Question

Statement-I : $\text{K}>\text{Mg}>\text{Al}>\text{B}$ metallic character order. Statement-II: Ionic radius of any element is less than its atomic radius.

• A. Both statements are true
• B. Statement I is false but statement II is true.
• C. Both statements are False.
• D. Statement I is true but statement II is false.

Hint:

Metallic character is highest on the left and bottom of the periodic table. Cations are smaller than their parent atoms, while anions are larger.

Answer 1

The Correct Option is D

To solve this question, we must analyze both Statement I and Statement II separately using our understanding of periodic table trends and ionic and atomic radii.

Analyzing Statement I: \( \text{K} > \text{Mg} > \text{Al} > \text{B} \) metallic character order.

Metallic character generally decreases as we move from left to right across a period and increases down a group in the periodic table.

• Potassium (\( \text{K} \)) is located in Group 1, Period 4. It has a highly metallic character due to its tendency to lose electron easily.
• Magnesium (\( \text{Mg} \)) is in Group 2, Period 3. It is less metallic than potassium but more metallic than elements to its right in the period.
• Aluminum (\( \text{Al} \)) is in Group 13, Period 3. Its metallic character is less than that of magnesium but still considerable.
• Boron (\( \text{B} \)) is a metalloid, located in Group 13, Period 2, with significantly lower metallic character compared to the others.

Based on these explanations, the order \( \text{K} > \text{Mg} > \text{Al} > \text{B} \) correctly represents the decreasing metallic character. Therefore, Statement I is true.

Analyzing Statement II: Ionic radius of any element is less than its atomic radius.

The ionic radius compared to the atomic radius differs based on whether the atom loses or gains electrons to form an ion:

• For cations (positively charged ions, formed by losing electrons), the ionic radius is smaller than the atomic radius because there are fewer electrons, leading to a reduced screening effect and greater effective nuclear charge pulling the remaining electrons closer.
• For anions (negatively charged ions, formed by gaining electrons), the ionic radius is larger than the atomic radius due to the addition of electrons causing increased electron-electron repulsion.

Thus, Statement II is false because it is not universally applicable to all elements. It only holds true for cations, not anions.

Conclusion:

Since Statement I is true and Statement II is false, the correct answer is Statement I is true but Statement II is false.