Answer : The magnitude of the orbital angular momentum for its most energetic electron is, [tex]\sqrt{6}\hbar[/tex]
Explanation :
The formula used for orbital angular momentum is:
[tex]L=\sqrt{l(l+1)}\hbar[/tex]
where,
L = orbital angular momentum
l = Azimuthal quantum number
As we are given the electronic configuration of Fe is, [tex][Ar]3d^64s^2[/tex]
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:
[tex]L=\sqrt{2(2+1)}\hbar[/tex]
[tex]L=\sqrt{6}\hbar[/tex]
Therefore, the magnitude of the orbital angular momentum for its most energetic electron is, [tex]\sqrt{6}\hbar[/tex]
