Vanadium has $Z = 23$ and so the electronic configuration of an isolated neutral atom in the gas phase would be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3}$.
Now, magnetic moment of 1.73 means 1 unpaired electron (calculated using the formula $M^2 = n(n + 2)$). So, vanadium will have to lose 4 electrons.
I worked out the ion's electronic configuration to be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$. But internet and other sources say it's $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1}$. Note the $\ce{3d^1}$ instead of $\ce{4s^1}$.
