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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)

best_answer

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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prateek

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