Which of the following ions has the maximum magnetic moment?

  • Option 1)

    Mn^{2+}\; 

  • Option 2)

    \; Fe^{2+}\;

  • Option 3)

    \; Ti^{2+}\;

  • Option 4)

    \; Cr^{2+}

 

Answers (2)

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)

 

 Mn^{2+}\Rightarrow 3d^{5},\:\:\mu = \sqrt{5\left ( 5+2 \right )}= \sqrt{35}

Fe^{2+}\Rightarrow 3d^{6}, \:\:\mu = \sqrt{4\left ( 4+2 \right )}= \sqrt{24}

Ti^{2+}\Rightarrow 3d^{2},\:\:\mu = \sqrt{2\left ( 2+2 \right )}= \sqrt{8}

Cr^{2+} \Rightarrow 3d^{4}, \:\: \mu = \sqrt{4\left ( 4+2 \right )}= \sqrt{24}


Option 1)

Mn^{2+}\; 

This option is correct.

Option 2)

\; Fe^{2+}\;

This option is incorrect.

Option 3)

\; Ti^{2+}\;

This option is incorrect.

Option 4)

\; Cr^{2+}

This option is incorrect.

S Sahil

FE2+

 

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