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# Magnetic moment of $A^{3+}$ ion is $5.48\times 10^{-23}J/T$.Find out the number of unpaired electrons in it.($9.27\times 10^{-24}J/T$=1BM)

$(a)\;1\qquad(b)\;2\qquad(c)\;4\qquad(d)\;5$

Magnetic moment =$5.48\times 10^{-23}J/T$
$\large\frac{5.48\times 10^{-23}}{9.27\times 10^{-24}}=$$5.92$BM
$\therefore \sqrt{n(n+2)}=5.92=\sqrt{35}$
$\therefore n=5$
$\therefore$ No of unpaired electrons =5
Hence (d) is the correct answer.