Question

The magnetic moment of an Ion is root 24 BM then that iron maybe fe+2 explain me how

Answer 1

Answer:

Explanation:

the electronic configuration of fe is=1s²2s²2p⁶3s²3p⁶3d⁶4s²

the electronic configuration of Fe²⁺=1s²2s²2p⁶3s²3p⁶3d⁶

unpaired e⁻=4

magnetic moment=√n(n+2)

so magnetic moment of fe =√4(4+2)=√24

Answer 2

Answer:

The magnetic moment of an ion is $\sqrt 24 BM$ is proved for $F e^{2+}$

Explanation:

Given: ion and chemical compound

To find: Magnetic moment of an ion

The electronic configuration of $F e$with atomic number of 26 is:

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}$

The electronic configuration of $F e^{2+}$ formed by loss of three electrons is:

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4s^{2} 3 d^{4} \text { containing } 4\text { unpaired electrons. }$

Formula used for magnetic moment :

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

where,

$\mu=$ magnetic moment

$n=$ number of unpaired electrons $=4$

Now put all the given values in the above formula, we get the unpaired electrons.

Therefore, the magnetic moment of $F e^{2+}$ is $\sqrt{24} BM$.

The magnetic moment of an ion is $\sqrt 24 BM$ is proved for $F e^{2+}$