Chapter 6, Problem 76A

Interpretation Introduction

Interpretation:

The ionic size of the given ions is to be predicted.

Concept Introduction :

An ion is formed by gaining or losing electrons in an atom. The number of electrons lost or gained by an atom becomes its charge. If an element loses electrons, it has a positive charge. In the symbol for an ion, the charge is represented as a superscript of the symbol of the atom for that element.

Answer to Problem 76A

  $S{c}^{3+}<C{a}^{2+}<K^{+}<C{l}^{-}<S^{2-}$

  $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}$

Explanation of Solution

When an atom loses one or more electrons, a positively charged ion or cation is formed. The cations are smaller in size than the atom from which it is formed. This is because they contain fewer electrons while having the same nuclear charge (no. of protons). The protons in the nucleus are holding the residual electrons more firmly and so, their radii are less than those of the parent atoms. Thus, the more the positive charge, the less will be the ionic radius. So, the order of the size will be:

  $S{c}^{3+}<C{a}^{2+}<K^{+}<C{l}^{-}<S^{2-}$

Similarly, for $O^{2-}$, $F^{-}$, $N{a}^{+}$, $M{g}^{2+}$, and $A{l}^{3+}$, the ionic size varies as shown:

  $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}$

Conclusion

The charge of the elements and their nuclear charge determines their ionic radius.