Question

In which of the following options, elements are correctly arranged in the increasing order of their atomic radius?

• A. Si < P < Na < N < F
• B. Na < Si < P < N < F
• C. F < N < P < Si < Na
• D. N < F < Si < P < Na

Hint:

Remember the general trend: Radius increases down-left. Helium is smallest, Francium/Cesium is largest.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Atomic radius trends in the periodic table: 1. Across a Period (Left to Right):
Atomic radius decreases due to increased effective nuclear charge. 2. Down a Group (Top to Bottom):
Atomic radius increases due to the addition of electron shells.
Step 3: Detailed Explanation:

Let's locate the elements: - Period 2: N (Group 15), F (Group 17). Trend: Radius decreases \( \text{N} \textgreater \text{F} \). So, \( \text{F} \textless \text{N} \). - Period 3: Na (Group 1), Si (Group 14), P (Group 15). Trend: Radius decreases \( \text{Na} \textgreater \text{Si} \textgreater \text{P} \). So, \( \text{P} \textless \text{Si} \textless \text{Na} \). - Comparing Period 2 vs Period 3: Elements in Period 3 generally have larger radii than those in Period 2 due to an extra shell. So, \( \{ \text{F, N} \} \textless \{ \text{P, Si, Na} \} \). Combining the orders: Smallest to Largest: Period 2: \( \text{F} \textless \text{N} \) Period 3: \( \text{P} \textless \text{Si} \textless \text{Na} \) Overall Order: \( \text{F} \textless \text{N} \textless \text{P} \textless \text{Si} \textless \text{Na} \)
Step 4: Final Answer:

The correct increasing order is \( \text{F} \textless \text{N} \textless \text{P} \textless \text{Si} \textless \text{Na} \).