Question

An orbital with $m=+3$ has an unpaired electron. If it is the only unpaired electron in the atom which of the following electronic arrangements corresponds to that atom?

• A. $\left[Kr\right]5d^{4}4f^{1}6s^2$
• B. $\left[Kr\right]4d^{2}5s^2$
• C. $\left[Kr\right]5d^{0}4f^{1}6s^2$
• D. $\left[Xe\right]4f^{2}5s^2$

Answer 1

Answer: (C)

$\left[Kr\right]5d^{0}4f^{1}6s^2$

Given that the magnetic quantum number $\left(m\right)=+3$, therefore the angular quantum number is $3$ corresponding to $f-$subshell and the principal quantum number is $4$. The one unpaired electron resides in the $4f$ subshell. Therefore the configuration is : $\left[Kr\right]5d^{0}4f^{1}6s^2$