Question

The correct order of atomic radii of B, Be, N and C is

• A. Be $<$ B $<$ C $<$ N
• B. N $<$ B $<$ C $<$ Be
• C. N $<$ C $<$ B $<$ Be
• D. Be $<$ C $<$ B $<$ N

Hint:

Atomic Radius Trend: Across a period → Decreases (nuclear charge ↑)
Down a group → Increases (new shells added)

Answer 1

The Correct Option is C

Across a period (left to right), atomic size decreases due to increased nuclear charge with the same shell. Order of atomic numbers: Be (4), B (5), C (6), N (7) Thus, decreasing radii: Be>B>C>N Hence, increasing order: N<C<B<Be This matches option (3).

Answer 2

Step 1: Understand the trend of atomic radii in the periodic table
Atomic radius generally decreases across a period from left to right due to increasing nuclear charge pulling the electrons closer.
Atomic radius increases down a group because of the addition of electron shells.

Step 2: Identify the elements and their positions
- Be (Beryllium): Group 2, Period 2
- B (Boron): Group 13, Period 2
- C (Carbon): Group 14, Period 2
- N (Nitrogen): Group 15, Period 2

Step 3: Apply periodic trends to these elements
All these elements are in the same period (Period 2). Moving from left to right:
Be → B → C → N
Atomic radius decreases because the number of protons increases, pulling electrons closer.

Step 4: Final order of atomic radii
Since Be is furthest left, it has the largest radius, followed by B, then C, and N has the smallest radius.
Therefore, the correct order is:
N < C < B < Be.