Question

The correct order of magnetic moments of the given compounds is (spin only values in $B.M$.)
(Atomic nos.: $Mn=25,Fe=26,Co=27)$

• A. ${\left[MnC{l}_4\right]}^{2-}>{\left[CoC{l}_4\right]}^{2-}>{\left[Fe(CN{)}_6\right]}^{4-}$
• B. ${\left[MnC{l}_4\right]}^{2-}>{\left[Fe(CN{)}_6\right]}^{4-}>{\left[CoC{l}_4\right]}^{2-}$
• C. ${\left[Fe(CN{)}_6\right]}^{4-}>{\left[MnC{l}_4\right]}^{2-}>{\left[CoC{l}_4\right]}^{2-}$
• D. ${\left[Fe(CN{)}_6\right]}^{4-}>{\left[CoC{l}_4\right]}^{2-}>{\left[MnC{l}_4\right]}^{2-}$

Answer 1

Answer: (A)

Number of unpaired electrons $=0$

Magnetic moment $=n\sqrt{n+2}$

where $n=$ number of unpaired electrons.

i.e. greater the number of unpaired electrons, greater will be the paramagnetic character.