Question

The number of valence electrons of caesium is:
A.1
B.2
C.6
D.3

Answer 1

Caesium belongs to the s-block elements whose last electron enters in $ns$-orbital. The atomic number of caesium is 55 and its penultimate shell contains 8 electrons.

Complete step by step answer:
Caesium (IUPAC name) is a chemical element with the symbol Cs and atomic number 55. It is soft, and has a golden yellow appearance with a melting point of $28.5^\circ {\text{C}}$, and it is one of only five metals that are liquid at or near room temperature. Caesium, rubidium and potassium share the same physical and chemical properties. It imparts violet colour flame. The reason for flame colouration is that the energy of the flame causes an excitation of the outermost electrons which on return to their original position gives out the energy so absorbed in the visible region. It is the most reactive among all metals, it is pyrophoric (liable to ignite spontaneously on exposure to air) in nature and reacts with water even at $- 116^\circ {\text{C}}$. It has the least electronegativity, having a value of 0.79 on the Pauling scale. It has only one stable isotope, caesium-133. Most of the time Caesium is mined from pollucite, while the radioisotope (caesium-137) which is a fission product, is extracted from waste produced by nuclear reactors.
Caesium exhibits $[Xe]6{s^1}$ configuration i.e. 2, 8, 18, 18, 8, 1$[1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^6}4{d^{10}},5{s^2},5{p^6},6{s^1}]$, the configuration of the valence shell is $6{s^1}$
Therefore the number of valence electrons of caesium is 1.
Thus, option A is the correct answer.

Note:
Caesium has the least ionization enthalpy value among IA group elements, i.e. it loses its outermost electron quite easily, but the 2nd ionization enthalpy is very high. Consequently it is not possible for alkali metals to form ${M^{2 + }}$ ions under ordinary conditions.