Calculate spin only magnetic moment for mn2+ and ni++

1.	Calculate spin only magnetic moment for mn2+ and ni++
Answer» $Mn^{2+}$ $<klux>NI</klux>^{2+}$ $5.916\text{ BM},2.828\text{ BM}$ Answer : The spin only magnetic moment for and is respectively. Explanation : $Mn^{2+}$ For 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 $Ni^{2+}$ For ion, Atomic number of Ni = 25 Electronic configuration of Ni = $1s^22s^22p^63s^23p^64s^23d^8$ Electronic configuration of $Ni^{2+}$ ion = $1s^22s^22p^63s^23p^63d^8$ The unpaired electrons in $Ni^{2+}$ ion = 2 Formula used for spin only magnetic moment : $\mu_s=\sqrt{n(n+2)}$ where, $\mu_s$ = spin only magnetic moment n = number of unpaired electrons $Mn^{2+}$ The spin only magnetic moment for is, $\mu_s=\sqrt{5(5+2)}=5.916\text{ BM}$ $Ni^{2+}$ The spin only magnetic moment for is, $\mu_s=\sqrt{2(2+2)}=2.828\text{ BM}$ $Mn^{2+}$ $Ni^{2+}$ $5.916\text{ BM},2.828\text{ BM}$ Therefore, the spin only magnetic moment for and is respectively.

Answer»

$Mn^{2+}$ $<klux>NI</klux>^{2+}$ $5.916\text{ BM},2.828\text{ BM}$ Answer : The spin only magnetic moment for and is respectively.