Question

Among the elements of Period 2, pick out the element with the lowest atomic size.

Answer 1

1. Atomic size of any atom is defined as the distance from the nucleus of that atom to its outermost shell. It is generally measured in angstroms $\left(\mathrm{A}°\right)$.
2. Atomic size decreases on moving left to right in the Period or row in the periodic table.
3. The reason for the decrease in size is due to an increase in the effective or net nuclear charge due to the addition of electrons in the same energy level. Due to the increase in nuclear charge felt by valence electrons, the valence electrons move towards the nucleus, and size decreases.
4. Period 2 consists of eight elements, Lithium (Li), Beryllium (Be), Boron (B), Carbon (C), Nitrogen (N), Oxygen (O), Fluorine (F), and Neon (Ne).
5. Exception: Since Neon has a stable octet configuration, the interelectronic repulsion between the electrons in Ne is more than that in F. As a result, the effective nuclear charge is less in Ne compared to that in F. Therefore, the Ne atom is larger than the F atom.
6. Since Fluorine (F) is present at the rightmost side in Period 2 excluding Neon, it is the smallest or has the lowest atomic size in Period 2.