Question

The magnetic movement of a compound of transition element is 3.9 B.M. Write the number of unpaired electrons in the element

Answer 1

no of unpaired electrons are 3

Answer 2

The number of unpaired electrons in the element are 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 magnetic moment.

$3.9=\sqrt{n(n+2)}$

Squaring both sides:

$15.21={n(n+2)}$

$n=3$

Thus the number of unpaired electrons in the element are 3

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