Question

The magnetic moment of an oxide of vanadium is $2\sqrt{2}$ BM. Find the number of electrons having azimuthal quantum number = 1 or 2 in one molecule of the oxide.

Answer 1

Magnetic moment =$\sqrt{n(n+2)}=2\sqrt{2}$
$\Rightarrow n=2$
Since the number of unpaired electrons is 2, so V is in (+3) oxidation state.
So, oxide is $V_2{O}_3$
$V^{3+}=\left[Ar\right]3d^{2}4s^0$
$O^{2-}=\left[He\right]2s^{2}2p^6$
When $l=1$, we are talking about the p subshell.
The 2p and 3p orbitals of $V^{3+}$ are completely filled so a total of 24 electrons in two vanadium atoms of the oxide.
Similarly, $O^{2-}$ contains 6 electrons in 2p orbitals so, there are a total of 18 electrons from p orbitals in three oxygen atoms.
When $l=2$, we are talking about the d orbital.
V has 4 electrons (2 from each) in its 3d orbitals.
Therefore, total no. of given electrons = 24 + 18 + 4 = 46