Question

The ${71}^{st}$ electron of an element $X$ with an atomic number of $71$ enters into the orbital:

• A. $4f$
• B. $6p$
• C. $5d$
• D. $6s$

Answer 1

The correct option is C $5d$

In the lanthanoid series, atomic number of fourteen $4f-$block elements ranges from $58\left(Ce\right)to71\left(Lu\right)$. Ytterbium, $Yb(Z=70)$ has electronic configuration :$\left[Xe\right]4f^{14}6s^2$. So, the ${71}^{nth}$ electron of lutetium, $Lu(Z=71)$ should enter into $5d$ orbital and its (here, $Lu$ is $‘X^{\prime })$ electronic configuration will be : $\left[Xe\right]4f^{14}5d^{1}6s^2$. It happens so, because f-block elements have general electronic configuration, $(n-2)f^{1-14}(n-1)d^{1-10}n{s}^2$. Therefore, option (c) is correct.