Magnetic moment of x^{n+} is sqrt {24} | vedclass

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Question

$\mathrm{X}^{\mathrm{n}+}$

$\mu=\sqrt{24}=4.92\,\mathrm{BM}$

$\mathrm{n}=4=\mathrm{No} .$ of unpaired $\mathrm{e}^{-}$

$\mathrm{Fe}^{+2} \Rig

Vedclass · Accepted Answer

$4 , 2$

Vedclass · Answer

$\mathrm{X}^{\mathrm{n}+}$

$\mu=\sqrt{24}=4.92\,\mathrm{BM}$

$\mathrm{n}=4=\mathrm{No} .$ of unpaired $\mathrm{e}^{-}$

$\mathrm{Fe}^{+2} \Rightarrow[\mathrm{Ar}] 4 \mathrm{s}^{2} 3 \mathrm{d}^{6}$

$n=2$

Magnetic moment of $x^{n+}$ is $\sqrt {24} $ $B.M.$ Hence No. of unpaired electron and value of $'n'$ respectively. (Atomic number $= 26$)

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