Explain which atom of the following pair have the smaller ionization energy: As, I

To determine which atom has the smaller ionization energy between As (Arsenic) and I (Iodine), we need to consider their respective positions on the periodic table.

Ionization energy refers to the amount of energy required to remove an electron from an atom or ion in its gaseous state. As we move across a period from left to right on the periodic table, the atomic radius decreases while the ionization energy generally increases. This is because the effective nuclear charge (the positive charge experienced by the outermost electrons) increases, making it more difficult to remove an electron.

In this case, As is located to the left of I on the periodic table. Arsenic (As) is in period 4, while Iodine (I) is in period 5. As we move from left to right across a period, the ionization energy generally increases. Therefore, Iodine (I) will have a higher ionization energy compared to Arsenic (As).

So, in conclusion, Arsenic (As) will have the smaller ionization energy compared to Iodine (I).