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The metal ion ( in gaseous state) with lowest spin-only magnetic moment value is
Option: 1
$\mathrm{V}^{2+}$

Option: 2
$\mathrm{Ni}^{2+}$

Option: 3
$\mathrm{Cr}^{2+}$

Option: 4
$\mathrm{Fe}^{2+}$

Answers (1)

best_answer

The metal ion with their electronic configuration

$\mathrm{V^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3},}$ $\mathrm{unpaired \: e^{-}=3}$

$\mathrm{Ni^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{8},}$ $\mathrm{unpaired \: e^{-}=2}$

$\mathrm{Cr^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{4},}$ $\mathrm{unpaired \: e^{-}=4}$

$\mathrm{Fe^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6},}$ $\mathrm{unpaired \: e^{-}=4}$

Lowest spin-only magnetic moment value will be for the ion having the least number of unpaired electron.

Thus, $\mathrm{Ni^{2+}}$ has the lowest spin only magnetic moment out of the given ions.

So, option (2) is correct.

Posted by

Pankaj

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