Magnetic moment of 25Mnx+ is 15 then the value of X is:

Magnetic moment of $_{25} {Mn}^{x +}$ is $\sqrt{15}$ then the value of X is:

A
1
B
2
C
3
D
4

Solution:

No. of unpaired $e^{-} = 3;$

Thus, given element can be written as,

$_{25} {Mn}^{4 +}$

On comparing we get the value of X as 4.

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