Answer: Ar > Cl > Na
To arrange the given set of atoms in order of decreasing ionization energy (IE), we need to consider their positions in the periodic table. Ionization energy generally increases as we move from left to right across a period and decreases as we move down a group.
The atoms given are:
Cl - Chlorine (Group 17, Period 3)
Ar - Argon (Group 18, Period 3)
Na - Sodium (Group 1, Period 3)
Now let's arrange them in order of decreasing IE:
Ar (highest IE): As a noble gas, argon has a full electron shell, making it highly stable and difficult to remove an electron from it. Thus, it has the highest ionization energy among the three atoms.
Cl: Chlorine is just one electron short of a full electron shell, so it requires a significant amount of energy to remove an electron. Its ionization energy is lower than argon but higher than sodium.
Na (lowest IE): Sodium is in Group 1, which means it has only one electron in its outermost shell. This electron is relatively easy to remove, so sodium has the lowest ionization energy among the three atoms.
In summary, the order of decreasing ionization energy is: Ar > Cl > Na.