Chapter 12, Problem 59A

Interpretation Introduction

Interpretation:

The given ions are to match with the pictures representing radius.

Concept introduction:

Atomic size depends on the placement of molecules in periodic table.

Left to right, atomic radius decreases and top to bottom, radius increases.

Same trend is applicable for ions.

Answer to Problem 59A

Let’ assign numbering to pictures:

  

Matching of ions with their radius is as:

Size of atom	ion
(i)	$K^{+}$
(ii)	$C{a}^{2+}$
(iii)	$S{c}^{3+}$
(iv)	$S^{2-}$
(v)	$C{l}^{-}$

Explanation of Solution

The ions which are to be arranged in increasing order of radius are isoelectronic.

Ions	$S{c}^{3+}$	$C{a}^{2+}$	$K^{+}$	$C{l}^{-}$	$S^{2-}$
Protons	21	20	19	17	16
Electrons	18	18	18	18	18

Thus, all ions have same number of electrons but different number of protons.

As seen from table, effective nuclear charge of $S{c}^{3+}$ is maximum so it has smallest radius and effective nuclear charge of $S^{2-}$ is minimum so it has largest radius

So the trend is:

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

Conclusion

Thus, in isoelectronic species, atomic radius increases with decrease in effective nuclear charge