Coordination Compounds 2 Question 9
####10. The correct order of the spin only magnetic moment of metal ions in the following low spin complexes, $\left[V(CN) _6\right]^{4-}$, $\left[Fe(CN) _6\right]^{4-},\left[Ru\left(NH _3\right) _6\right]^{3+}$, and $\left[Cr\left(NH _3\right) _6\right]^{2+}$, is
(a) $Cr^{2+}>Ru^{3+}>Fe^{2+}>V^{2+}$
(2019 Main, 8 April I)
(b) $V^{2+}>Cr^{2+}>Ru^{3+}>Fe^{2+}$
(c) $V^{2+}>Ru^{3+}>Cr^{2+}>Fe^{2+}$
(d) $Cr^{2+}>V^{2+}>Ru^{3+}>Fe^{2+}$
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Solution:
Key Idea In presence of strong field ligands, $\Delta _0>p$, for fourth electron it is more energetically favourable to occupy $t _{2 g}$ orbital with configuration $t _{2 g}^{4} e _g^{0}$ and form low spin complexes.
The correct order of the spin only magnetic moment of metal ions in the given low-spin complexes is $V^{2+}>Cr^{2+}>Ru^{3+}>Fe^{2+}$. All the given complexes possess strong field ligands $\left(CN, NH _3\right)$. Hence, readily form low spin complexes.
