Question

An ion $M{n}^{a+}$ has the magnetic moment equal to $4.9BM$. The value of $a$ is:

• A. $3$
• B. $4$
• C. $2$
• D. $5$

Answer 1

Answer: (A)

$3$

As we know,

Magnetic moment $=\sqrt{n(n+1)}$; where $n$ is no. of unpaired electron
here,

$4.9=\sqrt{n(n+2)}$

$\to n=4$ means this ion has 4 unpaired electrons.

electronic configuration of $Mn=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^{5}4s^2$ have 5 unpaired electron so to have a ion with 4 unpaired electron, it has to loose 3 electron and ion forms is

${}_{25}M{n}^{3+}:1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^4$;

$\therefore a=3$

Hence, the correct option is $A$