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

• Electronegativity trend: As you move across the 2nd period of the periodic table (from left to right), electronegativity increases. This happens because the nuclear charge increases, pulling the electrons closer to the nucleus, while the size of the atoms decreases.
• Specific Trends: In the second period, electronegativity increases as follows:
  • Li (Lithium): Being the first element in the period, it has a low electronegativity (0.98 on the Pauling scale), because it has a large atomic radius and is less effective at attracting electrons.
  • Be (Beryllium): Slightly higher than lithium, but still relatively low compared to the elements that follow. Its electronegativity is around 1.57.
  • B (Boron): As you move further right, the electronegativity increases, and boron has an electronegativity of 2.04.
  • C (Carbon): With an electronegativity of 2.55, carbon attracts electrons more effectively.
  • N (Nitrogen): Nitrogen has an electronegativity of 3.04, further increasing as we approach the more electronegative elements.
  • O (Oxygen): Oxygen has an electronegativity of 3.44, being one of the highest in the 2nd period.
  • F (Fluorine): Fluorine is the most electronegative element in the 2nd period, with an electronegativity of 3.98, making it the most effective in attracting electrons.

Conclusion: The correct trend in electronegativity for the second period elements is: Li < Be < B < C < N < O < F. This shows the general increase in electronegativity as you move across the period from left to right.