Question

which of the following atoms is the smallest?
be
o
li
b
ne

Explanation:

Step1: Recall atomic radius trend

In a period (row) of the periodic table, atomic radius decreases from left to right. This is because as atomic number increases, the number of protons in the nucleus increases, leading to a stronger effective nuclear charge that pulls the electrons closer to the nucleus.

Step2: Identify the period of the atoms

Be (beryllium), O (oxygen), Li (lithium), B (boron), and Ne (neon) are all in the second period of the periodic table.

Step3: Determine the order from left to right

In the second period, the order from left to right is: Li, Be, B, O, Ne.

Step4: Apply the atomic radius trend

Since atomic radius decreases from left to right in a period, the atom furthest to the right (with the highest atomic number among them, excluding Ne which is a noble gas with a different electron configuration and slightly larger radius due to van der Waals forces) will have the smallest radius. Among Li, Be, B, O, the order of atomic number is Li (3) < Be (4) < B (5) < O (8). So O has a smaller atomic radius than Li, Be, and B. Neon, although has a higher atomic number, has a filled valence shell and its atomic radius is measured differently (van der Waals radius) which is larger than the covalent radius of O. So among the given options, O is the smallest.

Answer:

O (the option with "O" as the atom, which was already selected in the provided image, but the correct atom is O)