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

$[Xe]5d^16s^2$

The correct answer is Option (1) → $[Xe]5d^16s^2$

To write configuration:

1. Noble gas core of Xenon (Xe) = 54 electrons

2. Remaining electrons = 57 - 54 = 3 electrons

Now we decide where these 3 electrons go.

According to energy order after Xe:

$6s→5d→4f$

For Lanthanum, 4f filling has not started significantly yet.

Electronic Distribution

First two electrons go into 6s:

$6s^2$

The third electron goes into 5d:

$5d^1$

So configuration:

$[Xe] 5d^1 6s^2$