Question

a chemist makes a graph showing the electronegativity of some of the elements in periods 2 and 3 of the periodic table. electronegativity of elements (pauling scale) 0 li b n f na al p cl element 2 3 li be 4 11 na 12 mg 13 b 14 c 15 n 16 o 17 f 18 ne 13 al 14 si 15 p 16 s 17 cl 18 ar which value is most likely the electronegativity for o? a. 2.5 b. 3.0 c. 3.5 d. 4.0

Explanation:

Step1: Analyze Electronegativity Trends

In the periodic table, electronegativity generally increases across a period (left to right) and decreases down a group (top to bottom). For period 2, elements Li, B, N, F: F has the highest electronegativity (~4.0), N is ~3.0, B ~2.0, Li ~1.0. For period 3, Na, Al, P, Cl: Cl is highest (~3.0), P ~2.0, Al ~1.5, Na ~1.0. Oxygen (O) is in period 2, group 16, between N (group 15) and F (group 17). N has ~3.0, F ~4.0, so O should be between them, closer to F? Wait, no—wait the bar graph: N (purple) is ~3.0, F (purple) ~4.0. O is in period 2, group 16, same period as N, O, F. So N (15) < O (16) < F (17) in period 2. So electronegativity: N (~3.0) < O < F (~4.0). Wait, but the yellow bars: Cl (period 3, group 17) is ~3.0? Wait no, the purple bars are period 2 (Li, B, N, F), yellow are period 3 (Na, Al, P, Cl). Wait Li (period 2, group 1) purple ~1.0, B (period 2, group 13) ~2.0, N (period 2, group 15) ~3.0, F (period 2, group 17) ~4.0. Yellow: Na (period 3, group 1) ~1.0, Al (period 3, group 13) ~1.5, P (period 3, group 15) ~2.0, Cl (period 3, group 17) ~3.0. So period 3, group 17 (Cl) has ~3.0, period 2, group 17 (F) has ~4.0. So period 2, group 16 (O) should be between N (group 15, ~3.0) and F (group 17, ~4.0). Wait but the options: A.2.5, B.3.0, C.3.5, D.4.0. N is ~3.0, F ~4.0, so O should be higher than N (3.0) and lower than F (4.0). So 3.5 (option C) is between 3.0 and 4.0, which fits O's position (between N and F in period 2, group 16). Wait but wait the bar for Cl (period 3, group 17) is yellow, ~3.0. So period 3, group 17: 3.0; period 2, group 17: 4.0. So period 2, group 16 (O) should be higher than period 3, group 16 (S, which would be lower than Cl). Wait S is in period 3, group 16, so S < Cl (group 17) in period 3. So O (period 2, group 16) should be higher than S (period 3, group 16) and lower than F (period 2, group 17). So O's electronegativity should be between N (3.0) and F (4.0), so 3.5 is reasonable. Wait but the options: B is 3.0 (same as N), C is 3.5, D is 4.0 (F). So O can't be 4.0 (F is higher), so 3.5 (C) is between 3.0 (N) and 4.0 (F), which makes sense.

Step2: Match with Options

Options: A.2.5 (too low, lower than N), B.3.0 (same as N, but O is more electronegative than N), C.3.5 (between N and F), D.4.0 (same as F, but F is more electronegative than O). So O's electronegativity is most likely 3.5 (option C).

Answer:

C. 3.5