Chemistry - f-block Question with Solution | TestHub

The atomic number of Neodymium (Nd) is 60. Its electronic configuration is $\left[\mathrm{Xe}\right] 4{\mathrm{f}}^4 6{\mathrm{s}}^2$.

Now, by removing two electrons from the outermost shell 6s, ${\mathrm{Nd}}^{2+}$ ion is formed. Hence, electronic configuration of ${\mathrm{Nd}}^{2+}$ is $\left[\mathrm{Xe}\right] 4{\mathrm{f}}^4$.