Question

an ion Mn​a+has the magnetic momrntum equal to 4.9 B.M the value of a is

Answer 1

Since Magnetic moment = n(n+2)−−−−−−−√n(n+2) = 4.9

∴∴ n= 4(where n is no. of unpaired electron)

Thus Mna+Mna+ ion has four unpaired electron 25Mn3+:1S2,2S22P6,3S2,3P63d425Mn3+:1S2,2S22P6,3S2,3P63d4

∴∴ a = 3

Answer 2

Answer: a = 3

Explanation:

Formula used for magnetic moment :

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

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

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

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

$n=4$

Therefore, the number of unpaired electrons is 4.

Manganese must contain 4 unpaired electrons.

Thus the element had lost 2 electrons from 4s shell and  1 electron from 3d shell.

$1s^22s^22p^63s^23p^63d^54s^2$

The electronic configuration of $Mn^{3+}$ is,  

$1s^22s^22p^63s^23p^63d^4$ containing 4 unpaired electrons.

Thus the value of a is 3.