Question

Valence shell electronic configuration of Barium is:
(A) $5{{s}^{2}}$
(B) $6{{s}^{2}}$
(C) $5{{s}^{1}}$
(D) $6{{s}^{1}}$

Answer 1

Hint: To obtain the valence shell occupancy of the given element, the filling of electrons based on the energy of the orbitals is done. The lower energy subshells are occupied first, followed with the higher energy subshells.

Complete step by step solution:
Barium has atomic number 56. It belongs to period 6 and group 2 of the periodic table. In group 2, it is the fifth element.
From the Aufbau principle, the filling of electrons from the lower energy shell to the higher energy shell, gives the electronic configuration as $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{2}}4{{d}^{10}}5{{p}^{6}}6{{s}^{2}}$or $\left[ Xe \right]6{{s}^{2}}$. The outermost or the valence shell is the 6s orbital with occupancy of two electrons.

Therefore, the valence shell electronic configuration is option (B)- $6{{s}^{2}}$.

Additional information:
Being present in group 2 of the periodic table, Barium is an alkaline earth metal with two valence electrons and $(+2)$ oxidation state.
-With the decrease in the electronegativity down the group, the hydride of barium is most ionic.
-The oxide and hydroxide of barium are the most basic in the group with the increase in the size of the atom down the group.
-The sulphates, nitrates, and chlorides of barium are the most soluble in the group.

Note: In the Aufbau principle used for the electronic configuration, the $\text{(n + l)}$ rule must be taken care of, where, when the $\text{(n + l)}$value is the same for two subshells. The higher n value is given preference. Thus, the 4s is filled before 3d and 5s before 4d.