1. Periodic trend to use

Atomic radius generally:

• decreases across a period from left to right,
• increases down a group from top to bottom.

We will check each option using these trends.

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

• In period 3: $\mathrm{Si} > \mathrm{P} > \mathrm{Cl}$ is correct because radius decreases left to right.
• $\mathrm{Cl}$ is in period 3, while $\mathrm{F}$ is in period 2. Since atomic radius increases down a group, $\mathrm{Cl} > \mathrm{F}$ is also correct.

So, Option A is correct.

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

• In period 3: $\mathrm{Mg} > \mathrm{Al}$ is correct.
• In period 2: $\mathrm{C} > \mathrm{O}$ is correct.
• Also, period 3 atoms are generally larger than period 2 atoms, so $\mathrm{Al} > \mathrm{C}$ is reasonable.

Thus, Option B is correct.

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

• $\mathrm{Al}$ lies below $\mathrm{B}$ in group 13, so $\mathrm{Al} > \mathrm{B}$ is correct.
• In period 2: $\mathrm{B} > \mathrm{N} > \mathrm{F}$ is correct because radius decreases left to right.

Thus, Option C is correct.

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

Check carefully:

• In group 2, radius increases down the group, so $\mathrm{Mg} > \mathrm{Be}$ not $\mathrm{Be} > \mathrm{Mg}$.
• In period 3: $\mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$ is correct.

Therefore the full decreasing order should begin as $\mathrm{Mg} > \mathrm{Be}$ or at least $\mathrm{Be} > \mathrm{Mg}$ is definitely wrong.

So, Option D is incorrect.

6. Final answer

The incorrect decreasing order of atomic radii is: $\boxed{\mathrm{D}}$

This matches the stored correct answer.