as we all know that if the value of $n+l$ is same then the orbital with lower value of $n$ will get filled first so in case of first member of lanthanide series the configuration is $(\ce{Xe})\mathrm{6s^2 5d^1 4f^1}$. Here, the $n+l$ value for $\mathrm{d}$ orbital and $\mathrm{f}$ orbital is same i.e. 7 and $\mathrm{4f}$ have a lower value of $n$. Then why are we not filling the electron in $\mathrm{4f}$ first?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.