Question

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

Explanation:

Step1: Recall Periodic Trend

Atomic radius decreases across a period (left to right) due to increasing nuclear charge (more protons) pulling electrons closer, with same energy level. These elements (Be, O, Li, B, Ne) are in period 2.

Step2: Order Elements Left to Right

Period 2 order: Li (leftmost), Be, B, O, Ne (rightmost).

Step3: Apply Trend to Find Smallest

Across period 2, atomic radius decreases left to right. Ne is noble gas, but O is to the right of B, Be, Li. Wait, Ne has van der Waals radius (larger for noble gases), but among non - noble gas period 2, O is rightmost (excluding Ne's special case? Wait, no—wait, the trend: for main - group elements, across period, atomic radius decreases. Li (1), Be (2), B (3), O (6), Ne (10) protons. O has more protons than B, Be, Li. So O's electrons are pulled more. Wait, Ne is noble gas, its atomic radius is measured as van der Waals, which is larger than covalent radius. So among these, O is the smallest (excluding Ne's abnormal measurement). Wait, let's check: Li (atomic number 3), Be (4), B (5), O (8), Ne (10). The atomic radius (covalent) for Li: ~152 pm, Be: ~112 pm, B: ~87 pm, O: ~66 pm, Ne: van der Waals ~154 pm. Wait, no, I made a mistake. Wait, B's atomic radius is ~87, O is ~66. So O is smaller than B, Be, Li. So the order from largest to smallest (atomic radius) in period 2 (main - group, excluding Ne's van der Waals) is Li > Be > B > O. Ne's atomic radius is larger due to noble gas configuration (electron repulsion in van der Waals). So among the options, O has the smallest atomic radius.

Answer:

O (the option with "o" - note: likely typo for "O", Oxygen)