Question

The incorrect decreasing order of atomic radii is:

• A. \( Mg>Al>C>O \)
• B. \( Al>B>N>F \)
• C. \( Be>Mg>Al>Si \)
• D. \( Si>P>Cl>F \)

Hint:

Remember, atomic radius decreases across periods and increases down groups in the periodic table.

Answer 1

The Correct Option is C

The atomic radius generally decreases across a period (from left to right) and increases down a group.

(1) Mg > Al > C > O:
Mg and Al are in Period 3 (Mg is in Group 2, Al is in Group 13). Mg has a larger atomic radius than Al due to the general trend of decreasing radius across a period.
C and O are in Period 2 (C is in Group 14, O is in Group 16). C has a larger atomic radius than O for the same reason.
However, comparing elements from different periods (e.g., Mg/Al with C/O) violates periodic trends because atomic size increases down a group. Hence, this order is not entirely correct.

(2) Al > B > N > F:
Boron (B) and Aluminum (Al) are in Group 13, with B in Period 2 and Al in Period 3. Al has a larger radius than B due to being in a lower period.
Nitrogen (N) and Fluorine (F) are in Period 2. Radius decreases from B to N to F across the period. So this sequence follows the trend correctly within their groups and periods.
This sequence is correct.

(3) Be > Mg > Al > Si:
Beryllium (Be) is in Period 2, Group 2.
Magnesium (Mg) is in Period 3, Group 2 — so Mg > Be (down a group, size increases).
Aluminum (Al) is in Period 3, Group 13 — so Mg > Al (across period, size decreases).
Silicon (Si) is in Period 3, Group 14 — so Al > Si (again, across period, size decreases).
This sequence follows the periodic trend correctly.

Conclusion:
Among the given options, the correct order of atomic radii based on periodic trends is Be > Mg > Al > Si.

Final Answer:
The final answer is $ Be > Mg > Al > Si $.