The “spin­-only” magnetic moment [ in units of Bohr magneton, (\mu_{} _{B}) ] of Ni^{2+} in aqueous solution would be (atomic number of Ni = 28)

  • Option 1)

    2.84

  • Option 2)

    4.90

  • Option 3)

    0

  • Option 4)

    1.73

 

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)

 

 The magnetic moment of = Ni^{2+} = \sqrt{2\left ( 2+2 \right )}

=\sqrt{8}= 2.84

 


Option 1)

2.84

This option is correct

Option 2)

4.90

This option is incorrect

Option 3)

0

This option is incorrect

Option 4)

1.73

This option is incorrect

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