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Question: Number of valence electrons in \({{{O}}^{2 - }}\) is: A. \(6\) B. \(8\) C. \(10\) D. \(4\)...

Number of valence electrons in O2−{{{O}}^{2 - }} is:
A. 66
B. 88
C. 1010
D. 44

Explanation

Solution

We know that valence electrons are the electrons which are present in the outermost shell of an atom. This is purely dependent on the atomic number and the electronic configuration of the atom. Oxygen lies in the group XVI and period II.

Complete step by step answer:
To find the number of valence electrons, first we have to know the atomic number and then the electronic configuration. As we know, an atomic number is obtained when we add the number of protons or number of electrons in an atom. Number of protons is equal to the number of protons. So the atomic number of oxygen is eight. It can be expressed in electronic configuration as 1s22s22p41{{{s}}^2}2{{{s}}^2}2{{{p}}^4}, i.e. the orbitals can be arranged as given below:

↑↓\uparrow \downarrow↑↓\uparrow \downarrow↑↓\uparrow \downarrow↑\uparrow↑\uparrow

1s2{{1}}{{{s}}^2} 2s2{{2}}{{{s}}^2} 2p4{{2}}{{{p}}^4}
Here, there are two unpaired electrons in the p orbital. Now let’s consider the O2−{{{O}}^{2 - }} ion. Generally ions are formed by either gaining or losing electrons. The given ion is an anion, i.e. it is formed by gaining electrons. So they have a negative charge. Cations are formed by losing electrons. They have a positive charge. Since it has −2 - 2 as charge, two electrons are gained. So the electronic configuration will be 1s22s22p61{{{s}}^2}2{{{s}}^2}2{{{p}}^6}. This electronic configuration is the same as that of neon noble gas. The orbitals are arranged as:

↑↓\uparrow \downarrow↑↓\uparrow \downarrow↑↓\uparrow \downarrow↑↓\uparrow \downarrow↑↓\uparrow \downarrow

1s2{{1}}{{{s}}^2} 2s2{{2}}{{{s}}^2} 2p6{{2}}{{{p}}^6}
There are six valence electrons in the valence shell in the oxygen atom. While in O2−{{{O}}^{2 - }}, the number of valence electrons in the outermost shell is 88.

Hence the correct option is B.

Note: When the fluorine and oxygen in anionic forms are considered, fluorine is more electronegative than oxygen. Since the electronegativity is very high in oxygen in anionic form, it is very stable than any other elements in the periodic table.