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Help me please, Magnetic moment 2.83 BM is given by which of the following ions?(At nos. Ti=22, Cr=24, Mn=25, Ni=28):-

Magnetic moment 2.83 BM is given by which of the following ions?
(At nos. Ti=22, Cr=24, Mn=25, Ni=28):-

  • Option 1)

    Ti^{3+}

  • Option 2)

    Ni^{2+}

  • Option 3)

    Cr^{3+}

  • Option 4)

    Mn^{2+}

 
Answers (1)
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As we learned in concept

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)

 

 magnetic movement, \mu =\sqrt{x(x+2)}

 

=> x(x + 2) = 8

=> x = 2

Electronic confiq of N{_{i}}^{2t}

                                  

Clearly, no of unpaired is = 2

therefore, solution is 2


Option 1)

Ti^{3+}

incorrect

Option 2)

Ni^{2+}

correct

Option 3)

Cr^{3+}

incorrect

Option 4)

Mn^{2+}

incorrect

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