Question

Which of the following is the correct order of the size of the ions?
A.$Na^{+} > Mg^{2+} > F^{-} > O^{2-}$
B.$O^{2-} > F^{-} > Na^{+} > Mg^{2+}$
C.$Mg^{2+} > Na^{+} > F^{-} > O^{2-}$
D.$O^{2-} > F^{-} > Mg^{2+} > Na^{{}+}$

Answer 1

Hint: In order to solve the question, first compare the number of shells in each species. If they all are equal in number, compare the positive charge and negative charge. We will get the correct option.

Complete answer:
- Initially comparing the number of shells, all species of ions are having the same number of shells. It is clear that each species is having two shells.
- As all given ions are having the same number of shells, we will now compare according to the charge.
- If positive charge increases, the size of the ion decreases. This is because, if positive charge is more, the effective nuclear charge is more which leads to the smaller size of the ion.
- If negative charge increases, the size of the ion increases. This is because if negative charge is more, the effective nuclear charge is less which leads to the larger size of the ion.
- Now we shall compare the options.
- Oxygen ion comes first as it is having negative charge when compared to the remaining ions and it is followed by fluorine ion as it has a negative charge whereas sodium and magnesium ions contain positive charge. So, among the negative signs, the oxygen will have larger size than the fluorine, i.e., ${{O}^{2-}} > {{F}^{-}}$
- Comparing sodium ion and magnesium ion, sodium ion is higher in size than magnesium ion because sodium ions contain less positive charge than magnesium ions. So, among the positive signs, sodium ion will have larger size than the magnesium ion, i.e., $N{{a}^{+}} > M{{g}^{2+}}$

Therefore, the correct answer is an option (b).

Note:
The reason for larger size of ions which contain more negative charge is an increase in positive charge also increases effective nuclear charge. Less effective nuclear charge is favorable for more size.