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Q)

An ion $Mn^{a+}$ has the magnetic moment equal to 4.9bm. What is the value of a ?

$(a)\;2\qquad(b)\;3\qquad(C)\;4\qquad(d)\;1$

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A)
Since Magnetic moment = n(n+2)=√n(n+2)= 4.9
$\therefore$ n= 4(where n is no. of unpaired electron)
Thus $Mn^{a+}$ ion has four unpaired electron $25 Mn^{3+} : 1S^2 , 2S^22P^6, 3S^2, 3P^6 3d^4$
$\therefore$ a = 3
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