Question

How many electrons are lost or gained in forming each $B{a^{2 + }}$?

Answer 1

Hint: The valence electrons are those electrons which are present in the outermost electronic configuration of the atom. So the atom loses its valence electron to form a cation and gains electrons to form anion.

Complete step by step answer:
The chemical elements except the noble gases, are not stable in nature as they do not follow octet rule which says that the atom should contain 8 electrons in the outermost orbital. Due to their instability they react with other chemical elements to form new products.
Thus, the atom loses its valence electrons or the outermost electrons to form a cation or a positively charged species. The atom gains the electrons from the other neighbouring chemical element to form an anion or negatively charged species. The atom does so to obtain the noble gas configuration.
The noble gas configuration means that the atoms by losing or gaining the electrons forms a configuration of the nearest noble gas.
The barium is the chemical element which is present in the group 2 of the periodic table also known as alkaline earth metal. The barium is denoted by the symbol Ba. The atomic number of the barium is 56 which denotes that the number of electrons present in the barium atom is 56. The electronic configuration of the barium atom is $[Xe]6{s^2}$. The barium atom will lose its two electrons from the valence shell to form a $B{a^{2 + }}$.

Therefore, 2 electrons are lost in forming each $B{a^{2 + }}$.

Note:You can now see that when the barium atom loses its two electrons, it gets the configuration of the xenon atom. So the electronic configuration of the xenon atom and the $B{a^{2 + }}$ ion is the same.