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Question: Which of the following arrangements represents the increasing order (smallest to largest) of ionic r...

Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species O2−,S2−,N3−,P3−{ O }^{ 2- },{ S }^{ 2- },{ N }^{ 3- },{ P }^{ 3- } ?
(a) O2−<N3−<S2−<P3−{ O }^{ 2- }<{ N }^{ 3- }<{ S }^{ 2- }<{ P }^{ 3- }
(b) O2−<P3−<N3−<S2−{ O }^{ 2- }<{ P }^{ 3- }<{ N }^{ 3- }<{ S }^{ 2- }
(c) N3−<O2−<P3−<S2−{ N }^{ 3- }<{ O }^{ 2- }<{ P }^{ 3- }<{ S }^{ 2- }
(d) N3−<S2−<O2−<P3−{ N }^{ 3- }<{ S }^{ 2- }<{ O }^{ 2- }<{ P }^{ 3- }

Explanation

Solution

The long form of periodic table or the modern periodic table is the current form of the periodic table. It consists of a total of 118 elements. We can predict the properties of an element belonging to a particular group because of the periodicity of properties. Take the charges into consideration while deciding the trend.

Complete step by step solution:
Before solving this question, we first need to know about the modern periodic table. In modern periodic tables the elements are arranged in increasing order of their atomic numbers. The modern periodic table has 18 vertical columns and 7 horizontal rows.
Properties of the Groups in the Modern Periodic Table. They are the vertical columns in the long form of the periodic table and the total number of these vertical columns is 18, with the first group denoted as Group 1. For each group, the elements belonging to a particular group will have the same type of outer shell electronic configuration where for every element the principal quantum number will increase as we move down the group. As we go down a group there are certain general trends such as the electronegativity will decrease, the atomic radii will increase, electropositive character will increase, and ionization energy will decrease. Properties of the Periods in the Modern Periodic Table. They are the horizontal rows in the long form of the periodic table and the total number of these horizontal rows is 7. The first period only has 2 elements (Hydrogen and Helium), the second and third period have 8 elements, the fourth and the fifth period have 18 elements while the sixth and the seventh period have 32 elements. With each successive period, the principal quantum number increases. As we move across a period there are certain general trends such as the electronegativity will increase, the atomic radii will decrease, electropositive character will decrease, and ionization energy will increase. Now let us solve the question. Sulphur and Phosphorus belong to the third period hence their atomic and ionic radii will be larger than that of Nitrogen and Oxygen since these elements belong to the second period of the periodic table. Now, In order to compare the size of P3−{ P }^{ 3- } and S2−{ S }^{ 2- } ion we need to first compare their atomic radius. Since the size of atoms decreases along a period, hence the atomic radius of sulphur atom will be less than that of phosphorus atom. Also in P3−{ P }^{ 3- } ion the ionic charge is greater than that for S2−{ S }^{ 2- }; this will lead to more electronic repulsions in P3−{ P }^{ 3- } ion and hence its size will increase. Therefore, P3−{ P }^{ 3- } ion will have the largest ionic radius.

Hence the correct answer is (a) O2−<N3−<S2−<P3−{ O }^{ 2- } <{ N }^{ 3- }<{ S }^{ 2- }<{ P }^{ 3- }

Note: Due to the periodicity of the properties, the unknown properties of an element can be predicted but they are not always true. Certain exceptions within a group do exist. For example, the electron affinity trend of group 17 where the electron affinity of Fluorine is less than that of Chlorine.