"The incorrect decreasing order of atomic radii is"

Question

Accepted Answer

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Let's analyze the options based on the periodic trends in atomic radii. Recall two key trends:

Atomic radii decrease as you move from left to right in a period due to increasing effective nuclear charge.

Atomic radii increase as you move down a group because additional electron shells are added.

We'll review each option:

Option A: $\mathrm{Si} > \mathrm{P} > \mathrm{Cl} > \mathrm{F}$

Silicon (Si), phosphorus (P), and chlorine (Cl) are in the same period, with a gradual decrease in atomic radius from left to right.

Fluorine (F) is in an earlier period (it has only 2 electron shells) and is much smaller.

Hence, the order follows the expected trend.

Option B: $\mathrm{Mg} > \mathrm{Al} > \mathrm{C} > \mathrm{O}$

Magnesium (Mg) and aluminum (Al) are period 3 elements, whereas carbon (C) and oxygen (O) belong to period 2.

Period 3 elements generally have larger radii than period 2 counterparts.

The order is consistent: Mg and Al are larger than C and O, and within each period, radii decrease from left to right.

Option C: $\mathrm{Al} > \mathrm{B} > \mathrm{N} > \mathrm{F}$

Aluminum (Al) is in period 3 and is expected to be larger than boron (B), nitrogen (N), and fluorine (F) from period 2.

Within period 2 (B, N, F), the atomic radii decrease from left to right.

Thus, this order is correct as well.

Option D: $\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$

Beryllium (Be) is in period 2, while magnesium (Mg), aluminum (Al), and silicon (Si) are in period 3.

Since atomic radii increase with the number of electron shells, Be (with only 2 shells) should be smaller than Mg.

This means that stating $\mathrm{Be} > \mathrm{Mg}$ is incorrect.

Thus, the incorrect decreasing order of atomic radii is given in Option D.

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The incorrect decreasing order of atomic radii is