Question

The set representing the correct order of ionic radius is:
(A) $L{{i}^{+}}>B{{e}^{2+}}>N{{a}^{+}}>M{{g}^{2+}}$
(B) $N{{a}^{+}}>L{{i}^{+}}>M{{g}^{2+}}>B{{e}^{2+}}$
(C) $L{{i}^{+}}>N{{a}^{+}}>M{{g}^{2+}}>B{{e}^{2+}}$
(D) $M{{g}^{2+}}>B{{e}^{2+}}>L{{i}^{+}}>N{{a}^{+}}$

Answer 1

Hint: The ionic radius depends upon the specific trends of the periodic table. The trend for the element’s atomic radius will be continual in the group and across the period as well unlike ionic radius. If asked for the discontinuous pattern of elements, we need to consider some numerical figures and empirical definitions.

Complete step by step solution:
Let us first see the trend of atomic radius and then focus on the ionic radius patterns.
Atomic radius- The distance between the nucleus of an atom and the outermost shell having an electron is known as the atomic radius of the same. The atomic radius of the atom decreases as we move from left to right in a period and increases as we move from top to bottom in a group. Now focusing on ionic radius, let us look forward to cations; When a neutral atom loses an electron, a positive charge is built and the atoms become cations. The resulting cations are smaller than that of the respective neutral atoms. This takes place because when an electron is lost, the repulsion between electrons decreases and protons more easily pull the electrons towards the nucleus. Each atom has an ability to either lose or gain electrons. So, the trend of ionic radius won’t be constant as that of the atomic radius. Thus, the trends must be specific regarding different groups.
Illustration- The alkali and alkali earth metals form cations which increase in size down the group. However, across the period there is no significant change in the ionic radius. The ionic radius is measured by calculating its proportion as space while comparing it with another ionic bond in crystal lattice. Also, this trait isn’t permanent for any atom. Ionic radius mostly depends on coordination number, spin state etc.
So, $N{{a}^{+}}=0.98A{}^\circ $
$L{{i}^{+}}=0.68A{}^\circ $
$M{{g}^{2+}}=0.65A{}^\circ $
$B{{e}^{2+}}=0.62A{}^\circ $
Hence, $N{{a}^{+}}>L{{i}^{+}}>M{{g}^{2+}}>B{{e}^{2+}}$ is the correct order of ionic radius.

Therefore, option (B) is correct.

Note: Do note that the ionic radius depends on many factors and won’t be continuous for the specific trends. So, numerical values are important to describe the trend.