Question

The correct order according to size is
A. $O>{{O}^{-}}>{{O}^{2-}}$
B. ${{O}^{-}}>{{O}^{2-}}>O$
C. ${{O}^{2-}}>{{O}^{-}}>O$
D. $O>{{O}^{2-}}>{{O}^{-}}$

Answer 1

Hint: The size of the atom is going to depend on the number of electrons present in it or number of electrons it donated or number of electrons accepted. The size of the atom is going to decrease while we are moving from left to right in the periodic table.

Complete answer:
- In the question it is asked to write the correct order according to size of among the given options.
- First, we have to check the atomic size of the given ions ${{O}^{2-}},{{O}^{-}}and\text{ }O$ .
- The given atom is oxygen.
- We have to find the correct order of the atomic size of the ions ${{O}^{2-}},{{O}^{-}}and\text{ }O$ and have to arrange them according to their atomic size.
- We know that if an atom is going to accept the electrons, the newly added electrons are going to enter into a vacant or new orbital.
- By the addition of new electrons to the outermost shell the size of the atom is going to increase.
- Means the atom which accepts more electrons is going to exhibit high atomic size when compared to the atoms which accept less number of electrons or the atoms which do not accept the electron.
- Therefore the atomic size of the ${{O}^{2-}}$ is high when compared to ${{O}^{-}}$ and the atomic size of the ${{O}^{-}}$ Is high when compared to ‘O’.
- Therefore, the correct order of atomic size of the oxygen and its derivatives is ${{O}^{2-}}>{{O}^{-}}>O$ .

So, the correct option is C.

Note:
If an atom is going to lose the electrons then the atomic radius of the particular atom is going to decrease. Means the atom which has large positive charge is going to have less atomic size when compared to the atoms which contain less charge or no charge on them.