Question

Among 2nd period elements, correct electronegativity trend is:

• A. B $<$ C $<$ N $<$ O $<$ F
• B. F $<$ O $<$ N $<$ C $<$ B
• C. N $<$ C $<$ O $<$ F $<$ B
• D. F $<$ N $<$ C $<$ B

Hint:

To determine the electronegativity trend across a period, remember that it increases from left to right due to increased nuclear charge and decreased atomic radius.

Answer 1

The Correct Option is A

The electronegativity trend is: B < C < N < O < F.

• Electronegativity Trend: Electronegativity increases across the 2nd period of the periodic table from left to right. This is due to increasing nuclear charge and decreasing atomic size, leading to a stronger pull on electrons.
• Specific Trends: Within the second period, electronegativity increases sequentially:
  • Li (Lithium): Electronegativity 0.98. Low due to large atomic radius and weak electron attraction.
  • Be (Beryllium): Electronegativity approximately 1.57. Slightly higher than lithium.
  • B (Boron): Electronegativity 2.04.
  • C (Carbon): Electronegativity 2.55. More effective electron attraction.
  • N (Nitrogen): Electronegativity 3.04.
  • O (Oxygen): Electronegativity 3.44. High within the period.
  • F (Fluorine): Electronegativity 3.98. The most electronegative element in the 2nd period, exhibiting maximum electron attraction.

Conclusion: The established electronegativity progression for second-period elements is: Li < Be < B < C < N < O < F. This confirms the general upward trend of electronegativity from left to right across the period.