Question

What is the magnetic moment for ${Mn^{2+}}$ ion?

• A. 5.92 B.M.
• B. 2.80 B.M
• C. 8.95 B.M.
• D. 9.55 B.M.

Answer 1

The Correct Option is A

${Mn^{2+}} : 3d^5,$ no. of unpaired electrons $(n) = 5$
Magnetic moment $(\mu) = \sqrt{n(n+2)} \, {B.M.}$
$= { \sqrt{5(5+2)} \, B.M. = \sqrt{35}}$
$= {= 5.92 \,B.M.}$

Answer 2

Firstly, Let us understand what exactly the magnetic moment means,

It is a tendency for a substance to align itself when it is placed in a magnetic field. It will align itself in such a way that the system has least energy. There is torque produced, which is called a restoring torque when the external magnetic field is removed and the system is back to its original state.

 The formula to calculate Magnetic moment is:

\(\mu = \sqrt{n(n+1)}\)

Where, the

\(\mu\) = Magnetic moment

\(n\) = Number of unpaired electrons

So, now let us look at our elements,

Manganese, Mn+2, it falls under the transition element and it has twenty-five electrons

Its configuration is:

1s²2s²2p⁶3s²3p⁶4s²3d⁵

And the configuration of the ion is:

1s²2s²2p⁶3s²3p⁶4s⁰3d⁵

As, the number of unpaired electrons are five.

Therefore, n=5

After substituting these values in the above equation we get,

\(\mu = \sqrt{5(5+1)}\)⇒5.92 B.M.

Therefore the correct option is ‘A’.