Question

Lanthanide having atomic number $57$ is
A. s-block element
B. p-block element
C. d-block element
D. f-block element

Answer 1

The block of the element can be identified based on the outermost orbitals. The valence electrons of the lanthanide enter into the d-orbital.

Complete answer:
p, d and f are four orbitals in which the electron enters. All these orbitals contain electrons in an element but the orbitals that receive the last electrons are known as valence orbitals and the valence orbital indicates the block type of which the elements belong.The atomic number $57$ is of lanthanum.

The electronic configuration of the lanthanum element is as follows: $\left[ {{\text{Xe}}} \right]\,5{{\text{d}}^1}\,{\text{6}}{{\text{s}}^{\text{2}}}$

The s-orbitals are completely filled and the d-orbital has one electron. The energy of $6s$ orbitals is less than the $5d$ orbital, so the electron first enters into $6$s orbitals, and then the remaining valence electrons enter into $5d$ orbital. The valence electrons of the lanthanum element enter into d-orbital so the lanthanum is a d-block element. So, lanthanide having atomic number $57$ is a d-block element.

Additional information: The period of the element tells the principal quantum number of the valence shell of the element. The group number tells the number of valence electrons entered in the outermost orbital. F-block elements from atomic numbers $58 - 71$ are known as lanthanides and elements from atomic numbers $90 - 103$ are known as actinides.

Therefore, the correct answer is option (C) d-block elements.

Note: The block of the element tells the type of outermost orbital which receives the last electrons. The atomic number of xenon is $54$ so, if an atom contains electrons more than a noble gas its electronic configuration is shown by the symbol of noble gas and valence electronic configuration. As the $57$ stands for lanthanum so all forward elements from $58 - 71$ are known as lanthanides. Valence shell electrons of lanthanides enter into f-orbitals.