Question

calculate magnetic moment of Fe2+ ion

Answer 1

Answer : The magnetic moment is 4.9 BM

Explanation:

The metal is Fe and the electronic configuration of Fe is, $1s^22s^22p^63s^23p^63d^64s^2$

The oxidation state of Fe is +2, thus the configuration is:

$1s^22s^22p^63s^23p^63d^6$

The number of unpaired electrons in Fe = 4

Formula used for magnetic moment :

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

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

$\mu=\sqrt{4(4+2)}=4.9BM$

Therefore, the magnetic moment is 4.9 BM

Answer 2

Answer:

$\sqrt{24}$

Explanation:

Explanation:

The metal is Fe and the electronic configuration of Fe is, 1s^22s^22p^63s^23p^63d^64s^21s

2

2s

2

2p

6

1. 3s

2

3p

6

3d

6

4s

2

The oxidation state of Fe is +2, thus the configuration is:

1s^22s^22p^63s^23p^63d^61s

2

2s

2

2p

6

3s

2

3p

6

3d

6

The number of unpaired electrons in Fe = 4

Formula used for magnetic moment :

\mu=\sqrt{n(n+2)}μ=

n(n+2)

where,

\muμ = magnetic moment

n = number of unpaired electrons

\mu=\sqrt{4(4+2)}=4.9BMμ=

4(4+2)

=4.9BM

Therefore, the magnetic moment is 4.9 BM