Question

Calculate the spin only magnetic moment of mn2+ ion

Answer 1

For, Mn2+ ion

Z= 27

Electronic Configuration of Mn2+= [Ar] 3d7

Number of unpaired electrons = 3

Spin only magnetic moment == 3.87 B.M

Answer 2

Answer : The spin only magnetic moment is 5.916 BM

Explanation:

For $Mn^{2+}$ ion,

Atomic number of Mn = 25

Electronic configuration of Mn=$1s^22s^22p^63s^23p^64s^23d^5$

Electronic configuration of $Mn^{2+}$ ion = $1s^22s^22p^63s^23p^63d^5$

The unpaired electrons in $Mn^{2+}$ ion = 5

Formula used for magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = spin only magnetic moment

n = number of unpaired electrons

Now put all the given values in the above formula, we get the magnetic moment.

$\mu=\sqrt{5(5+2)}=5.916BM$

Therefore, the spin only magnetic moment is, 5.916 BM