Question

Which of the following sequences is correct for decreasing order of ionic radius?
A.$S{e^{2 - }} > {I^ - } > B{r^ - } > {O^{2 - }} > {F^ - }$
B.${I^ - } > S{e^{2 - }} > {O^{2 - }} > B{r^ - } > {F^ - }$
C.$S{e^{2 - }} > {I^ - } > B{r^ - } > {F^ - } > {O^{2 - }}$
D.${I^ - } > S{e^{2 - }} > B{r^ - } > {O^{2 - }} > {F^ - }$

Answer 1

Ionic radius increases down the group and decreases across the period. When an electron is added to the neutral atom or ion, then the ionic radius will increase. Thus, the ion with more negative charge has more ionic radius. The ion with more charge has more ionic radius.

Complete answer:
Halogens are non-metals that belong to group $17$ . Fluorine, chlorine, bromine, and iodine belong to halogens. Iodine ion ${I^ - }$ has more ionic radius as it is the chemical element with atomic number $53$ . The ion $S{e^{2 - }}$ has ionic radius followed by ${I^ - }$ as selenium is the element with atomic number $34$ . Bromide ion has the atomic number $35$ due to the one negative charge it has less ionic radius than $S{e^{2 - }}$ .
Oxygen ion ${O^{2 - }}$ has more ionic radius than fluoride ion ${F^ - }$ as oxygen has negative charge of $- 2$ whereas fluoride ion has single negative charge. Oxygen is the element with atomic number $8$ and fluorine has atomic number $9$ .
${O^{2 - }}$ and ${F^ - }$ are the same isoelectronic series. But the oxygen ion has more negative charge whereas fluoride ion has less negative charge.
Thus, the order is ${I^ - } > S{e^{2 - }} > B{r^ - } > {O^{2 - }} > {F^ - }$

Option D is the correct answer.

Note:
The ionic radius down the group increases as the electron is added to another shell. When the elements in the same period are taken, the elements that have more negative charge have more ionic radius. Isoelectronic series are the ions that have the same electrons. When isoelectronic series are taken, the ion with more negative charge is considered.