Question

Which one of the following ions will be smallest in size?

• A. Na$^+$
• B. Mg$^{2+}$
• C. F$^-$
• D. $O^{-}_2$

Answer 1

The Correct Option is B

To determine which ion is the smallest in size among the given options, we need to consider the effective nuclear charge and the number of electrons in each ion.

1. Na⁺: The sodium ion is formed by losing one electron from the neutral sodium atom. As a result, it has 10 electrons and 11 protons, leading to a higher effective nuclear charge compared to the neutral atom. The reduced electron cloud results in a smaller ionic size compared to the original atom.
2. Mg²⁺: The magnesium ion is formed by losing two electrons from the neutral magnesium atom. It has 10 electrons and 12 protons. The loss of two electrons increases the effective nuclear charge even more significantly than in Na⁺, making Mg²⁺ even smaller.
3. F^-: The fluoride ion gains an electron, resulting in 10 electrons and 9 protons. This excess of electrons reduces the effective nuclear charge experienced per electron, leading to an increase in size.
4. O^2-: The oxide ion gains two electrons, resulting in 10 electrons and 8 protons. The substantial increase in electron repulsion causes an even larger increase in size than for F^-.

Given the above analysis:

• Na⁺ is smaller than Na neutral atom.
• Mg²⁺ is smaller than Mg and Na⁺ due to even higher effective nuclear charge.
• F^- is larger than F due to adding an electron.
• O^2- is larger than O due to adding two electrons.

Therefore, the smallest ion by size is Mg²⁺ due to its high effective nuclear charge after losing two electrons, which compacts the electron cloud.