The atomic number of Lanthanum is 57. Its electronic configuration will be ……………………

The correct option is [Xe] 5d¹ 6s².

Lanthanum (atomic number, $Z=57$) is the first element of the lanthanide series. To find its electronic configuration, we build upon the electronic configuration of the preceding noble gas, Xenon ($\text{Xe}$), which has an atomic number of 54.

The electronic configuration of Xenon ($\text{Xe}$) is:
$\left[\text{Xe}\right]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^6$

This accounts for 54 electrons. For Lanthanum ($Z=57$), we must distribute the remaining 3 electrons ($57-54=3$) into the next available orbitals: $6s$, $5d$, and $4f$.

According to the Aufbau principle, the $6s$ orbital is lower in energy than both the $5d$ and $4f$ orbitals. Therefore, 2 electrons fill the $6s$ orbital first:
$6s^2$

The remaining 1 electron enters the $5d$ orbital instead of the $4f$ orbital. This is because the energy of the $5d$ orbital is slightly lower than that of the $4f$ orbital at this point in the periodic table, providing a more stable configuration. Consequently, the last electron is placed in the $5d$ orbital:
$5d^1$

Combining these, the complete valence shell configuration for Lanthanum is:
$\left[\text{Xe}\right]5d^{1}6s^2$