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
$$Cr^{2+} > Ru^{3+} > Fe^{2+} > V^{2+}$$
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.Complex | Configuration | 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 |