Q. The correct order of magnetic moments (spin only values in BM) among the following is (Atomic no. Mn=25, Fe=26, Co=27)
A) $\left\lbrack {MnCl}_{4} \right\rbrack^{2 -}\ > \ \left\lbrack {CoCl}_{4} \right\rbrack^{2 -}\ > \left\lbrack Fe(CN)_{6} \right\rbrack^{4 -}$
B) $\left\lbrack {MnCl}_{4} \right\rbrack^{2 -}\ > \left\lbrack Fe(CN)_{6} \right\rbrack^{4 -}\ > \ \left\lbrack {CoCl}_{4} \right\rbrack^{2 -}$
C) $\left\lbrack Fe(CN)_{6} \right\rbrack^{4 -}\ > \left\lbrack {MnCl}_{4} \right\rbrack^{2 -}\ > \left\lbrack {CoCl}_{4} \right\rbrack^{2 -}$
D) $\left\lbrack Fe(CN)_{6} \right\rbrack^{4 -}\ > \left\lbrack {CoCl}_{4} \right\rbrack^{2 -}\ \ > \left\lbrack {MnCl}_{4} \right\rbrack^{2 -}$
Explanation
(a)
Number of unpaired electrons in $\left\lbrack Fe(CN)_{6} \right\rbrack^{4 -}$ is zero.
Thus, magnetic moment
=$\sqrt{n(n + 2)}$=0 BM
(n=unpaired electrons)
n in $\left\lbrack {MnCl}_{4} \right\rbrack^{2 -}$=5, $\sqrt{35}$ BM
n in $\left\lbrack {CoCl}_{4} \right\rbrack^{2 -}$=3, $\sqrt{15}$ BM
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.