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Question

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:

A
$$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$$
B
$$Cr^{2+} > V^{2+} > Ru^{3+} > Fe^{2+}$$
C
$$V^{2+} > Ru^{3+} > Cr^{2+} > Fe^{2+}$$
D
$$Cr^{2+} > Ru^{3+} > Fe^{2+} > V^{2+}$$
Solution
Verified by Toppr

Correct option is A. $$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$$
Solution:- (A) $$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$$

According to equations, all the complexes are low spin.
ComplexConfiguration No. of
unpaired
electrons
$$[V(CN)_6]^{4-}$$$$t_{2g} ^3e_g^0$$3

$$[Cr(NH)_3)_6]^{2+}$$

$$t_{2g} ^4e_g^0$$

2

$$[Ru(NH_3)_6]^{3+} $$

$$t_{2g} ^5e_g^0$$

1

$$[Fe(CN)_6]^{4-}$$
$$t_{2g} ^6e_g^0$$ 0

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