Question

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

Answer 1

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

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

• For $Ni^{2+}$ 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

The spin only magnetic moment for $Mn^{2+}$ is,

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

The spin only magnetic moment for $Ni^{2+}$ is,

$\mu_s=\sqrt{2(2+2)}=2.828\text{ BM}$

Therefore, the spin only magnetic moment for $Mn^{2+}$ and $Ni^{2+}$ is $5.916\text{ BM},2.828\text{ BM}$ respectively.

Answer 2

"The spin-only magnetic moment of $\mathrm{Mn}^{2+} is 4 BM and \mathrm{Ni}^{++} is 1.73 BM.$

The spin-only magnetic moment formula can be written in two different ways;

One is the based on the number of unpaired electrons (n), and another is based on the total number of electron spins (S).

For $Mn^{2+}$,

Atomic number of Mn = 27

Electronic configuration$=[\mathrm{Ar}] 3 \mathrm{d}^{7} 4 \mathrm{s}^{2}\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 \mathrm{d}^{7}$

It has 3 unpaired electrons.

Therefore, n = 3

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

$\mu=\sqrt{3(3+2)}$

$\mu =\sqrt{15} \cong 4 \mathrm{BM}$

For $\mathrm{Ni}^{++},$

Electronic configuration $=[\mathrm{Ar}] 3 \mathrm{d}^{6} 4 \mathrm{s}^{2}\mathrm{Ni}^{++}=[\mathrm{Ar}] 3 \mathrm{d}^{6}$

It has 1 unpaired electrons.

Therefore, n = 1

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

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

$\mu =\sqrt{3} \cong 1.73 \mathrm{BM}$"