The correct order of magnetic moments (spin only values in B.M.) among is

(Atomic nos.: Mn = 25, Fe = 26, Co = 27)

  • Option 1)

    [MnCl_4]^{2-}>[CoCl_4]^{2-}>[Fe(CN)_6]^{4-}

  • Option 2)

    [MnCl_4]^{2-}>[Fe(CN)_6]^{4-}>[CoCl_4]^{2-}

  • Option 3)

    [Fe(CN)_6]^{4-}>[MnCl_4]^{2-}>[CoCl_4]^{2-}

  • Option 4)

    [Fe(CN)_6]^{4-}>[CoCl_4]^{2-}>[MnCl_4]^{2-}

     

 

Answers (1)

As we learnt in

Calculation of spin only magnetic moment -

\mu =\sqrt{n(n+2)}

\mu =magnetic\:moment\:in(BM)

where n = no. of unpaired electron 

n=1......5(no.\:of\:unpaired\:e^{-})

- wherein

V^{2+}\rightarrow3d^{3}=\sqrt{3(3+2)}

=\sqrt{15}

=3.87\:\:\:BM

3+.9\rightarrow 3.9(BM)

 

1.  \left [ MnCl_{4} \right ]^{2-} has Mn^{2+} ion and its magnetic moment is \sqrt{5\left ( 5+2 \right )} = \sqrt{35}

2. \left [ CoCl_{4} \right ]^{2-} has Co^{2+} ion and its magnetic moment is \sqrt{3\left ( 3+2 \right )} =\sqrt{15}

3. \left [ Fe\left ( CN \right )_{6} \right ]^{4-} has Fe^{2+} ion with strong field ligand and its magnetic moment is zero.

Order  of magnetic moment

 \Rightarrow \left [ MnCl_{4} \right ]^{2-} > \left [ CoCl_{4} \right ]^{2-}> \left [ Fe\left ( CN \right )_{6} \right ]^{4-}


Option 1)

[MnCl_4]^{2-}>[CoCl_4]^{2-}>[Fe(CN)_6]^{4-}

This option is correct

Option 2)

[MnCl_4]^{2-}>[Fe(CN)_6]^{4-}>[CoCl_4]^{2-}

This option is incorrect

Option 3)

[Fe(CN)_6]^{4-}>[MnCl_4]^{2-}>[CoCl_4]^{2-}

This option is incorrect

Option 4)

[Fe(CN)_6]^{4-}>[CoCl_4]^{2-}>[MnCl_4]^{2-}

 

This option is incorrect

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