Question

Fe has the electron configuration [Ar]3d64s2. What is the magnitude of the orbital angular momentum for its most energetic electron? √2h Võh - V12h V12h 20h

Answer 1

Answer : The magnitude of the orbital angular momentum for its most energetic electron is, $\sqrt{6}\hbar$

Explanation :

The formula used for orbital angular momentum is:

$L=\sqrt{l(l+1)}\hbar$

where,

L = orbital angular momentum

l = Azimuthal quantum number

As we are given the electronic configuration of Fe is, $[Ar]3d^64s^2$

Its most energetic electron will be for 3d electrons.

The value of azimuthal quantum number(l) of d orbital is, 2

That means, l = 2

Now put all the given values in the above formula, we get:

$L=\sqrt{2(2+1)}\hbar$

$L=\sqrt{6}\hbar$

Therefore, the magnitude of the orbital angular momentum for its most energetic electron is, $\sqrt{6}\hbar$