Chemical Bonding: Question 1

Worked solution

Step 1: Recall what bond polarity depends on

A covalent bond is polar when the two bonded atoms have different electronegativities. The shared electron pair is pulled towards the more electronegative atom, giving it a small negative charge ($\delta^-$) and leaving the other atom with a small positive charge ($\delta^+$). The bigger the electronegativity difference between the two atoms, the more polar the bond.

Step 2: Calculate the electronegativity difference for each bond

Each option is a bond between hydrogen ($\text{EN} = 2.1$) and one other element from the table. Taking the magnitude of the difference for each:

$\Delta\text{EN}(\text{C–H}) = |2.5 - 2.1| = 0.4$

$\Delta\text{EN}(\text{N–H}) = |3.0 - 2.1| = 0.9$

$\Delta\text{EN}(\text{O–H}) = |3.5 - 2.1| = 1.4$

$\Delta\text{EN}(\text{F–H}) = |4.0 - 2.1| = 1.9$

Step 3: Identify the greatest difference

The differences increase steadily: $0.4 < 0.9 < 1.4 < 1.9$. This matches the trend in electronegativity across period 2 (C $<$ N $<$ O $<$ F, since electronegativity increases left to right across a period).

The largest value, $1.9$, belongs to the F–H bond, so this bond is the most polar of the four.

Why the other options are wrong

• A (C–H): this has the smallest electronegativity difference ($0.4$) of the four, so it is actually the least polar bond, not the most polar.
• B (N–H): a difference of $0.9$ is bigger than C–H but still smaller than both O–H and F–H.
• C (O–H): a difference of $1.4$ is the second-largest, but F has an even higher electronegativity than O, so F–H is more polar still.

Final answer

• Greatest bond polarity: F–H ($\Delta\text{EN} = \boxed{1.9}$), option D.