Question

both vanadium and its 3+ ion are paramagnetic. use electron configurations to explain why this is so.

Explanation:

1. First, recall the electron configuration of vanadium (V). Vanadium has an atomic number of 23. The electron configuration of a neutral vanadium atom is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3$ (or can be written as $[Ar] 4s^2 3d^3$ using the noble gas core notation).
2. Now, consider the vanadium(III) ion ($V^{3+}$). When vanadium forms a $3+$ ion, it loses 3 electrons. Electrons are removed from the outermost shell first, so it loses the 2 electrons from the $4s$ orbital and 1 electron from the $3d$ orbital. The electron configuration of $V^{3+}$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^2$ (or $[Ar] 3d^2$).
3. Paramagnetism is caused by the presence of unpaired electrons. In the neutral vanadium atom, the $3d$ subshell has 3 electrons. According to Hund's rule, these electrons will occupy separate orbitals within the $3d$ subshell with parallel spins, so there are 3 unpaired electrons.
4. In the $V^{3+}$ ion, the $3d$ subshell has 2 electrons. Again, by Hund's rule, these 2 electrons will occupy separate $3d$ orbitals with parallel spins, resulting in 2 unpaired electrons.
5. Since both the neutral vanadium atom and the $V^{3+}$ ion have unpaired electrons (3 in V, 2 in $V^{3+}$), they are both paramagnetic.

Answer:

• Electron configuration of V (atomic number 23): $\boldsymbol{[Ar] 4s^2 3d^3}$ (has 3 unpaired $3d$ electrons).
• Electron configuration of $\boldsymbol{V^{3+}}$: Loses 3 electrons (2 from $4s$, 1 from $3d$), so $\boldsymbol{[Ar] 3d^2}$ (has 2 unpaired $3d$ electrons).
• Paramagnetism requires unpaired electrons. Both V (3 unpaired $3d$ electrons) and $V^{3+}$ (2 unpaired $3d$ electrons) have unpaired electrons, so they are paramagnetic.