $Co^{3+}$ $(3d^6)$: Weak field ligand,$t_{2g}^4 e_g^2$,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.9 	ext{ BM}$.$(iii)$ $Co^{2+}$ $(3d^7)$: Tetrahedral complex,$e^4 t_2^3$,$n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 	ext{ BM}$.$(iv)$ $Co^{2+}$ $(3d^7)$: Square planar complex,$d_{xz}^2 d_{yz}^2 d_{z^2}^2 d_{xy}^1$,$n = 1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 	ext{ BM}$.Comparing the values: $ii (4.9) > iii (3.87) > iv (1.73) > i (0)$.Thus,the decreasing order is $ii > iii > iv > i$.

The magnetic moments of the following,arranged in decreasing order,will be:-
$(i)$ $Co^{3+}$ (Octahedral complex with a strong field ligand)
$(ii)$ $Co^{3+}$ (Octahedral complex with a weak field ligand like $F^{-}$)
$(iii)$ $Co^{2+}$ (Tetrahedral complex)
$(iv)$ $Co^{2+}$ (Square planar complex)

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The magnetic moments of the following,arranged in decreasing order,will be:-$(i)$ $Co^{3+}$ (Octahedral complex with a strong field ligand)$(ii)$ $Co^{3+}$ (Octahedral complex with a weak field ligand like $F^{-}$)$(iii)$ $Co^{2+}$ (Tetrahedral complex)$(iv)$ $Co^{2+}$ (Square planar complex)