Question

calculate the number of unpaired electrons in Ti3+ , Mn2+ and calculate the spin only magnetic moment?​

Answer 1

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Answer 2

The number of unpaired electrons and the spin only magnetic moment in $Ti^{3+}$ are 1 and 1.7 BM.

The number of unpaired electrons and the spin only magnetic moment in $Mn^{2+}$ are 5 and 5.9 BM.

Explanation:

1) Electronic configuration of $Ti:22:1s^22s^22p^63s^23p^64s^23d^2$

Electronic configuration of $Ti^{3+}$ ion = $1s^22s^22p^63s^23p^63d^1$

The unpaired electrons in $Ti^{3+}$ ion = 1

Formula used for spin only magnetic moment :

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

where, $\mu$ = spin only magnetic moment

n = number of unpaired electrons

$\mu=\sqrt{1(1+2)}=1.7BM$

2) 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 spin only magnetic moment :

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

where, $\mu$ = spin only magnetic moment

n = number of unpaired electrons

$\mu=\sqrt{5(5+2)}=5.9BM$

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