Question

Arrange Se^2-, I^-, Br^-, O^2- and F^- in decreasing order of their ionic radius.

Answer 1

A negatively charged ion is known as an anion.

Anions have a larger ionic radius than their parent element (the element from which they are formed).

Ionic radius trend for anions in the periodic table:

1. The ionic radius increases down the group (from top to bottom).
2. The ionic radius decreases across the period from left to right.

For Selenium anion (Se^2-)

• The atomic number is $34$.
• Periodic position: ${16}^{\mathrm{th}}$ group, $4^{\mathrm{th}}$ period.
• The number of electrons in the anion will be $34+2=36$.

For Iodine (I^-)

• The atomic number is $53$.
• Periodic position: ${17}^{\mathrm{th}}$ group, $5^{\mathrm{th}}$period
• The number of electrons in the anion will be $53+1=54$.

For Bromine (Br^-)

• The atomic number is $35$.
• Periodic position: ${17}^{\mathrm{th}}$ group, $4^{\mathrm{th}}$ period
• The number of electrons in the anion will be $35+1=36$.

For Oxygen (O^2-)

• The atomic number is $8$
• Periodic position: ${16}^{\mathrm{th}}$ group, $2^{\mathrm{nd}}$ period
• The number of electrons in the anion will be $8+2=10$.

For Fluorine (F^-)

• The atomic number is 9
• Periodic position: 17th group, 2nd period
• The number of electrons in the anion will be $9+1=10$.

Considering both the period and group trends for the anion, the decreasing order of ionic radius is

I^- > Se^2- > Br^- > O^2- > F^-